Hubert Schaetzel

Hilbert's tenth problem. Asymptotic Diophantine enumerations. Hypervolumes method

We build up a general method to evaluate the number of solutions of Diophantine equations with asymptotic branches that we call hypervolumes method by analogy to Hardy-Littlewood famous circle method. The estimates are based on integral calculus of volumes associated with an asymptotic sieve resting on Hasse local-global principle in order to get a right volumes' weighting. Matrices with remarkable properties emerge from this weighting process and are certainly beyond everything else, except the notion of representative of a variable, the most essential contribution of our study. Indeed, to an independent expression within an equation will correspond a specific matrix, a basic brick reusable in another context, namely for other equations solutions counting. Among these properties, replacing a variable of integers by a variable of prime numbers amounts to a simple subtraction by the identity matrix while the elevation of the degree of the variable meets some simple rules of additivity to be discovered in this article. The usefulness of these objects is obvious for recurrent terms equations, such as Waring sums, but also for any other heteroclite assemblage since the matrix of change of basis of a given rank happens to be unique. The ultimate evaluations lead therefore to mere products of eigenvalues. No calculation becomes vain as soon as it is completed, as now it will be reusable in another equation configuration. Is the "diophantine problem" then only the art of composing matrices eigenvalues ? The reader can make his opinion himself. Besides of that, we cross-check our estimates with either proven results (Vinogradov, Iwaniec/Friedlander, Terence Tao/Ben Green) or Number Theory famous conjectures (Goldbach, De Polignac, Hardy-Littlewood, Fermat-Catalan, Pillai, twin primes) and we submit many additional more complex peculiar or general results demonstrating the ease of use and the prolific nature of this tool.

Dixième problème de Hilbert. Dénombrements asymptotiques d'équations diophantines. Méthode des hypervolumes

Résumé

Nous construisons une méthode générale de dénombrement des solutions d'équations diophantines à branches asymptotiques dite des hypervolumes par analogie à la célèbre méthode du cercle de Hardy-Littlewood. L'estimation repose sur le calcul de volumes par intégrations multiples associé à des corrections de ces volumes par un crible asymptotique reposant sur le principe local-global de Hasse. Les matrices mises en oeuvre pour l'évaluation de ces corrections ont des propriétés remarquables et sont ici au-delà de tout le reste, à part la notion de représentant d'une variable, la contribution essentielle de notre étude. En effet, à une expression indépendante dans une équation va correspondre une matrice spécifique, brique de base réutilisable dans un autre contexte, c'est-à-dire pour une autre équation. Parmi ces propriétés, le remplacement d'une variable de nombres entiers par une variable de nombres premiers revient à une simple soustraction par la matrice identité tandis que l'élévation du degré de la variable répond à de simples règles d'additivité à découvrir dans cet article. L'utilité de ces objets est manifeste pour les équations à termes récurrents, telles les sommes de Waring, mais aussi dans tout autre assemblage hétéroclite, car la matrice de passage d'un rang donné s'avère unique. Ainsi les évaluations ultimes mènent à de simples produits de valeurs propres de matrices. Aucun calcul n'est désormais vain sitôt terminé car réutilisable dans d'autres configurations d'équations. Le « problème diophantien » ne serait-il alors que l'art de composer des valeurs propres entre elles ? Le lecteur peut se faire sa propre opinion. Par ailleurs, nous vérifions le recoupement des dénombrements établis ici avec des résultats effectivement démontrés (Vinogradov, Iwaniec/Friedlander, Terence Tao/Ben Green) ou des conjectures de la Théorie des Nombres (Goldbach, De Polignac, Hardy-Littlewood, Fermat-Catalan, Pillai, nombres premiers jumeaux) et nous proposons de nombreux autres résultats particuliers ou généraux plus complexes, démontrant la facilité d'utilisation et le caractère prolifique de cet outil.

KEYS FOR ASYMPTOTIC ENUMERATION

Preamble

David Hilbert stated in 1900, during the 2 nd international congress of mathematics, a list of 23 problems, the tenth being to find a general and systematic method to solve diophantine equations. In 1970, Youri Matiiassevitch showed the impossibility of a single algorithm giving solutions of any diophantine equation [3]. However what about :

-a general method of enumeration of solutions ? -aa anticipation of the bounded or infinite character of the number of solutions ? -the determination of an asymptotic expression in the case of infinite number of solutions ?

For the first question, one might doubt that the problem is still opened, given Youri Matiiassevitch result. The two other problems are explored here through literal examples where the hereby suggested method can be applied.

Generally, the enumerations of diophantine equations (Vinogradov, De Polignac, Hardy Littlewood…) give, in the case of infinite numbers of solutions, formulas which can be divided into two parts :

-one dealing with the infinity or not of the solutions, in connection with the peculiar form, and in particular the degree, of the studied diophantine equation : a function mixing polynomials (in numerator generally) and logarithms (in denominator), which we will call the function volume, -the other depending on criteria of divisibility related to the constants of the diophantine equation : a limit coefficient of an infinite product of terms (Euler product) which we will call the (normalized) abundance factor (fudge factor in English, séries singulières in French).

The object of this article is to give an expression in the form of integrals and polynomials of the function volume for several general equations types which include famous conjectures, a systematic method of determination of the abundance factors, a matrix method applying to many situations and an exposure of precautions for use.

Our intention will not be, in this text, to prove all what follow, since on one hand the subject is too vast and on the other hand some demonstrations are entirely beyond our understanding. We will rather seek to gather as many as possible elements of reflexion around our approach whose advantage is to be at the same time elementary and rich in results.

If this article opens new horizons of resolution in the enumeration field, we will be satisfied with it. Besides, let us note that it comes up with as many new conjectures as it may close old ones, which is a positive thing.

The levels of difficulty and quality between the various parts of the article may seem heterogeneous. This is due firstly to the need sometimes for pointing out or establishing some elementary results, secondly to the time passed between the first stoke of the pen and the present text.

The choice of the article's title 'hypervolumes method' is quite old. It reflects a lack of retrospect on the mathematical literature for hereby subject. The method was developed originally with a minimal and relatively trivial background. Today, to replace the article in its mathematical context, more appropriate titles could be "local-global method for n variables", 'Euler products in n dimensions" or "Hasse principe in n-volumes" even though none of these title does completely match.

Foot-note

Fudge factor and normalized abundance factor are only one and only thing. The first term is used when expressions result from the mathematical literature, the second when we make our own evaluations, thus avoiding an ambiguity between our and other contributions.

Heart of the method

The adopted approach is geometric. We count the number of solutions points in an n dimensions volume, not necessarily bounded (quite to the contrary), delimited by an integer parameter, the so called target c. This volume cumulates the solutions for c = 0, c = 1, c = 2,…, c = c.

We seek then the number of solutions on the surface of the volume while incrementing (by 1) the target. The variation of volume is obtained by derivation of the function volume. The number of solutions is obtained by a evaluation based on matrices of proportions between c, c+1, etc.

We get then the laws governing the enumerations :

-a geometrical law of evaluation of the volume, as a law of average, to which is subjected the number of solutions,

Asymptotic sieve

The method of determination of the abundance factors which we propose is not a sieve according to usual definition (see for example a general definition in [11]). We chose the term of sieve because it allows, like the standard sieves, to achieve the same ends, namely to count the solutions of asymptotic diophantine equations. In addition, we use the concept of sequence (see definition page 19) which usually exists in a sieve's case.

The sieve consists in substituting mathematical objects, which we called variables of integers x i and variables of prime numbers y j , by equivalent tools {{x i }} and {{y j }}. We build two objects {{x i }} and {{y j }} with respect to weightings and which replace the variables of one or the other types in the diophantine equations. If it acts on R(x 1 , x 2 , …,x k , y 1 , y 2 , …, y m ) = c, the equation becomes R({{x 1 }}, {{x 2 }}, …,{{x k }}, {{y 1 }}, {{y 2 }}, …, {{y m }}) = c. We will see in 1 the detail of this process by developing the concept of representatives, fundamental to this approach.

Concept of asymptotic representatives

We will see the "equivalence" of a variable of positive integers with the set {0, 1, 2, 3,…, p-1} modulo p and of a variable of prime numbers positive with the set {1, 2, 3,…, p-1} modulo p at a sequence p. The approach modulo p is then extended to an approach modulo p i (and p ∞ by extension). The concept of (asymptotic) representative is thus the essential modus operandi here. For a construction modulo p : {{x i }} p = [0,1,2,…,p-1] and {{y i }} p = [1,2….,p-1] For the approach modulo p i δ , the representatives are respectively for the variables of integer and of prime numbers : {{x i }} p δ = [0,1,2,…,p δ -1] et {{y i }} p δ = [g 0 ,g 1 ,g 2 ,…,g φ(p^δ)-1 ]

The use of g a primitive of p makes it possible to simply eliminate all multiples of p (including 0). Let us note that the adequate representative of a variable is obtained modulo p δ when δ → +∞.

Implementation of the formalism modulo p

The enumeration of the t-uples (x 1 , x 2 , …, x k , y 1 , y 2 , …, y m ) satisfying the diophantine equality R(x 1 , x 2 , …, x k , y 1 , y 2 , …, y m ) = c amounts drawing a multidimensional table (or a succession of two-dimensional tables) having {{x 1 pi }}, {{x 2 pi }}, …, {{x k pi }}, {{y 1 pi }}, {{y 2 pi }}, …, {{y m pi }} as vectors (or axis) generators of the target c. The abundance of each target c according to integer values 0, 1,…, p i is collected inside the table. Set {0, 1, 2… p i -1} is the tool at the step p i for handling of integers' variables x and set {1, 2… p i -1} for prime numbers variables y. The abundance of c obtained at sequence p i is

Normalization

Normalization is an operation of weighting of the abundance factors resulting from the preceding formalism. This normalization gives the occurrences of the targets c to their actual proportion. This normalization is not an artifice which allows coherence between surface and volume's data, but results from arithmetic arguments. The procedure of normalization modulo p and modulo p k is given at exercise 2. The Euler product of the normalized abundance factors fan(c,p i ) at each sequence p i is called the normalized abundance factor of the target c and is noted fan(c).

Well posed problem

We are never short of self-evident argument : the method of the asymptotic sieve applies only if the studied problem is indeed asymptotic. This follows from the fundamental nature of the sieve we use. Indeed, the mathematical objects {{x i }} and {{y j }} are declined on two infinite sets (N and P). The enumeration equals only if the objects {{x i }} and {{y j }} are actually extended to all their whole domain of definition. Let us take examples. The diophantine equation y 1 +y 2 = c, c a value fixed in advance is not an admissible problem. But y 1 +y 2 = 2n+c is an asymptotic problem (c is still a given constant).

Previously, the problem related to only one value and could not be solved within our framework. Now, it relates to all the even numbers and admits an asymptotic formula (we could also replace 2n by n). Let us notice however, and that is essential, that the asymptotic sieve is useful and can be used if the problem is well posed or not. We regularly do that in the exercises which follow, in particular for the study on Waring sums. Indeed, a badly posed problem may be transformed into a problem well posed by the simple addition of a variable and former calculations reused. We will see how to implement this operation.

Field of application

We want simply to reconsider here two examples not to leave ambiguity to our field of application within the asymptotic framework. It is the conjecture of Goldbach and the conjecture of the prime numbers twins. The first problem is written y 1 +y 2 = 2n+c and the second y 1 -y 2 = c. The target c being fixed, when y 1 and y 2 describe all the possible solutions, we can show infinity from the solutions in (y 1 , y 2 ) or not. However, we cannot say which values variable n will take. In other words, we can solve the conjecture of the prime numbers but not that of Goldbach because the fact of observing a particular value of n makes leave the problem the asymptotic framework. However, we can give the behaviour of the solutions in arithmetic progression of the latter with precision without particular difficulty. Recently, we have shown an adaptation of the method allowing its use in a non-asymptomatic framework, i.e. in the case of a finite number of solutions. This is the subject of another article. However, the presence of certain differences of evaluation remains to be addressed (and explained) when the number of solutions is zero (while the evaluation sometimes seems to indicate 1 or 2 solutions).

Personal point of view on a peculiar case

The example of the asymptotic formula of Hardy Littlewood for the diophantine equation x 1 n + x 2 n +…+ x k n = c is famous. Obtained by the circle method, the evaluation of this formula holds at the same time of the mathematical feat and the miracle. Indeed for the miracle, the major arcs, although of negligible extent in front of the minor arcs, have a majority contribution which, by chance, can be calculated. For the feat, the reader can for example refer to [11] where we find a method based on the Weyl inequality and the Hua lemma and combining mathematical prowess and tricks. In the broad outline, we can recognize those developments referring to the geometrical law (volume) and those referring to the impact of the arithmetic law (surface). However, the restrictions being born in the process are not clearly ascribable to one or the other and unfortunately the restrictions remain drastic at the end. Let us point out these restrictions. The asymptotic formula is presented as such, that means it is not true except when c tends towards infinite (thus non verifiable if a bad behaviour shows close to the origin). Moreover, a condition is imposed on the number of variables relative to the degree of the expression. Initially, the validity of the formula resulting from the method of the circle imposed k > 2 n +1. Thanks to K.B.Ford (see [7] and [11]), the condition was reduce to k ≥ n 2 .(ln(n)+ln(1n(n))+O (1)). This gives for example the following table : To simplify, we give here O(1) value 1 for the data simulation. Koch improvement thus takes effect only for relatively high degrees n. The number of variables remains important compared to the degrees of the equations. By a matter of fact, for as many variables, the abundance factors discriminate little or not the targets c. Let us take the example of n = 2 (and k = 6).

The normalized abundance factors are given below for the targets c ranging between 0 and 10 : Excepting c = 0, the targets have relative abundances varying between 1% and 6% compared to average 1. For n > 2, the discriminating effects are even weaker and the abundance factors tend quickly towards 1 with the increase in n. In fact, this table is a true consent of impotence. It is an illusion to believe that S(c)(Г(1+1/n)) k /Г(k/n).c (k/n-1) gives the fine enumeration of the solutions of surface x 1 n +x 2 n +…+x k n = c with such constraints between n and k. We can observe only the average behaviour imposed by the geometrical law of volume x 1 n +x 2 n +…+x k n < c, behaviour accessible without the least calculation of S(c). The restrictions which are currently imposed on the asymptotic formula of Hardy Littlewood make only revisit one multiple integral very accessible by standard calculations (which we give besides). We propose an alternative to this dilemma with our asymptotic sieve. This sieve makes it possible to separate the asymptotic problem of enumeration in two parts, one giving geometrical law, the other the arithmetic law. This can make believe in a loss of mathematical rigour, but it is not so. Moreover, this separation is profitable to simplicity of the developments and less reducing constraints if any.

Types of partitions

The decomposition 8 = 5+3 is similar to the mirror decomposition 8 = 3+5. It will be counted here for two occurrences. On the other hand, the decompositions 6 = 3+3 and 6 = 3+3 are perfectly identical and are counted for only one. We accordingly place ourselves in standard redundant partitions throughout this article

Enumeration in a limited volume

We will examine a simple case to arrive gradually at our objective of enumeration of asymptotic diophantine equations. Let us have :

x 1 + x 2 +…+ x k ≤ c

Here the variables x 1 , x 2 ,…, x k describes the natural integers N *+ (c>0). The solutions are here, like in the continuation, the points of the mesh of the first « quadrant » of the hypervolume based on the unit vectors of the axis. The exact number of solutions of this inequality is given by :

x 1 = c x 2 = c -x 1 x k = c -x 1 -x 2 -…-x k-1 ∑ ∑ … ∑ 1 x 1 = 1 x 2 = 1 x k = 1
We get a good approximation of this expression using the volume defined by inequality x 1 + x 2 +…+ x k ≤ c, where x 1 , x 2 , …, x k take positive values. We then get a formula which will be easier to handle, that is :

x 1 = c x 2 = c -x 1 x k = c -x 1 -x 2 -…-x k-1 ∫ ∫ … ∫ 1.1…1.dx 1 dx 2 …dx k-1 dx k x 1 = 0 x 2 = 0 x k = 0
The reader will find theoretical arguments and profusion of calculations of volumes for double and triple integrals in [13]. These arguments spread (beyond the triple integrals developed in [13]) and enable to calculate the preceding integral. In this one, the elementary volume of the mesh is the product 1.1… 1, that is 1. The exact evaluation of this integral (see exercise 16) is possible owing to the fact that the variables are separable. This condition of separation is essential with the explicit resolution of the problem. Of course, it is not always sufficient. Let us consider now the inequality :

y 1 + y 2 +…+ y m ≤ c
Here the variables y 1 + y 2 +…+ y m describes the prime numbers. The approximate number of solutions of this equation depends on the density of the prime numbers locally and the volume defined by the equation y 1 + y 2 +…+ y m ≤ c. Previously, each point of the unit mesh of the hypervolume was solution. Here, it is necessary to balance each point (y1, y2,…, ym) of this mesh by the approximate density :

1 . 1 … 1

Ln(y 1 ).Ln(y 2 )…Ln(y m ) resulting from the density of solutions 1/Ln(p) of the prime numbers at a given point p.

We get the estimate by using the integral directly : Foot-note :

The lower boundaries of the integrals are taken equal to 2 to avoid divergence. It would be absurd to integrate here from 0 or 1. The upper delimiters are subject to same adaptation (c must be replaced by c-2) in order to (possibly) avoid a divergence (inner loops). The expression is an approximation getting much better as the volume increases (although eventually with oscillations).

By extension, for a set of polynomials, all increasing functions, let us consider the following inequality now, taking account of the preceding notations for integers x i on the one hand and prime numbers y j on the other hand :

P 1 (x 1 ) + P 2 (x 2 ) + … + P k (x k ) + Q 1 (y 1 ) + Q 2 (y 2 ) + … + Q m (y m ) ≤ c
The number of solutions of such an inequality will answer to :

x 1 = P 1 -1 (c) x 2 = P 2 -1 (c -P 1 (x 1 )) y m = Qm -1 (c -2 -P 1 (x 1 ) -…-Q m-1 (y m-1 ))

∫ ∫ … ∫ dy 1 dy 2 … dy m-1 dy m dx 1 dx 2 …dx k-1 dx k Ln(y 1 ).Ln(y 2 )… Ln(y m-1 )Ln(y m ) 0 0 2

It is reasonable to think that such integration is not trivial :

-the reciprocal functions P 1 -1

, P 2 -1 …P k -1 , Q 1 -1
, Q 2 -1 …Q m -1 cannot be obtained in a literal form for a great number of cases (but literal approximations are possible), -the presence of logarithmic under integrals is a difficulty to overcome, -the convenience of its resolution may depend on the order of variables whereby we choose to integrate.

However, we will propose asymptotic solutions to the traditional problems (and undoubtedly most interesting). In exercise 13, we give ways and justifications for extracting logarithms from integrals. We carry out, starting at exercise 16, the decomposition of multiple integrals in a product of simple integrals for separable variables.

Enumeration in a boundless volume

In the preceding paragraph, the finite value of the parameter c delimits, within the framework of the inequality, a finite volume. We have to deal, within the asymptotic framework, with another category of inequalities generating infinite volumes. That can be, for a set of polynomials all increasing functions :

i j k m Σ P n (x n ) + Σ Q n (y n ) ≤ c + Σ R n (z n ) + Σ S n (t n ) n = 1 n = 1 n = 1 n = 1
Earlier, c introduced a constraint for monomials or polynomials located on the left of the inequality. Here this constraint is not exerted on the right member of the inequality (out c). The number of meshes in boundless volume is written then, x n and z n being variables of integers and y n and t n variables of prime numbers : k m k m i j-1 z 1 z k t 1 t m x 1 = P 1

Convergence or divergence of the asymptotic enumerations

We affirm the following statements :

-if the volume integral, asymptotic approximate cardinal of solutions, diverges the number of solutions then diverges, -if the volume integral is finite, the cardinal of the solutions is finite. When the integral is finite, the limit does not have actual value. Hence, if the integral converges towards 0, the number of solutions can be different from 0.

In addition, let us consider the case of a target with abundance factor null or converging towards zero. Asymptotically, we are in a borderline case 0.x , which requires a finer study to decide convergence or divergence of the number of solutions. It is the case, for example, of y 1 ²+y 2 ² = p+c, where we seek the cardinal of prime numbers, except for a translation, generated by the sum of two squares. For the targets 0 mod 2 and 2 mod 3, the abundance factor is 0 for an enumeration in p/ln 3 (p) if we regard the problem as concerning two variables y 1 and y 2 , whereas it is different from 0 for an enumeration in p 1/2 /ln 2 (p) if we regard it as concerning one variable y 1 and a constant y 2 (y 2 = 2 for the targets 0 mod 2 and y 2 = 0 for the targets 2 mod 3).

Variables definition domain

The method to get {{z}} representatives shows that the variables z definition domain are either positive values, either positive and negative values, that is we have the choice of intervals like [0,∞[ or like ]-∞,∞[. In the first instance, bound 0 may be replaced by any finite value since an asymptotic contribution between 0 and a (finite value) is finite asymptotically, hence negligible (unchanged abundance factor  a). When the choice of interval is ]-∞,∞[, we are no longer in the case of integration in the first quadrant presented until now and we must take this into account : in the case of integers numbers variables, we integrate simply between -∞ and +∞ but in the case of primes variables, we have to consider the sum of two integrals between -∞ and -2 and between 2 and +∞ to avoid any parasitic divergence.

Parasitic system of equations

Equalities

The hypervolumes method is not designed for a system of equations.

When we say that, we draw the reader's attention towards unwanted production of such system during numerical evaluations. For example, let us consider the integral : k m k m i j-1 z 1 z k t 1 t m x 1 = P 1 -1 (c+Σ R n (z n )+Σ S n (t n ))

y j = Q j -1 (c-2+Σ R n (z n )+Σ S n (t n ))-Σ P n (x n )-Σ Q n (y n )))
∫… ∫ ∫… ∫ ∫ … ∫ dy 1 …dy j .dt 1 …dt m . dx 1 …dx i .dz 1 …dz k Ln(y 1 )…Ln(y j ).Ln(t 1 )...Ln(t m ) 0 0 2 2 0 2

When we evaluate this volume for a numerical verification, we are led to assumptions on boundaries z 1 to z k and t 1 to t m interdependence (for example z 1 = … = z k = t 1 = … = t m ). Doing so, we are introducing additional equations to our initial diophantine equation. As hypervolumes method is not compatible for equations systems, we inject these relationships in the initial equation, creating an equation with fewer variables.

We are facing the paradox of a method that allows any situations evaluation as long as we do not actually seek to numerically evaluate it. This is analogous to the observer inference in quantum physics.

Thus, the volumes V(c) and V'(c) assessment should be "natural" by the mere fact of the studied diophantine equation. The artificial imposition of some shape of volume to evaluate V(c) is prohibited. Indeed, to a given shape of volume is corresponding a given number of targets at its surface (between V(c) and V(c+1)) and therefore a given abundance factor. If the volume is not only imposed by the diophantine equation, there is no reason such factor abundance, we compute by our method, matches as it should to correct in a right way the said volume.

Inequalities

Previous item being clarified, what about the introduction of inequalities ? When, we ask this question, we refer to evaluations with restrictions on the definitions of variables domains: The first chart shows, within the four quadrants defined by the axis in blue colour, a total asymmetry of the "volumes" induced by 2x 2 -3z < c. However, if we step back, the second chart gives us another perspective showing the asymptotic situation of the problem. We see the quasi-symmetry to the y axis. Thus an asymptotic enumeration in the first quadrant (above the curve) is equal to half of the evaluation in all quadrants (above the curve). This remark is quite general including in dimension greater than 2.

We are therefore entitled to introduce inequalities on variables during asymptotic evaluations (by the balance sheet of the quadrants with volume zero and quadrants with identical volumes).

Literal expressions of the normalized abundance factors

We get literal expressions in cases of polynomials with separable variables, some appearing already in the literature. The suggested method allows generalizing some of these expressions. Thus for example, the expression of the abundance factor for a 1 y 1 + a 2 y 2 + … + a n y n = c with n prime numbers variables (see exercise 5) is :

∞ ∞ ∞ П (1- (-1) m-2 ) П (1- (-1) m-1 ) П (1- (-1) m ) (p i -1) m-2
(p i -1) m-1 (p i -1) m p i \c, p i \a i or(and(p i ∤c,p i \a i ),and(p\c,p∤a i )) p i ∤c, p i ∤a i

This example and others relatively more complex are presented in the exercises :

-Waring enumeration (or Fermat enumeration pending on reader preference) (if boundless volume), -Hardy-Littlewood enumeration (or Terence Tao and Ben Green enumeration), -Fermat-Catalan enumeration (Pillai conjecture, Iwaniec and Friedlander enumeration), -quadratic enumeration,…

Opposite

This operation is an academic case. Let us write {{x i }} = [0,1,2,…,p i -1]. Then -{{x i }} = -[0,1,2,…,(p i -1)] = [-0,-1,-2,…,-(p i -1)] = [p i -0, p i -1, p i -2,…, p i -(p i -1)] mod p. Any permutation of the elements of an asymptotic representative does not affect at all the targets c. Hence [p i -0, p i -1, p i -2,…, p i -(p i -1)] ≡ [0,1,2,…,p i -1]. Thus :

-{{x i }} = {{x i }} then -{{x i 2n+1 }} = {{x i 2n+1 }}
The reader will check without pain that the preceding relation remains valid in the case of a prime number variable {{yi}} = [1,2,…, p i -1], like in a reasoning modulo p i δ .

Thus, when a diophantine equation includes variables with odd exponents, the abundance factors remain identical by replacing the negative signs in front of these variables by positive signs.

Examples

Fan(-x 7 -y 5 -z 4 .t = c) = Fan(x 7 +y 5 +z 4 .t = c) Fan(-x 7 -y 5 -z 4 = c) = Fan(x 7 +y 5 -z 4 = c) ≠ Fan(x 7 +y 5 +z 4 = c)

Matrix operators

A method must be associated with powerful operators to support his success. The asymptotic sieve leads to the use of matrices having remarkable geometrical properties, matrices which show a generous flexibility for the literal evaluation of the fudge factors. We show how to such diophantine expression is related such matrix and how the addition of a new variable to the initial expression involves a simple multiplication by a new matrix with the existing one (or deducted of the existing matrix).

The reader will also note that the behaviour of some targets versus the present concept is versatile modulo p δ . The reference is of course the δ → +∞ limit behaviour. However, it is useful to note that a target can be poor or rich (and possibly standard) according to the studied δ value. For example, let us consider the diophantine equation (and target c = 0) :

c = x 1 3 +2x 2 3
We can establish the following table : Rich targets  ε > 0, a.F(z 1 , …, z i ) << # asymp {R = c} < (F(z 1 , …, z i )) 1+ε Poor targets  ε > 0, (F(z 1 , …, z i )) 1-ε < # asymp {R = c} << a.F(z 1 , …, z i ) P 13/390

Keys for asymptotic enumeration Let us note that the versatility concept intervenes here as c = x 1 3 +2x 2 3 is an asymptotic problem only when c tends towards infinite, which has no practical application (infinite c cannot be reached).

Other examples

The study of the generation of squares by a quadratic equation gives the opportunity of many likewise rich targets (the phenomenon is not uncommon).

Ambiguous targets

Concept

The concept is as much ambiguous as its name. The concept of ambiguous targets includes rich or poor targets and gives some explanation to their being. We choose here to be relatively fuzzy on our explanation having no total control of the concept. Ambiguous targets are singularities which reveal a diophantine equation "boundaries". Indeed, target c is a parameter of given equation. At the said boundary, the given equation passes, implicitly or explicitly, from an irreducible equation to a reducible equation as for the diophantine enumeration point of view (and we think in general within the usual meaning of equations reducibility). We find ourselves then in the case of systems of equations for which the proposed method is not suitable. The asymptotic behaviour of an ambiguous target cannot be certified, a priori, by the study of the factors of abundance. Examples of contradictions can be produced. In some ways, consistency within the irreducible domains is ensured by the loss of consistency in reducible boundaries.

Example

The following example, we came about by Roger Heath-Brown as a challenge to our method, has being leading to the addition of the preceding paragraph : This equation can be put in more general form c = 5x 1 3 +9x 2 3 -(10x 3 3 +12x 4 3 ) and we can seek abundance factors for c = 0 or any other integer value c. However for c = 0, we have an ambiguity phenomenon since we write 5(x 1 3 -2x 3 3 ) = 3(3x 2 3 -4x 4

3

), which means the following system of two diophantine equations :

x 1 3 -2x 3 3 = 3r 3x 2 3 -4x 4 3 = 5r
When the study is done omitting this point, the normalized abundance factor of equation c = 5x 1 3 +9x 2 3 -(10x 3 3 +12x 4 3 ) tends to a nonzero value (approximating 0.5 in a study modulo p as well as modulo p ∞ ), which makes believe an infinity of solutions when in fact the equation has (a priori) only trivial solution (0,0,0,0). However, a certain part of truth remains here in abundance factors, but in a hidden way. Indeed, when we replace the equation in variables of integers by an equation in prime numbers variables 5y 1 3 +9y 2 3 = 10y 3 3 +12y 4 3 we then observe an interdiction at sequence p = 7, involving a null product of the factors of abundance (and a priori finite number of corresponding equation solutions). This finite value is somewhat inherited by former variables of integers equation (how?).

Geometry of numbers

We will see that the examination of a small amount of numbers suffices sometimes to understand the whole studied problem. These numbers are usually 0, 1, 2, 3 and 5. The integer 0 is foremost because it admits any none zero divisors while remaining an integer (i.e. equal to itself). Integer 1 is special because it is prime and not. For the other integers, let us have p one of them and let us consider then (-1) (p-1)/2 , that is ±i (for p = 2), -1 (for p = 3), 1 (for p = 5). These numbers are at the four cardinal points of the unit circle (of which 0 is the centre and 1 the « principal point »). We will see the central role of the unit circle in the next course. Pending on the case, this circle can undergo deformations : translation of the centre, eccentricity change, and so on. Exponent (p-1)/2 shows also the importance of p-1 and (p-1)/2 terms, that we will so often find, what the reader certainly already suspected.

Level of proof

The reader will judge the relevance of presented work. The geometrical law, based on multiple integrals and the exploitation of logarithms, is quite obvious. Concerning the arithmetic law, our arguments rest on the substitution in the studied diophantine equation of variables by equivalent objects. This last adjective is appropriate to us, but may be perhaps disputed. The union of the two laws is settled during the substitution operation.

In any case, if simplicity of the principles, systematic alignment on proved (or conjectured) results, universality of employment and richness of new research axis are positive signs, the reader will have leisure to check that.

Opening on the exercises

The set of problems raised by the diophantine enumeration is very rich what makes its exceptional interest. It passes by arithmetic (congruence, residues…), analysis (function of several variables, multiple integrals, convergence,…), algebra (groups' theory, matrices…), geometry (representation in space, hypervolume and hypersurface…). We can only complain about the astonishing harmony of some simplifications after long developments. The axes of research are innumerable.

Exercises which follow give some illustrations. The here chosen term "exercise" does not want to be reducing. Quite to the contrary, after what constitutes above a general guide of the method, the exercises are the place where proofs are elaborated, certainly sometimes summarily, and of concrete examples. Exercise 1 is number one by its importance while exposing the asymptotic sieve. Exercise 2 supplements it with an algorithmic point of view. Exercises 3 and 4 present the fundamental bases necessary to the literal evaluation of the abundance factors. The literal expressions of these factors for Waring sums are given in exercise 6 thanks to exercise 5 preludes. Exercises 7 to 10 give the abundance factors for traditional diophantine equations, such as Fermat-Catalan equations, polynomials and quadratic equations, equations with overlapping variables… In particular, Fermat-Catalan equations provide an explosion of conjectures and a strong potential of research. The foremost environment founding matrices, constructive of sequence matrices and increment of sequence matrices are approached there. Exercise 11 shows the universality of the method but also its limits. Exercise 12 gives ways for getting expressions generating many prime numbers. Exercise 13 has a great practical utility for enumerations with prime numbers variables (Golbach conjectures like). Exercise 14 is used primarily for comprehension of non-monotonous enumerations close to origin. The beginning of exercise 15 is a wink and a contribution of energy to continue with the reading. Then, this exercise gives a global approach to the problem of differences between prime numbers. It is almost out of subject here and is like a bottle launched to the sea. The later exercises offer heterogeneous partitions, but always captivating, by finally giving traditional or new asymptotic formulas. The abundance factors generally show as Euler products what allows their approximate evaluation without difficulty. The determination of their exact value is another science which is approached only for some favourable cases in the exercise 22. Our text finishes as a fugue (synonymous of run away in French) with the presentation of the research axis.

We hope that the reader will find among many awkward formulas, that of his choice, of a pure mathematical beauty.

Your right to main topics

We list here some of our own favourite topics : pages -asymptotic representative of a diophantine variable (of integer or prime numbers) 19 -approximate iterative generation of the prime numbers set 324 -primitive roots according to Legendre 43 -primitive roots modulo 4p 251 -quadratic and polynomial reciprocal classes 255 -rho function for the enumeration of the classes p such as x n = c mod p, given c 257 -inclusive and exclusive frequencies of a given divisor in the infinite set p-1 260 -corollary of the second theorem of Mertens and function ΓP 381 -structure in double helix of the elementary cardinal matrices 66 -theorem of decomposition of an elementary cardinal matrix (matrix eigen-elements) 72 -theorem of existence of a common environment matrix 311 -decomposition of the cardinal matrices (general case)

Minkowski approach

According to Hermann Minkowski, the diophantine equation R(x, y…) = c has a solution if and only if R(x, y…) = c mod k m has a solution whatever p and k, p positive prime, k positive integer and equation R(x, y…) = c has a real solution. ( [5] gives only the result and we do not have the proof). Let us stop on the part relating to R(x, y…) = c mod p k for any p and k as defined. It appears natural to write : {(x, y, …) / R(x, y, …) = c}  {(x, y, …) / k i ,  p i , R(x, y, …) = c modulo 2 k2 .3 k3 …p i ki …} and :

#{(x, y, …) / R(x, y, …) = c}  #{(x, y, …) / k i ,  p i , R(x, y, …) = c modulo 2 k2 .3 k3 …p i ki

…}

From Chinese theorem, it follows (see also relations (3) and ( 4 The type of variables, of integer or of prime numbers, does not have any importance here. These writings are the base of diophantine equation asymptotic enumeration. To comprehend the behaviour of R(…) = c, it is necessary and sufficient to indicate the behaviour for any prime number p i and to carry out an infinite product.

Product of cardinals relative to integers

We give at this paragraph and at the next one two numerical examples for good comprehension. We have, according to the Chinese theorem, the relation : #{n / P(n) = c modulo m i .m j } = #{n / P(n) = c modulo m i }. #{n / P(n) = c modulo m j } (3) provided that (m i ,m j ) = 1. Let us take an example to clarify things : P(n) = n², m i = 2², m j = 3².

Product of cardinals relative to prime numbers

We have : #{p / P(p) = c modulo m i .m j } = #(p / P(p) = c modulo m i ) . #{p / P(p) = c modulo m j } (4) provided that gcd(m i ,m j ) = 1.

Here the use of p instead of n as generator of the residues means that gcd(p,m i .m j ) = 1. Let us take again an example P(p) = p², m i = 5, m j = 7. We crossed the solutions which are eliminated by the condition gcd(p,m i .m j ) = 1. If this condition is not observed then either P(p) = 0 mod m i , or P(p) = 0 mod m j .

Concept of representatives

We specified the rule of multiplication of the cardinals modulo p i (and modulo p i ki indirectly). This rule is effective only if the first column of tables 1 and 2 is well selected. It is the choice of adequate representatives for the variables n or p (first column of the tables in examples) which induces the aforementioned rule of the cardinals' products. The concept of representatives is upstream of the rule of multiplication of the cardinals modulo p i ki .

Let us look further into this concept.

A variable of integers is the list of infinite natural integers (0, 1, 2,…, +∞). Each element of this list is balanced of a uniform weight of value 1. A variable of prime numbers is the list of infinite natural integers (0, 1, 2,…, +∞), but each element of this list is balanced of a uniform weight of value 1 for the prime numbers and 0 for the other numbers. We build the two objects {{x i }} and {{y j }} by respecting these weightings. As the case of integers' variables may seem too selfevident, we start with prime numbers variables.

Variable of prime numbers

Let us have the table :

sequence \ y j 0 1 2 3 4 5 6 7 8 9 10 11 … ∞ 2 ε 0 1-ε 1 ε 1 ε 2 1 ε 1 ε 3 1 ε 1 … … 3 ε 0 1-ε 1 1 ε 1 1 ε 1 1 ε 2 1 1 … … 5 ε 0 1-ε 1 1 1 1 ε 1 1 1 1 ε 1 … … 7 ε 0 1-ε 1 1 1 1 1 1 ε 1 1 1 1 … … 11 ε 0 1-ε 1 1 1 1 1 1 1 1 1 1 ε … … … … … … … … … … … … … … … … … p ε 0 1-ε 1 1 1 1 1 1 1 1 1 1 1 … … Π ε 0 t (1-ε 1 ) t ε ε ε 2 ε ε 2 ε ε 3 ε 2 ε 2 ε … … Π/ε ε 0 t /ε (1-ε 1 ) t /ε 1 1 ε 1 ε 1 ε 2 ε ε 1 … …
We seek to represent a variable of prime numbers y j by its decomposition into a succession of lines whose ordinate is the prime numbers set. The elements of this ordinate are called the sequences of the asymptotic sieve. When the sequence r divides y j , let us fix the weighting of y j at this sequence with ε k , where k is the multiplicity of r in y j , if not let us fix the weighting of y j at 1. As y j = 1 has a little particular statute among the numbers (it is not really a prime number), we give him a weight 1-ε 1 . In the same way for zero which receives weighting ε 0 . The product of the various weights is given before the last line of the table. We are particularly interested by the relative weight between these various numbers. Accordingly, we see that we can build the last line where each prime number receives a weighting equalizes to 1, where 1 receives weighting (1-ε 1 ) t /ε and where the other numbers are multiples of ε. Let us supplement the table until infinity. Let us make tend ε towards 0. We then get on the last line for y j the weighting of a variable of prime numbers. We observe more closely the case of y j = 1. Its weighting (1-ε 1 ) t /ε is lower than ε as soon as t > 2.ln(ε)/ln(1-ε 1 ). The parameter t being free, we choose it to answer the preceding expression. For y j = 0, we can easily choose ε 0 such as ε 0 t /ε < ε. Under these conditions, a variable of prime numbers tends to the sequence p towards: {{y j }} p ≡ [1,2,…,p-1] ≡ [g 0 ,g 1 ,g 2 ,…,g φ(p)-1 ] mod p (5) Moreover, if we consider the variable y j for all sequences, we can write symbolically : {{y j }} ≡ Π/ε ≡ cf. Π{{y j }} p (6) p Here cf is the factor of adjustment giving the bond between the representative of the aggregate variable and the representatives at each sequence. For a given ε (i.e. for a given precision), cf is a constant whatever the target c.

Let us go back to the table and make ε = 0, δ = 0.

P 20/390

Asymptotic sieve sequence \ y j 0 1 2 3 4 5 6 7 8 9 10 11 … ∞

2 0 1 0 1 0 1 0 1 0 1 0 1 … … 3 0 1 1 0 1 1 0 1 1 0 1 1 … … 5 0 1 1 1 1 0 1 1 1 1 0 1 … … 7 0 1 1 1 1 1 1 0 1 1 1 1 … … 11 0 1 1 1 1 1 1 1 1 1 1 0 … … … … … … … … … … … … … … … … … p 0 1 1 1 1 1 1 1 1 1 1 1 … … Π 0 1 0 0 0 0 0 0 0 0 0 0 … … limit Π/ε 0 1 1 1 0 1 0 1 0 0 0 1 … …
The two last lines, different, show the need for limit evaluation to achieve our aims. Without this artifice, it is impossible to distinguish the qualities of fundamental representation contained in [1,2,…, p-1] mod p when p describes prime numbers. Thus, our matter does not hold out of an asymptotic framework.

Variable of integers

On the same principle, we have immediately {{x i }} p ≡ [0,1,2,…,p-1] ≡ [0,g 0 ,g 1 ,g 2 ,…,g φ(p)-1 ] mod p (7) Foot-note

In the practice of enumerations, we observe, between the two types of variables, significant differences due to the only element {0}. This is foremost at the small values of sequences p : an object such [1] mod 2 (for a variable of prime numbers) and very different from [0,1] mod p (for a variable of integers).

Generalization of the concept of representatives

We reasoned modulo p in the preceding case. This is not generally sufficient. We must proceed modulo p δ . In this case, we have :

For variables x i of integers modulo p δ : {{x i }} ≡ [0,1,2,…,p δ -1] mod p δ (8)

For the variables y j of prime numbers modulo p δ : {{y i }} ≡ [1,2,…,p-1, p+1, p+2,…, 2.p-1, 2.p+1, 2.p+2,…, 3.p-1, …, p δ -p+1, p δ -p+2,…, p δ -1] mod p δ

The use of g, a primitive root of p, makes it possible to eliminate without effort all multiples of p (including 0). {{y i }} ≡ [g 0 ,g 1 ,g 2 ,…,g φ(p^δ)-1 ] mod p δ (9) These alternatives, and in particular when δ → +∞, are examined with Waring sums at exercise 6 and more generally at exercise 11. We will verify that interpretation δ = 1 is in certain cases identical to interpretation δ → +∞ and that in other cases a finite δ suffices.

Operations on the representatives

Let us return to our sieve and replace the first line y j by y j n . That is to say the

1-ε 1 ε 1 ε 2 1 ε 1 ε 3 1 ε 1 … … 3 ε 0 1-ε 1 1 ε 1 1 ε 1 1 ε 2 1 1 … … 5 ε 0 1-ε 1 1 1 1 ε 1 1 1 1 ε 1 … … 7 ε 0 1-ε 1 1 1 1 1 1 ε 1 1 1 1 … … 11 ε 0 1-ε 1 1 1 1 1 1 1 1 1 1 ε … … … … … … … … … … … … … … … … … p ε 0 1-ε 1 1 1 1 1 1 1 1 1 1 1 … … Π ε 0 t (1-ε 1 ) t ε ε ε 2 ε ε 2 ε ε 3 ε 2 ε 2 ε t … … Π/ε ε 0 t /ε (1-ε 1 ) t /ε 1 1 ε 1 ε 1 ε 2 ε ε 1 … …
The reasoning on y j applies in all points to y j n . (by replacing ε in the initial table by ε 1/n ). More generally, the polynomial operations R on entity {{}} is translated by the same operation R on each component of this entity because of the bijection between y j and R(y j ). That is : R({{x i }}) ≡ {{R (0),R (1),…,R(p i δ -1)}} mod p δ (10) and R({{y i }}) ≡ {{R(g 0 ),R(g),R(g 2 ),…,R(g φ(pi^δ)-1 )} mod p δ (11)

Limit of the reasoning

The free choice of t such as ε 0 t /ε tends towards 0 is contestable. It is disputed by exceptional cases which we called ambiguous when referring to the targets. It is often so with target c = 0 in a monomial type equation.

Definition domain of variables

We can build the previous two-dimensional tables with choice of x-coordinates not [ [0, +∞[[ but ]]-∞,+∞[[ with the same conclusions. The use of one or the other has no incidence in a modulo approach.

Decomposition and rebuilding of the diophantine equation

Let us reason modulo p i . The evaluation of the occurrences of a given target c is carried out for each sequence p i . For t a variable, {{t pi }} its representative modulo p i and {{t}} its representative modulo 2.3… p i , we write (the arrow → meaning "transforms in"): t → {{t pi }} => R(t) = c → R({{t pi }}) = c mod p i and R({{t}}) = c mod 2.3…p i where #{t / R(t) = c mod 2.3…p i } = Π pj  {2, 3, pi} #{t / R(t) = c mod p j } For several variables, we write successively :

x 1 → {{ x 1 pi }} => R(x 1 , x 2 , …, y m ) = c → R({{x 1 pi }}, x 2 , …, y m ) = c mod p i and R({{x 1 }}, x 2 , …, y m ) = c mod 2.3…p i where #{(x 1 ) / R(x 1 , x 2 , …, y m ) = c mod 2.3…p i } = Π pj  {2, 3, pi} #{(x 1 ) / R(x 1 , x 2 , …, y m ) = c mod p j } x 2 → {{ x 2 pi }} => R(x 1 , x 2 , …, y m ) = c → R({{x 1 pi }}, {{x 2 pi }}, …, y m ) = c mod p i and R({{x 1 }}, {{x 2 }}, …, y m ) = c mod 2.3…p i where #{(x 1 ,x 2 ) / R(x 1 , x 2 , …, y m ) = c mod 2.3…p i } = Π pj  {2, 3, pi} #{(x 1 ,x 2 ) / R(x 1 , x 2 , …, y m ) = c mod p j } … y m → {{ y m pi }} => R(x 1 , x 2 , …, y m ) = c → R({{x 1 pi }}, {{x 2 pi }}, …, {{y m }})= c mod p i and R({{x 1 }}, {{x 2 }}, …, {{y m }}) = c mod 2.3…p i where #{(x 1 ,x 2 ,…,y m ) / R(x 1 , x 2 , …, y m ) = c mod 2.3…p i } = Π pj  {2, 3, pi} #{(x 1 ,x 2 ,…,y m ) / R(x 1 , x 2 , …, y m ) = c mod p j } With R(x 1 , x 2 , …,x k , y 1 , y 2 , …, y m ) = c, the equation transforms at the sequence p, thanks to the concept of representatives, into R({{x 1 }} p , {{x 2 }} p , …,{{x k }} p , {{y 1 }} p , {{y 2 }} p , …, {{y m }} p ) = c.

Stochastic approach of the concept of representatives

The representatives are arithmetic operators. The asymptotic enumerations are obtained by passage to the limit. We show here the footbridge with the theory of probability which is useful for the evaluation, a priori, of such enumerations. We show why an approach by this theory is filled of success.

We remind the equiprobability of the prime numbers in the arithmetic progressions to find the concept of representative. Let us note [[a,b]] the set of integers of an interval [a,b]. We call this set a comb. If this comb is affected with some constraint, we express it with an index in the following way [[a,b]] C . Let us consider a succession of integers [[x,y]]. Let us consider a random pointer P A on this interval. Let us record the value pointed modulo t on the "comb" [[0,t-1]]. If y-x is a multiple of t the probability of recording on each notch of the comb is equivalent. Table 1 presents a random test of this type using a random pointer of a standard spread sheet. The points which interest us here are represented in green. This comb is not discriminating if we consider a discontinuous succession of integers and of suitable size (multiple of t) as we did; it is not the same if we consider different sets on an interval [x, y] such those from the even numbers (recording on the even notches), from the odd numbers (partial recording on the odd notches) or from the prime numbers (recording in quasiequal proportion on certain odd notches). We can adopt the following formalism:

Let it be {{n}} the average result of P A,mod(t) (n), n  [[x,y]], on a great number of tests then [[0,t-1]]  {{n}} (12) We can translate this result as : The comb [ [0,t-1]] is a representative modulo t of a random selected integer. We get an encapsulation property "being an integer" for the proposed numbers set.

We evoked the case of the prime numbers above. We can affirm, and it is a result due to Gustav Dirichlet [9], the equidensity of the prime numbers in the arithmetic progression a+k.m, a and m suitably choices. More precisely, it is necessary and it suffices that gcd(a,m) = 1. Thus in the formalism with constraint adopted previously, we can write now : [[0,t-1]] (x,t)=1  {{p}} where x is a suitable element ((x,t) =1) of the comb. As (0,t) ≠ 1, we can (and must) withdraw 0 within the list [[0,t-1]] :

P 22/390

Asymptotic sieve

[ [1,t-1]] (x,t)=1  {{p}} (13) So the comb[ [1,t-1]] (x,t)=1 is a representative modulo t of a random prime number. We get an encapsulation property "being a prime number" for the proposed numbers set.

The reader will find in table 1 an elementary example of enumeration concerning the prime numbers. They are the points represented in red. We observe a fluctuation of the results around the awaited average right-hand side. This results from the fact that the number of primes in a progression c + k.t, limited to the interval [1,65536], varies few units according to c. We will also observe that the odd divisors of t (here 210=2*3*5*7) do not obey the rule what is awaited. We can then normalize the occurrences according to these remarks to get the points in blue.

Table 1 Marginal variation in the case modulo 210 = 2.3.5.7 (carried out on x> 2)

Arithmetic prolongation of the stochastic reasoning Expressions ( 12) and ( 13), raising here of statistical arguments, find their origin in the exclusively arithmetic approach when using the asymptotic sieve. Let us suppose that the pointer P is no more one random pointer, but contains information. The result will be translated into relative abundance on the various notches of the considered comb. Such an operation is difficult to decipher in this form. It is much more convenient to point in a levelling way on the notches of the comb and to collect relative abundances defining the "characteristics" of the pointer. This pointer which may have several independent entries. So that : R(P 1 (x 1 ), P 2 (x 2 ), …, P k (x k ), Q 1 (y 1 ), Q 2 (y 2 ), … , Q l (y l )) = c Then R(P 1 ({{x 1 }}), P 2 ({{x 2 }}), …, P k ({{x k }}), Q 1 ({{y 1 }}), Q 2 ({{y 2 }}), … , Q l ({{y l }}) = c mod p j , p j  {2, 3, p i } (14) becomes the iterative form of our enumerations.

Sets approach of the concept of representatives

We refer to the extract of a lesson [8] on the contributions of Cantor and logicians to the mathematical corpus : « Depuis Cantor, tout ceci a été mieux formulé, en particulier, par Von Neumann qui a trouvé la bonne formulation : un ordinal est l'ensemble de ses prédécesseurs. C'est recycler un peu une idée de Shopenhauer qui disait "qu'un entier présuppose tous les précédents". C'est la même chose avec les ordinaux. L'ordinal 0, c'est l'ensemble vide ∅. L'ordinal 1, c'est l'ensemble dont l'unique élément est l'ensemble vide : {∅}. L'ordinal 2, c'est l'ensemble dont les deux éléments sont l'ordinal 0 et 1 : {∅, {∅}}. L'ordinal 3, c'est : {∅, {∅}, {∅, {∅}}}… On a une échelle d'ensembles sur lesquels l'appartenance coïncide avec l'inclusion : un ensemble appartient à un autre si et seulement s'il est inclus dans l'autre. Alors, qu'est-ce que w 0 , le premier ordinal infini ? Eh bien, c'est l'ensemble de tous ses prédécesseurs, c'est-`a-dire w 0 := {0, 1, 2, …} »

In this text (left in its original language and) that we discovered lately, we find on one side the expression of the integers representatives and on the other side the asymptotic concept. It recalls that Von Neumann and Shopenhauer "revealed" that an ordinal is the set of its predecessors. It is amazing to note that this set writing can be used without modification in diophantine enumeration problems as we do write {{x}} ≡ {0, 1, 2, …}. Of course, the generalisation to the prime numbers case is not referred to as it was not a preoccupation at that time.

Random occurrences of modulo classes

(modulo 210) (5 000 000 tests)

Reciprocal

It seems to us here that the concept of representative implies the equidensity of the prime numbers in the arithmetic progressions.

Introspection

As we underlined at the end of the preamble, we can stop here. All the following is repetition or technics of calculation. Let us continue however.

Normalization

Target c delimits a volume V(c) as we specified in the exposure of the article (Keys for asymptotic enumeration). Within V(c)-V(c-1), the number of actual solutions is proportional to the abundance factors product deduced from the asymptotic sieve. Then for neighbour targets c : #{R(…) = c} ≡ (V(c)-V(c-1)).a. Π Fae(c,p) p Here, a is a constant of proportionality, which depends only the number and the type of variables on which operates relation R. Normalization consists in finding a such as :

#{R(…) = c} ≡ Fan(c).(V(c)-V(c-1))

The sum on all cardinals #{R(…) = c} must restore the whole of volume V(c), that is ∑ Fan(c).(V(c)-V(c-1)) = V(c). We cannot solve this equality precisely because the reasoning does not hold "close" to the origin, but we can note that this part is negligible compared to infinite (whatever its size). On a large domain of targets (ad infinitum), the average weight of Fan(c) must be 1 which can be done only by assigning a unit weighting to the variables.

Hence, let us place at sequence p with the collection of occurrences of a given target c. The entities {{x i }} p ≡ [0,1,2,…,p-1] ≡ [0,g 0 ,g 1 ,g 2 ,…,g φ(p)-1 ] mod p and {{y j }} p ≡ [1,2,…,p-1] ≡ [g 0 ,g 1 ,g 2 ,…,g φ(p)-1 ] mod p are in fact tables with two entries. The first line is occupied by the components and the second by weightings:

{{x i }} p ≡ 0 1 2 … p-1 1 1 1 1 1 {{y i }} p ≡ 1 2 3 … p-1 1 1 1 1 1
Besides, the collection of target c abundances is given by :

{{c}} p ≡ 0 1 2 … p-1 #{0} #{1} #{2} #{…} #{p-1}
When R(…) = x = c, we have a bijection between the two weightings.

{{x i c }} p ≡ 0 1 2 … p-1 1 1 1 1 1
However, when R(…) = y = c, a correction is necessary (with still the bijection) : Here, the weighting (on the second line) will increase with the number of variables. The only means to avoid this resulting divergence is to give an unit weight to the whole entity when a new variable is introduced. Hence : Thus, on the basis of the abundance factors evaluated at the sequences p, we get the normalized abundance factors as follows :

P 24/390 Asymptotic sieve

Case modulo p i :

The targets are in the interval [0,p i -1] -for a variable of integers , the cardinal must be divided by p i at sequence p i -for a variable of prime numbers, the cardinal must be divided by p i -1 at sequence p i These ratios apply for each entry of variable : k variables of integers → ratio 1/(p i ) k , m variables of prime numbers → ratio 1/(p i -1) m . To restore the sum to p i , it is necessary and sufficient to carry out a final multiplication of this ratios by p i . Hence the rules (relations 1) : k variables of integers → ratio 1/(p i ) k-1 m variables of prime numbers → ratio p i /(p i -1) m k variables of integers and m variables of prime numbers → ratio p i /((p i ) k (p i -1) m ) Case modulo p i δ :

The targets are in the interval [0,

p i δ -1]
-for a variable of integers , the cardinal be divided by p i δ at sequence p i -for a variable of prime numbers, the cardinal must be divided by p i δ-1 (p i -1) at sequence p i These ratios apply for each entry of variable : k variables of integers → ratio Here, the multiples of p are prohibited with the prime numbers variables, hence the instructions "if" which appear in the loops (alternative to the use of primitive roots).

Comments

The computing times "explode" here with the increase of δ, the more so as the correct solution is given for this parameter ad infinitum. The recourse to a more effective algorithm is essential. It is still the case with Waring sums which we present at exercise 6.

Universality of the asymptotic sieve

The asymptotic sieve is universal for algorithmic usage according to the preceding comments :

-sieve indifferent to the types of variables (of integers or prime numbers), -sieve indifferent to the complexity of the expression (case of the overlapping variables).

The only restrictions, which appeared, are :

-one practical : exponential growth with the desired precision (in the absence of matrix operators) of the computing time of normalized abundance factor, -one essential : the possible absence of a convergence of normalized abundance factor (recurrent oscillations).

For the second point, according to the diophantine equations and/or the choice of the targets, the abundance factor may have no limit : divergence, oscillations (without damping), collapse… The divergence often appears for c = 0 which is a rather particular target. The oscillations are a phenomenon generally due to the distribution of the prime numbers inside the natural integers. Their width falls with the number of variables. The collapse is observed, for example, for a Waring sum with a number of variables lower or equal to the degree of the expression, where we note a sudden diminution of the abundance factor value at some sequence p.

Algorithm of construction of matrices

We will develop, later in this article, a method allowing to overcome the time problem for Waring sums. We present here an algorithmic translation of this method. For the initial data p, n and δ, this algorithm gives a matrix of transformation, centre point of our method. The disadvantage is always the individual treatment. Thus, as for the overlapping loops algorithm, its main interest is to check general results. The computing time of the matrix is about that of the first loop. The matrix allows thus the "instantaneous" calculation of a quasi-infinite number of overlapping loops (after determination of eigenvalues and eigenvectors). We do not give more explanation for the moment because it is necessary to clarify the subject before that. We invite the reader to return later on to the following algorithm after having taken knowledge of the global solution with p, n and δ in literal form.

Let us note that this algorithm is only one outline and can be improved and optimized in computing times. For k = 0 To wi -1 Range("A1").Offset(s, k+2) = Range("A1").Offset(s-1,k+2) *g-pdelta * Int((Range("A1").Offset(s-1,k+2) * g) / pdelta) Range("A1").Offset(s -1000, 0) = Range("A1").Offset(s -1000, 0) & "," & Range("A1").Offset(s, k + 2) Range("A1").Offset(s -1000, 1) = "." Next k Next j Suit6: Next i For j = 1 To wi -1 fac = gdi * fac -pdelta * Int((gdi * fac) / pdelta) t = t + 1 facpi = fac * (p ^ i) Range("A1").Offset(s, t) = facpi -pdelta * Int(facpi / pdelta) Range("A1").Offset(s -1000, 0) = Range("A1").Offset(s -1000, 0) & "," & Range("A1").Offset(s, t) Next j Suit7: If di <= 1 Then GoTo Suit6

If gi = 1 Then GoTo Suit2 Next i Suit1:

'Find deltan deltan = Int((delta -1) / n) pdelta = p ^ delta 'Base line vector (cardinals and targets values) Range("A1").Offset(502, 0) = p ^ (delta -deltan -1) Range("A1").Offset(502, 2) = 0 For i = deltan To 0 Step -1 phidin = (p ^ (delta -i * n -1)) * (p -1) a = n b = phidin Macrochdi din = x gdin =g^ din -pdelta * Int((g ^ din) / pdelta) cardi = (p ^ (i * (n -1))) * din Range("A1").Offset(i + 503, 0) = cardi win = phidin / din fac = 1 For j = 0 To win -1 fac = fac * gdin -pdelta * Int((fac * gdin) / pdelta) Range("A1").Offset(i + 503, j + 2) = fac*p ^ (i *n) Next j Next i 'Base columns vector (« false-cardinls » and targets values) Range("A1").Offset(1002, 2) = 0 Range("A1").Offset(1002, 0) = 1 Range("A1").Offset (2, 0) = ",0" s = 1002 For i = delta -1 To 0 Step -1 t = 1 phidi = (p ^ (delta -i -1)) * (p -1) a = n b = phidi Macrochdi di = x gdi =g^ di -pdelta * Int((g ^ di) / pdelta) wi = phidi / di t = t + 1 s = s + 1 Range("A1").Offset(s, 0) = wi gdipi = gdi * (p ^ i) Range("A1").Offset(s, t) = gdipi -pdelta * Int(gdipi / pdelta) Range("A1").Offset(s -1000, 0) = Range("A1").Offset(s -1000, 0) & "," & Range("A1").Offset(s, t) Range("A1").Offset(s -1000, 1) = "." If wi <= 1 Then GoTo Suit7 fac = gdi For j = 1 To di -1 s = s + 1 Range("A1").Offset(s, 0) = wi 'Matrix elaboration Range("A1").Select ActiveCell.FormulaR1C1 = "=Counta(R [2]C:R[450]C)" x = Range("A1") Range("A1") = "" For i = 2 To x + 1 For j = 2 To x + 1 Range("A1").Offset(i, j) = 0 Next j Next i i = 1001 Suit8: i = i + 1 j = 1 If Range("A1").Offset(i, 2) = "" Then GoTo Suit9 Suit11: j = j + 1 If Range("A1").Offset(i, j) = "" Then GoTo Suit8 x = Range("A1").Offset(i, j) k = 501 Suit10: k = k + 1 q = 1 If Range("A1").Offset(k, 2) = "" Then GoTo Suit11 Suit14: q = q + 1 If Range("A1").Offset(k, q) = "" Then GoTo Suit10 y = Range("A1").Offset(k, q) z = x + y -p ^ delta * Int((x + y) / p ^ delta) r = 1001 Suit12: r = r + 1 s = 1 Suit13: s = s + 1 If Range("A1").Offset(r, s) = "" Then GoTo Suit12 If z <> Range("A1").Offset(r, s) Then GoTo Suit13 Range("A1").Offset(r -1000, i -1000) = Range("A1").Offset(r -1000, i -1000) + Range("A1").Offset(k,0)/Range("A1").Offset(r, 0) GoTo Suit14 Suit9: End Sub Sub Macrochdi() Suit101: x = max(a,b) y = min(a,b) 'Find gcd(a,b), a and b positive given numbers If x = y Then GoTo Suit100 a = x -y b = y GoTo Suit101 Suit100: ' This is a relatively complex case as p divides n. We will see at exercise 6 that the general form of the matrices is simpler when p does not divide n.

Algorithm of Fermat-Catalan

This algorithm allows to calculate Fermat-Catalan matrices described in exercise 9. The algorithm is led modulo p (the approach modulo p δ→∞ can be explored by the reader). A small number of lines of the algorithm are to be adapted to a particular studied case. When d = cm, the matrix is the "standard" matrix.

Expression of the solutions x

Let g j be a solution. The other solutions result immediately and are g j .g k(p-1)/d , k = 0 to d-1. Indeed, g j .g k(p-1)/d is a solution (immediate verification by replacing x in the equation) and all the values from k = 0 to d-1 are distinct by definition of g. The first value of k, after k = 0, such as g k(p-1)/d = 1 is k = d. g j.n =g i.d mod p  j.n = i.d mod p-1  j(n/d)d = i.d mod p-1 j = i(n/d) -1 mod p-1 satisfies this equation. Let us have β = (n/d) -1 mod p-1. Thus x = g j = g i.β mod p. As g i.d = c mod p, g i mod p is necessarily d th root of c and we can write g i = c (1/d) mod p and this root is unique as if g i = g j then i = j mod p-1 by definition of g. Then x = g j = c (1/d) β mod p. Let us pose in addition α = c (1/d) mod p. Then x = α β mod p. Let us rewrite these relations : α = c (1/d) mod p α d = c mod p β = (n/d) -1 mod p-1 β.(n/d) = 1 mod p-1 x = α β mod p ( 9)

LEMMA 1

Let c be a residue of x n = c mod p. Solutions of equation x n = c mod p (c ≠ 0), numbering d = (n,p-1), are : {α β mod p, α β g (p-1)/d mod p , , α β g 2(p-1)/d mod p , , …, α β g (d-1)(p-1)/d mod p , } where α = c (1/d) mod p and β = (n/d) -1 mod p-1 (10)

Euler criterion

If c = g i.d then c (p-1)/d = g i(p-1) = 1 mod p.

If not c = g j g id , then c (p-1)/d = g j(p-1)/d mod p where j is an integer in interval [1,d-1]. By definition of g, p-1 is the first exponent k (≠0) such as g k = 1 mod p. So g j(p-1)/d is necessarily different from 1 mod p.

LEMMA 2

x n = c mod p, c ≠ 0, has d solutions if and only if c (p-1)/d = 1 mod p where d = (n,p-1) (11) 8. Legendre notation Primitive roots and residues

(
If n = 2 then for any odd p, d = 2 and we find the usual forms of the Euler criterion and Legendre notation. Indeed, let us write c = g j mod p (according to notations and remarks made higher). Then :

(p-1) j.(p-1) ( c ) = c 2 = g 2 mod p p 2 Let us have relative primes with 22 (up to p) : {1, 3, 5, 7, 9, 13, 15, 17, 19, 21}. φ(φ(23)) = φ(22) = φ (2).φ(11) = 10.

We have then the On each column (and line besides) figure one and only one primitive roots of 23.

P 33/390

Primitive roots and residues Property 2

If g is primitive root for the prime number p, then g -1 is too.

Indeed, let us consider g 0 a primitive root of p. Then, if x is relative prime with p-1, then p-1-x is also relative prime with p-1 and thus g 0 x and g 0 p-1-x are primitive roots of p. For p = 23, we have the following couples (g,g -1 ) : (5, 14), (7,10), (11,21), (15,20), (17,19). The primitive roots join in φ(φ (p))/2 couples whose product equals 1.

Property 3

No primitive root can be the product modulo p of two other primitive roots. However, common factors are usual, including for a couple (g,g -1 ). (such as for example (15,20) mod 23).

Indeed, let us have g a primitive root. Let us consider the set {1, np 1 , np 2 , …, np k } of the relative primes with p-1 and the set of primitive roots {g 1 , g np1 , g np2 , …, g npk } of p. The exponents npi are all odd and thus never follow themselves in the natural order of integers. Then, let us consider the product of two primitive roots h = g 1 .g 2 . There is a value nps such as g 1 = g 2 nps as g 2 is primitive root. Then h = g 2 nps+1 . As nps+1 cannot be relative prime with p-1 (because an even integer), it follows immediately that h cannot be a primitive root.

n th root of the unit modulo p δ

We examine only the case where p does not divide n.

Case odd p (p > 2)

We like to solve : ω > 0 \ g ω.n = 1 mod p δ

We seek the smallest positive integer ω = ω g,p,n,δ,0 answering the preceding equation. Integer w does not depend on the particular choice of g, primitive root as we will see below. Thus, we can write ω g,p,n,δ,0 = ω p,n,δ,0 .

Let us put in parallel view the same successive terms : {g 0 , g 1 , g 2 , …, g (Φ(δ)-1) }, {g 0 =g Φ(δ) , g 1 , g 2 , …, g (Φ(δ)-1) }, …,{g 0 =g k.Φ(δ) , g 1 , g 2 , …, g (Φ(δ)-1) } called first set and {g 0.n , g 1.n , g 2.n , g 3.n , …, g i.n } called second set. The first terms, respectively of each set, are self-evidently equal. The elements of the first set are then, within the n th step, identical with the terms of the second set. We seek the smallest value ω p,n,δ,0 (ω p,n,δ,0 > 0) for which g ω [p,n,δ,0].n equals 1, that is g k.Φ(δ) . This coincidence is got, thanks to the Chinese theorem, by considering the common divisor d with n and Φ(δ), when the following condition is realised :

ω p,n,δ,0 = Φ(δ)/d n,δ , k = n/d n,δ,0 , d n,δ,0 = (n,Φ(δ)) LEMMA 5 (fundamental lemma 1)
The minimal value of positive integer ω in equation g ω.n = 1 mod p δ equals Φ(δ)/(n,Φ(δ)).

g ω.n = 1 mod p δ => ω p,n,δ,0 = Φ(δ)/(n,Φ(δ)) (23)
Let us have m la multiplicity of factor p in n. Then n = r.p m and as Φ(δ) = p δ-1 .(p-1), d = p min(m,δ-1) . (n.p -min(m,δ-1) ,p δ- 1-min(m,δ- 1) .(p-1)), that is : 1-min(m,δ-1) .(p-1) (24) gcd (n.p -min(m,δ-1) ,p δ- 1-min(m,δ-1) .(p-1))

ω p,n,δ,0 = p δ-
The preceding result applies as long as the elements of p j.n .{g 0.n , g 1.n , g 2.n , …, g (Φ(δ-j)-1).n }mod p δ are not multiple of p δ and thus, since a primitive root g cannot generate multiples of p, as long as p j.n < p δ , that is δ > j.n. Beyond that, the elements are null modolo p δ , i.e. for j ≥ δ/n. Let us pose δn = int((δ-1)/n). The preceding condition is written :

j > δn (34)
The repetition of the residues c = 0 is obtained by counting all the families answering this inequality, that is to say the families : p (δn+1).n .{g 0.n , g 1.n , g 2.n , …, g (Φ(δ-δn-1)-1).n }mod p δ-δn-1 , p (δn+2).n .{g 0.n , g 1.n , g 2.n , …, g (Φ(δ-δn-2)-1).n }mod p δ-δn-2 , … p (δ-1).n .{g 0.n , g 1.n , g 2.n , …, g (Φ(1)-1).n }mod p} mod p δ which cardinal is : Φ(δ-δn-1)+Φ(δ-δn-2)+…+Φ(1) = (p-1).p (δ-δn-2) +(p-1).p (δ-δn-3) +…+(p-1) = p (δ-δn-1) -1

It is necessary to add one unit for target 0. Thus : #{0} = p δ-δn-1 (35) P 36/390

Primitive roots and residues LEMME 7

Let us have d i = (n,Ф(δ-i)) where Ф(δ-i) = p δ-i-1 .(p-1). Equation x n = c mod p δ , p prime odd number, admits d i.n .p i.(n-1) solutions for c like p i.n .g i.d[i.n] and i ≤ δn, p δ-δn-1 solutions for c = 0, if not it has no solution.

We can thus draw the picture of the residues cardinals (relations 36) :

Line x x n = c mod p δ #{c} δn+1 0 p δ-1 .{g 0 , g 1 , … , g Φ(1)-1 } p δ-2 .{g 0 , g 1 , … , g Φ(2)-1 } … p δn+1 .{g 0 , g 1 , … , g Φ(δ-(δn+1))-1 } 0 p δ-δn-1
δn p δn .{g 0 , g 1 , … , g Φ(δ-δn)-1 } p δn.n .{g 0.d[δn.n] , g 1.d[δn.n] , … , g (Φ(δ-δn.n 0.d[i.n] , g 1.d[i.n] , … , 0.d[n] , g 1.d[n] , … , g (Φ(δ-n)/d[n]-1).d[n] } d n .p (n-1) 0 p 0 .{g 0 , g 1 , … , g Φ(δ)-1 } p 0 .{g 0.d[0] , g 1.d[0] , … , g (Φ(δ)/d[0]-1).d [0] } d 0

)/d[δn.n]-1).d[δn.n] } d δn.n .p δn.(n-1) … … … … i p i .{g 0 , g 1 , … , g Φ(δ-i)-1 } p i.n .{g
g (Φ(δ-i.n)/d[i.n]-1).d[i.n] } d i.n .p i.(n-1) … … … 1 p 1 .{g 0 , g 1 , … , g Φ(δ-1)-1 } p n .{g
We adopt the rule of writing v i or v[i] which means that term v i varies between 0 and ω i -1and where ω i = Φ(δ-i)/d i and

d i = (n, Φ(δ-i))
The useful part of this table for the method of the two-dimensional tables is: 0 .gd [0]-1 .g v [0].d [0] } 0

#{p δ } = #{0} p δ-δn-1 #{p δn.n .g 0 .g v[δn.n].d[δn.n] } d δn.n .p δn.(n-1) #{p δn.n .g 1 .g v[δn.n].d[δn.n] } 0 … … #{p δn.n .gd [δn.n]-1 .g v[δn.n].d[δn.n] } 0 ... … #{p j.n .g 0 .g v[j.n].d[j.n] } d j.n .p j.(n-1) #{p j.n .g 1 .g v[j.n].d[j.n] } 0 … … #{p j.n .gd [j.n]-1 .g v[j.n].d[j.n] } = 0 (37) … … #{p 1.n .g 0 .g v[n].d[n] } d n .p n-1 #{p 1.n .g 1 .g v[n].d[n] } 0 … … #{p 1.n .gd [n]-1 .g v[n].d[n] } 0 #{p 0 .g 0 .g v[0].d[0] } d 0 #{p 0 .g 1 .g v[0].d[0] } 0 … … #{p
This general case simplifies if n has no factor equal to p :

p ∤ n (38) Then : d i = (n,Ф(δ-i)) = (n,p δ-1-i .(p-1)) = (n,p-1) = d 0 = d (
39) So that : {g 0.n , g 1.n , g 2.n , …, g (Φ(δ-j)-1).n } ≡ U d.(p^j.(n-1)) times {g 0.d , g 1.d , g 2.d , …, g (Φ(δ-j.n)/d-1).d } mod p δ-j.n (40) and P 37/390

Primitive roots and residues

#{p δ } = #{0} p δ-δn-1 #{p δn.n .g 0 } d.p δn.(n-1) #{p δn.n .g 1 } 0 … … #{p δn.n .gd -1 } 0 ... … #{p j.n .g 0 } d.p j.(n-1) #{p j.n .g 1 } = 0 (41) … … #{p j.n .gd -1 } 0 … … #{p 0 .g 0 } d = #{1} #{p 0 .g 1 } 0 … … #{p 0 .gd -1 } 0
For each line, it is implied that the cardinal of p j.n .g i is also that of p j.n .g i .g u.d . Let us note that if d = 1, then the exponent d-1 equals 0 and lines corresponding to cardinals #{p j.n .g j.n] } do not exist. We have then #{p j.n .g 0 } = #{p j.n .g 1 } = … = #{p j.n .g u(j.n)-1 } = p j.(n-1)

1 .g v[j.n].d[j.n] } to #{p j.n .gd [j.n]-1 .g v[j.n].d[
Case even p (p = 2)

We proceed with the specific characteristic of this case :

Φ(δ) = 2 δ-1 (2-1) = 2 δ-1 d i = (n,Ф(δ-i)/2) = (n,2 δ-i-2 ) for 0≤i≤δ-2
Let us write n = r.2 m where r does not contain factor 2. Then :

d i = (n,Ф(δ-i)/2) = (2 m ,2 δ-i-2 ) = 2 min(m,δ-i-2)
Let us seek the roots x (0 < x < 2 δ -1) of :

x n = 1 mod 2 δ

According to lemma 6, the minimal value of ω (≥0) of the equation (5 ω ) n = 1 mod 2 δ equals 2 δ-2-min(m,δ-2) . It results the following table (relations 42) :

x conditions on n conditions on k,i c #{c} = #{1} 1 odd n 1 1 (-1) i +k.2 δ-min(m, δ-2) even n i = 0, 1 k = 0, 1, …, min(2 m ,2 δ-2 )-1

1 min(2 m+1 ,2 δ-1 )
We use the multiplicity m of factor 2 for even numbers. In case of odd numbers, the value to be retained for m is not 0, but -1 in case of generalization with a single literal formula #{1} = min(2 m+1 ,2 δ-1 ). While recalling this particular point of values jump of m when passing from odd value to even value, a physico-philosophical sentence rushes to our mind :

There is no quantum jump but the antagonism of even and odd numbers.

Let us continue. We have (relation 43

) : {0, 1, 2,… 2 δ -1} ≡ {0, 2 0 .{5 0 , 5 1 , 5 2 , …, 5 Φ(δ)/2-1 }, 2 0 .{-5 0 , -5 1 , -5 2 , …, -5 Φ(δ)/2-1 }, 2 1 .{5 0 , 5 1 , 5 2 , …, 5 Φ(δ-1)/2-1 }, 2 1 .{-5 0 , -5 1 , -5 2 , …, -5 Φ(δ-1)/2-1 }, … 2 j .{5 0 , 5 1 , 5 2 , …, 5 Φ(δ-j)/2-1 }, 2 j .{-5 0 , -5 1 , -5 2 , …, -5 Φ(δ-j)/2-1 }, … 2 δ-2 .{5 0 }, 2 δ-2 .{-5 0 }, 2 δ-1 } mod 2 δ
The reader will note pairing (by opposite values in the equivalence second member) in intermediate lines by opposition to the extreme lines.

Then let us study the residues c cardinals for

x n = c mod 2 δ while writing first, by simple exponentiation of each term of {0, 1, 2,… 2 δ -1} :

{0 n , 1 n , 2 n ,… (2 δ -1) n } ≡ {0, 2 0.n .{5 0.n , 5 1.n , 5 2.n , …, 5 (Φ(δ)/2-1).n }, (-1) n .2 0.n .{5 0.n , 5 1.n , 5 2.n , …, 5 (Φ(δ)/2-1).n }, 2 1.n .{5 0.n , 5 1.n , 5 2.n , …, 5 (Φ(δ-1)/2-1).n }, (-1) n .2 1.n .{5 0.n , 5 1.n , 5 2.n , …, 5 (Φ(δ-1)/2-1).n }, … 2 j.n .{5 0.n , 5 1.n , 5 2.n , …, 5 (Φ(δ-j)/2-1).n }, (-1) n .2 j.n .{5 0.n , 5 1.n , 5 2.n , …, 5 (Φ(δ-j)/2-1).n }, … 2 (δ-2).n .{5 0.n }, (-1) n .2 (δ-2).n .{5 0.n }, 2 (δ-1).n } mod 2 δ

Let us study the line 2 j.n .{5 0.n , 5 1.n , 5 2.n , …, 5 (Φ(δ-j)/2-1).n } mod p δ . Thanks to cyclic group {5 0 , 5 1 , 5 2 , …, 5 Φ(δ-j)/2-1 } mod 2 δ-j , expression 2 j .{5 0 , 5 1 , 5 2 , …, 5 Φ(δ-j)/2-1 } mod 2 δ and then 2 j.n .{5 0.n , 5 1.n , 5 2.n , …, 5 (Φ(δ-j)/2-1).n } mod 2 δ forms complete cycles.

The study of 2 j.n .{5 0.n , 5 1.n , 5 2.n , …, 5 (Φ(δ-j)/2-1).n } mod 2 δ and of {5 0.n , 5 1.n , 5 2.n , …, 5 (Φ(δ-j)/2-1).n } mod 2 δ-j.n is equivalent here. Let us consider the set of numbers {5 0 , 5 1 , 5 2 , 5 3 , …, 5 i } mod 2 δ-j.n . The smallest integer i (i > 0) such as 5 i = 1 mod 2 δ-j.n is, according to lemma 6, i = Φ(δ-j.n)/2. Let us then consider the two sets of successive terms : {5 0 , 5 1 , 5 2 , …, 5 (Φ(δ-n.j)/2-1) }, {5 0 =5 Φ(δ-n.j)/2 , 5 1 , 5 2 , …, 5 (Φ(δ-n.j)/2-1) }, {5 0 =5 2.(Φ(δ-n.j)/2) , 5 1 , 5 2 , …, 5 (Φ(δ-n.j)/2-1) }, …, {5 0 =5 k.(Φ(δ-n.j)/2) , …} mod 2 δ-j.n and {5 0.n , 5 1.n , 5 2.n , …, 5 (Φ(δ-j)/2-1).n },{ 5 (Φ(δ-j)/2).n , …}mod 2 δ-j.n

Still according to lemma 6, we have 5 ω.n = 1mod 2 δ-j.n when ω = (Φ(δ-j.n)/2)/(n,Φ(δ-j.n)/2). Set {5 0.n , 5 1.n , 5 2.n , …, 5 (Φ(δ-j)/2- 1).n } mod 2 δ-j.n contains then (Φ(δ-j)/2)/(Φ(δ-j.n)/2/(n,Φ(δ-j.n)/2)) repetitions, that is : 1) (44) and : {5 0.n , 5 1.n , 5 2.n , …, 5 (Φ(δ-j)/2-1).n } ≡ U d[j.n].(2^j.(n-1)) times {5 0.n , 5 1.n , 5 2.n , …, 5 (ω[j.n]-1).n } mod 2 δ-j.n where d j.n = (n,Φ(δ-j.n)/2) and ω j.n = Φ(δ-j.n)/(2d j.n ) (45) Let us pose n = k.d j.n , then : {5 0.n , 5 1.n , 5 2.n , …, 5 (Φ(δ-j)/2-1).n } ≡ U d[j.n].(2^j.(n-1)) times {5 0.k.d[j.n] , 5 1.k.d[j.n] , 5 2.k.d[j.n] , …, 5 (Φ(δ-j.n)/2d[j.n]-1).k.d[j.n] } mod 2 δ-j.n From (44), we deduce :

#{5 i.n } = d j.n .2 j.(n-
 n, d j.n = (n,Φ(δ-j.n)/2),  i > 0 \ 5 i.n = 5 d[j.n] mod 2 δ-j.n
Then {5 0.k.d[j.n] , 5 1.k.d[j.n] , 5 2.k.d[j.n] , …, 5 (Φ(δ-j.n)/2d[j.n]-1).k.d[j.n] } is the repetition (possibly only once) of a cyclic group which contains necessarily 5 d[j.n] . It follows immediately: 0.d[j.n] , 5 1.d[j.n] , 5 2.d[j.n] , …, 5 (ω[j.n]-1).d[j.n] } mod 2 δ-j.n (46)

{5 0.n , 5 1.n , 5 2.n , …, 5 (Φ(δ-j)/2-1).n } ≡ U d[j.n].(2^j.(n-1)) times {5
We have also :

{(-5) 0.n , (-5) 1.n , (-5) 2.n , …, (-5) (Φ(δ-j)/2-1).n } ≡ U d[j.n].(2^j.(n-1)) times {(-5) 0.d[j.n] , (-5) 1.d[j.n] , (-5) 2.d[j.n] , …, (-5) (ω[j.n]-1).d [j.n] } mod 2 δ-j.n 16 We notice to have not only c = 0 but also c = 2 δn.n as peculiar targets to study. To confirm it, let us return to d j.n and develop its expression :

= 2 δn = 7 x target c #(c) #(k) #(c). #(k) 2 0 +2 15 k, -2 0 +2 15 k 2 0 +2 3 k 2 2 2 13 32768 2 1 +2
n = 3 δn = 5 x target c #(c) #(k) #(c). #(k) (2 0 +2 1 k) 1/3 or 1+2k' 2 0 +2 1 k 2 0 2 15 32768 (2 3 +2 4 k) 1/3 +16384k' 2 3 +2
d j.n = (n,Φ(δ-j.n)/2) = 2 δ-j.n-2 (48)
This term has any meaning only as long as :

δ-j.n-2 ≥ 0 that is j ≤ (δ-2)/n
Hence, it is necessary for us to examine the particular cases where :

j > (δ-2)/n = int((δ-1)/n)+(δ-1)/n-int((δ-1)/n)-1/n = δn+((δ-1)/n-int((δ-1)/n)-1/n) > δn-1 that is j ≥ δn (49)
This result is to be compared with j > δn when we study the cases p > 2.

The repetition of the residues c = 0 is obtained by counting all the families answering to (49), that is : 2 (δn+1).n .{5 0.n , 5 1.n , 5 2.n , …, 5 (Φ(δ-δn-1)/2-1).n }, 2 (δn+2).n .{5 0.n , 5 1.n , 5 2.n , …, 5 (Φ(δ-δn-2)/2-1).n }, … 2 (δ-2).n .{5 0.n } mod 2 δ and 2 (δn+1).n .{(-5) 0.n , (-5) 1.n , (-5) 2.n , …, (-5) (Φ(δ-δn-1)/2-1).n }, 2 (δn+2).n .{(-5) 0.n , (-5) 1.n , (-5) 2.n , …, (-5) (Φ(δ-δn-2)/2-1).n }, … 2 (δ-2).n .{(-5) 0.n } mod 2 δ so also 2 (δ-1).n mod 2 δ and {0} mod 2 δ with cardinal 2.(Φ(δ-δn-1)/2+Φ(δ-δn-2)/2+…+Φ(2)/2)+1+1 = 2 (δ-δn-2) +2 (δ-δn-3) +…+2 1 +1+1 = 2 (δ-δn-1) -1+1 Thus :

#{0} = 2 δ-δn-1 (50)

From ( 49), it remains us to consider the particular case j = δn for the families : 2 j.n .{5 0.n , 5 1.n , 5 2.n , …, 5 (Φ(δ-j)/2-1).n } mod 2 δ 2 j.n .{(-5) 0.n , (-5) 1.n , (-5) 2.n , …, (-5) (Φ(δ-j)/2-1).n } mod 2 δ whose quantity of terms is :

#{2 δn..n } = 2.(Φ(δ-δn)/2) = 2 δ-δn-1 (51) P 40/390
Primitive roots and residues LEMMA 8

Let us have d i = (n,Ф(δ-i)/2) where Ф(δ-i) = 2 δ-i-1 . Equation x n = c mod 2 δ admits d i.n .2 i.(n-1) solutions for c of form 2 i.n .5 i.d[i.n] and i ≤ δn, 2 δ-δn-1 solutions for c = 0, 2 δ-δn-1 solutions for c = 2 δn.n , if not it has no solution.

We can thus draw the picture of the residues cardinals (relations 52): 0.d[i.n] , 5 1.d[i.n] , … , 5 (Φ(δ-i.n)/d[i.n]-1).d[i.n] } 2 i.n .{(-5) 0.d[i.n] , (-5) 1.d[i.n] , … , (-5 0.d[n] , 5 1.d[n] , … , 5 (Φ(δ-n)/d[n]-1).d[n] } 2 n .{(-5) 0.d[n] , (-5) 1.d[n] , … , (-5 0.d[0] , 5 1.d[0] , … , 5 (Φ(δ)/d[0]-1).d [0] } 2 0 .{(-5) 0.d[0] , (-5) 1.d[0] , … , (-5

Line x x n = c mod p δ #{c} ≥δn+1 2 δ-1 .{5 0 } 2 δ-2 .{5 0 , 5 1 , … , 5 Φ(2)-1 } 2 δ-2 .{(-5) 0 , (-5) 1 , … , (-5) Φ(2)-1 } … 2 δn+1 .{5 0 , 5 1 , … , 5 Φ(δ-(δn+1))-1 } 2 δn+1 .{(-5) 0 , (-5) 1 , … , (-5) Φ(δ-(δn+1))-1 } 0 2 δ-δn-1 δn 2 δn .{5 0 , 5 1 , … , 5 Φ(δ-δn)-1 } 2 δn .{(-5) 0 , (-5) 1 , … , (-5) Φ(δ-δn)-1 } 2 δn.n .{5 0 , 5 1 , … , 5 (Φ(δ-δn.n)/d[δn.n]-1) } 2 δn.n .{(-5) 0 , (-5) 1 , … , (-5) (Φ(δ-δn.n)/d[δn.n]-1) } 2 δ-δn-1 … … … … i 2 i .{5 0 , 5 1 , … , 5 Φ(δ-i)-1 } 2 i .{(-5) 0 , (-5) 1 , … , (-5) Φ(δ-i)-1 } 2 i.n .{5
) (Φ(δ-i.n)/d[i.n]-1).d[i.n] } d i.n .2 i.(n-1) … … … 1 2 1 .{5 0 , 5 1 , … , 5 Φ(δ-1)-1 } 2 1 .{(-5) 0 , (-5) 1 , … , (-5) Φ(δ-1)-1 } 2 n .{5
) (Φ(δ-n)/d[n]-1).d[n] } d n .2 (n-1) 0 2 0 .{5 0 , 5 1 , … , 5 Φ(δ)-1 } 2 0 .{(-5) 0 , (-5) 1 , … , (-5) Φ(δ)-1 } 2 0 .{5
) (Φ(δ)/d[0]-1).d[0] } d 0
We adopt the rule of writing v i or v[i] which means that term v i varies between 0 and ω i -1 and where ω i = Φ(δ-i)/d i and

d i = (n, Φ(δ-i)/2)
The useful part of this table in the method of two-dimensional tables is : [0] } 0

#{2 δ } = #{0} 2 δ-δn-1 #{2 δn.n .5 0 .5 v[δn.n].d[δn.n] } or #{2 δn.n .(-5) 0 .(-5) v[δn.n].d[δn.n] } 2 δ-δn-1 #{2 δn.n .5 1 .5 v[δn.n].d[δn.n] } or #{2 δn.n .(-5) 1 .(-5) v[δn.n].d[δn.n] } 0 … … #{2 δn.n .5 d[δn.n]-1 .5 v[δn.n].d[δn.n] } or #{2 δn.n .(-5) d[δn.n]-1 .(-5) v[δn.n].d[δn.n] } 0 ... … #{2 j.n .5 0 .5 v[j.n].d[j.n] } or #{2 j.n .(-5) 0 .(-5) v[j.n].d[j.n] } d j.n .2 j.(n-1) #{2 j.n .5 1 .5 v[j.n].d[j.n] } or #{2 j.n .(-5) 1 .(-5) v[j.n].d[j.n] } 0 … … #{2 j.n .5 d[j.n]-1 .5 v[j.n].d[j.n] } or #{2 j.n .(-5) d[j.n]-1 .(-5) v[j.n].d[j.n] } = 0 (53) … … #{2 1.n .5 0 .5 v[n].d[n] } or #{2 1.n .(-5) 0 .(-5) v[n].d[n] } d n .2 n-1 #{2 1.n .5 1 .5 v[n].d[n] } or #{2 1.n .(-5) 1 .(-5) v[n].d[n] } 0 … … #{2 1.n .5 d[n]-1 .5 v[n].d[n] } or #{2 1.n .(-5) d[n]-1 .(-5) v[n].d[n] } 0 #{2 0 .5 0 .5 v[0].d[0] } or #{2 0 .(-5) 0 .(-5) v[0].d[0] } d 0 #{2 0 .5 1 .5 v[0].d[0] } or #{2 0 .(-5) 1 .(-5) v[0].d[0] } 0 … … #{2 0 .5 d[0]-1 .5 v[0].d[0] } or #{2 0 .(-5) d[0]-1 .(-5) v[0].d
Let us note that if d j.n = 1, then the exponent d j.n -1 equals 0 and lines corresponding to null cardinals do not exist. This case corresponds to n not having any factor equal to 2 :

2 ∤ n (54)
Let us note #{1} the cardinal of solutions of x n = 1 mod 2 δ :

#{1} = #{x / x n = 1 mod 2 δ , x = 0,1, …, 2 δ -1} (55)
Let us note c a residue mod 2 δ , δn = int((δ-1) /n) and m the multiplicity of factor 2 in n. Hence, we get the summary 1) .(#{1}) 2 δ-1-i.n /(#{1})

k = 0, 1, …, 2 δ-δn-1 -1 0 2 δ-δn-1 1 2 δn .(1+2k) k = 0, 1, …, 2 δ-δn-1 -1 2 δn.n 2 δ-δn-1 1 2 i .(1+2.(#{1}).k) 1/n +2 δ-i.(.n-1) /(#{1})k' k = 0, 1, …, 2 δ-1-i.n /(#{1})-1 i = 0 to δn-1 k' = 0 to 2 i.(n-1) .(#{1})-1 2 i.n (1+2.#{1}.k) 2 i.(n-
Making then the sum ∑ #{c}.#{alternatives of c}, we find it equal to 2 δ , which shows that all the solutions are described. The peculiarity of case p = 2 is in the second line of preceding table data (#{2 δn.n } = 2 δ-δn-1 ) which does not exist in the odd p case.

LEMMA 9 (alternative of lemma 8) Equation x n = c mod 2 δ admits 2 i. (n-1) .(#{1}) solutions for c different from 0 and 2 δn.n , 2 δ-δn-1 solutions for c = 0 and c = 2 δn.n , if not it has no solution.

Legendre notation

(modulo p δ )
In p odd case (p > 2), a necessary and sufficient condition so that c is a residue of equation x n = c mod p δ is that c is of form p

i.n .g v[i.n].d[i.n] . Then c/p i.n = g v[i.n].d[i.n]
Let us have d j.n = (n,Φ(δ-i.n)) and i ≤ δn and pose :

Φ(δ) ( c ) = (c/p i.n ) d i.n mod p δ (57) p,δ n,i Then : ( c ) = 0  c = k.p i mod p δ and i > δn p,δ n,i ( c ) = 1 and i ≤ δn  c is a n-residue mod p δ (58) p,δ n,i ( c ) ≠

Displacement modulo p

Case p > 2

We have the bijection (with equality between components) : {1, 2, 3, …, p-1} ≡ {g 0 , g 1 , g 2 , …, g p-2 } mod p It follows, the first equivalence being self-evident and the second being deduced immediately of the aforesaid bijection (relation 59) : {g 0 , g 1 , g 2 , …, g Φ(δ)-1 }mod p δ ≡ {1, 1+p, 1+2p, …, 1+(p δ-1 -1).p}mod p δ {2, 2+p, 2+2p, …, 2+(p δ-1 -1).p}mod p δ {3, 3+p, 3+2p, …, 3+(p δ-1 -1).p}mod p δ … {p-1, p-1+p, p-1+2p, …, p-1+(p δ-1 -1).p}mod p δ ≡ {g 0 , g 0 +p, g 0 +2p, …, g 0 +(p δ-1 -1).p}mod p δ {g 1 , g 1 +p, g 1 +2p, …, g 1 +(p δ-1 -1).p}mod p δ {g 2 , g 2 +p, g 2 +2p, …, g 2 +(p δ-1 -1).p}mod p δ … {g p-2 , g p-2 +p, g p-2 +2p, …, g p-2 +(p δ-1 -1).p}mod p δ Let us take a little different approach for the same result :

P 42/390
Primitive roots and residues {g 0 , g 1 , g 2 , …, g Φ(δ)-1 } ≡ {g 0 , g 0 .g p-1 , g 0 .g 2(p-1) , …, g 0 .g (p^δ-1).(p-1) } {g 1 , g 1 .g p-1 , g 1 .g 2(p-1) , …, g 1 .g (p^δ-1).(p-1) } {g 2 , g 2 .g p-1 , g 2 .g 2(p-1) , …, g 2 .g (p^δ-1).(p-1) } … {g p-2 , g p-2 .g p-1 , g p-2 .g 2(p-1) , …, g p-2 .g (p^δ-1).(p-1) } mod p δ

The second member of the equivalence is read column by column and is self-evidently equal to the first member. In addition, g is not a divisor of p, and according to the Fermat theorem, we have g p-1 -1 = k.p, k an integer. We draw immediately from it relation (59).

By exponentiation of the terms, we have (relation 60) : {g 0.n , g 1.n , g 2.n , …, g (Φ (δ)-1).n } ≡ {g 0.n , g 0.n +p, g 0.n +2p, …, g 0.n +((p δ-1 -1)/n).p} {g 1.n , g 1.n +p, g 1.n +2p, …, g 1.n +((p δ-1 -1)/n).p} {g 2.n , g 2.n +p, g 2.n +2p, …, g 2.n +((p δ-1 -1)/n).p} … {g (p-2).n , g (p-2).n +p, g (p-2).n +2p, …, g (p-2).n +((p δ-1 -1)/n).p}mod p δ Indeed, let us consider (g i +k1.p) n = g i.n +k1'.p and (g i +k2.p) n = g i.n +k2'.p mod p δ . If k1 ≠ k2 then necessarily k1' ≠ k2'. Indeed, let us reason by absurdity. Let us pose k1 ≠ k2 and k1' = k2'. Then, by the remarkable identity, (g i +k1.p) n -(g i +k2.p) n = (k1-k2).p.∑(g i +k1.p) 2n-x .(g i +k1.p) x , integer x varying 0 to 2n, so that after few rearrangements (g i +k1.p) n -(g i +k2.p) n = (k1-k2).p.(g 2n.i +k.p), k being a composite expression of k1, k2 and p. With primitive root g, g 2n.i is not a multiple of p, nor g 2n.i +k.p. Hence, this term cannot be null what suffices for our argument. It rises immediately.

{(g i ) n , (g i +p) n , (g i +2.p) n , ..., (g i +(u(k1)-1).p) n } ≡ {(g i ) n , (g i ) n +p, (g i ) n +2.p, ..., (g i ) n +(u(k1)-1).p}

This proves relation (60).

Let us consider the equivalence we got higher : {g 0.n , g 1.n , g 2.n , …, g (Φ(δ-j)-1).n } ≡ U d[j.n].(p^(n-1)) times {g 0.d[j.n] , g 1.d[j.n] , g 2.d[j.n] , …, g (Φ(δ-j.n)/d[j.n]-1).d [j.n] } mod p δ-j.n

We have then (relation 61) :

LEMMA 10 {g 0.d[j.n] , g 1.d[j.n] , g 2.d[j.n] , …, g (Φ(δ-j.n)/d[j.n]-1).d[j.n] } ≡ { g 0.d[j.n] , g 0.d[j.n] +p, g 0.d[j.n] +2p, …, g 0.d[j.n] +(p δ-1-j.n p} { g 1.d[j.n] , g 1.d[j.n] +p, g 1.d[j.n] +2p, …, g 1.d[j.n] +(p δ-1-j.n p} { g 2.d[j.n] , g 2.d[j.n] +p, g 2.d[j.n] +2p, …, g 2.d[j.n] +(p δ-1-j.n p} … {g (p-2).d[j.n] +p, g (p-2).d[j.n] +2p, …, g (p-2).d[j.n] +(p δ-1-j.n p}mod p δ-j.n Let us see that with an example. Let us have p = 3,g=2 and δ = 4 (d = 2) : {g 0 , g 1 , g 2 , …, g Φ( 4)-1 } mod 3 The second members show the importance of the heads of lines {1, 2, …, p-1} and {1 n , 2 n , …, (p-1) n }. We will use this result at exercise 6.

Case p = 2

We can proceed in the same way but make it short here (to shorten our text).

Decomposition of p

We seek a condition on the couple (i,j) when p is broken up into two terms as follows (g primitive root) : p = g i+u.d +g j+v.d mod p (62)

Here, d is a constant (integer), u and v arbitrary integers.

The preceding relation is equivalent to 0 = g i+u.d +g j+v.d mod p, then g i+u.d = -g j+v.d mod p, that is also g i-j+(u-v).d = -1 mod p. Let us replace u-v by integer w and -1 by g (p-1)/2 . Thus : We will examine other decompositions of p, which are essential within our article, in particular at exercises ( 5) and ( 9).

g i-j+w.d = g (p-

Alternative definition of the concept of primitive root mod p

We use Legendre notation for this elegant and fundamental definition. Let us have g a primitive root. As g (p-1)/2 = -1, we thus have :

LEMMA 11 g primitive root  ( g ) = -1 (64) p 2
The reciprocal is immediate. This lemma is used at exercise 10 page 252.

Generalization of the concept of primitive root

This part of our article does not affect the other chapters. We however report it, not having found trace of it in the mathematical literature for the most general cases (n unspecified). The reader will retain of it only what it will think useful to other ends.

Let us have a positive integer with a ∤ n

There is an integer exponent m > 0 such as : a m = 1 mod n

In function of a, the smallest number m is given by the table :   n m p Ω(a,p) p 2 Ω(a,p).if(a = g p mod p 2 , 1, p) … … p k Ω(a,p).if(a = g p mod p 2 , 1, p).if(a = g p^2 mod p 3 , 1, p)…if(a = g p^(k-1) mod p k , 1, p)

Here g is a primitive root of p. If a is a primitive root of p, we have Ω(a,p) = p-1, if not the expression divides p-1 :

Ω(a,p) \ p-1 (65)
P 44/390

Primitive roots and residues

The behaviour of Ω as function of a and p is a great "unknown" of Number Theory. However, if we admit to have elucidated it, the remainder follows.

Let us pose : Ψ(a,p k ) = if(a = g p mod p 2 , 1, p).if(a = g p^2 mod p 3 , 1, p)…if(a = g p^(k-1) mod p k , 1, p) (66) Ψ(a,p k ).Ω(a,p) is the smallest number such as a m = 1 mod p k . It is maximum for Ψ(a,p k ).Ω(a,p) = (p-1).p k-1 = Φ(p), traditional result. The number a could be described as a primitive root in this case (m maximum).

Let us suppose n of the form : n = ∏ p i ki i = 1 to r Then, the smallest exponent such as a m = 1 mod n is given, using the lowest common multiple of a list, by : lcm(Ω(a,p 1 ), Ω(a,p 2 ), …, Ω(a,p r )) . ∏ Ψ(a,p i ki ) (67) i = 1 to r Number a could be described as primitive root mod n when m, checking a m = 1 mod n, is maximum, i.e. if and only if :

m min = lcm(p 1 -1, p 2 -1, …, p r -1). ∏ p i ki-1 (68) i = 1 to r

Test of a prime number

By successive products of random integer a > 1, we get a m = 1 mod n. We have equivalence :

m \ n-1  n is prime
This rises intuitively from the relation ( 67), but the proof is not undertaken here.

Factorisation

In the case of a product of two prime factors n = p.q, we have for any integer : a lcm(Ω(a,p), Ω(a,q)) = 1 mod n (69)

As Ω(a, p) divides p-1, if we find quickly (few tests), values of m (and/or a) such as a m = 1 mod n, we can reconstitute by products of the factors potentially p-1 or q-1, which by tests (and errors) makes it possible to factorize n. This problem remains difficult.

EXERCISE 4 : CIRCULANT MATRICES

Circulant matrices are the essential tools to the study of the abundance factors. Complex exponentials, which rule among them, will rapidly become our friends in that exercise. Circular matrices are fundamental to Number Theory although this fact is rarely (or never) viewed and described as such. Nevertheless, complex exponentials showing-up is not without remembering the "great sieve".

Types of matrices

The mathematical literature distinguishes two types of matrices : the group of the right circulant matrices (indice d) and groups it left circulant matrices (indice g), respectively :

c 0 c n-1 … c 1 [CI d ] = [CI d (c 0 ,c 1 , …,c n-1 )] = c 1 c 0 … c 2 (1) … … … … c n-1 c n-2 … c 0 and c 0 c 1 … c n-1 [CI g ] = [CI g (c 0 ,c 1 , …,c n-1 )] = c 1 c 2 … c 0 (2) … … … … c n-1 c 0 … c n-2

Eigen-elements

Right circulant matrices

Circulant matrices are diagonalisable (non-zero matrices). Indeed, the determinant of right circulant matrices, which is calculated thanks to Vandermonde determinant, is indicated for example in [16] p 262. We have :

n-1 n-1 Det([CI d ]) = ∏ ∑ c t .e -2πi.t.v/n (3) v = 0 t = 0
Hence, we deduce eigenvalues σ of these matrices [CI d, t,1 ] = [C(c 0 = 0, …, c t-1 = 0, c t = 1, c t+1 = 0, …, c n-1 = 0)], where only one diagonal is non-null and equal to 1 :

n-1 Det([CI d,t,1 ]-σ[I]) = ∏ (e -2πi.t.v/n -σ) = 0 v = 0
Hence the eigenvalues, with integer v varying 0 to n-1, and t a constant for the moment :

σ v, t = e -2πi.t.v/n (4)
The eigenvalues of the circulant matrices [CI Let us consider (X) the eigenvectors (x 1 , x 2 , …, x n ), column vector. We have :

([CI d,t,1 ]-σ v, t .[I])(X) = 0 Hence, it results immediately : x t+1 -e -2πi.t.v/n .x 1 = 0 x t+2 -e -2πi.t.v/n .x 2 = 0 x t+k -e -2πi.t.v/n .x k = 0 (5) ... x t+n -e -2πi.t.v/n .x n = 0
Indices of x are modulo n here (that is within 0 and n-1). Let us form symmetric matrix (in which we will recognize a Discrete Fourier Transform) : 1.(n-1) [P] = (1/n Lines and columns of matrix [P] are indexed (r,s) starting with 0.

1 1 1 ... 1 1 (e 2πi/n ) .1.1 (e 2πi/n ) .1.2 ... (e 2πi/n ) .
We have then :

[P(r,s)] = (1/n 1/2 ).[(e 2πi/n ) .r.s ]
Hence the transposed of the conjugated matrix :

[ t P*(t,s)] = (1/n 1/2 ).[(e -2πi/n ) .t.s ]
The product of these matrices is then :

n-1 [P(r,s)].[ t P*(t, s)] = (1/n). [ ∑ e (2πi/n).s(r-t) ] s = 0
If r = t, the sum ∑e (2πi/n).s(r-t) equals n. When r ≠ t, e (2πi/n) on one hand and e (2πi/n).s on the other hand (for any integer s) are geometrically uniformly distributed around the origin. The sum ∑e (2πi/n).s(r-t) is then equal to 0. Thus [P(r,s)].

[ t P*(t, s)] = [I]
where [I] is the identity matrix of dimension n.

Thus [P], symmetric matrix, is also unitary matrix :

[P -1 ] = [ t P*] (7)
The component in (r,s) of [P] after multiplying by n 1/2 is :

x r, s = e (2πi/n).r.s
We have also : x r+k, s = e (2πi/n).(r+k).s = e (2πi/n).r.s .e (2πi/n).k.s = e (2πi/n).r.s .x k, s

We deduce relation ( 5) when r = t and s = v. Thus :

The matrix [P] is common change of basis matrix to all right circulant matrices [C d,t,1 ] whatever the choice of t.

Having written [CI d,t,1 ] = [P].[σ t ].[P -1 ], we have then :

[CI d ] = ∑ c t .[CI d,t,1 ] = [P].( ∑c t .[σ t ] ).[P -1 ] (8) t = 0 to n-1 t = 0 to n-1
The eigenvalues of [CI d,t,1 ], the below matrix [ ] being a diagonal matrix, are thus (first component v = 0, second v = 1, etc.) :

∑c t .[e -2πi.t.v/n ] (9)
t = 0 to n-1

Let us develop expression (8) to find the component in (r,s) :

CI d (r,s) = ∑ P(r,w) .( ∑c t .σ t (w,w) ) . P -1 (w,s)

w = 0 to n-1 t = 0 to n-1
That is using [P ).e (-2πi/n).w.s w = 0 to n-1 t = 0 to n-1

So that : CI d (r,s) = ∑ ∑c t .e (2πi/n).w.(-t+r-s) (10) w = 0 to n-1 t = 0 to n-1

Left circulant matrices

We can get a left circulant matrix (or right-hand side) starting from a right circulant matrix (or left respectively) by the transformation : 0 0 0 1 … 0 0 0 0 1 0 … 0 0 0 1 0 0 … 0 0 With [σ m ] the matrix of the eigenvalues and [P] a matrix of eigenvectors of the circulant matrix [CI m ], we get :

1 0 0 0 … 0 0 0 0 0 0 … 0 1 0 0 0 0 … 1 0 [CI g (1,0, …,0)] = … … … … … … … (11)
[CI m ] = [P].[σ m ].[P -1 ] Thus : [CI m ].[CI g (1,0, …,0)] = [P].[σ m ].[CI g (1,0, …,0)].[P -1 ]
We preserve the change of basis matrix [P]. However, the components of the "eigenvalues' matrix" are not aligned on the principal diagonal in this diagram. We will call this matrix, if it is necessary, the left eigenvalues matrix.

EXERCISE 5 : MODULO P i ABUNDANCE FACTORS FOR WARING SUMS

We seek here the abundance factors of Waring (or Fermat) sums x 1 n + x 2 n + … + x k n = c and y 1 n + y 2 n + … + y k n = c in the case of a modulo p i approach. The correction to be brought between the modulo p i approach and the modulo p i δ approach is studied in exercise 6. For the reader, this exercise constitutes a first "big piece". If we get Vinogradov result for three variables, with only few lines, the general case offers a crowd of developments associating conjectures and unquestionable results.

Case of the hyperplanes unit mesh

x 1 + x 2 + … + x k = c
(1) Here x i are positive integers. Let us start with :

x 1 + x 2 + … + x k = c mod p i p i describes the set of prime numbers and each variable is replaced by its representative [0, 1, 2,…, p i -1]. We are thus brought to build a table with k dimensions and of size p i following each axis of freedom. The elements of this table are obtained by modulo p i sums in accordance with the operators (+) present in the equality (1). We then count the occurrences of each number (between 0 and p i -1) in the table. We find afterwards, by normalization, the abundance factor for each target c. In present case, the enumeration of occurrences is self-evident : each number [0, 1, 2,…, p i -1] appears the same number of times and, whatever k and c, conduct to only one result : Fan(c) = 1.

Case of linear hyperplanes unit mesh

a 1 x 1 + a 2 x 2 + … + a k x k = c (2) Starting from : a 1 x 1 + a 2 x 2 + … + a k x k = c mod p i
With a j one of the coefficients, let us write a 1 x 1 + a 1 x 1 + a 2 x 2 + … + a k x k = c -a j x j mod p i , if p j \a i , then a j x j « disappears » in mod p i . Thus, in case p j \a i the targets multiple of p i alone have non-null cardinals (equal to p i ). Factor Fan(c) is written (p i describing 2,3, 5, 7 , 11, +∞) : П p i . П 0 (3) p i \c, p i \a i p i \c p i ∤a i , These terms of abundance are conditional. If the condition is not realised, the term equals 1.

Case of the hyperplanes logarithmic mesh

y 1 + y 2 + … + y m = c (4)
Here y i are positive prime numbers. Proceeding in the same way, the n-dimensional arrays are built starting from the representatives [1, 2,…, p i -1]. The elements of the tables are obtained by modulo p i sums in accordance with the operators (+) present in the equality (4). Then, we count the occurrences of each number (numbers ranging between 0 and p i -1).

Let us start with some examples.

Case p i = 2, m = 1 Matrix (1dimension) 1 Elements Occurrences 1 0 0 1 1 Case p i = 2, m = 2 Matrix (2 dimensions) 1 Elements Occurrences 1 0 0 1 1 0 Case p i = 2, m = 3 Matrix (3 dimensions) +1 1 Elements Occurrences 1 1 0 0 1 1 Case p i = 3, m = 1 Matrix 1 2 Elements Occurrences 1 2 0 0 1 1 2 1 Case p i = 3, m = 2 Matrix 1 2 Elements Occurrences 1 2 0 0 2 2 0 1 1 1 2 1 Case p i = 3, m = 3 Matrix +1 1 2 1 0 1 Elements Occurrences 2 1 2 0 2 1 3 +2 1 2 2 3 1 1 2 2 2 0
These examples make us consider the following hypothesis as well for odd p i as for p i = 2 :

For m odd, if element 0 appears x times then the other elements are present x + 1 times; for m even, if element 0 appears x+1 times then the other elements are present x times. As the matrix of dimension m at sequence p i contains exactly (p i -1) m elements, it follows :

For m odd : x+(p i -1)(x+1) = (p i -1) m , so that x = ((p i -1) m -(p i -1))/p i and x + 1 = ((p i -1) m +1)/p i For m even : x+1+(p i -1)(x) = (p i -1) m , so that x = ((p i -1) m -1)/p i and x + 1 = ((p i -1) m +(p i -1))/p i These two combined cases are written : Occurrence of element 0 : ((p i -1) m +(-1) m (p i -1))/p i (i) Occurrence of other elements : ((p i -1) m -(-1) m )/p i (ii) These is the recurrence hypothesis we will demonstrate now.

Proof :

For m = 1, the occurrence of element 0 is ((p i -1) 1 +(-1) 1 (p i -1))/p i = 0 and that of the other elements is ((p i -1) 1 -(-1) 1 )/p i = 1 in conformity with representative [1, 2, …, p i -1] construction. Let us pose y the occurrence of element 0 and x the occurrence of other elements. The account of the elements of the multi-dimensional table is got according to the following diagram :

Matrix Vector dim m+1 0 1 2 … p i -2 p i -1 Counting at m y x x x x x Counting at m+1 Increment by 1 x y x x x x Increment by 2 x x y x x x … … … … … … … Increment by p i -2 x x x x y x Increment by p i -1 x x x x x y
The number of occurrences of element 0 is thus (p i -1)x, that is (p i -1)((p i -1) m -(-1) m )/p i from (ii), that is also ((p i -1) m+1 +(-1) m+1 (p i -1))/p i what confirms (i).

The number of occurrences of the other elements is (p i -2)x+y, that is (p i -2)((p i -1) m -(-1) m )/p i +((p i -1) m +(-1) m (p i -1))/p i from (i) and (ii), that is also ((p i -1) m+1 -(-1) m+1 )/p i what confirms (ii). The abundance factors are obtained for c by a product according to whether p i divides c (case corresponding to element 0 or not (case corresponding to the other elements).

Normalization is obtained by division by the number of elements of the matrix (that is (p i -1) n ) and multiplication by the number of elements, or p i . All made operations, we get :

Fan(c) = П (1- (-1) m ) П (1- (-1) m-1 ) (5) (p i -1) m (p i -1) m-1 p i ∤ c p i ∖ c
(c) = П (1-(-1) m ) П (1- (-1) m-1 ) (6) (p i -1) m-1 (p i -1) m (1- (-1) m ) p i p i ∖ c (-1) m
In the case m = 2, corresponding to Goldbach conjecture, we get :

Fan(c) = = П (1- 1 ) П (1+ 1 ) П (1- 1 ) П ( p i -1 ) (7) (p i -1) 2 (p i -1) (p i -1) 2 p i -2 p i ∤ c p i ∖ c p i p i ∖ c
The second form of the result is that given, for example, in [2] p 13 for "Goldbach conjecture" (the form was not developed by Goldbach) and in [3] for De Polignac conjecture. This interests p+q = c and p-q = c. The same form of results for the two conjectures is explained simply by the arguments concerning the signs evoked in the pages of introduction of our article.

In case m = 3, corresponding to Vinogradov result, we get with (5) :

Fan(c) = = П (1+ 1 ) П (1- 1 ) П (1+ 1 ) П (1- 1 ) (8) (p i -1) 3 (p i -1) 2 (p i -1) 3 p i 2 -3p i +3 p i ∤ c p i ∖ c p i p i ∖c
This result is found, in the second form, in [2] (page 17) and confirms the suggested approach. In [2] (page 66), we find an expression of Fan(c) for m positive integer in the form :

S m (c) = ∑ μ m (p i ) c pi (c) φ m (p i ) p i
Here φ is Euler totient function, μ Mobius function and c pi Ramanujan function defined by :

c pi (c) = ∑ e 2πifc/pi (f,p i )=1
In [2], it is proposed to seek the preceding expression in the form of product of Euler, what we do here. We have according to [11] p 213 :

c pi (c) = { p i -1 if p\c -1 if p i ∤ c
From [11] p 325, we get in the case where ∑ f(n) converges absolutely, the equality :

∞ +∞ ∑ f(n) = ∏ (1+f(p)+f(p²)+ …) = ∏ (1+∑ f(p k )) n=1 p p k=1
Absolute convergence of S m (c) is proven in [11] for m = 3, method which spreads without particular difficulty. Using the above equality, we have then :

+∞ S m (c) = ∏ (1+∑ μ m (p i k ) c pi (c) ) φ m (p i k ) p i k=1
By definition, Mobius function is 1 for 1, 0 for numbers divisible by a perfect square and (-1) w(n) for integers n with w(n) distinct prime factors. Thus, for any prime number p i , we have :

μ(p i ) = -1 μ(p i k ) = 0 if k > 1
(c) = ∏ (1+ μ m (p i ) c pi (c) ) = ∏ (1+ (-1) m if(p i \c,p i -1,-1) ) φ m (p i ) (p i -1) m p i p i Hence: S m (c) = П (1- (-1) m ) П (1- (-1) m-1 ) = Fan(c) (9) (p i -1) m (p i -1) m-1 p i ∤ c p i ∖ c
We thus get a coincidence between the normalized abundance factor Fan(c) and the singular series S m of target c. Let us note also that, when n increases, we attend progressive equalization of normalized factors, which is an awaited and natural result. Indeed, the more the number of independent variables in ( 4) is large, larger will be the "degree of freedom" towards target c. All normalized factors tend then towards 1.

Numerical applications

The abundance factor varies according to the finite number of the divisors of c. Expression П(1-(-1) m /(p i -1) m ) can be considered as the fixed part of this factor. We give an approximation of it pending on the sequence to which we develop the calculations (p = 2 excluded). Except traditional case m = 2, the other cases converge quickly towards a limiting value. One thus needs little calculations to get a good numerical approximation. When m increases, the asymptotic value 1 is alternatively approximate by the top or bottom better and better.

Case of linear hyperplanes logarithmic mesh

a 1 y 1 + a 2 y 2 + … + a m y m = c (10) This case admits many common points with the preceding one.

Vectors [1, 2,…, p i -1], supports of construction of the multidimensional tables, identical previously, are replaced by vectors a

i .[1, 2,…, p i -1]. If p i does not divide a i , then a i [1, 2, …, p i -1] = [a i , 2a i , …, (p i -1)a i ] is an elementary permutation of [1, 2,…, p i -1] (if not one
of the elements of [a i , 2a i , …, (p i -1)a i ] would be twice, then there would be n and m positive integers inferior to p i such as n.a i -m.a i = 0 mod p i , then there would be n positive integer inferior to than p i such as n.a i = 0 mod p i , that is to say still n = 0 mod p i or a i = 0 mod p i , the first proposition being contradictory to conditions on n, the second proposition to hypothesis on a i ) : the account is identical that in absence of coefficient a i . The reader will refer then to the preceding paragraph giving the result for y 1 +y 2 +…+y m = c.

If p i divides a i then a i [1, 2, …, p i -1] = [0, 0, …, 0] mod p i : at sequence p i , the vector [0, 0… 0] does not modify the value of deduced elements. The number of occurrences of element 0 is thus ((p i -1) m -(-1) m (p i -1) 2 )/p i and of the other elements is ((p i -1) m +(-1) m (p i -1))/p i . Proof of this result is done by recurrence without difficulty by a same reasoning used higher. We get immediately the general expression of the normalized abundance factors Fan(c) :

П (1- (-1) m ) П (1- (-1) m-1 ) П (1- (-1) m-2 ) П (1- (-1) m-1 ) (p i -1) m (p i -1) m-1 (p i -1) m-2 (p i -1) m-1 p i ∤c, p i ∤a i p i \c, p i ∤a i p i \c, p i \a i p i ∤c, p i \a i
That is also :

П (1- (-1) m-2 ) П (1- (-1) m-1 ) П (1- (-1) m ) (11) (p i -1) m-2 (p i -1) m-1 (p i -1) m p i \c, p i \a i or(et(p i ∤c,p i \a i ),et(p\c,p∤a i )) p i ∤c, p i ∤a i
In case m = 2, corresponding to generalized Goldbach conjecture, we get :

П p i (p i -2) П ( p i ) П (0) (12) (p i -1) 2 p i -1 p i ∤c , p i ∤a i p i \c p i ∤c , p i \a i
The last multiplicative term is conditional. If the condition is not satisfied, the term equals 1.

Case of the integer coefficients not all positive, unit or logarithmic mesh

Change of sign of a coefficient a i does not affect the abundance factor. Indeed a i [1, 2, …, p

i -1] = -a i [-1, -2, …, -p i -1] = - a i [p i -1, p i -2, …, 2, 1].
A change of sign being equivalent to a simple permutation, this does not affect the enumeration of the occurrences of the elements in set [1, 2, …, p i -1] modulo p i .

Case of the hypersurfaces unit mesh

x 1 n + x 2 n + … + x k n = c ( 13
)
The enumeration is carried out modulo p i according to the now familiar preceding method. p i describes the set of prime numbers and each variable x i n is replaced by its representative [0 n , 1 n , 2 n , …, (p i -1) n ]. We are thus brought to a table with k dimensions and of size p i following each axis. The elements of these tables are obtained by modulo p i sums in accordance with the operators (+). Then, we count in the table the number of occurrences of each number (between 0 and p i -1). Then, by normalization, we get the abundance factor for target c.

We explore first the case p i = 2 which is particular. This case amounts forming multidimensional tables with a generating vector {0,1} for each axis. The reader will verify easily that we get :

#{0} = 2 k-1 and #{1} = 2 k-1 if p i = 2 (14)
The use of the method of the overlapping loops exposed to exercise 2 shows the restricted number of distinct cardinals that we get. Distinct cardinals are no more than d+1 where d = gcd (n,p-1). To arrive at this result, we start with the argument developed in exercise 3 on residues of x n mod p i giving immediately the table of the number of distinct cardinals for k = 1, that is 3 distinct cardinals :

Target c Cardinal (for k = 1) {0} 1 {gd, g 2d , g 3d , …, g (pi-1) } mod p i d = (n,p i -1) Others among {0, 1, 2, …, (p i -1)} 0

Note : It would be more rigorous to indice d (d i ) like p i . It is not done in general for simplification of writing.

In exercise 3, we recalled the equivalence, except for a permutation, between {1, 2, …, (p i -1)} and {g 0 .{g d , g 2d , g 3d , …, g (pi- 1) }, g 1 .{g d , g 2d , g 3d , …, g (pi-1) }, g 2 .{g d , g 2d , g 3d , …, g (pi-1) }, …, g (d-1) .{g d , g 2d , g 3d , …, g (pi-

… card d-1 0 g 0 .g d g 0 .g 2d … g 0 .g (pi-1) g 1 .g d g 1 .g 2d … g 1 .g (pi-1) g d-1 .g d g d-1 .g 2d … g d-1 .g (pi-1) card = 1 0 card = d g d g 2d … g (pi-1)
Our recurrence hypothesis, implicit in the table above, is that a set of type g r .{g d , g 2d , g 3d , …, g (pi-1) } = {g r .g d , g r .g 2d , g r .g 3d , …, g r .g (pi-1) } has, for each one of its cardinals, the same value card r+1 . It is necessary then to verify the property while passing from k to k+1 recalling that it is indeed true for k = 1 (self-evident case). We give first an example to clarify the proof. Let us have p = 13 and n = 9. The values which follow are modulo 13. d = (9,12) = 3. g= 2 is a primitive root of p = 13. Then, we get table 2 : card 0 card 1 card 2 card 3 0 g 0d .g 0 g 1d .g 0 g 2d .g 0 g 3d .g 0 g 0d .g 1 g 1d .g 1 g 2d .g 1 g 3d .g 1 g 0d .g 2 g 1d .g 2 g 2d .g 2 g 3d .g 2 Thus, there is a matrix of transformation which allows calculating by iteration, to order k, the cardinals of each target modulo p i -1. This matrix is:

card' 0 1 12 0 0 card 0 card' 1 = 3 1 3 6 card 1 card' 2 0 3 7 3 card 2 card' 3 0 6 3 4 card 3
In general case, we observe that table 1 is formed of the following parts: the first line, the first column except the element of the first line, the first square (under card 1 ) and the other squares (under card 2 to card d-1 ). We are thus brought to consider the cases corresponding to each part of the table, in order to get corresponding contributions for given target c, contributions which we must then add. We call generating vector of table 1 following x, respectively y, the elements 0 and g u.d located on the left first column of this table and the elements 0 and g y .g v.d located above the first line of this table :   Case 1 : First line of table 1. This line, where we add 0, simply reproduces, with cardinal 1, each element of the generating vector following y. The contribution to the cardinals is the identity : card' r = card r (15) Case 2 : First column of table 1, except line 1, already mentioned above.

The contribution is d.card 0 for each target of the form g u.d and is 0 for each target of the form g x .g u.d with x ≠ 0. With adopted rule on cardinals' indices, we act on card' 1 , each target g u.d being concerned in an identical way, as :

card' 1 = d.card 0 card' i = 0, i ≠ 1 (16)
Case 3 : First square of table 1 right of the first column and following squares.

Here a target is written c = g u.d + g y .g v.d mod p i and we seek, for y given, the couples (u,v) answering that equation. That is to say s(c) = #{(u,v)}. Then, for the target c.g d , we have c.g d = g (u+1).d + g y .g (v+1).d mod p i and the number of couples (u,v) solutions is identical to the number of preceding couples and this with the same contributions by each square of the table. Thus, it suffices to solve respectively each case, possibly distinct, of type c = g r , r = 0 to d-1, to get respectively the identical cases c = g r .g d , c = g r .g 2d , …, g r .g ((pi-1)/d-1).d mod p i , with cardinal (p i -1)/d (the case c = g r included). Hence the contribution :

d-1 card' r+1 = ∑ d.(#{(u,v) / g r = g u.d + g y .g v.d mod p i }.card y+1 ) (17) y = 0
The argument is identical for c = 0 with multiplicity equal to (p i -1)/d to be multiplied by d, that is p i -1 what imposes the existence of only one non-null component on the first line of the matrix. We verify this point on the example. The equation of problem is g u.d + g y .g v.d = 0, u and v varying independently from 1 to (p i -1) /d, that is also g y .g v.d = -g u.d , then g y+(v-u).d = -1 mod p i . However as g is a primitive root of the sequence p i , we have necessarily g (pi-1)/2 = -1 mod p i . It follows y+(v-u).d = (p i -1)/2 mod p i -1, then y = (p i -1)/2+(v-u).d mod p i -1. However d divides p i -1. We deduce immediately y = (p i -1)/2 mod d. This means that matrix column n° y+2 carries on the first line all the solutions and the cardinal equals p i -1 at indicated position y+2. Hence the contribution (indice is y+1 here) : card' 0 = (p i -1).card (pi+1)/2 (18)

Addition of contributions 1, 2 and 3 gives the following matrix. This one, noted [A], is written in a "shortened" way according to the numbers of lines and columns (x,y) of the matrix (relation 19) as :

[A(x,y)] = [if((x,y) = (2,1), d, if((x,y) = (1,(p i +1)/2 mod d), p i -1, if((x,y) > (1,1), #(u,v).d / g x-2 = g u.d + g y-2 .g v.d mod p i , 0)))]+[ I ]
where x is indice of line 1 to d+1, y is indice of column 1 to d+1, {u,v} integers describing [0,(p

i -1)/d-1] 2 , d = (n,p i -1), [I] is identity matrix dimension d+1.
x axis is directed to the bottom, y axis to the right. By convention also, (x,y) > (1,1) means x > 1 and y > 1.

Writing the expression of the cardinals as (u k , v k , w k , x k ), we get :

k u k 1 v k 0 (20) w k = A 0 … … x k 0
As the new cardinals express versus the old ones by means of a dimension d+1 matrix evolves that d+1 is the maximum number of distinct cardinals. For k = 1, the result of ( 19) is the first column of A which, referring to the construction of A, is the column vector generator of A. This is in accordance with the result of the first step recurrence hypothesis.

The correspondence between the cardinals values results from matrices construction (relations 21) : 1) .g 0 g d .g 1 g 2d .g 1 … g (pi-1) .g 1 g d .g d-1 g 2d .g d-1 … g (pi-1) .g d-1

# u k = card 0 v k = card 1 w k = card 2 … x k = card d c 0 g d .g 0 g 2d .g 0 … g (pi-
We draw this table at selected sequence k and carry out ascending sorting of the second line (having care to attach the first line to it). We will note T(u k , v k , w k , x k ) the function corresponding to this sorting and T(c, u k , v k , w k , x k ) the value corresponding to the target c. This notation will be useful to us later in our matter.

The determination of abundance factors for the hypersurfaces x 1 n + x 2 n + … + x k n = c can then be carried out with an algorithm based on the contributions developed previously. The important fact is that there is here no more exponential increase in the processing time with k (unlike in the case of overlapping loops method).

Case of the hypersurfaces, logarithmic mesh

y 1 n + y 2 n + … + y m n = c (22)
Here y i are positive prime numbers.

The case is very similar to the previous one. The representative is no more [0 n , 1 n , 2 n , …, (p i -1) n ] but [1 n , 2 n , …, (p i -1) n ].

The absence of {0} makes that the first column of previous example table disappears (column corresponding to card 0 ).

Thus : {B] = [A]-[I] (23)

where [A] is previously obtained matrix, [B] is the matrix which interests us here and [I] the identity matrix of dimension d+1. Thus (relation 24) :

[B(x,y)] = [if((x,y) = (2,1), d, if((x,y) = (1,(p i +1)/2 mod d), p i -1, if((x,y) > (1,1), #(u,v).d / g x-2 = g u.d + g y-2 .g v.d mod p i , 0)))]
If we are interested only in the reduced matrix [B'] which is matrix [B] out of first column and first line, we also have (with lines and columns being numbered from 1 to d) :

[B'(r,s)] = [#(u,v).d / g r-1 = g u.d + g s-1 .g v.d mod p i ] (25)
The cardinals, at order m, are given by :

m u m 1 v m 0 (26) w m = B 0 … … x m 0
Again d+1 is the maximum number of distinct cardinals.

As previously, the correspondence between cardinals values results from construction of the matrices (relation 27) : 1) .g 0 gd.g 1 g 2d .g 1 … g (pi-1) .g 1 gd.g d-1 g 2d .g d-1 … g (pi-1) .g d-1

# u m = card 0 v m = card 1 w m = card 2 … x m = card d c 0 g d .g 0 g 2d .g 0 … g (pi-
We draw this table at selected sequence k and carry out ascending sort of the second line (having cared to attach the first line to it). We will note T(u k , v k , w k , x k ) the function corresponding to the sorting and T(c, u k , v k , w k , x k ) the value corresponding to the target c. This notation will be useful to us later in our matter.

Abundance factors determination for the hypersurfaces y 1 n + y 2 n + … + y m n = c, relative to prime numbers variables, can then be carried out like previously without exponential increase in the processing time with m.

Cardinal matrix

We propose to call the precedent matrices of transformation as cardinal matrices since they interest the cardinality of the targets. Matrix [A] is the cardinal matrix of the integers' variables. The matrix [B] is the cardinal matrix of the prime numbers variables. When the context allows it, we speak about cardinal matrix without noting the type of variable. The cardinal matrices (modulo p) are the angular stones of abundance factors construction. It is important to study them in detail because their utility goes largely beyond the scope of Waring sums.

Dependence on primitive root

It is essential to notice that the cardinal matrix depends on the choice which is made for the primitive root g. Indeed, pending on g and g', the sets of numbers {g u.d .g i } and {g' v.d .g' i } may not be the same (but permutations) as u and v vary in their field of definition from 0 to (p-1)/d-1. Nevertheless, and it is self-evident, application to a target c, whatever choice of g, leaves unchanged the abundance factor of c. In addition, as g 0 is indifferent to the choice of g, value of the component x 1 at (x=2, y=2) is same whatever the choice of g.

We must remember this dependence to g when we give a literal expression of a cardinal matrix. According to the level of literal expression parameter setting, the expression may not be single. We are not able unfortunately to give a literal correspondence between the particular shape of the matrix according to an, a priori, choice of g not having a function (like g(p,d)) appropriate to each situation (can there be some ?). We can however anticipate this one under the following conditions : let us have g a primitive root, then every primitive root g' such as g' = g k.d (k an integer) lead to the same cardinal matrix, result which rises self-evidently from the equations being used to build the cardinal matrices. Thus, there exists φ(d) distinct shapes of matrices corresponding to φ(d) choice of families of targets {g, g 2+k1.d , g 3+k2.d ,…, gd -1+kd.d } (where choice of k1, k2,…, kd involves indeed that g 2+k1.d , g 3+k2.d ,…, gd -1+kd.d are primitive roots). Indeed, g being primitive root, g r is also primitive when, as we have seen at exercise 3, (r,d) = 1. Let us have family {g r , g r(2+k1.d) , g r(3+k2.d) ,…, g r (d- 1+kd.d) }. Then {r, 2r, 3r, …, r(d-1)} mod p-1 describes the whole set {1, 2, 3, …, d-1} (complete system of residues) for φ(d) distinct values of r. The table below thus gives the number of distinct cardinal matrices for some values of d : d # matrices forms Admissible permutations 1 1 2 1 3 2 {g 0 ,g 1 ,g 2 }.g 3u {g 0 ,g 2 ,g 1 }.g 3u 4 2 {g 0 ,g 1 ,g 2 ,g 3 }.g 4u {g 0 ,g 3 ,g 2 ,g 1 }.g 4u 5 4 {g 0 ,g 1 ,g 2 ,g 3 ,g 4 }.g 5u {g 0 ,g 2 ,g 4 ,g 1 ,g 3 }.g 5u {g 0 ,g 3 ,g 1 ,g 4 ,g 2 }.g 5u {g 0 ,g 4 ,g 3 ,g 2 ,g 1 }.g 5u 6 2 {g 0 ,g 1 ,g 2 ,g 3 ,g 4 ,g 5 }.g 6u {g 0 ,g 5 ,g 4 ,g 3 ,g 2 ,g 1 }.g 6u 7 6 {g 0 ,g 1 ,g 2 ,g 3 ,g 4 ,g 5 ,g 6 }.g 7u {g 0 ,g 2 ,g 4 ,g 6 ,g 1 ,g 3 ,g 5 }.g 7u {g 0 ,g 3 ,g 6 ,g 2 ,g 5 ,g 1 ,g 4 }.g 7u {g 0 ,g 4 ,g 1 ,g 5 ,g 2 ,g 6 ,g 3 }.g 7u {g 0 ,g 5 ,g 3 ,g 1 ,g 6 ,g 4 ,g 2 }.g 7u {g 0 ,g 6 ,g 5 ,g 4 ,g 3 ,g 2 ,g 1 }.g 7u 8 4 {g 0 ,g 1 ,g 2 ,g 3 ,g 4 ,g 5 ,g 6 ,g 7 }.g 8u {g 0 ,g 3 ,g 6 ,g 1 ,g 4 ,g 7 ,g 2 ,g 5 }.g 8u {g 0 ,g 5 ,g 2 ,g 7 ,g 4 ,g 1 ,g 6 ,g 3 }.g 8u {g 0 ,g 7 ,g 6 ,g 5 ,g 4 ,g 3 ,g 2 ,g 1 }.g 8u 9 6 {g 0 ,g 1 ,g 2 ,g 3 ,g 4 ,g 5 ,g 6 ,g 7 ,g 8 }.g 9u {g 0 ,g 2 ,g 4 ,g 6 ,g 8 ,g 1 ,g 3 ,g 5 ,g 7 }.g 9u {g 0 ,g 4 ,g 8 ,g 3 ,g 7 ,g 2 ,g 6 ,g 1 ,g 5 }.g 9u {g 0 ,g 5 ,g 1 ,g 6 ,g 2 ,g 7 ,g 3 ,g 8 ,g 4 }.g 9u {g 0 ,g 7 ,g 5 ,g 3 ,g 1 ,g 8 ,g 6 ,g 4 ,g 2 }.g 9u {g 0 ,g 8 ,g 7 ,g 6 ,g 5 ,g 4 ,g 3 ,g 2 ,g 1 }.g 9u … d φ(d)

Consequences for numerical applications

As soon as φ(d) > 1, as we explained previously, matrix form is related to the choice of g. Any abundance factor evaluation passes by such a choice. Literal matrix expressions introduce this dependence with respect to g inevitably. Under these conditions (d > 2), (approximate) abundance factor evaluation at sequence p always resort to basic equation g x-2 = g u.d + g y- 2 .g v.d mod p i .

Properties of cardinal matrices

Property 1

A and B being bound by A = B+I, the dimension of these associated matrices are identical; this dimension is equal to 1+d i where d i = gcd(n,p i -1), n being the exponent of the studied monomial.

Property 2

First column is written (1, d i , 0, …, 0) and (0, d i , 0, …, 0) respectively for A and B. Indeed, this column corresponds to the first column of the two-dimensional table used to work out the matrices (column card 0 ) and is thus identical to column vector generator (addition with 0). What involves the proposition.

Property 3

The sum of each line is respectively p i and p i -1 for matrices A and B. Indeed, the set of values 0 to p i -1 appear one and only once in the line vector generator of The sum of the components of the reduced matrix B' (we got by removing the first line and column of the matrix) on a diagonal, parallel with the principal diagonal is equal to p-1. Indeed, we start from g x-2 = g u.d + g y-2 .g v.d mod p i . The two components of the couple (x,y) vary together by increment of a unit. The sum, following a diagonal, equals ∑ (x,x+t) #{(u,v) \ g x-2 = g u.d + g y-2 .g v.d mod p i }, t a constant integer depending on the chosen diagonal, x = 2 to d+1 (mod d). For y = x+m, the local equation is g x+m-2 = g u.d + g y+m-2 .g v.d mod p i , that is also g x-2 = g -m mod d .g u.d + g y-2 .g v.d mod p i . When m varies in the field of definition 2 to d+1 mod d, the required sum is thus given by #{(u,v,n) \ g x-2 = g n .g u.d + g y-2 .g v.d which, by construction, is self-evidently p-1.

Property 5

The sum of each column is respectively p i and p i -1 for the matrices A and B, except for first column and column whose first line is non-null. Indeed, in the case of B, the sum of a column is equal to the cardinal of the elements of

Property 6

First line comprises only one non-null component, except the first column (which is 1 or 0 for A and B respectively). This component equals, according to the preceding property, p i -1 for matrices A and B. The component appears in column ((p i -1)/2 mod d i )+2 (first column starts as number 1) as we showed higher. Let us precise this locus r. Let us write 2+((p i -1)/2 mod d i ). We have d i = (p i -1,n), and thus p i -1 = k.d i , k an integer, so that (p i -1)/2 = k.d i /2. If d i is odd, then (p i -1)/2 = k.d i /2 is even, k must be even and thus (p i -1)/2 = k'.d i , k' integer, hence r = 2 mod d i = 2, which corresponds to only one position of p i -1and only one shape of matrix.

If d i is even, then (p i -1)/2 = k.d i /2
where k is either even, or odd, so r = 2 mod d i or r = 2+d i /2, with only two corresponding locus of p i -1 in matrix first line, and two forms of associated matrices. For r = 2, k is even and p i = 1 mod 2d i . For r = d i /2, k is odd and p i = 1+d i mod 2d i . Let us summarize : d i = (n,p i -1) odd p i = 1 mod 2d i matrix type 1 p i -1 in position 2 in first line d i = (n,p i -1) even p i = 1 mod 2d i matrix type 1 p i -1 in position 2 in first line d i = (n,p i -1) even p i = 1+d i mod 2d i matrix type 2 p i -1 in position 2+d i /2 in first line

The odd case d i and p = 1+d i mod 2d i does not exist.

Properties 7

We seek for B the identities which exist between certain elements of the matrix apart from the first line and first column. We note B' the restriction on these elements. Position (x,y) in B' is not located compared to the position in B, but in B' (x ≥ 1 and y ≥ 1). The equation being used for the enumeration in line x and column y of B' is then :

B' =[B'(x,y)] = [ #(u,v) / {g x-1 = g u.d + g y-1 .g v.d mod p i }]
Our research consists in finding permutations of exponents involving systematically B'(x',y') = B'(x,y) We start with two expressions :

B'(x,y) = #(u,v) / {g x-1 = g y-1 .g v.d + g u.d }] B(x',y')] = #(u',v') / {g x'-1 = g y'-1 .g v'.d + g u'.d }]
Six arrangements (including the initial one) are possible for the first equation and leaves it unchanged : P 59/390 Modulo pi abundance factors for Waring sums

B'(x,y) = #(u,v) / {g x-1 = g y-1 .g v.d + g u.d }] B'(x,y) = #(u,v) / {g x-1 = g u.d + g y-1 .g v.d }] B'(x,y) = #(u,v) / {-g u.d = g y-1 .g v.d -g x-1 }] B'(x,y) = #(u,v) / {-g u.d = -g x-1 + g y-1 .g v.d }] B'(x,y) = #(u,v) / {-g y-1 .g v.d = -g x-1 + g u.d }] B'(x,y) = #(u,v) / {-g y-1 .g v.d = g u.d -g x-1 }]
The first arrangement corresponding to the identity is self-evident and is not useful in the next course. Using property g (p-1)/2 = -1, it follows for the five remaining expressions :

B'(x,y) = #(u,v) / {g x-1 = g u.d + g y-1 .g v.d }] B'(x,y) = #(u,v) / {g (p-1)/2 .g u.d = g y-1 .g v.d + g x-1+(p-1)/2 }] B'(x,y) = #(u,v) / {g (p-1)/2 .g u.d = g x-1+(p-1)/2 + g y-1 .g v.d }] B'(x,y) = #(u,v) / {g y-1+(p-1)/2 .g v.d = g x-1+(p-1)/2 + g u.d }] B'(x,y) = #(u,v) / {g y-1+(p-1)/2 .g v.d = g u.d + g x-1+(p-1)/2 }]
Putting all the last terms on the right to form g r.d , by adequate multiplication by g s.d , it follows:

B'(x,y) = #(u,v) / {g -y+x = g -y+1 .g u.d + g v.d }] B'(x,y) = #(u,v) / {g -x+1 .g (u+w).d = g y-x-(p-1)/2 .g (v+w).d + g w.d }] B'(x,y) = #(u,v) / {g -y+1+(p-1)/2 .g u.d = g -y+x+(p-1)/2 + g v.d }] B'(x,y) = #(u,v) / {g y-1+(p-1)/2 .g v.d = g x-1+(p-1)/2 + g u.d }] B'(x,y) = #(u,v) / {g y-x .g (v+w).d = g -x+1-(p-1)/2 .g (u+w).d + g w.d }]
The order of the permutations (u,v) do not matter, provided that all these permutations are actually carried out. An exponent incrementing by (given) w does not influence the permutations. We can thus write :

B'(x,y) = #(u,v) / {g x-1 = g y-1 .g v.d + g u.d }] B'(x,y) = #(u,v) / {g -y+x = g -y+1 .g v.d + g u.d }] B'(x,y) = #(u,v) / {g -x+1 = g y-x-(p-1)/2 .g v.d + g u.d }] B'(x,y) = #(u,v) / {g -y+1+(p-1)/2 = g -y+x+(p-1)/2 .g v.d + g u.d }] B'(x,y) = #(u,v) / {g y-1+(p-1)/2 = g x-1+(p-1)/2 .g v.d + g u.d }] B'(x,y) = #(u,v) / {g y-x = g -x+1-(p-1)/2 .g v.d + g u.d }]
We then get by identification of each expression B(x',y') and B'(x,y) the couples of conditions : Here y, y', x, x' are numbers corresponding to locus of columns and lines lying between 1 and d+1. Thus, the preceding equalities are written precisely :

x' = 1+((x-y) mod d) y' = 1+((-y + 1) mod d) x' = 1+((-x+1) mod d) y' = 1+((-x+y-(p-1)/2) mod d) x' = 1+((-y+1+(p-1)/2) mod d) y' = 1+((x-y+(p-1)/2) mod d) x' = 1+((y-1+(p-1)/2) mod d) y' = 1+((x-1+(p-1)/2) mod d) x' = 1+((-x+y) mod d) y' = 1+((-x+1-(p-1)/2) mod d)
We keep in sight this point (mod d for x ≥ 1) while simplifying the writing of these equalities. We note that (p-1)/2 or -(p-1)/2 is without effect modulo d (systematic sign + below). Thus :

x' = x-y+1 y' = -y + 2 x' = -x+2 y' = -x+y+1+(p-1)/2 x' = -y+2+(p-1)/2 y' = x-y+1+(p-1)/2 x' = y+(p-1)/2 y' = x+(p-1)/2 x' = -x+y+1 y' = -x+2+(p-1)/2
Let us have : X = (x,y), X' = (x',y').

We can write the preceding equalities in the form of linear transformations :

X' = T(X) = R.X + S P 60/390

Modulo pi abundance factors for Waring sums R square of dimension 2, S vector of dimension 2. This gives respectively :

R1 = 1 -1 S1 = 1 0 -1 2 R2 = -1 0 S2 = 2 -1 1 1+(p-1)/2 R3 = 0 -1 S3 = 2+(p-1)/2 1 -1 1+(p-1)/2 R4 = 0 1 S4 = (p-1)/2 1 0 (p-1)/2 R5 = -1 1 S5 = 1 -1 0 2+(p-1)/2
We get, knowing that p-1 = 0 mod d : The reader will be able to verify without pain that any other couple (Ti,Tj), i≠j, is also appropriate, except for (T3,T5).

T1 R1 2 = I T1 2 = I Involution T2 R2 2 = I T2 2 = I Involution T3 R3 2 = R5, R3 3 = I T3 2 =
I = T3 x T5 T1 ≠ Π (T3,T5) T2 ≠ Π (T3,T5) T3 = T5 x T5 T4 ≠ Π (T3,T5) T5 = T3 x T3
Here Π (T3, T5) is any product of T3 and T5. It is not possible to generate T1, T2 and T4 starting from a couple (Ti,Tj) which does not shelter at least an involution.

With T2 and T4 transforms and properties called upon higher, we can, using some program code lines, get the form of the matrices for given d and limit the number of different components.

We have two cases : If (p-1)/2 = 0 mod d, then :

R2 = -1 0 S2 = 2 -1 1 1 P 61/390
Modulo pi abundance factors for Waring sums

R4 = 0 1 S4 = 0 1 0 0 If (p-1)/2 = d/2 mod d, then : R2 = -1 0 S2 = 2 -1 1 1+d/2 R4 = 0 1 S4 = d/2 1 0 d/2
12. Forms of matrices We deduce this form directly from properties 2 and 6. Thus B is written in the form :

0 0 … 0 p-1 0 … 0 d 0 … U V … 0
Two cases arise : If p = 1 mod 2d, then U is empty and V is square of dimension d. If p = 1+d mod 2d, then U and V have identical dimension (each d/2 columns, d lines).

Property 9 ('of symmetry')

Let us show that the square matrix formed by using U and V, in the following order, is symmetrical (if U is empty, the property interests of course only V) :

V U As :

B' = U V = d.(#(u,v)/{g x-1 = g y-1 .g v.d + g u.d })
it is necessary and it suffices to show that :

#(u,v)/{g x-1 = g u.d + g (p-1)/2+y-1 .g v.d } = #(u,v)/{g y-1 = g u.d + g (p-1)/2+x-1 .g v.d }
We can demonstrate this result either directly, or by using transform T4, what we do here :

Preceding equation at second member : y

'-1 = (p-1)/2+x-1 x'-1 = y-1 Transform T4 : y'' = 1+((x'-1+(p-1)/2) mod d) x'' = 1+((y'-1+(p-1)/2) mod d)
That is, with p-1 = 0 mod d, for the first member of the equation :

y''-1 = 1+((y-1+(p-1)/2) mod d) x''-1 = (x-1) mod d
This is the sought identity.

Let us note here that T4 transform governs the symmetry of matrix B.

Detailed form

Detailed form Let us start with some examples. We build the cardinal matrices for d = 14. We get, while applying transformation T2 and T4, the following forms.

For p = 1 mod 28 (that is (p- Thus, the matrices of the type (p-1)/2 = 0 mod d are built geometrically, a priori, thanks to six helix branches as follows (geometrical construction n°1) :

0 p-1 0 0 0 0 0 0 … 0 0 0 0 0 0 d x1-d 0 0 0 0 0 xmax … xmax xmax 0 0 0 0 0 symmetric 0 0 x2+d-1 0 x2
In the same way, the matrices of the type (p-1)/2 =d/2 mod d are built by a completely similar method :

We form initially three vortex branches like previously. We make then an overall exchange of lines 2 to d/2+1 and lines d/2+2 to d+1, except for the first column, to find the appropriate matrix.

If the direction of the arrows above is reversed, we can check that another geometrical construction can be also proposed (geometrical construction n°2) : In the geometrical construction, "entries" of matrix are in positions which apply in a systematic way (1 entry on the left at column 2, 1 entry on the right bottom following trace, 1 entry just above the preceding one in the same direction for the first construction, etc.). The "exits" depend on the value modulo 3 of d and the value modulo 2d of p. Thus, we have the configurations of positioning of x max and x max-1 as follows (which can be proved from T2 and T4 transforms, but left here as non-critical proof) :

P
For the first geometrical construction : If p = 1 mod 2d d = 0 mod 3 d = 1 mod 3 d = 2 mod 3

x max-1 x max-1 x max-1 x max x max-1 x max-1 x max x max-1 x max x max x max x max x max-1

x max x max-1 x max-1 If p = 1+d mod 2d d = 0 mod 3 d = 1 mod 3 d = 2 mod 3

x max-1 x max-1 x max-1 x max-1 x max x max x max x max x max-1

x max-1 x max x max x max-1 x max x max-1 x max-1

The figures in cases p = 1+d mod 2d are in a mirror figures of cases p = 1 mod 2d. If we use the second geometrical construction we get the following situations :

If p = 1 mod 2d d = 0 mod 3 d = 1 mod 3 d = 2 mod 3 x max-1 x max-1 x max-1 x max x max-1 x max x max-1
x max x max x max-1 x max x max x max-1

x max x max-1 x max-1 If p = 1+d mod 2d d = 0 mod 3 d = 1 mod 3 d = 2 mod 3

x max-1 x max-1 x max-1 x max x max-1

x max x max x max x max-1

x max x max x max-1 x max-1 x max x max-1 x max-1

We can give a correspondence between the classifications (with the proviso of keeping writing conventions for the entries x1) of the two geometrical constructions : Trace in red and first line in green correspond by T2 and T4 transforms :

Geometrical system 1 Geometrical system 2 Sum of position numbers (i-1)d-3(i-1)(i-2)/2+1 (i-1)d-3(i-1)(i-2)/2+1 2((i-1)d-3(i-1)(i-2)+2 (i-1)d-3(i-1)(i-2)/2+2 to (i-1)d-3(i-1)(i-2)/2+d-3(i-1) d+(i-1)d-3i(i-1)/2 to (i-1)d-3(i-1)(i-2)/2+2 d+2(i-1)d-3(i-1)²/2+2
x' = -x+2 mod d y' = -x+y+1+(p-1)/2 mod d x'' = y+(p-1)/2 mod d y'' = x+(p-1)/2 mod d Indeed, here (p-1)/2 = 0 mod 2d and for the trace x = y, that is thus x' = -x+2 and y' = 1, then by symmetry y'' = -y+2 and x'' = 1 what is the sought correspondence : (y,y'') = (2,14), that is y+y'' = 16 = 2 mod 14, (y,y'') = (3,13), that is y+y'' = 2 mod 14, etc. We verify, in the same way, for in blue oblique row above the trace, the correspondence with the in red elements of the third line, etc. We get by successively withdrawing the elements which correspond on the two tables : For the elements remaining on top of the trace in the first table, it is necessary to set out again with complete initial matrix.

The in green last column corresponds to blue oblique part above the trace. Indeed, let us take again the T2 transform :

x' = -x+2 mod d y' = -x+y+1+(p-1)/2 mod d

For y = 14 and (p-1)/2 = 0 mod d, it follows x' = -x+2 and y' = -x+15, that is x' = -x+2 and x'-y' = -13 = 1 mod 14, what demonstrates our point. We proceed in the same way for the other columns on the left of that studied previously. Thus, all the part above matrix trace, trace included, is taken into account. The symmetrical part is taken into account by T4 transform immediately.

If we consider then the (p-1)/2 = d/2 mod d case, using example d = 14 and T2 and T4 transforms, we can describe, by same elementary arguments, the whole set of correspondences. We can proceed in the same way with second geometrical construction method.

Let us come to the second table presented above. We note, with each coupled withdrawal of elements in correspondence from the first table, the reduction in length of the new paired segments by exactly three elements. That is due to the fact that starting from a horizontal line (respectively oblique, respectively vertical), one precisely needs three steps of withdrawal to return to horizontal (respectively oblique, respectively vertical). This implies also exactly three entry points in matrix and a three blades vortex shape.

To finish, let us notice that the T2 and T4 transforms apply whatever the number d. Thus our argument applies to any matrix B. Hence it follows :

Property 10 ('geometrical construction')

We proceed with at the same time biological and nautical analogy. The cardinal matrices [B] admit a geometrical construction with (symmetrical ) double-three blades vortices. The vortices are not oriented and can be described either in a trigonometric or reverse way. For matrices of type 1 (where (p-1)/2 = 0 mod d), construction succeeds immediately. For matrices of type 2 (where (p-1)/2 = d/2 mod d), an overall exchange, following the horizontal, out of first line and first column, of the constituent halves is necessary (the vortices are broken before the exchange of the components) to reveal the two vortices. In the first case, we will thus speak of a double helix without crack, in the second, of double helix with crack.

Reduction of unknowns

We showed that the matrices B are built geometrically using a double three blades vortices. We will draw conclusions, from the set of symmetries, concerning identical elements enumeration for these matrices.

For that, we consider successively the three "branches" of the vortex on the part above the trace, trace included. For the first branch, classification starts at 1 and finishes with max, max being possibly common to the other branches. However, until d, the trace is common to two branches. For the second, it starts at 2 and finishes with max (max possibly common to the other branches). For the third, classification starts at d+1 and finishes with max (let us recall that if the first branch is actually numbered by increment of 1, classifications of the other branches are given by the T2 and T4 transforms). It follows immediately for the described part :

0 2d-1 1 2d-1 p-1 1 1 1 d 1 1 1 x1(-d) 1 1 1 x 2 to x d 3 d-1 3(d-1) x d+1 to x max (= x (d²+3d+2)/6 ) 6 (d-1).(d-2)/6 (d-1).(d-2) Total (d+1) 2
Our work of reduction of the number of variables is not completed here. Indeed, we also know that the sums of matrices lines and columns are equal to p-1. We thus have d+d additional equations. However, matrix B' being symmetric, we have in the 2d preceding equations d redundant equations. We consider then T2 transform : From these two remarks, it follows that we do not dispose now of d equations, but only of 1+int(d/2) residual equations, the integer value resulting from a last redundancy in the middle of the table in regard of parity of d.

Property 12 ( 'additional reduction of cardinals')

r ad = 1+int(d/2)
The table below gives the value, for given d, of three things :

-the number of different elements after the use of involutive T1 (or T2) and T4 transforms, -the number of additional equations resulting from lines and columns sums, -the remaining number of cardinals to be found using [#(u,v) / {g x-1 = g y-1 .g The term "unknown factors" must, of course, being taken with its peculiar meaning since the components of the matrices described here are actually perfectly defined.

Matrices production algorithm

Such an algorithm is very simple to produce thanks to relation B'(x,y) = #(u,v)/{g x-1 = g u.d + g y-1 .g v.d mod p i } and use of properties 2 and 6. The speed of execution will be improved when d increases using property 7 but the algorithm then requires more sophistication. -1 , all results which rise from the linear algebra. card'(i) = ∑ #(i-j mod p).card(j) (29) j = 0 to p-1

Matrices diagonalisation

Cardinal matrices eigenvalues and eigenvectors

Here #(i-j) are the components at locus (i,j) of matrix [CI] which is a right circulant matrix since #(i-j) = #(i+t-(j+t)). Hence, for the quoted example, we have : d = (n,p-1) = (3,12) = 3 g primitive root of p (g = 2 here for p = 13) #( 0 For more clarity of the writing, let us pose c t = c (t). We have c (t) = #(t), that is : We observe the degenerescence of the eigenvalues because of the periodicity of the coefficients c t . When v' = v.g u.d , u integer, we have indeed σ v' = σ v .

) = 1 #(g u.d ) = #(2 4u ) = 3 #(g.g u.d ) = #(2 4u+1 ) = 0 #(g 2 .g u.d ) = #(2 4u+2 ) = 0
c(0) = #(0) = if(variable of integers = vi, 1, 0) c(g k .g u.d ) = #(g k .g u.d ) = if(k = 0, d, 0) So that σ v = if(vi,1,0)+ d.
In

1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
0,89 0,57 0,12 -0,4 -0,7 -1 -1 -0,7 -0,4 0,12 0,57 0,89 1 0 0,46 0,82 0,99 0,94 0,66 0,24 -0,2 -0,7 -0,9 -1 -0,8 -0,5 2 0,57 -0,4 -1 -0,7 0,12 0,89 0,89 0,12 -0,7 -1 -0,4 0,57 2 0 0,82 0,94 0,24 -0,7 -1 -0,5 0,46 0,99 0,66 -0,2 -0,9 -0,8 3 0,12 -1 -0,4 0,89 0,57 -0,7 -0,7 0,57 0,89 -0,4 -1 0,12 3 0 0,99 0,24 -0,9 -0,5 0,82 0,66 -0,7 -0,8 0,46 0,94 -0,2 -1 4 -0,4 -0,7 0,89 0,12 -1 0,57 0,57 -1 0,12 0,89 -0,7 -0,4 4 0 0,94 -0,7 -0,5 0,99 -0,2 -0,8 0,82 0,24 -1 0,46 0,66 -0,9 5 -0,7 0,12 0,57 -1 0,89 -0,4 -0,4 0,89 -1 0,57 0,12 -0,7 5 0 0,66 -1 0,82 -0,2 -0,5 0,94 -0,9 0,46 0,24 -0,8 0,99 -0,7 6 -1 0,89 -0,7 0,57 -0,4 0,12 0,12 -0,4 0,57 -0,7 0,89 -1 6 0 0,24 -0,5 0,66 -0,8 0,94 -1 0,99 -0,9 0,82 -0,7 0,46 -0,2 7 -1 0,89 -0,7 0,57 -0,4 0,12 0,12 -0,4 0,57 -0,7 0,89 -1 7 0 -0,2 0,46 -0,7 0,82 -0,9 0,99 -1 0,94 -0,8 0,66 -0,5 0,24 8 -0 ,7 0,12 0,4 0,57 0,7 0,8 0,24 0,9 0,2 0,7 0,89 0,57 0,12 0,9 0,66 0,24 0,2 0,7 0,94 10 0,4 0,89 0,7 0,57 0,2 0,94 0,7 0,66 0,9 0,24 0,99 11 0,7 0,12 0,89 0,89 0,4 0,2 0,66 0,99 0,7 0,24 0,94 0,82 12 0,89 0,57 0,4 0,12 0,57 0,2 0,24 0,66 0,94 0,99 0,82 0,46 We then carry out a rearrangement of generating elements different from 0 modulo p (first line and first column out of our table) while starting with g u.d , u = 0 to (p-1)/d-1, then g u.d+1 , then g u.d+2 , etc. This gives a new matrix: The same rearrangement is made on [P] (and [P -1 ]). Thus : 0,89 -0,7 0,89 -0,7 0,57 0,12 0,57 0,12 -0,4 -1 -0,4 -1 1 0 0,46 -0,7 -0,5 0,66 0,82 0,99 -0,8 -1 0,94 0,24 -0,9 -0,2 8 -0,7 0,89 -0,7 0,89 0,12 0,57 0,12 0,57 -1 -0,4 -1 -0,4 8 0 -0,7 -0,5 0,66 0,46 0,99 -0,8 -1 0,82 0,24 -0,9 -0,2 0,94 12 0,89 -0,7 0,89 -0,7 0,57 0,12 0,57 0,12 -0,4 -1 -0,4 -1 12 0 -0,5 0,66 0,46 -0,7 -0,8 -1 0,82 0,99 -0,9 -0,2 0,94 0,24 5 -0,7 0,89 -0,7 0,89 0,12 0,57 0,12 0,57 -1 -0,4 -1 -0,4 5 0 0,66 0,46 -0,7 -0,5 -1 0,82 0,99 -0,8 -0,2 0,94 0,24 -0,9 2 0,57 0,12 0,57 0,12 -0,4 -1 -0,4 -1 -0,7 0,89 -0,7 0,89 2 0 0,82 0,99 -0,8 -1 0,94 0,24 -0,9 -0,2 -0,7 -0,5 0,66 0,46 3 0,12 0,57 0,12 0,57 -1 -0,4 -1 -0,4 0,89 -0,7 0,89 -0,7 3 0 0,99 -0,8 -1 0,82 0,24 -0,9 -0,2 0,94 -0,5 0,66 0,46 -0,7 11 0,57 0,12 0,57 0,12 -0,4 -1 -0,4 -1 -0,7 0,89 -0,7 0,89 11 0 -0,8 -1 0,82 0,99 -0,9 -0,2 0,94 0,24 0,66 0,46 -0,7 -0,5 10 0,12 0,57 0,12 0,57 -1 -0,4 -1 -0,4 0,89 -0,7 0,89 -0,7 10 0 -1 0,82 0,99 -0,8 -0,2 0,94 0,24 -0,9 0,46 -0,7 -0,5 0,66 4 -0,4 -1 -0,4 -1 -0,7 0,89 -0,7 0,89 0,12 0,57 0,12 0,57 4 0 0,94 0,24 -0,9 -0,2 -0,7 -0,5 0,66 0,46 0,99 -0,8 -1 0,82 6 -1 -0,4 -1 -0,4 0,89 -0,7 0,89 -0,7 0,57 0,12 0,57 0,12 6 0 0,24 -0,9 -0,2 0,94 -0,5 0,66 0,46 -0,7 -0,8 -1 0,82 0,99 9 -0,4 -1 -0,4 -1 -0,7 0,89 -0,7 0,89 0,12 0,57 0,12 0,57 9 0 -0,9 -0,2 0,94 0,24 0,66 0,46 -0,7 -0,5 -1 0,82 0,99 -0,8 7 -1 -0,4 -1 -0,4 0,89 -0,7 0,89 -0,7 0,57 0,12 0,57 0,12 7 0 -0,2 0,94 0,24 -0,9 0,46 -0,7 -0,5 0,66 0,82 0,99 -0,8 -1

Hence it results eigenvalues rearrangement : ] [P' -1 (r,s)] = (1/p 1/2 ). [(e (-2πi/p).if(r=0,0,(g^int((r-1)/((p-1)/d))).(g^d.((r-1) mod (p-1)/d))).if(s=0,0,(g^int((s-1)/((p-1)/d))).(gd.^((s-1) mod (p-1)/d))) ] [σ'(r,s)] = [if(r≠s,0,σ if(r=0,0,(g^int((r-1)/((p-1)/d))).(g^d.((r-1) mod (p-1)/d))) ] Progression of the indices is here of the form : {{0}, {g 0.d , g 1.d , …, g ((p-1)/d-1).d }, {g.g 0.d , g.g 1.d , …, g.g ((p-1)/d-1).d }, …, {gd -1 .g 0.d , gd -1 .g 1.d , …, gd -1 .g ((p-1)/d-1).d }} It was represented only for matrix of eigenvalues but applies to all matrices.

[σ']
We proceed then to the "merger" per blocks of the matrices.

0 0 1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 0,89 -0,7 0,89 -0,7 0,57 0,12 0,57 0,12 -0,4 -1 -0,4 -1 8

1 -0,7 0,89 -0,7 0,89 0,12 0,57 0,12 0,57 -1 -0,4 -1 -0,4 12 1 0,89 -0,7 0,89 -0,7 0,57 0,12 0,57 0,12 -0,4 -1 -0,4 -1 51 -0,7 0,89 -0,7 0,89 0,12 0,57 0,12 0,57 -1 -0,4 -1 -0,4 2 2 1 0,57 0,12 0,57 0,12 -0,4 -1 -0,4 -1 -0,7 0,89 -0,7 0,89 3 1 0,12 0,57 0,12 0,57 -1 -0,4 -1 -0,4 0,89 -0,7 0,89 -0,7 11 1 0,57 0,12 0,57 0,12 -0,4 -1 -0,4 -1 -0,7 0,89 -0,7 0,89 10 1 0,12 0,57 0,12 0,57 -1 -0,4 -1 -0,4 0,89 -0,7 0,89 -0,7 3 4 1 -0,4 -1 -0,4 -1 -0,7 0,89 -0,7 0,89 0,12 0,57 0,12 0,57 61 -1 -0,4 -1 -0,4 0,89 -0,7 0,89 -0,7 0,57 0,12 0,57 0,12 91 -0,4 -1 -0,4 -1 -0,7 0,89 -0,7 0,89 0,12 0,57 0,12 0,57 71 -1 -0,4 -1 -0,4 0,89 -0,7 0,89 -0,7 0,57 0,12 0,57 0,12

The merger gives for the example:

[P' -1 (r+1 mod d, s+1 mod d)] = [P' -1 (r mod d, s mod d)]
The sum of a block ((p-1)/d,(p-1)/d) divided by (p-1)/d is thus the sum of a line (1,(p-1)/d). Thus, the sum bearing on r = 0 to (p-1)/d-1: (u≠v,0, if(vi,1,0)+d.∑e (-2πi/p).(g^(u-1)).(g^(r.d)) ]

[P''(u,v)] = [if(v = 0, 1, if(u = 0, (p-1)/d, (1/p 1/2 ).∑e (2πi/p).(g^(u-1)).(g^(r.d)) ] [P'' -1 (u,v)] = [if(v = 0, 1, if(u = 0, (p-1)/d, (1/p 1/2 ).∑e (-2πi/p).(g^(u-1)).(g^(r.d)) ] [σ''(u,v)] = [if
We get for u > 0 and v > 0 (in block ((p-1)/d,(p-1)/d) :

P''(u,v) = P''(u-1,v+1) P'' -1 (u,v) = P'' -1 (u-1,v+1) The inner part of these matrices is left circulant. This results (σ'' i are replaced by σ i for a better homogeneity of writing) in the following form: THEOREM: Decomposition of cardinal matrices

Cardinal matrix [M], relative to operation x n (with ve = 1), respectively y n (with ve = 0), at sequence p, of dimension d+1 where d = (n, p-1), is diagonalisable according to :

1 λ 0 */d λ 0 */d … λ 0 */d σ 0 0 0 … 0 1 λ 0 /d λ 0 /d … λ 0 /d 1 λ 1 */d λ 2 */d … λ d */d 0 σ 1 0 … 0 1 λ 1 /d λ 2 /d … λ d /d [M] = (1/p) 1 λ 2 */d λ 3 */d … λ 1 */d 0 0 σ 2 … 0 1 λ 2 /d λ 3 /d … λ 1 /d (32) … … … … … … … … … … … … … … … 1 λ d */d λ 1 */d … λ d-1 */d 0 0 0 … σ d 1 λ d /d λ 1 /d … λ d-1 /
(34) r = 0 to (p-1)/d-1
This writing is to be privileged by its simplicity, although we used it only seldom for historic reasons (calendar of drafting). Other writings are possible. Thus, while multiplying on the left and on the right, the matrix of eigenvalues by :

1 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0
We have firstly We use to pass from a form to the other the matrix whose square is identity :

1 0 0 0 0 … 0 0 0 1 0 0 0 … 0 0 0 0 0 0 0 … 0 1 [CC] = 0 0 0 0 0 … 1 0 … … … … … … … … 0 0 0 0 1 … 0 0 0 0 0 1 0 … 0 0 0 0 1 0 0 … 0 0 P 73/390
Modulo pi abundance factors for Waring sums

Multiplication on the right and on the left of the matrix of eigenvalues by [CC] reverse initial order of eigenvalues starting from σ 2 . Multiplication on the right of the matrix of eigenvectors by [CC] leaves unchanged first and second column, with remainder of the circulant block becoming right circulant. Multiplication on the left of the matrix P'' -1 (u,v) by [CC] leaves unchanged first column, permute elements of second column starting from second line with remainder of the circulant block becoming right circulant. Hence the form of the decomposition in eigen-elements :

1 λ 0 */d λ 0 */d … λ 0 */d σ 0 0 0 … 0 1 λ 0 /d λ 0 /d … λ 0 /d 1 λ 1 */d λ d-1 */d … λ 2 */d 0 σ 1 0 … 0 1 λ 1 /d λ 2 /d … λ d /d [M] = (1/p) 1 λ 2 */d λ 1 */d … λ 3 */d 0 0 σ d … 0 1 λ d /d λ 1 /d … λ d-1 /d (35) … … … … … … … … … … … … … … … 1 λ d */d λ d-1 */d … λ 1 */d 0 0 0 … σ 2 1 λ 2 /d λ 3 /d …
1 λ 0 */d λ 0 */d … λ 0 */d σ 0 0 0 … 0 1 λ 0 /d λ 0 /d … λ 0 /d 1 λ 1 */d λ 2 */d … λ d */d 0 σ 1 0 … 0 1 λ 1 /d λ d /d … λ 2 /d [M] = (1/p) 1 λ d */d λ 1 */d … λ d-1 */d 0 0 σ 2 … 0 1 λ 2 /d λ 1 /d … λ 3 /d (36) … … … … … … … … … … … … … … … 1 λ 2 */d λ 3 */d … λ 1 */d 0 0 0 … σ d 1 λ d /d λ d-1 /d … λ 1 /d
We will use for reasons of drafting calendar the former right circulant change of basis matrix. However, in this precise case, we must be vigilant on indices of λ u and σ u :

Positions in column 2 to λ u and σ u 1

P' = (1/p 1/2 ). λ 1 */d λ 2 */d … λ d */d P' -1 = (1/p 1/2 ). λ 1 /d λ d /d … λ 2 /d λ d */d λ 1 */d … λ d-1 */d λ 2 /d λ 1 /d … λ 3 /d … … … … … … … … λ 2 */d λ 3 */d … λ 1 */d λ d /d λ d-1 /d … λ 1 /d
Thus P' is right circulant and unitary : P' -1 = t P'* Eigenvectors matrix is written :

1 λ 0 /d λ 0 /d … λ 0 /d 1 p 1/2 .P' [P] = (1/p 1/2 ). 1 … 1
We name this kind of matrix intra-unitary.

The following notations will be useful for us in other exec ices :

P 74/390

Modulo pi abundance factors for Waring sums -1+u.d )) u = 0 to (p-1)/d-1

[λ'] = λ 1 λ d … λ 2 [λ'*] = λ 1 * λ 2 * … λ d * λ 2 λ 1 … λ 3 λ d * λ 1 * … λ d-1 * (38) … … … … … … … … λ d λ d-1 … λ 1 λ 2 *
We have d = (n, p-1). Thus p-1 = m.d. In addition, p-1 is even. Therefore, m is even, then p = 1 mod 2d. Thus (p-1)/2d is an integer. Let us pose u' = u+(p-1)/2d. Then, for all u, there is u' such as g u'.d = g u.d+(p-1)/(2d).d = -g u.d .. Thus, the first sum bearing on the indices u = 0 to (p-1)/d-1 and the second on the indices u = 0 to (p-1)/(2d)-1, we get ∑sin((2π/p).g k+u.d

) = ∑sin((2π/p).g k+u.d )-sin((2π/p).g k+u.d ) = 0.

Then :

σ k = if(vi,1,0)+d. ∑ cos((2π/p).g k-1+u.d ) (39) u = 0 to (p-1)/d-1
All the eigenvalues are real numbers.

Case even d

We have two alternatives : Case p = 1 mod 2d

We are somewhat in the preceding case. Indeed, g (u+(p-1)/(2d)).d = g (p-1)/2 .g u.d = -g u.d mod p. So that cos((2π/p).g k-1+(u+(p-1)/(2d)).d ) = cos((2π/p).g k-1+u.d ) and sin((2π/p).g k-1+(u+(p-1)/(2d)).d ) = -sin((2π/p).g k-1+u.d ). It follows :

σ k = if(ve,1,0)+d. ∑ cos((2π/p).g k-1+u.d ) = if(ve,1,0)+2d. ∑ cos((2π/p).g k-1+u.d ) (40) u = 0 to (p-1)/d-1 u = 0 to (p-1)/(2d)-1
All the eigenvalues are real numbers.

Case p = 1 + d mod 2d

We have to use the general expression. Thus :

μ k = 1+d.( ∑ cos((2π/p).g k-1+u.d )-i.sin((2π/p).g k-1+u.d )) (41) u = 0 to (p-1)/d-1 λ k = d.( ∑ cos((2π/p).g k-1+u.d )-i.sin((2π/p).g k-1+u.d )) (42) u = 0 to (p-1)/d-1 As d/2 = (p-1)/2 = mod d, we have g d/2 = g (p-
(λ k+d/2 ) = (λ k ) (43)  (λ k+d/2 ) = -(λ k )
The eigenvalues are imaginary numbers (or real, case that we reject without proving it here).

Particular cases studies.

We established cardinal matrices eigenvalues and eigenvectors expressions previously. These expressions, however, have the disadvantage of being sums whose number of elements increases with sequence p. The purpose of the following particular cases studies are to get, when possible, "simpler" expressions, not sums in any case.

Dimension 2 matrices

We establish a general result here for d = (n,p-1) = 1 P 75/390 Modulo pi abundance factors for Waring sums Matrices A and B are then dimension 1 + d = 2. From the general form of the matrices, we deduce immediately :

A = 1 p-1 1 p-1 B = 0 p-1 1 p-2 We have : Det(A-μI) = μ(μ-p) Det(B-λI) = (λ+1)(λ-p+1) The eigenvalues of matrix A are : μ 0A = p μ 1A 0
and are associated to the eigen-vectors matrix (λ 0B = λ 0B * = μ 0A -1, λ 1B = λ 1B * = μ 1A -1) :

P A = (1/p 1/2 ). 1 p-1 = (1/p 1/2 ). 1 μ 0A -1 = (1/p 1/2 ). 1 μ 0A *-1 1 -1 1 μ 1A -1 1 μ 0A *
#{0} =(1/p) 1 p-1 p k 0 1 p-1 1 #{g u } 1 -1 0 0 k 1 -1 0
That is also :

#{0} = p k-1 (44) #{g u } p k-1
The eigenvalues of matrix B are shifted of a unit :

λ 0B = p-1 λ 1B -1
and are associated with the eigenvectors matrix identical to that of A :

P B = (1/p 1/2 ). 1 p-1 = (1/p 1/2 ). 1 λ 0B = (1/p 1/2 ). 1 λ 0B * 1 -1 1 λ 1B 1 λ 1B *
and to the reverse matrix (identical to P -1 A ) P -1 B = P B Then :

#{0} =(1/p) 1 p-1 (p-1) k 0 1 p-1 1 #{g u } 1 -1 0 (-1) k 1 -1 0
That is also : We use relation A = B + I to determine cardinal matrices forms which we give directly here. The reader will refer to next paragraph for the evaluations of B. Thus :

#{0} = (1/p).((p-1) k + (p-1).(-1) k ) (45) #{g u } (1/p).((p-1) k -(-1) k ) Hence the table in example : k = 1 k = 2 k = 3 k = 4 k = 5 #{0} 0 p-1 (p-1
For p = 2 A = 1 1 (47) 1 1 For p = 1 mod 4 1 p-1 0 A = 2 (p-3)/2 (p-1)/2 (48) 0 (p-1)/2 (p+1)/2 Det(A-μI) = -μ 3 +p.μ 2 +p.μ-p 2 = -(μ-p)(μ 2 -p) = -(μ-p)(μ 2 +√p)(μ-√p) For p = 3 mod 4 1 0 p-1 A = 2 (p-1)/2 (p-3)/2 (49) 0 (p+1)/2 (p-1)/2 Det(A-μI) = -μ 3 +p.μ 2 -p.μ+p 2 = -(μ-p)(μ 2 +p) = -(μ-p)(μ 2 +i√p)(μ-i√p)
The enumerations are then (except case p

= 2) k #{0} 1 #{g 0 .g 2u } = A 0 #{g 1 .g 2u } 0
Having established cardinal matrix A, the evaluation of A k imposes to get eigenvalues and eigenvectors.

For p = 2, deduced from case d = (n,p-1) = 1 already studied, eigenvalues matrix A are :

μ 0A = 2 μ 1A 0
and are associated with eigenvectors matrix :

P A = (1/√2)
P -1 A = (1/p 1/2 ). 1 (-1+i√p)/2 (-1-i√p)/2 1 (-1-i√p)/2 (-1+i√p)/2
We may notice the simplicity of passage from case p = 1 mod 4 to case p = 3 mod 4 thanks to the imaginary number -i. Then :

#{0}

-i) k )) #{g 0 .g 2u } = (1/p).(p k -(1/2).p k/2 .(i k +(-i) k +p 1/2 .(i k+1 +(-i) k+1 ))) #{g 1 .g 2u } (1/p).(p k -(1/2).p k/2 .(i k +(-i) k -p 1/2 .(i k+1 +(-i) k+1 )))
The preceding values being cardinals do not have imaginary parts. We find an equivalent expression by replacing i k +(-i) k by (-1) k/2 (1+(-1) k ) : Cases p = 1 mod 4 and p = 3 mod 4 can be put together in only one, while using (-1) (p-1)/2 : #{0}

#{0} (1/p).(p k +(1/2).(p-1).p k/2 .(-1) k/2 (1+(-1) k ) #{g 0 .g 2u } = (1/p).(p k -(1/2).p k/2 .((-1) k/2 (1+(-1) k )+p 1/2 . (-1) (k+1)/2 (1+(-1) k+1 )) (51) #{g 1 .g 2u } (1/p).(p k -(1/2).p k/2 .((-1) k/2 .(1+(-1) k ))-p 1/2 .(-1) (k+1)/2 .(1+(-1) k+1 )) Hence the table in example: k = 1 k = 2 k = 3 k = 4 k = 5 #{0} 1
(1/p).(p k +(1/2).(p-1).p k/2 .(-1)

k(p-1)/4 (1+(-1) k ) #{g 0 .g 2u } = (1/p).(p k -(1/2).p k/2 .((-1) k(p-1)/4 (1+(-1) k )+p 1/2 . (-1) (k+1)(p-1)/4 (1+(-1) k+1 ))

(52) #{g 1 .g 2u } (1/p).(p k -(1/2).p k/2 .((-1)
k(p-1)/4 .(1+(-1) k ))-p 1/2 .(-1) (k+1)(p-1)/4 .(1+(-1) k+1 ))

Foot-note :

P 78/390
Modulo pi abundance factors for Waring sums The reader will be able to verify that the expressions obtained above give the cardinals in awaited order (#{0}, #{g 0 .g 2u }, #{g 1 .g 2u }). This is not obvious, a priori, because it depends on a good correspondence during attribution of eigenvalues and eigenvectors. By tests and errors, it is easy to find this order.

Beyond the literal expressions, certainly interesting, it is essential to check if these expressions cannot lie within a more broad scope. What we do now. Let us have μ i eigenvalues of A and λ i eigenvalues of B. Let us have μ i *et λ i * complex conjugate of μ i and λ i . Then, according to the results obtained higher, we have well :

1 λ 0 */2 λ 0 */2 P A = (1/p 1/2

Primitive roots and the prime numbers test

There is equivalence between p is prime and there is g such as -(i) (p-1)/2 .p 1/2 = 1+2 ∑e (-2πi/p).g^(2r) (57) r = 0 to (p-1)/2-1 Not knowing g a priori, its choice is critical :

-If, for given g (2 ≤ g ≤ p-2), relation (57) is verified, then p is prime, -If, for given g, relation ( 57) is verified on the real part or the imaginary part, but not on the whole set of this relation (imaginary and real together), then g is not primitive root of p (and another value is then tested), -If, for given g, relation ( 57) is verified neither on the real part, nor on the imaginary part, p is not a prime number.

Warning

This test is based on cyclic groups modulo p. It does not bring anything new as test of prime compared to what rises from a direct study of these groups.

Proof

The relation is true for p prime number. The eigenvalues expression is obtained in two ways each actually demonstrated. The reciprocal rises from little Fermat theorem and Chinese theorem. When p is not prime the right member of (57) presents a surplus of terms hampering equality.

Example : p = 5,g= The two last lines of last column contribute to a sum on sines different from zero. (-0,94722 = -0,75419-0,19303).

The two columns of right-hand side do not have null sums since all the cycles on primitive roots are not complete. There is failure on this completeness as soon as q is not any more a prime number. Indeed, the recurrent part is sum in the complex plan of points distributed uniformly around the origin. When truncation of part of the points is not symmetrical, the sum locus is isolated from origin which is the case here.

Foot-note

For great prime numbers, it is essential to detect the recurrent part (in blue) of the table, to deduce the not-cyclic part from it and thus to save computing time for eigenvalues evaluation.

Case of hyperspheres, logarithmic mesh

y 1 2 + y 2 2 + … + y k 2 = c (58)
Here y i are positive prime numbers. For p = 2, B is written according to the case general

B = 0 1 1 0
For odd p, d = (n,p-1) = (2,p-1) = 2. Thus matrix B is square of dimension 2+1 = 3. We write it :

0 p-1 0 0 p-1 0 B = 2 #{(0,0)} #{(1,0)} = 2 B' 0 #{(0,1)} #{(1,1)} 0
The already registered values rise from the already demonstrated general relations. Here #{(x,y)} corresponds to the cardinals of the solutions modulo p of equations g 2i +g 2j .g y-2 = g x-2 (i and j varying in the usual field of definition), that is :

B' = #{g 2i +g 2j = g 2k } #{g 2i +g 2j .g = g 2k } #{g 2i +g 2j = g 2k g} #{g 2i +g 2j .g = g 2k g} B' being the part of matrix B which interests us for the moment. After rearrangement of terms, and with g primitive root of p (and necessarily g (p-1)/2 = -1)

B' = B'1 B'3 = #{g 2i -g 2k = -g 2j } #{g 2i -g 2k = -g 2j g} B'2 B'4 #{g 2i -g 2j .g (p-1)/2 = -g 2k .g.g (p-1)/2 } #{g 2j -g 2k = -g 2i g -1 }
Here (B'3) and (B'4) are clearly equivalent by permutation i → j, k → k, j → i-1. If p = 1 mod 4, let us have (p-1)/2 = 2r, then (B'2) and (B'3) are equivalent by permutation i → i, k → j+r, j → -k. If p = 3 mod 4, let us have (p-1)/2 = 2r+1, then (B'1) and (B'4) are equivalent by permutation i → k, k → j, j → i+r.

Thus B is written :

Case 1 mod 4 0 p-1 0 B = 2 x 1 -2 x 2 0 x 2 x 2 Case 3 mod 4 0 p-1 0 B = 2 x 2 x 2 0 x 1 x 2
The same result is obtained by geometrical construction (cf. property 10). It suffices then to remember that each line has as sum p-1 to lead to the sought relations :

For p = 1 mod 4 0 p-1 0 B = 2 (p-5)/2 (p-1)/2 0 (p-1)/2 (p-1)/2 For p = 3 mod 4 0 0 p-1 B = 2 (p-3)/2 (p-3)/2 0 (p+1)/2 (p-3)/2
The enumerations are then (except case p = 2) (after having taken care to verify the exactitude of result for k = 1)

k #{0} 1 #{g 0 .g 2u } = B 0 #{g 1 .g 2u } 0
Then, we seek, for B k evaluation, the eigenvalues and eigenvectors of B.

For p = 2 λ 0B = -1 λ 1B 1 For p = 1 mod 4 : λ 0B p-1 λ 1B = -1-√p λ 2B -1+√p
For p = 3 mod 4 :

λ 0B p-1 λ 1B = -1+i.√p λ 2B -1-i.√p
Again, we notice the simplicity of the passage of case p = 1 mod 4 to case p = 3 mod 4 thanks to imaginary number -i. We have μ 2 = μ 1 *.

For p = 2, thus :

#{0} = (1/2).(1+(-1) k ) (59) #{g u } (1/2).(1-(-1) k )
For p = 1 mod 4 :

#{0} 1 (p-1)/2 (p-1)/2 (p-1) k 0 0 1 (p-1)/2 (p-1)/2 1 #{g 0 .g 2u } = (1/p) 1 (-1-√p)/2 (-1+√p)/2 0 (-1) k (p 1/2 +1) k 0 1 (-1+√p)/2 (-1-√p)/2 0 #{g 1 .g 2u } 1 (-1+√p)/2 (-1-√p)/2 0 0 (p 1/2 -1) k 1 (-1-√p)/2 (-1+√p)/2 0 All calculations done : #{0} (1/p).((p-1) k +(1/2).(p-1).((p 1/2 -1) k +(-1) k .(p 1/2 +1) k )) #{g 0 .g 2u } = (1/p).((p-1) k +(1/2).((p 1/2 -1) k+1 +(-1) k+1 .(p 1/2 +1) k+1 )) (60) #{g 1 .g 2u } (1/p).((p-1) k -(1/2).(p-1).((p 1/2 -1) k-1 +(-1) k-1 .(p 1/2 +1) k-1 )
)

Writing e x = (p 1/2 +1)/(p-1) 1/2 , we get e -x = (p 1/2 -1)/(p-1) 1/2 and x = Ln((p 1/2 +1)/(p-1) 1/2 ). Then the preceding expressions give two cases:

If k = 0 mod 2 : #{0}

(1/p).((p-1) k +(p-1) (k/2+1) .ch(k.Ln((p 1/2 +1)/(p-1) 1/2 )))

#{g 0 .g 2u } = (1/p).((p-1) k -(p-1) ((k+1)/2) .sh((k+1).Ln((p 1/2 +1)/(p-1) 1/2 ))) #{g 1 .g 2u } (1/p).((p-1) k +(p-1) ((k+1)/2) .sh((k-1).Ln((p 1/2 +1)/(p-1) 1/2 )))

If k = 1 mod 2 : #{0}

(1/p).((p-1) k -(p-1) (k/2+1) .sh(k.Ln((p 1/2 +1)/(p-1) 1/2 )))

#{g 0 .g 2u } = (1/p).((p-1) k +(p-1) ((k+1)/2) .ch((k+1).Ln((p 1/2 +1)/(p-1) 1/2 ))) #{g 1 .g 2u } (1/p).((p-1) k -(p-1) ((k+1)/2) .ch ((k-1).Ln((p 1/2 +1)/(p-1) 1/2 )))

All of these expressions are of course integers.

As examples, we constituted the following table :

k = 1 k = 2 k = 3 k = 4 k = 5 #{0} 0 2(p-1) (p-1).(p-5) (p-1).(p 2 -2p+9) (p-1).(p 3 -4p²+p-14) #{g 0 .g 2u } 2 p-5 p 2 -2p+9 p 3 -4p²+p-14 p 4 -5p 3 +11p 2 +5p+20 #{g 1 .g 2u } 0 p-1 (p-1).(p-3) (p-1).(p 2 -3p+6) (p-1).( p 3 -4p 2 +5p-10)
For 3 mod 4 :

#{0} 1 (p-1/2 (p-1)/2 (p-1) k 0 0 1 (p-1)/2 (p-1)/2 1 #{g 0 .g 2u } = (1/p) 1 (-1-i√p)/2 (-1+i√p)/2 0 (i.p 1/2 -1) k 0 1 (-1+i√p)/2 (-1-i√p)/2 0 #{g 1 .g 2u } 1 (-1+i√p)/2 (-1-i√p)/2 0 0 (-1) k (i.p 1/2 +1) k 1 (-1-i√p)/2 (-1+i√p)/2 0
All calculations done : #{0}

(1/p).((p-1) k +(1/2).(p-1).((-1+i.p 1/2 ) k +(-1-i.p 1/2 ) k ))

#{g 0 .g 2u } = (1/p).((p-1) k +(1/2).(p+1).((-1+i.p 1/2 ) k-1 +(-1-i.p 1/2 ) k-1 )))

(61)

#{g 1 .g 2u } (1/p).((p-1) k +(1/2).((-1+i.p 1/2 ) k+1 +(-1-i.p 1/2 ) k+1 )))
As (1/2).(-1+i.p 1/2 ) k +(-1-i.p 1/2 ) k = (p+1) k/2 .cos(k.arcos (-1/(p+1) There again, all of these expressions are integers.

Foot-note :

The reader will be able to check that above expressions give cardinals in awaited order (#{0}, #{g 0 .g 2u }, #{g 1 .g 2u }).

As examples, we constituted the following table :

k = 1 k = 2 k = 3 k = 4 k = 5 #{0} 0
λ 0 /2 λ 0 /2 P -1 B = 1/p 1/2 1 λ 1 /2 λ 2 /2 1 λ 2 /2 λ 1 /2 then #{0} = (1/p) [λ 0 n 1 + 1/2 (λ 1 k λ 0 * + λ 2 k λ 0 * )] #{g 0 .g 2u } 1 λ 1 * λ 2 * #{g 1 .g 2u } 1 λ 2 * λ 1 *
We have also : ∑ μ i = 0 and ∑ μ i μ j = -(-1) (p-1)/2 p and ∑ μ i 2 = (-1) (p-1)/2 2p i ≠ 0 i ≠ 0 j ≠ 0 i ≠ 0 i ≠ j ∑ λ i = -2 and ∑ λ i λ j = 1-(-1) (p-1)/2 p and ∑ λ i 2 = 2((-1) (p-1)/2 p+1) i ≠ 0 i ≠ 0 j ≠ 0 i ≠ 0 i ≠ j

Case of 3_hypersurfaces, unit mesh

Generality

Let us have :

x 1 3 + x 2 3 + … + x k 3 = c (63)
Here x i are positive integers. The analysis is carried out modulo p i . Sequence p i describes the prime numbers set and each variable is replaced by its representative [0 3 , 1 3 , 2 3 , …, (p i -1) 3 ] when building a k dimensions table. Table elements are obtained by modulo p i sums.

Matrices form

Like previously, we use relation A = B + I to determine the cardinal matrices. The reader will refer to next paragraph for evaluations of B from which we deduce the continuation immediately.

Matrix A is written : For p = 2, p = 3 mod 6 and p = 5 mod 6, we gat d = (n,p-1) = (3,p) =1 (hence dim(A) = 2) and :

A = 1 p-1 1 p-1
The corresponding enumeration was studied previously and we have :

P 83/390
Modulo pi abundance factors for Waring sums

#{0} = p k-1 #{g u } p k-1
For p = 1 mod 6 (according to B in the following paragraph)

1 p-1 0 0 A = 3 x 1 -2 x 2 x 3 0 x 2 x 3 +1 x 1 0 x 3 x 1 x 2 +1

Determinant

We have by simple development of the terms :

Det(A-μI) = μ 4 -(x 1 +x 2 +x 2 +1)μ 3 +(x 1 x 2 +x 2 x 3 +x 3 x 1 -x 1 2 -x 2 2 -x 3 2 +3x 1 -3p)μ 2 +(x 1 3 +x 2 2 +x 3 3 -3x 1 x 2 x 3 -2x 1 x 2 -2x 1 x 3 +x 2 x 3 - x 1 2 +2x 2 2 +2x 3 2 -3x 1 +3x 2 +3x 3 +3(p-1)(x 2 +x 3 )+6p-1)μ+p².(3x 1 -p-1)
Using x 1 +x 2 +x 3 = p-1, we get :

Det(A-μI) = μ 4 -p.μ 3 -(3(x 1 2 +x 1 x 2 +x 2 2 -px 1 -(p-1)x 2 )+(p-1) 2 +3p).μ 2 +p(3(x 1 2 +x 1 x 2 +x 2 2 -(p+1)x 1 +p 2 +2p+2).μ+p².(3x 1 -p-1) Conjecture 1 From x 1 +x 2 +x 3 = p-1, it follows (x 1 +x 2 +x 3 ).(x 2 -x 3 ) -(p-1).(x 2 -x 3 ) = 0 then, 3(x 1 2 +x 1 x 2 +x 2 2 )-3px 1 -3(p-1)x 2 +(p-1) 2 = 3(x 1 2 +x 1 x 3 +x 3 2 )-3px 1 -3(p-1)x 3 +(p-1) 2
We can also write

(x 1 +x 2 +x 3 ).(x 1 -x 2 ) -(p-1).(x 1 -x 2 ) = 0 then, 3(x 1 2 +x 1 x 3 +x 3 2 )-3px 1 -3(p-1)x 3 +(p-1) 2 = 3(x 2 2 +x 2 x 3 +x 3 2 )-3(p-2)(x 2 +x 3 )+(p-1)(p-4)
We observed that the members of the preceding equality cancel themselves at each test carried out on p (checking up to p = 8677). Hence, a priori, remarkable identities between the couples of integers (x 1 ,x 2 ), (x 1 ,x 3 ), (x 2 ,x 3 ) :

3(x 1 2 +x 1 x 2 +x 2 2
)-3px 1 -3(p-1)x 2 +(p-1) 2 =0 3(x 1 2 +x 1 x 3 +x 3 We note that x 2 and x 3 are not differentiable in this system of equations (same symmetries), peculiar point that we will find again further. Hence the general form x 2 +xy+y 2 -p/3 = 0 (66) alternatively associated with couples :

(x,y) = (x 1 -(p+1)/3,x 2 -(p-2)/3) (x,y) = (x 1 -(p+1)/3,x 3 -(p-2)/3) (67) (x,y) = (x 2 -(p-2)/3,x 3 -(p-2)/3) Let us pose X = x/√p, Y = y/√p, then : X 2 +XY+Y 2 -1/3=0 (68)
After translations of (p-1)/3 or (p-2)/3 following x axis and (p+2)/3 following y axis, solutions locus are on an ellipse of large axis 2.√(2p/3) tilted with angle -π/4 and small axis 2.(√(2p))/3 tilted with angle +π/4. Couples of integers (x 1 ,x 2 ), (x 1 ,x 3 ), (x 2 ,x 3 ) are on conic curves (and two on top on the same one).

In polar coordinates, these conics can be written also :

x 1 = (p+1+2.√p.cos(θ))/3 (69) x 2 = (p-2+2.√p.cos(-2π/3+θ))/3 x 3 = (p-2+2.√p.cos(2π/3+θ))/3 with the proviso of posing : θ = if(x 3 >x 2 ,-1,1

1 = 2f (71) s 1 = 2f 1 = 3x 1 -p-1 s 2 = 2f 2 = 3x 2 -p+2 s 3 = 2f 3 = 3x 3 -p+2
We will reconsider the noted expressions s i = 2f i when we approach the concept of constructive matrix of sequence at exercise (9) while noticing already here (d = 3) that :

s i = 1 mod d (72)
Indeed, replacing in matrix A the triplet (x 1 ,x 2 ,x 3 ) by the triplet (s 1 ,s 2 ,s 3 ), the previous formula becomes somewhat obvious: The reason of this indiscernibility is elementary. We know that cardinal matrix depends on choice of primitive root g. For d = 3, there exists φ (d=3) = 2 configurations according to whether we choose family {g 0 , g 1 , g 2 }.g 3u or {g 0 , g 2 , g 1 }.g 3u to build the cardinal matrix. This can be taken into account by a condition on permutation of x 2 and x 3 (such as x 2 > x 3 ).

1 p-1 0 0 A = 3 (p-

Determinant evaluation

Conjecture 1 allows us to draw a very simple expression of the determinant.

From Det(A-μI) = μ 4 -p.μ 3 -(3(x 1 2 +x 1 x 2 +x 2 2 -px 1 -(p-1)x 2 )+(p-1) 2 +3p).μ 2 +p(3(x 1 2 +x 1 x 2 +x 2 2 -(p+1)x 1 +p 2 +2p+2
).μ+p².(3x 1 -p-1) and 3(x 1 2 +x 1 x 2 +x 2 2)-3px 1 -3(p-1)x 2 +(p-1) 2 = 0, we get :

det(A-μI) = μ 4 -p.μ 3 -3pμ 2 +p(-3x 1 +4p+1)μ +p².(3x 1 -p-1)
Then :

det(A-μI) = μ 4 -p.μ 3 -3pμ 2 +p(3p-2f)μ +2p².f = (μ-p)(μ 3 -3p.μ-2p.f) Hence (taking μ = 0) : det(A) = p 2 .(2f) = p 2 .s 1 (73)

Remarkable identities

From e -i.θ +e -i.(θ-2π/3) +e -i.(θ+2π/3) = 0 and cos²(θ)+sin²(θ) = 1, it is easy to deduce : cos(θ)+cos(-2π/3+θ)+cos(2π/3+θ) = 0 (74) cos²(θ)+cos²(-2π/3+θ)+cos²(2π/3+θ) = 3/2 (75)

The first identity gives (p+1+2√p.cos(θ))/3+(p-2+2√p.cos(-2π/3+θ))/3+(p-2+2√p.cos(2π/3+θ))/3 = p-1, hence the earlier formula :

x 1 +x 2 +x 3 = p-1

The second identity yields : p = ((3x Starting with p = r 2 +3s 2 ,we can deduce 3 « trivial » solutions associated with 4p = v 2 +3w 2 noticing that :

4p = (2r) 2 +3(2s) 2 = (v 0 ) 2 +3(w 0 ) 2 4p = (r-3s) 2 +3(r+s) 2 = (v 1 ) 2 +3(w 1 ) 2 4p = (r+3s) 2 +3(r-s) 2 = (v 2 ) 2 +3(w 2 ) 2
The first identity is obvious and will not interest us. The interest of the two other decompositions is in some possibilities of writing simplifications as we will see later (using alternatively the couples (r,s) and (v,w)).

Classification into families

We may find groups of pertinent results by one or the other choice of parameters (t 1 ,t 2 ), (r,s), (v 1 ,w 1 ) or (v 2 ,w 2 ). The most efficient approach seems to be to consider first the following families (relations 82) . The distinction between « primed » and « non-primed » families is related to the indiscernibility of x 2 and x 3 (relations 83) :

Families

Families Order 1, 2, 3, 4, 5 or 6

x 2 > x 3 1', 2', 3', 4', 5' or 6' x 3 > x 2

Conjecture 2

Let us come back to the expression of the matrix A determinant : det(A) = p 2 .(2f) = p 2 .s 1 .

The sign of this determinant is that of f, that is also that of 3x 1 -p-1.

That sign, noted underneath s (A), is related to the belonging to one of the 12 previous families according to the conjecture : Families s(A) 3, 4, 6, 3', 4', 6' 1 (84) 1, 2, 5, 1', 2', 5' -1

Conjecture 3

We now observe that : Parameter t is also related to the parameters v 1 and v 2 in the earlier given decomposition of 4p. It identifies with the combination ±or(0,1).v 1 ±or(0,1).v 2 which is positive and a multiple of 3, so that :

x 1 = (p+1+s(A).(4p-3t 2 ) 1/2 )/3 x 2 = (p-1-x 1 +(-1) m .t)/2 (85) x 3 = (p-1-x 1 -(-1) m .t)/2 where (x 2 ≠x 3 ) m = 0 if x 2 > x 3, m = 1 if x 2 <
P 86/390
Modulo pi abundance factors for Waring sums

t = if(v 1 mod 3 = 0, v 1 , if(v 2 mod 3 = 0, v 2 , if(v 1 +v 2 mod 3 = 0, v 1 +v 2 , abs(v 1 -v 2 ))) (87)
This parameter t allows a rapid research of the x 1 value without using the primitive roots of p and enumeration of the couples (u,v) such as 1 = g 3u + g 3v modulo p. Indeed, the domain of definition of t in (4p-3t 2 ) 1/2 is limited by (generating already the square root economy) :

Df(t) = ]0, 2(p/3) 1/2 [
Collecting integer values (4p-3t 2 ) 1/2 by varying t by integer increment within that domain, we get at most three solutions to our diophantine problem. Then reducing the admissible values of t to 0 modulo 3 (as expressed by relations 86), there remains a unique solution. Going back to x 1 = (p+1+s (A).(4p-3t 2 ) 1/2 )/3, we get two values for x 1 when we alternatively attribute to s(A) the value 1 and -1. The alternative which yields an integer is the good one and is the unique solution to remember here.

This conjecture is remarkable since it establishes a footbridge between the field of enumeration and the field of arithmetic resolution. We get across from a diophantine enumeration of several equations (with exponents and primitive root) -i.e. #(u,v) / g i = g 3u +g j .g 3v mod p -to a "standard" diophantine equation (without exponent nor primitive root) -i.e. {(t 1 ,t 2 ) \ p = t 1 2 +t 1 t 2 +t 2 2 } -.

Let us note, to finish with, that by the method of the primitives, the machine time to get x 1 up to p = 37 243 (for example) is several hours, while one can calculate these values up to p = 821 551 (for example) in a few seconds by hereby method. We get x 2 and x 3 at the same time, but we cannot discern which is which.

Conjecture 4

It is a variant of the preceding conjecture.

Let us bring together two earlier equations :

p = ((3x 1 -p-1) 2 +(3x 2 -p+2) 2 +(3x 3 -p+2) 2 )/6 p = ((t 1 +2t 2 ) 2 +(t 1 -t 2 ) 2 +(-(2t 1 +t 2 )) 2 )/6
We get then the Bringing together x 1 = (p+1+2.√p.cos(θ))/3 and x 1 = (p+1+s (A).(4p-3t 2 ) 1/2 )/3, we get two relations, the second one taking into account the choice made at (70) :

cos(θ) = s(A).(1-3t 2 /4p) 1/2 (89) sin(θ) = if(x 3 >x 2 ,-1,1).(1/2).(3/p) 1/2 .t
A numerical simulation (up to p = 821551) gives an equidistribution of the values collected with arcsine((√3/2).(t/√p)) inside ]0, π/2[, that is also with θ in the first quadrant. Let us have θ i one sample of the angles θ and #(θ p ) the number of angles obtained for our trials between 7 and p (that is about half of the prime numbers between 7 and p). The probability to get an angle between θ i and θ i +π/#(θ p ) is thus :

Prob(θ \ θ i ≤ θ < θ i +π/#(θ p ), p → +∞) = 1/#(θ p ) (90)
For the underneath chart, we collected the angles θ i for p ≤ 821551 with increasing values and we numbered them from n = 1 to n = 32712 (p = 3 and p = 5 mod 6 being exclude from the list). We can see that the dispersion about a straight line is no longer visible and so will it certainly be as p tends towards infinity.

P 87/390

Modulo pi abundance factors for Waring sums This equidistribution is certainly connected with the Sato-Tate conjecture, which usual form is given by the relations a p (E) = p+1-#E(F p ) and cos(θ p (E)) = a p (E)/(2√p) and lim p→∞ #{E ϵ E p : α ≤ θ p (E) ≤ β}/#(E p ) = (2/π).∫ α,β sin²(θ).dθ. It shows here in a much simpler and bare form as it is downstream of the generating procedure involved in elliptic curves. It would impact so far only the elliptic curves like y 2 = x 3 +c, the effect of x 3 being studied here (and not that of x 3 +a.x). The Sato-Tate conjecture is thus closely linked to the decomposition of p in integers (as t 1 2 +t 1 t 2 +t 2 2 or r 2 +3s 2 here). We will get more general decompositions of p at exercise 9 (starting at &21) using specific matrices.

The study of the distributions of (r,s) and (t 1 ,t 2 ) enables to get another point of view for the origin of that equidistribution. Let us start from 1 = sin 2 (α)+cos 2 (α), it follows :

1 = sin 2 (α)+sin 2 (-α+π/2)

The last term shows the symmetry of the two squares. Varying α between 0 and π/2, we get values of sin(α) between 0 and 1. Random drawings, without constraints here, give a set of equidistributed values α in interval ]0, π/2[.

In addition, the decomposition p = r 2 +3s 2 write :

1 = (r/(p) 1/2 ) 2 +(s/(p/3) 1/2 ) 2 By trivial identification, we get the two relations : r/√p = sin(α) s/√p = (1/3) 1/2 sin(-α+ π/2 mod 2π)

Our search being only on positive values of r and s. The trigonometric identity brings no constraint and the statistic phenomena can prevail fully. Our random drawings are done by the choice of prime numbers p. Numerous distinct drawings should a priori result in an equidistribution of the angles α between 0 and π/2. We adopt of course systematic drawings of p (here all p ≤ 821551). The ratios (r/√p, s/√p) and (t 1 /√p, 2 /√p) are given underneath with the same presentation by increasing values :

The horizontal axis is referred as fp instead of p because, as we said earlier, the points obtained in ordinates are submitted in increasing values. Let us have n the index of p (that is p 0 = 3, p 1 = 5, p 2 = 7, p 3 = 11…). It follows then : r ≈ p 1/2 .sin(n/n max ) ≈ p 1/2 .sin((p/ln(p))/(p max /ln(p max ))) s ≈ (1/3) 1/2 .p 1/2 .sin(n/n max ) ≈ (1/3) 1/2 .p 1/2 .sin((p/ln(p))/(p max /ln(p max )))

The numerical results are those awaited by statistics.

We get the second graph for (t 1 ,t 2 ) by studying the different (modulo) cases relating the couples (t 1 ,t 2 ) and the couples (r,s). We leave to the interested reader the leisure of this peculiar task.

Of course, the curves are not linear, but we observe a yielding at higher values, the effect (of sinus) being less marked for the couples (t 1 ,t 2 ) than for the couples (r,s). Let us note also the similitude of the curves concerning s/√p and r/√p with a ratio of ordinates equal to (1/3) 1/2 , ratio that we mentioned effectively just above.

As t 1 > t 2 > 0, no value of t 1 is never inferior to (p/3) 1/2 (noticing that (1/3) 1/2 ≈ 0,577). When we write p = t 1 2 +t 1 t 2 +t 2 2 , so long we don't impose t 1 > t 2 > 0, the equation is perfectly symmetrical in t 1 and t 2 . In a heuristic point of view, t 1 and t 2 are not discernible. This means that if the distribution of the values (t 1 /√p,t 2 /√p) is statistical, the slopes on the right of t 1 /√p and on the left of t 2 /√p are the same on the preceding graphic. That is what we observe. Conjecture 6. Generalisation of Sato-Tate conjecture.

It is again only an analogy to the said conjecture. Let us have a and c two distinct positive prime relative integers, (a,c) = 1, given in advance. Let p a prime number. The diophantine equation (r > 0, s > 0) p = a.r 2 +c.s 2 (91) has 0 or 1 unique solution. When one solution exists, we get by identification :

sin(α) = r/(p/a) 1/2 sin(-α+π/2) = s/(p/c) 1/2
Then taking values among the first quadrant :

α = arsine(r/(p/a) 1/2 ) (92) -α+π/2 = arsine(s/(p/c) 1/2 )
All numerical test that we made give an equidistribution of the angle α within the domain of definition ]0, π/2[ except a unique case, that is (a,c) = (1,1), where equality of a and c induces to make a choice on the order of r and s (r ≥ s or reverse) and equidistribution is than on ]0, π/4[ or on ]π/4, π/2[. Note also that we lose this equidistribution if we look at p = a.r 2 +c.s 2 +b and sin(α) = r/((p-b)/a) 1/2 and sin(-α+π/2) = s/((pb)/c) 1/2 where b is a non-null constant. The failure is also patent with p = a.r u +c.s v and sin(α) = (r u /(p/a)) 1/2 and sin(-α+π/2) = (s v /(p/c)) 1/2 where u or v are distinct from 2.

Conjecture 7 : Evaluation of x 1 -(p-1)/3

The initial goal was here to find a still faster method to determine x 1 than that proposed at conjecture 3. Our success is very partial, but the incongruity, in some way, of the heuristic distribution that follows gives the explanation. Let us pose :

Δx 1 = x 1 -(p-1)/3 and pm = (p mod 27 -1)/3 Then :

Δx 1 mod 9 = 2.pm+if(pm mod3=2;-3;0)+if(pm>5;-9;0)

This gives more information on (Δx 1 -(Δx 1 mod9))/9 constituting a first stage for evaluation of relation Δx The precise evolution of m = (Δx 1 -(Δx 1 mod9))/9 escape our understanding to this day (except for what follows). Nevertheless, its sign depends only on the modulo 3 values of t 1 and t 2 . We observe that (relations (94)) :

1 = f(p). p Δx 1 Δx 1 mod9 (Δx 1 -Δx 1 mod9)/9 p Δx 1 Δx 1 mod9 (Δx 1 -Δx 1 mod9)/9 p Δx 1 Δx 1 mod9 (Δx 1 -Δx 1 mod9)/9
Families Sign ( 4,6,3',4',6' Positive or null 1,2,5,1',2',5' Negative or null

The sign of (Δx 1 -(Δx 1 mod9))/9 is thus the sign of the determinant of A. Indeed, we have : m = (s(A).v+2-3.(Δx 1 mod9))/27 where 4p = v 2 +3w 2 (95)

Let us recall that only one of the three solutions of 4p = v 2 +3w 2 is actually the right solution here. It is where m is an integer. We get also with v = or(v 1 , v 2 , v 0 = 2r) :

m = s(A).(int(v/27)+or(0,-1)) (96)
For a given p, the maximal absolute value of m, equal to s(A).m, is obtained for w = 1 at best. So we have (s (A).m max +or(0,1)) 2 = (int(v max /27)) 2 = (int((4p-3.w min 2 ) 1/2 /27)) 2 ≤ (int((4p-3) 1/2 /27)) 2 ≤ (4p-3)/27². So that also : p ≥ ( 27 The p m = f(m) curve has approximately a parabola shape in the range displayed with significant gaps near the origin (even if the graph scale hides such gaps). We considered here that or(0,1) gives a mean value of 1/2 and we plotted thus the test curve p = 182,25.(m+1/2) 2 +0,75.

The regularity of the points near such a curve is not a surprise, but this is not the end of the story.

Representing the evolution of the populations at constant m, we obtain for a sample of the said parameter m the graph which follows :

P 90/390

Modulo pi abundance factors for Waring sums

We observe, at constant m, three parts :

-the zone without of solutions -the zone of germination and upgrading of populations (main stream between p m et p m+1 ) -the zone of compromise (all staffs homogenize gradually)

The set of curves seems tending at infinity to the same point.

Taken one after the other, the populations near the origin are typically:

#(p) ≈ if(p < p m , 0, c m .(p-p m ) rm ) (98) m = 0 m = 18
The r m coefficient increases with m while c m decreases. The ascent into the germination zone must be increasingly steep as m grows to allow numbers to catch up.

Of course, we do assume nothing here concerning asymptotic behaviour (neither shape of the curve like c m .(p-p m ) rm which is indeed incompatible with the general trend of the curves to point towards the same locus at infinity, nor even infinity of cardinals). We can also question the order of appearance of p m : have we always p m < p m+1 ? This is actually observed in the studied range. The chart below is an illustration of it.

If we classify the values of p m following opposite signs, that is, as p 0,p 1,p 2,p 3,p 4,etc., the curve of gaps like p next -p previous presents a much greater dispersal than in the previous graphic (hence to our opinion of little interest). We can P 91/390

Modulo pi abundance factors for Waring sums even notice a reversal for (p 2, p -3 ) = (919, 877).

Let us then look at the distribution of values θ based on the 1 to 6 families. To do this, let us write t 2 = α.t 1 . From t 1 > t 2 > 0, we get 0 < α < 1. As p = t sin 2 (θ) (3/4).(α) 2 /(1+α+α 2 ) (3/4)/(1+α+α 2 ) (3/4).(1+α) 2 /(1+α+α 2 ) sin 2 (θ) with 0 < α < 1 ]0,1/4[ ]1/4,3/4[ ]3/4,1[ θ mod π/2 ]0,π/6 There is no reason to believe that the said blending does not imply families 1, 1' and 2, 2' (for higher values p than those examined here). The interest of the previous chart is to show that the process of emergence of new values m takes place preferentially according to low values of θ (in each domain of definition) for the first samples of m (without total order with respect to p as does not indicate the chart however), phenomenon which is more pronounced for families 5, 5' and 6, 6'.

As we it reported above, this distribution is otherwise equiprobable for abs(θ) in its domain of definition ]0, π/2[. In particular, populations in the six families tend to the same order of magnitude which is not either suggested by the graph.

The equidistribution is also suggested within each family by statistical trials.

The study of m is not complete here, even statistically, as one can look now at the second, than the third, and so on, instance of m,, one of the a priori easy results being p ≥ (27 2 .(s (A).m w,max +or(0,1)) 2 +3.w 2 )/4 for the w th instance of m.

Eigenvalues and eigenvectors

We got already an expression of these eigenvalues and eigenvectors. We seek an alternative expression here. Let us start with previously found expression :

det(A-μI) = (μ-p)(μ 3 -3p.μ-2p.f)
Research of eigenvalues of [A] amounts cancelling det (A-μI), that is :

μ-p = 0 or μ 3 -3p.μ-2p.f = 0
Resolution of the last equation is carried out, according to Horner method [16] (knowing that all solutions are real). We get :

μ = t.z (t.z) 3 -3p.t.z-2p.f = 0 z = cos(θ) t = 2/√p
with cos(3θ) = f/√p and using cos(3θ) = 4cos 3 (θ) -3cos(θ).

Let us note that the solutions are indeed real if |f| < √p, condition which rises (also) immediately from ( 89), more exactly owing to the fact that the solutions (x 1 ,x 2 ), (x 1 ,x 3 ), (x 2 ,x 3 ) are on the ellipses described in (69).

Then we get eigenvalues : p 2.√p.cos((1/3).(arcos(f/√p))) 2.√p.cos((1/3).(-2π+arcos(f/√p))) 2.√p.cos((1/3).(2π+arcos(f/√p)))

We have to solve here the question of the order of these eigenvalues because all permutations do not give an eigenvectors matrix satisfying the adopted notations (coherent with order of λ i in matrices P B and P -1 B ). The following order is appropriate (what is checked by substitution). Let us pose :

θ = if(x 3 >x 2 ,-1,1).arcos(f/√p)+k.2π μ 0A = .p (99) μ 1A 2.√p.cos(θ/3) μ 2A 2.√p.cos((2π+θ)/3) μ 3A 2.√p.cos((-2π+θ)/3)
The three following circulant shifts on indices 1, 2 and 3 are also appropriate :

(0,1,2,3) ≡ (0,2,3,1) ≡ (0,3,1,2))
These permutations correspond to the three possible choices of k :

( k = 0 mod 3 ) or ( k = 1 mod 3 ) or ( k = 2 mod 3 )
But the three following ones (which is not circulant shifts are to be drawn aside (but only because of adopted notation choice). (0,1,3,2) ≡ (0,3,2,1) ≡ (0,2,1,3)

The eigenvalues of [B] = [A]-[I] are then :

λ 0A = .p-1 (100) λ 1A -1+2.√p.cos(θ/3) λ 2A -1+2.√p.cos((2π+θ)/3) λ 3A -1+2.√p.cos((-2π+θ)/3)
We also have for the associated eigenvectors matrix (P A = P B ) :

1 λ 0 /3 λ 0 /3 λ 0 /3 P B = (1/p 1/2 ). 1 λ 1 /3 λ 2 /3 λ 3 /3 1 λ 3 /3 λ 1 /3 λ 2 /3 1 λ 2 /3 λ 3 /3 λ 1 /3
Let us note here, with λ i * conjugate of eigenvalue λ i , and λ i all real, that this expression is identical to :

1 λ 0 */3 λ 0 */3 λ 0 */3 P B = (1/p 1/2 ). 1 λ 1 */3 λ 2 */3 λ 3 */3 1 λ 3 */3 λ 1 */3 λ 2 */3 1 λ 2 */3 λ 3 */3 λ 1 */3
It rises from trigonometric functions properties (of cosine function), that the reader may verify it by a simple evaluation of P -1 B .P B using afore suggested expressions, that inverse matrix of P B can be written :

1 λ 0 /3 λ 0 /3 λ 0 /3 P -1 B = (1/p 1/2 ). 1 λ 1 /3 λ 3 /3 λ 2 /3 1 λ 2 /3 λ 1 /3 λ 3 /3 1 λ 3 /3 λ 2 /3 λ 1 /3
P B and P -1 B show a remarkable simplicity. Let us note that the evaluation of P -1 B , except the first column, is not really useful to this stage of our enumerations, since P -1 B is multiplied on the right by the vector column (1,0,0,…, 0). It will be later.

All calculations done, the enumerations are then :

k #{0} 1 #{g 0 .g 3u } = A 0 #{g 1 .g 3u } 0 #{g 2 .g 3u } 0 That is : #{0} = (1/p) [μ 0 k 1 + 1/3 (μ 1 k λ 0 + μ 2 k λ 0 + μ 3 k λ 0 )] (101) #{g 0 .g 3u } 1 λ 1 λ 2 λ 3 #{g 1 .g 3u } 1 λ 3 λ 1 λ 2 #{g 2 .g 3u } 1 λ 2 λ 3 λ 1
All of these expressions are of course integers.

Alternative expressions of angle θ

There are three alternatives according to choice of x 1 , x 2 or x 3 :

θ = if(x 3 >x 2 ,-1,1).arcos((3x 1 -p-1)/(2.√p)) mod 2π θ = 2π/3+if(x 1 >x 3 ,-1,1).arcos ((3x 2 -p+2)/(2.√p)) mod 2π θ = -2π/3-if(x 1 >x 2 ,-1,1).arcos ((3x 3 -p+2)/(2.√p)) mod 2π
According to modulo 2π residue, eigenvalues order is not same from one to the other of these alternative choices. However, the order is well that of a circulant shift and is appropriate.

Alternative expression of the components

To find another expression of triplet (x 1 ,x 2 ,x 3 ), let us return to the identity B = P B .D B .P B -1 :

0 p-1 0 0 1 λ 0 /3 λ 0 /3 λ 0 /3 λ 0 0 0 0 1 λ 0 /3 λ 0 /3 λ 0 /3 3 x 1 -3 x 2 x 3 = 1/p 1 λ 1 /3 λ 2 /3 λ 3 /3 0 λ 1 0 0 1 λ 1 /3 λ 3 /3 λ 2 /3 0 x 2 x 3 x 1 1 λ 3 /3 λ 1 /3 λ 2 /3 0 0 λ 2 0 1 λ 2 /3 λ 1 /3 λ 3 /3 0 x 3 x 1 x 2 1 λ 2 /3 λ 3 /3 λ 1 /3 0 0 0 λ 3 1 λ 3 /3 λ 2 /3 λ 1 /3 That is : x 1 λ 0 2 +λ 1 λ 2 λ 3 x 2 = 1/3p λ 0 2 +(λ 1 2 λ 3 +λ 2 2 λ 1 +λ 3 2 λ 2 )/3 (102) x 3 λ 0 2 +(λ 1 2 λ 2 +λ 2 2 λ 3 +λ 3 2
λ 1 )/3 and in a condensed form :

3 x i = 1/(3.p) (λ 0 2 + Σ λ k λ k+2-i λ k+2 /3) (103) k=1
We can check, in addition, by direct calculation the remarkable identities for eigenvalues (relations (104) :

∑ μ i = 0 and ∑ μ i μ j = -3p and ∑ μ i 2 = 6p i ≠ 0 i ≠ 0 j ≠ 0 i ≠ 0 i ≠ j ∑ λ i = -3 and ∑ λ i λ j = -3(p-1) and ∑ λ i 2 = 3(2p+1) i ≠ 0 i ≠ 0 j ≠ 0 i ≠ 0 i ≠ j

Enumeration by only one column

Let us seek some simplifications with the evaluation of x 1 , x 2 and x 3 . For that purpose, let us return on expressions of type g i .g 3u +g j .g 3v , we get for u = 1 to (p-1)/3 andv = 1 to (p-1)/3

: #(u,v) / {g 3u +g 3v =0} = (p-1)/3 #(u,v) / {g 3u +g.g 3v =0} = 0 #(u,v) / {g 3u +g 2 .g 3v =0} = 0 and x 1 /3 = 1+#(u,v) / {g 3u +g 3v =1} = #(u,v) / {g 3u +g.g 3v =g 2 } = #(u,v) / {g 3u +g 2 .g 3v =g} x 2 /3 = #(u,v) / {g 3u +g 3v =g} = #(u,v) / {g 3u +g.g 3v =1} = #(u,v) / {g 3u +g 2 .g 3v =g 2 } x 3 /3 = #(u,v) / {g 3u +g 3v =g 2 } = #(u,v) / {g 3u +g.g 3v =g} = #(u,v) / {g 3u +g 2 .g 3v =1}
Three last expressions result from general relations given higher in the text. For first expression, (p-1)/3 is self-evidently a minimal enumeration. As it is maximum, there is an equality. The two other relations result as there are no more solutions to allot in the two-dimensional tables. At (x,y), we have value x+y mod p. At (x+1,y+1), we have x.g 3 +y.g 3 = (x+y).g 3 mod p. Each "diagonal" is thus a set of type g i .g 3u mod p, i fixed, u describing increments of 1. As g p-1 = 1 mod p, we have g 3u +g 3(v+(p-1)/3) = g 3u +g 3v mod p, a "diagonal" admits its full set (p-1)/3 further diagonal. The diagonal of zeros is given by g 3(u+(p-1)/6) +g 3u , that is a starting point at column (p-1)/6+1 = (p+5)/6 (= 6 here).

We are able then to make the enumeration of the elements of a two-dimensional table of type g i .gd .u +g j .gd .v by the analysis of only one column (or line). Indeed, let us consider last column. It is set 1+g 3u , u = 1 to (p-1)/3. These elements are representative of the table .  As

x 1 = 3+3.#(u,v)/{g 3u +g 3v =1} x 2 = 3.#(u,v)/{g 3u +g 3v =g} x 3 = 3.#(u,v)/{g 3u +g 3v =g 2 }
it suffices to count the occurrences of g 3u ,g.g 3v and g 2 .g 3w in set 1+g 3u , u = 1 to (p-1)/3. That is also : [1,(p-1)/3] In our example : 28 = g.g 3(v=5) , 17 = g.g 3(v=2) , 30 = g 3(v=5) , 9 = g 2 .g 3(v=10) , 5 = g 2 .g 3(v=6) , 16 = g 3(v=2) , 3 = g.g 3(v=10) , 24 = g.g 3(v=4) , 2 = g 3(v=8) That is 3 occurrences of type g 3u , 4 of type g.g 3v and 2 of type g 2 .g 3w , so that x 1 = 3.(1+3) = 12, x 2 = 3.(4) = 12, x 3 = 3.(2) = 6.

x 1 = 3+3.#(u)/{1+g 3u =g 3v } x 2 = 3.#(u)/{1+g 3u =g.g 3v } x 3 = 3.#(u)/{1+g 3u =g 2 .g 3v } u integer of
In the same way, two-dimensional tables g 3u +g.g 3v mod p and g 3u +g 2 .g 3v mod p give : [1,(p-1)/3] andx

x 1 = 3.#(u)/{g+g 3u =g 2 .g 3v } x 2 = 3.#(u)/{g+g 3u =g 3v } x 3 = 3.#(u)/{g+g 3u =g.g 3v } u integer of
1 = 3.#(u)/{g 2 +g 3u =g.g 3v } x 2 = 3.#(u)/{g 2 +g 3u =g 2 .g 3v } x 3 = 3.#(u)/{g 2 +g 3u =g 3v } u integer of [1,(p-1)/3]
Here, x 1 is obtained without adding 3 as there is no diagonal with zero value.

Case of the 3_hypersurfaces, logarithmic mesh

Let us have y

1 3 + y 2 3 + … + y k 3 = c (105)
Here y i are positive prime numbers.

For p = 2, p = 3 mod 6 and p = 5 mod 6, we have d = (n,p-1) = (3,p-1) =1 (hence dim(A) = 2)

B = 0 p-1 1 p-2
The corresponding enumeration was studied previously and leads to :

#{0} = (1/p).((p-1) k + (p-1).(-1) k ) #{g u } (1/p).((p-1) k -(-1) k ) If p = 1 mod 6 then (n,p-1) = (3,6k) = 3 and square matrix B is of dimension 1+3 = 4.
The sum of each line being p-1, we get by our standard geometrical construction :

0 p-1 0 0 B = 3 x 1 -3 x 2 x 3 0 x 2 x 3 x 1 0 x 3 x 1
x 2 where :

x 1 -3 #{ g 3i +g 3j .g = g 3k g 2 } x 2 = #{ g 3i +g 3j .g 2 = g 3k g 2 } x 3 #{ g 3i +g 3j = g 3k g 2 } These calculations done, enumerations are then given by : k

#{0} 1 #{g 0 .g 3u } = B 0 #{g 1 .g 3u } 0 #{g 2 .g 3u } 0
Locus as of the these values in abundance factors generating table is carried out according to the standard procedure of sorting. The eigenvalues of B were obtained previously :

λ 0B = p-1 λ 1B -1+2.√p.cos((1/3).(arcos (f/√p))) (106) λ 2B -1+2.√p.cos((1/3).(2.π+arcos (f/√p))) λ 3B -1+2.√p.cos((1/3).(-2.π+arcos (f/√p)))
and are associated to eigenvectors matrix (P A = P B )

1 λ 0 /3 λ 0 /3 λ 0 /3 P B = (1/p 1/2 ). 1 λ 1 /3 λ 2 /3 λ 3 /3 1 λ 3 /3 λ 1 /3 λ 2 /3 1 λ 2 /3 λ 3 /3 λ 1 /3 and inverse matrix 1 λ 0 /3 λ 0 /3 λ 0 /3 P -1 B = (1/p 1/2 ). 1 λ 1 /3 λ 3 /3 λ 2 /3 1 λ 2 /3 λ 1 /3 λ 3 /3 1 λ 3 /3 λ 2 /3 λ 1 /3
Finally, returning to expression B k = P B D B k P B -1 (and to B k .(1,0,0,0)), we get :

P 96/390
Modulo pi abundance factors for Waring sums

#{0} = (1/p) [λ 0 k 1 + 1/3 (λ 1 k λ 0 + λ 2 k λ 0 +λ 3 k λ 0 )] (107) #{g 0 .g 3u } 1 λ 1 λ 2 λ 3 #{g 1 .g 3u } 1 λ 3 λ 1 λ 2 #{g 2 .g 3u } 1 λ 2 λ 3 λ 1
There again, all of these expressions are integers, expressions which can be compare with those obtained for the hypersphere.

Case of 4_hypersurfaces, unitary and logarithmic meshes

General form

With equivalent permutations on (x,y) in g 4i +g 4j .g y-2 = g x-2 , we find relations of equality between the components of the matrices.

For prime p, two cases are possible : p = 1 mod 4 and p = 3 mod 4.

For p = 3 mod 4, we have d = (n,p-1) = ( 4,2) = 2, completely previously solved case.

For d = 4 and p = 1 mod 4 two types of matrices present :

If (p-1)/2 mod 4 = 0 ( p = 1 mod 8 ) d = (n,p-1) = (4,8k) = 4 0 p-1 0 0 0 4 x 1 -4 x 2 x 3 x 4 B(p,4) = 0 x 2 x 4 x 5 x 5 0 x 3 x 5 x 3 x 5 0 x 4 x 5 x 5 x 2 If (p-1)/2 mod 4 = 2 ( p = 5 mod 8 ) d = (n,p-1) = (4,4+8k) = 4 0 0 0 p-1 0 4 x 3 x 5 x 3 x 5 B(p,4) = 0 x 4 x 5 x 5 x 2 0 x 1 x 2 x 3 x 4 0 x 2 x 4 x 5 x 5
Then, from relations on lines and columns (that is p-1 for each one except for the first two columns whose sums are worth respectively 4 and 2(p-1)-4), it follows immediately :

If (p-1)/2 mod 4 = 0 ( p = 1 mod 8 ) x 1 +3x 3 -(p-1) = 0 x 1 -3x 5 +(p-1)/2 = 0 2x 1 +3x 2 +3x 4 -2(p-1) = 0 So that : 0 p-1 0 0 0 4 x 1 -4 x 2 (p-x 1 -1)/3 (2p-2x 1 -3x 2 -2)/3 B(p,4) = 0 x 2 (2p-2x 1 -3x 2 -2)/3 (p+2x 1 -1)/6 (p+2x 1 -1)/6 (108) 0 (p-x 1 -1)/3 (p+2x 1 -1)/6 (p-x 1 -1)/3 (p+2x 1 -1)/6 0 (2p-2x 1 -3x 2 -2)/3 (p+2x 1 -1)/6 (p+2x 1 -1)/6 x 2 If (p-1)/2 mod 4 = 2 ( p = 5 mod 8 ) x 1 +3x 3 -(p-5) = 0 x 1 -3x 5 +(p-5)/2 = 0 2x 1 +3x 2 +3x 4 -2(p+1) = 0 So that : 0 0 0 p-1 0 4 (p-x 1 -5)/3 (p+2x 1 -5)/6 (p-x 1 -5)/3 (p+2x 1 -5)/6 B(p,4) = 0 (2p-2x 1 -3x 2 +2)/3 (p+2x 1 -5)/6 (p+2x 1 -5)/6 x 2 (109) 0 x 1 x 2 (p-x 1 -5)/3 (2p-2x 1 -3x 2 +2)/3 0 x 2 (2p-2x 1 -3x 2 +2)/3 (p+2x 1 -5)/6 (p+2x 1 -5)/6

Conjectures about the components

We observe the following conjectures :

If p = 1 mod 8, 5x 1 2 +6x 1 x 2 +9x 2 2 -4(p+2)x 1 -6(p-1)x 2 +(5p 2 -34p+29)/4 = 0 (110) If p = 5 mod 8, 5x 1 2 +6x 1 x 4 +9x 4 2 -4(p+1)x 1 -6(p+1)x 4 +(5p 2 -26p+5)/4 = 0 (111)
Let us express x 2 in function of x 1 : For p = 1 mod 8,

x 2 = (p-1-x 1 )/3±(-4x 1 2 +2(p+5)x 1 -(p-1)(p-25)/4) 1/2
The expression under the root -4x 1 2 +2(p+5)x 1 -(p-1)(p-25)/4 must be positive. Its reduced discriminant is 36p. Hence :

(p-11)-6√p ≤ x 1 -4 ≤ (p-11)+6√p 4 4 and (p-3)-2√p ≤ x 2 ≤ (p-3)+2√p 4 4 If p = 5 mod 8, x 2 =3(p+1-x 1 )/3±(-4x 1 2 +2(p+1)(x 1 )-(p 2 -34p+1)/4) 1/2
The expression under the root -4x 1 2 +2(p+1)(x 1 )-(p 2 -34p+1 must be positive. Its reduced discriminant is again 36p. Hence :

(p+1)-6√p ≤ x 1 ≤ (p+1)+6√p 4 4 and (p-1)-2√p ≤ x 2 ≤ (p-1)+2√p 4 4
This confines (x 1 ,x 2 ) to a reduced number of couples of integers likely, a priori, to answer the problem (there is of course only one which is appropriate). In general, the integer solutions obtained a priori in this manner are not unique.

Synoptic of relations between matrix components

We find linear or quadratic forms for all relations between components of the dimension 4 matrices (with dependence to x 1 here) as it was already in case d = 3 and d = 2 but with larger complexity : 2x 3 x 4 +5x 4 2 -2(p-9)x 3 -4x 4 +(p-5)(p-13)/4 = 0

If p = 1 mod 8, 5x 1 2 +6x 1 x 2 +9x 2 2 -4(p+2)x 1 -6(p-1)x 2 +(5p 2 -34p+29)/4 = 0 or 5(x 1 -4) 2 +6(x 1 -4)x 2 +9x 2 2 -4(p-8)(x 1 -4)-6(p-5)x 2 +(5p-13)(p-17)/4 = 0 x 1 +3x 3 -(p-1) = 0 or (x 1 -4)+3x 3 -(p-5) = 0 5x 1 2 +6x 1 x 4 +9x 4 2 -4(p+2)x 1 -6(p-1)x 4 +(5p 2 -34p+29)/4 = 0 or 5(x 1 -4) 2 +6(x 1 -4)x 4 +9x 4 2 -4(p-8)(x 1 -4)-6(p-5)x 4 +(5p-13)(p-17)/4 = 0 x 1 -3x 5 +(p-1)/2 = 0 or (x 1 -4)-3x 5 +(p+7)/2 = 0 x 2 2 -2x 2 x 3 +5x 3 2 -2(p-3)x 3 +(p-1)(p-9)/4 = 0 5x 2 2 +6x 2 x 4 +5x 4 2 -4(p-3).(x 2 +x 4 )+(p-1)(p-9) = 0 x 2 2 +2x 2 x 5 +5x 5 2 -(p-1)x 2 -(3p+1)x 5 +(p-1) 2 /2 = 0 x 3 2 -2x 3 x 4 +5x 4 2 -2(p-3)x 3 +(p-1)(p-9)/4 = 0 x 3 +x 5 -(p-1)/2 = 0 x 4 2 +2x 4 x 5 +5x 5 2 -(p-1)x 4 -(3p+1)x 4 +(p-1) 2 /2 = 0 If p = 5 mod 8, 5x 1 2 +6x 1 x 2 +9x 2 2 -4(p+1)x 1 -6(p+1)x 2 +(p-5)(5p-1)/4 = 0 x 1 +3x 3 -(p-5) = 0 5x 1 2 +6x 1 x 4 +9x 4 2 -4(p+1)x 1 -6(p+1)x 4 +(p-5)(5p-1)/4 = 0 x 1 -3x 5 +(p-5)/2 = 0 x 2 2 -2x 2 x 3 +5x 3 2 -4x 2 -2(p-9)x 3 +(p-5)(p-13)/4 = 0 5x 2 2 +6x 2 x 4 +5x 4 2 -4(p+1).(x 2 +x 4 )+(p-1) 2 = 0 x 2 2 +2x 2 x 5 +5x 5 2 -(p-1)x 2 -(3p-7)x 5 +(p-1)(p-5)/2 = 0 x 3 2 -
x 3 +x 5 -(p-5)/2 = 0 x 4 2 +2x 4 x 5 +5x 5 2 -(p-1)x 4 -(3p-7)x 5 +(p-1)(p-5)/2 = 0

Determinant and eigenvalues

It is again a question of finding alternative expression to the general kind. Relations ( 110) and ( 111) are useful in the research of eigenvalues of matrices A(p,4) and B(p,4) in the two kind of p congruencies. By seeking the solutions of det(A-μI) = 0, we are led to two very long expressions in μ, x 1 and x 2 (that we do not give literally here for lack of place).

The first case p = 1 mod 8 leads to a set of real eigenvalues and eigenvectors, the second case p = 5 mod 8 to a set of imaginary eigenvalues and eigenvectors. Indeed :

For p = 1 mod 8, characteristic equation of eigenvalues μ i becomes after use of the relation ( 110), some pages of calculations and other ups and downs :

-(μ-p)(μ 4 -6p.μ 2 -4p.f.μ+p.(p-f 2 )) = 0 where f = (4x 1 -p-5)/3 (112)
Resolution of equation μ 4 -6p.μ 2 -4p.f.μ+p.(p-f 2 ) = 0, described for example in [16], uses Ludovici Ferrari method (for degree 4), then Horner method (for degree 3) (in absence of obvious solution) :

Let us add each side of the preceding equation (m.μ+b) 2 and identify expression obtained to the left with square μ 4 +2hμ 2 +h 2 . Then we get by identification of coefficients m

2 -6p = 2h, h 2 = p(p-f 2 )+b 2 , -4p.f+2m.b = 0. It follows b = 2p.f/m and h 2 = p.(p-f 2 )+4p 2. f 2 /m 2 = p.(p-f 2 )+2p 2. f 2 /(h+3p).
Hence the equation of degree 3 resolvent :

h 3 +3.p.h 2 -p.(p-f 2 ).h-p 2 (3p-f 2 ) = 0
This equation does not require use of Horner method for resolution, as we can easily verify by substitution that one of equation solutions is h = -p. Hence we deduce by Euclidean division h 2 +2p.h+p(f 2 -3p) = 0 and :

h = -p h = -p+(p.(4p-f 2 )) 1/2 h = -p-(p.(4p-f 2 )) 1/2
Any of these solutions gives the initial characteristic solutions of the equation by Ferrari method. Here, with h = -p, we have μ 2 +h ± (m.μ+b) = 0, m = (2(h+3p)) 1/2 = 2p 1/2 , b = p 1/2 .f, so that :

sol 1 = (-m-(m 2 -4(h+b)) 1/2 )/2 sol 2 = (m-(m 2 -4(h-b)) 1/2 )/2 sol 3 = (-m+(m 2 -4(h+b)) 1/2 )/2 sol 4 = (m+(m 2 -4(h-b)) 1/2 )/2 Then : sol 1 = p 1/2 .(-1-(2-f/p 1/2 ) 1/2 ) = p 1/2 .(-1-(2+(p+5-4x 1 )/(3.p 1/2 )) 1/2 ) sol 2 = p 1/2 .(1-(2+f/p 1/2 ) 1/2 ) = p 1/2 .(1-(2-(p+5-4x 1 )/(3.p 1/2 )) 1/2 ) (113) sol 3 = p 1/2 .(-1+(2-f/p 1/2 ) 1/2 ) = p 1/2 .(-1+(2+(p+5-4x 1 )/(3.p 1/2 )) 1/2 ) sol 4 = p 1/2 .(1+(2+f/p 1/2 ) 1/2 ) = p 1/2 .(1+(2-(p+5-4x 1 )/(3.p 1/2 )) 1/2 )
Order of solutions is to be established to still ensure an adapted solution within the framework of general adopted notations. We know that there exists φ(d=4) = 2 matrices forms according to such or such choice of the primitive root g at sequence p. Choosing {g 0 , g 1 , g 2 , g 3 }.g 4u , families {g 0r , g 1r , g 2r , g 3r }.g 4u are acceptable when (r,4) = 1, that is r = 1 (identity) or r = 3. Thus we have two acceptable families {g 0 , g 1 , g 2 , g 3 }.g 4u and {g 0 , g 3 , g 2 , g 1 }.g 4u . Permutation of g 1 and g 3 amounts giving a condition on the respective values of x 2 and x 4 :

μ 1 = p 1/2 .(-1-(2-f/p 1/2 ) 1/2 ) μ 2 = p 1/2 .(1-if(x 4 >x 2 ,-1,1).(2+f/p 1/2 ) 1/2 ) (114) μ 3 = p 1/2 .(-1+(2-f/p 1/2 ) 1/2 ) μ 4 = p 1/2 .(1+if(x 4 >x 2 ,-1,1).(2+f/p 1/2 ) 1/2 )
The circulant shifts of these eigenvalues are acceptable solutions. (We can build an eigenvectors matrix adapted to any order of eigenvalues but notations would evolve consequently losing generality).

We get, as in case d = 3, two cases x 4 >x 2 and x 2 >x 4 which are related to primitive root choice.

In a similar way, for p = 5 mod 8, characteristic equation of eigenvalues μ i is written :

μ 4 +2p.μ 2 -4p.f.μ+ p(9p-f 2 ) = 0 where f = (4x 1 -p-1)/3 (115)
The solution of this equation is again obtained using Ludovici Ferrari method by adding on each side of preceding equation (m.μ+b) 2 and by identifying the expression obtained on the left with a square μ 4 +2hμ 2 +h 2 . We get then m

2 +2p = 2h, h 2 = p(9p-f 2 )+b 2 , -4p.f+2m.b = 0. It follows b = 2p.f/m, so that h 2 = p(9p-f 2 )+4p 2. f 2 /m 2 = p(9p-f 2 )+2p 2. f 2 /(h-p).
Hence the equation of degree 3 resolvent :

h 3 -p.h 2 -p(9p-f 2 ).h+3p 2 (3p-f 2 ) = 0
We can verify by substitution that one of the solutions of equation is h = 3p. Hence we deduce by Euclidean division h 2 +2p.h+p(f 2 -3p) = 0 and :

h = 3p h = -p+(p.(4p-f 2 )) 1/2 h = -p-(p.(4p-f 2 )) 1/2
Any of these solutions gives the initial characteristic solutions of the equation by Ferrari method. We have μ 2 +h ± (m.μ+b) = 0,b = p 1/2 .f, so that :

sol 1 = p 1/2 .(-1-i(2+f/p 1/2 ) 1/2 ) = p 1/2 .(-1-i(2+(4x 1 -p-1)/(3p 1/2 )) 1/2 ) sol 2 = p 1/2 .(1-i(2-f/p 1/2 ) 1/2 ) = p 1/2 .(1-i(2-(4x 1 -p-1)/(3p 1/2 )) 1/2 ) (116) sol 3 = p 1/2 .(-1+i(2+f/p 1/2 ) 1/2 ) = p 1/2 .(-1+i(2+(4x 1 -p-1)/(3p 1/2 )) 1/2 ) sol 4 = p 1/2 .(1+i(2-f/p 1/2 ) 1/2 ) = p 1/2 .(1+i(2-(4x 1 -p-1)/(3p 1/2 )) 1/2 )
Let us note again, λ i and μ i the eigenvalues of matrices B(p,4) and A(p,4)+I (λ i = μ i -l, λ 0 = μ 0 -l = p-1). In addition, we have μ 3 = μ 1 * and μ 4 = μ 2 *.

The order of solutions is to be established to ensure an adapted solution. The condition (>) on relative values of x 2 and x 4 allows that, all circulant shifts being also solutions :

μ 1 = p 1/2 .(-1-i(2+f/p 1/2 ) 1/2 ) μ 2 = p 1/2 .(1-if(x 2 >x 4 ,-1,1).i(2-f/p 1/2 ) 1/2 ) (117) μ 3 = p 1/2 .(-1+i(2+f/p 1/2 ) 1/2 ) μ 4 = p 1/2 .(1+if(x 2 >x 4 ,-1,1).i(2-f/p 1/2 ) 1/2 )

Eigenvectors

We have then : Foot-note : Again, the equalities are verified only with correct order of (μ 1 , μ 2 , μ 3 , μ 4 ) and (λ 1 , λ 2 , λ 3 , λ 4 ) among a choice of 4! = 30 possible permutations.

P 100/390

Modulo pi abundance factors for Waring sums

Conjecture about f

As in case n = 3, we observe a dependence of f to a unique integer decomposition of p following a quadratic equation. Here p is 1 mod 8 or 5 mod 8 (thus 1 mod 4) and decomposition into two squares is a natural track to be explored. According to Leonard Euler, if p = 1 mod 4, there exists unique positive integers (a,b) such as p = a 2 +b 2 . As a and b are different parities, there are two integers α > 0 and β odd > 0 such as : p = (2α) 2 +β 2 Cas p = 1 mod 8

In this case, we have moreover α even. We get the conjecture : f = (4x 1 -p-5)/3 = (-1) (β+1)/2 .2.β Thus :

μ 1 = p 1/2 .(-1-(2.(1-(-1) (β+1)/2 .β/p 1/2 )) 1/2 ) μ 2 p 1/2 .(1-if(x 4 >x 2 ,-1,1).(2.(1+(-1) (β+1)/2 .β/p 1/2 )) 1/2 ) (118) μ 3 p 1/2 .(-1+(2.(1-(-1) (β+1)/2 .β/p 1/2 )) 1/2 ) μ 4 p 1/2 .(1+if(x 4 >x 2 ,-1,1).(2.(1+(-1) (β+1)/2 .β/p 1/2 )) 1/2 ) and x 1 (p+5)/4+(-1) (β+1)/2 .(3/2).β x 2 (p-3)/4+2α.if(x 4 >x 2 ,-1,1)-(-1) (β+1)/2 .(1/2).β x 3 = (p-3)/4-(-1) (β+1)/2 .(1/2).β (119) x 4 (p-3)/4-2α.if(x 4 >x 2 ,-1,1)-(-1) (β+1)/2 .(1/2).β x 5 (p+1)/4+(-1) (β+1)/2 .(1/2).β Case p = 5 mod 8
In this case, we have moreover α odd. Again, we can use :

f = (4x 1 -p-1)/3 = (-1) (β+1)/2 .2.β Thus : μ 1 = p 1/2 .(-1-i(2.(1+(-1) (β+1)/2 .β/p 1/2 )) 1/2 ) μ 2 p 1/2 .(1-if(x 2 >x 4 ,-1,1).i(2.(1-(-1) (β+1)/2 .β/p 1/2 )) 1/2 ) (120) μ 3 p 1/2 .(-1+i(2.(1+(-1) (β+1)/2 .β/p 1/2 )) 1/2 ) μ 4 p 1/2 .(1+if(x 2 >x 4 ,-1,1).i(2.(1-(-1) (β+1)/2 .β/p 1/2 )) 1/2 ) and x 1 (p+1)/4+(-1) (β+1)/2 .(3/2).β x 2 (p+1)/4+2α.if(x 4 >x 2 ,-1,1)-(-1) (β+1)/2 .(1/2).β x 3 = (p-7)/4-(-1) (β+1)/2 .(1/2).β (121) x 4 (p+1)/4-2α.if(x 4 >x 2 ,-1,1)-(-1) (β+1)/2 .(1/2).β x 5
(p-3)/4+(-1) (β+1)/2 .(1/2).β

Case of 5_hypersurfaces, unitary and logarithmic meshes

General points

If p = 3 mod 10, p = 5, p = 7 mod 10 or p = 9 mod 10 , then d = (5,p-1) = 1, completely solved case (matrix of dimension 2). If p = 1 mod 10, then d = (5,p-1) = 5 and characteristic matrices are of dimension 6. From equivalent permutations on (x,y) in g 5i +g 5j .g y-2 = g x-2 , it is easy to find equalities between some components of the characteristic matrices. Thus :

If (p-1)/2 mod 5 = 0 ( p = 1 mod 10 ) 0 p-1 0 0 0 0 5 x 1 -5 x 2 x 3 x 4 x 5 B(p,5) = 0 x 2 x 5 x 6 x 7 x 6 0 x 3 x 6 x 4 x 7 x 7 0 x 4 x 7 x 7 x 3 x 6 0 x 5 x 6 x 7 x 6 x 2
Then starting from the relations on lines and columns (that is sum p-1 for each one except for first two columns whose sums are respectively 5 and 2(p-1)-5), it immediately follows x 1 +x 2 +x 3 +x 4 +x 5 = p-1, x 2 +x 5 +2x 6 +x 7 = p-1 and x 3 +x 4 +x 6 +2x 7 = p-1, then x 6 = 2x 1 /3-x 3 -x 4 -(p-1)/3 and x 7 = -x 1 /3-x 3 -x 4 +2(p-1)/3, so that :

0 p-1 0 0 0 0 5 x 1 -5 x 2 x 3 x 4 -x 1 -x 2 -x 3 -x 4 +p-1 B(p,5) = 0 x 2 -x 1 -x 2 -x 3 -x 4 +p-1 2x 1 /3-x 3 -x 4 -(p-1)/3 -x 1 /3-x 3 -x 4 +2(p-1)/3 2x 1 /3-x 3 -x 4 -(p-1)/3 0 x 3 2x 1 /3-x 3 -x 4 -(p-1)/3 x 4 -x 1 /3-x 3 -x 4 +2(p-1)/3 -x 1 /3-x 3 -x 4 +2(p-1)/3 0 x 4 -x 1 /3-x 3 -x 4 +2(p-1)/3 -x 1 /3-x 3 -x 4 +2(p-1)/3 x 3 2x 1 /3-x 3 -x 4 -(p-1)/3 0 -x 1 -x 2 -x 3 -x 4 +p-1 2x 1 /3-x 3 -x 4 -(p-1)/3 -x 1 /3-x 3 -x 4 +2(p-1)/3 2x 1 /3-x 3 -x 4 -(p-1)/3 x 2

Cardinal matrix

Let us note, as previously, λ i and μ i the eigenvalues of matrices B(p,5) and A(p,5) + I (λ i = μ i -l, λ 0 = μ 0 -l = p-1).

Then we write for eigenvectors matrix : 5 λ 0 /5 λ 0 /5 λ 0 /5 λ 0 /5 1 λ 0 /5 λ 0 /5 λ 0 /5 λ 0 /5 λ 0 /5 1 λ 1 /5 λ 2 /5 λ 3 /5 λ 4 /5 λ 5 /5 1 λ 1 /5 λ 5 /5 λ 4 /5 λ 3 /5 λ 2 / A (et B) has 4 independent parameters at most. Indeed x 1 +x 2 +x 3 +x 4 +x 5 = p-1. Thus, there is a function g such as g(μ 1 ,μ 2 ,μ 3 ,μ 4 ,μ 5 ) = 0 and variables μ i are controlled by 4 independent parameters at most. We write this, a priori, in form of a parameter of module and of three parameters of angle a priori, that is :

1 λ 0 /
μ i+1 = ρ.cos(θ 1 +2πi/5).cos(θ 2 +2πi/5).cos(θ 3 +2πi/5), i = 0 to 4 (122) where ρ =ρ(p), θ 1 =θ 1 (p), θ 2 =θ 2 (p), θ 3 =θ 3 (p)
We note two conjectures ρ = 8√p and θ 2 = θ 1 +π/2 hence, we get immediately : μ i+1 = 4√p.sin(2(θ 1 +2πi/5)).cos(θ 3 +2πi/5), i = 0 to 4 (123) λ i+1 = 4√p.sin(2(θ 1 +2πi/5)).cos(θ 3 +2πi/5)-1, i = 0 to 4 (124)

This led after some developments and terms simplifications to :

5 Σ μ i = 0 i=1 5 Σ μ i 2 = 20p and Σ μ i .μ j = -5p i=1 i ≠ 0, j ≠ 0, i ≠ j 5 Σ μ i 3 = 30.p 3/2 .(sin(6θ 1 -θ 3 )-sin(2θ 1 +3θ 3 )) = 60.p 3/2 .sin(2(θ 1 -θ 3 )).cos(4θ 1 +θ 3 ) i=1 From equality B(p,5) = P B .[λ].P B -1
, we deduce for second line and second column component x 1 -5 = (1/5p).(λ 0 2 +(λ 1 3 + λ 2 3 +λ 3 3 + λ 4 3 +λ 5 3 )/5), so that :

5 Σ λ i 3 = 5(5p(x 1 -5)-(p-1) 2 ) i=1
Hence, using relations on sums and sums of squares of μ i , we get : , we deduce for second line and second column component :

5 Σ μ i 3 = 5p.(5x 1 -p-11) i=1 Then at least : sin(2(θ 1 -θ 3 ).cos(4θ 1 +θ 3 )) = (5x 1 -p-11)/(12√p) (125)
x 2 = (1/5p).( λ 0 2 +(λ 5 .λ 1 2 +λ 1 .λ 2 2 +λ 2 .λ 3 2 +λ 3 .λ 4 2 +λ 4 . λ 5 2 )/5).
Using λ i = μ i -l, Σμ i = 0, Σμ i .μ j = -5p, after some elementary handling, we get :

μ 5 .μ 1 2 +μ 1 .μ 2 2 +μ 2 .μ 3 2 +μ 3 .μ 4 2 +μ 4 .μ 5 2 = 5p(5x 2 -p+4)
Moreover, for the same expression, we have by developing functions of θ 1 and θ 3 : sin 2 (2(θ 1 +2π.i/5)).cos 2 (θ 3 +2π.i/5).sin(2(θ 1 +2π.(i-1)/5)).cos(θ 3 +2π.(i-1)/5) (126) i=1 i=1

5 5 Σ μ i 2 .μ i-1 = Σ 64.p 3/2 .
This expression does not have unfortunately the simplicity of (125). By bringing closer equations ( 125) and ( 126), we can, in theory, find θ 1 and θ 3 functions of x 1 and x 2 only (independence with x 3 , x 4 and x 5 ). This means that vector (μ 1 , μ 2 , μ 3 , μ 4 , μ 5 ) may be expressed only with x 1 and x 2 (without inevitably by an explicit literal expression as for equations of degree equal to or higher than 5).

The couple of the values of the angles (θ 1 ,θ 3 ) mod 2π is not unique solution. Indeed θ 1 ' = θ 1 +3π/10.i and θ 3 ' = θ 3 -2π/10.i (i an integer) are appropriate also, that is 20 distinct expressions mod 2π.

Injecting again this result in the cardinal matrix, we necessarily have that x 3 , x 4 and x 5 are expressed according to x 1 et x 2 (there still without inevitably by an explicit literal expression).

Research that we carried out to find simple algebraic dependence (possibly quadratic), without use of primitive roots, between x 1 and x 2 as in cases d = 2, d = 3 and d = 4 was unfruitful (what does not mean that it does not exist). Expression ( 106) is given here to facilitate the task of enough courageous lad or girl to further in search for explicit solutions (the literal problem being possibly insoluble as for degree 5 equations).

As examples, we can give to the following table for prime numbers lower than 300 (with d = 5) :

p θ 1 θ 2 θ 3 μ 1 μ 2 μ 3 μ 4 μ 5
x 1 x 2 x 3 x 4 x 5 x 6 x 7 11 1,96899183 3,53978816 0,12231286 -8,59492974 -0,42314838 -5,54860734 5,15415014 9, 41253532 5 5 0 0 0 0 5 31 5,42497645 0,71258747 4,48452492 19,80194466 3,28288712 15 0 0 10 5 10 5 41 5,48108018 0,76869120 3,70473978 3,23050715 16,97933412 7,08143762 5 15 10 10 0 10 5 61 5,41052168 0,69813270 3,85686998 24,47344816 3,2258973 4,33153986 15 15 0 20 10 10 15 71 5,63382773 0,92143875 3,28527692 12,9876689 17,4420362 0,78600551 5 25 20 10 10 10 15 101 1,56268869 3,13348502 5,7501906 17,693203 4,44714944 16,56953663 5 15 30 20 30 20 15 131 5,78724905 1,07486007 0,16203602 37,46278244 8,92922816 35 25 10 40 20 30 25 151 5,95919365 1,24680467 5,77743553 25,6427931 27,8929505 0,18404478 30 30 30 40 20 40 20 181 6,26144130 1,54905232 6,15607240 2 ,32067239 -14,3050398 -37,7493303 44,2091345 5,52456324 45 15 30 40 50 40 35 191 5,36445573 0,65206675 0,20106121 52,2529591 -20,4851358 -3,90261783 -25,4764801 -2,38872543 65 40 20 40 25 40 45 211 4,72627276 0,01388378 1,45036853 -29,7819181 37,2625092 27,106761 -34,7811546 0,19380256 45 45 40 30 50 30 55 241 4,68952002 6,26031635 1,22272437 16,0751071 -0,9684057 49,6037186 -34,1447613 -30,5656587 60 50 60 40 30 60 40 251 2,13869055 3,70948688 2,2038577 24,6018212 43,48474101 50 70 50 40 40 40 60 271 1,96141378 3,53221011 2,64943499 7 ,23914379 -40,8644203 -11,5078631 58,2510405 -13,1179008 75 50 30 70 45 60 55 281 0,66011873 2,23091506 2,45693314 50,08660571 22,8279138 65 40 50 70 55 70 45 Let us note however that condition θ 2 = θ 1 +π/2 is not mandatory.

Case of d_hypersurfaces, d odd prime, unitary and logarithmic meshes

We are interested in eigenvalues μ i geometry. Let us start with the following observations :

-In case d = 3 (dim = d-1 = 2), we place ourselves in the plane. Matrix A eigenvalues are projections, on an axis passing by the origin, axis varying according to p (angle θ = 1/3.arcos (f(2√p))), of three vectors of same modulus directed at 120° one of the other whose vector sum is null.

-In case d = 5 (dim = d-1 = 4), we place ourselves in the plane. Matrix A eigenvalues are projections, on an axis passing by the origin, axis varying according to p, of five vectors of same modulus and equal relative angles whose vector sum is null.

By analogy, in the case of d any odd prime, we can conjecture the following rule :

We place ourselves in a plane. Matrix A eigenvalues are projections, on an axis passing by the origin, axis varying according to p, of d vectors of same modulus and equal relative angles whose vector sum is null. Construction of such vectors is not complex except determination of axis of one of them.

P 103/390

Modulo pi abundance factors for Waring sums Polar reference makes possible to locate any point in this space :

x 0 = ρcos(θ 1 )cos(θ 2 )cos(θ 3 ) cos(θ 4 ) … cos(θ d-3 ) cos(θ d-2 ) x 1 = ρsin(θ 1 )cos(θ 2 )cos(θ 3 )cos(θ 4 ) … cos(θ d-3 ) cos(θ d-2 ) x 2 = ρsin(θ 2 )cos(θ 3 )cos(θ 4 ) … cos(θ d-3 ) cos(θ d-2 ) x 3 = ρsin(θ 3 )cos(θ 4 ) … cos(θ d-3 ) cos(θ d-2 ) … x d-3 = ρsin(θ d-3 )cos(θ d-2 ) x d-2 = ρsin(θ d-2 )
We seek d vectors ( x 0(1) ,x 1(1) ,x 2(1) ,(x 0(2) ,x 1(2) ,x 2(2) ,(x 0(d) ,x 1(d) ,x 2(d) ,of form (x 0 ,x 1 ,x 2 ,), vector null-sum, that is :

d Σ x j(i) = 0, j = 1 to d-2 i = 1 Let us write : x j(i) = ρsin(θ j +2πi/d)cos(θ j+1 +2πi/d) … cos(θ d-3 +2πi/d) cos(θ d-2 +2πi/d)
We did not develop this expression which is certainly a valid subject of exercise. The interested reader will have in particular to prove and use trigonometric remarkable identities such as :

d Σ sin(2πi/d)cos j (2πi/d) = 0, j integer (positive, null or negative) i = 1 d Σ sin k (2πi/d)cos j (2πi/d) = d.x/2 i+j-1 , j integer (positive, null or negative), (conjecture x integer) i = 1 d Σ sin k (2πi/d)cos j (2πi/d) = 0, j integer (positive, null or negative), k positive odd number i = 1
to prove that vectors (x 0(j) , x 1(j) , x 2(j) , … x d-3(j) , x d-2(j) ) answer the question. μ = x 0 = ρcos(θ 1 )cos(θ 2 )cos(θ 3 ) … cos(θ d-2 ) cos(θ d-2 ) is projection on a given axis of the initial point (x 0 , x 1 , …, x d-2 ).

The vector (μ 1 , μ 2 , … μ d-1 , μ d ) where

μ i+1 = ρcos(θ 1 +2πi/d)cos(θ 2 +2πi/d)cos(θ 3 +2πi/d)…cos(θ d-3 +2πi/d)cos(θ d-2 +2πi/d) (127)
represent the whole set of eigenvalues resulting from the vectors, with null sum, evoked previously.

The above conjecture is checked for d = 3 and d = 5 by earlier literal formulas (which are still conjectures).

We propose for ρ the following expression :

ρ = 2 d-2 .√p (128)
We can illustrate our matter by a short table concerning the case n = 7 : 614355 1,236948 2,735251 5,417584 -282,47643 -70,06005 -17,22707 19,56917 68,61039 71,27661 210,30737 3361 -0,938912 0,231923 1,308717 1,616279 2,739032 -334,68227 -6,53937 11,55847 55,66224 70,67597 91,42730 111,89767 To get the values of the angles θ i is relatively difficult and requires perseverance (of the computer), hence little of results presented above. Indeed, the geometry of cos(θ

p θ 1 θ 2 θ 3 θ 4 θ 5 μ 1 μ 2 μ 3 μ 4 μ 5 μ 6 μ 7 3319 0,279854 0,
1 +2πi/d)cos(θ 2 +2πi/d)cos(θ 3 +2πi/d)…cos(θ d-3 +2πi/d)cos(θ d-2 +2πi/d
) is that of a sea urchin. The solutions μ i are at the bottom of "valleys". The mesh of research for the d-2 parameters of angles must be sufficiently fine obliging many tests, each one be followed of an evaluation of minima (which succeeds only when the minima is indeed 0) which must be shortened when probably unfruitful. The order of magnitude of the mesh seems to be about 0,1 to succeed in "falling" into all the distinct valleys what means (2π/0,1) d-2 potential tests. We see the interest of simplifications of type θ 2 = π/2+θ 1 (obtained in the case n = 5) as well on the practical level as on the theoretical level for the production of possible literal formulas. The case n = 11, which follows the case n = 7, "explodes" the potential number of values of angles to examine (which passes from 10 9 to more than 10 16 ).

Case of d_hypersurfaces, d = p n , unitary and logarithmic meshes

In case d = 2, we place ourselves on a line (dim = d-1 = 1). Matrix eigenvalues are projections, on an axis passing by origin with angle θ (pending on p), of two opposite vectors (same modulus directed 180° one with the other), thus whose vector sum is null. For p = 1 mod 4, we are on a real line. The focus is 0 and the only possible projection is θ = 0. For p = 3 mod 4, it are necessary to be on the imaginary line. The focus is 0.i (= 0) and projection is θ = 0 (or θ =π/2 in the complex plane).

In case d = 4, we return in a space of dimension 1 (= (d-1) /2). For p = 1 mod 8, we are on a real line. We have two focus p 1/2 and -p 1/2 (of null sum), two circles around these focus of different radius and the solutions are the intersections with axis θ = 0. For p = 5 mod 8, the focus are still p 1/2 and -p 1/2 . The equations of two preceding circles are taken again but the solutions are this time the intersections with axis of imaginary (θ =π/2).

For eigenvalues, the passage of p to p n seems to correspond geometrically to the passage of one collection of vectors with one focus (before projection on an given axis), to collections of vectors with distinct focus. We speculate on "sea urchins" with p elements with equal modulus and relative angles and p n-1 focus (in a space of adapted size).

General properties of eigenvalues and eigenvectors

For the eigenvectors matrix, we place ourselves in the particular case of the intra-unitary change of basis matrices choice.

Property 13 d d.(#(u,v) / { g x-2 = g u.d + g y-2 .g v.d mod p i }) = (1/p).((p-1) 2 /d + Σ λ k .λ k+d+2-x *.λ k+d+2-y /d 2 ) (129) k=1
The first member of this identity has some number of properties (1 to 7) that we exposed previously and who apply to matrices A and B. We will establish later the coincidence of these properties with those of the second member subjugated to the choice of the aforesaid particular shape of eigenvectors matrices and their inverse matrices. We start from the result (i.e. the aforementioned form). We deduce properties on eigenvalues.

Property 14

Numbers p i , respectively p i -1, are always eigenvalues (among the other eigenvalues) of matrices A and B, associated eigenvectors (columns) being (1, 1,…, 1). Indeed, the sum of each lines equals p i , respectively p i -1, for matrices [A] and [B]. This involves the property self-evidently.

Property 15

(description of the decomposition of the matrix)

Let us have a d+1-uplet (μ 0 , μ 1 , μ 2 , … μ d-1 , μ d ) = (λ 0 , λ 1 , λ 2 , … λ d-1 , λ d ) + (1, 1, 1, … 1, 1). Let us have also the d+1-uplet conjugate of the precedent (μ 0 *, μ 1 *, μ 2 *, … μ d-1 *, μ d *) = (λ 0 *, λ 1 *, λ 2 *, … λ d-1 *, λ d *) + (1, 1, 1, … 1, 1).
We build, starting from these elements, four matrices [A]

, [P A ], [D A ], and [

Q A ] such as [A] = [P A ].[D A ].[Q A ]
and it follows :

1 λ 0 */d λ 0 */d … λ 0 */d μ 0 0 0 … 0 1 λ 0 /d λ 0 /d … λ 0 /d 1 λ 1 */d λ 2 */d … λ d */d 0 μ 1 0 … 0 1 λ 1 /d λ d /d … λ 2 /d A = (1/p) 1 λ d */d λ 1 */d … λ d-1 */d 0 0 μ 2 … 0 1 λ 2 /d λ 1 /d … λ 3 /d … … … … … … … … … … … … … … … 1 λ 2 */d λ 3 */d … λ 1 */d 0 0 0 … μ d 1 λ d /d λ d-1 /d … λ 1 /d The factor 1/p = (1/√p).(1/√p) is divided equitably between [P A ] and [Q A ]. We have immediately, with A = B + I 1 λ 0 */d λ 0 */d … λ 0 */d λ 0 0 0 … 0 1 λ 0 /d λ 0 /d … λ 0 /d 1 λ 1 */d λ 2 */d … λ d */d 0 λ 1 0 … 0 1 λ 1 /d λ d /d … λ 2 /d B = (1/p) 1 λ d */d λ 1 */d … λ d-1 */d 0 0 λ 2 … 0 1 λ 2 /d λ 1 /d … λ 3 /d … … … … … … … … … … … … … … … 1 λ 2 */d λ 3 */d … λ 1 */d 0 0 0 … λ d 1 λ d /d λ d-1 /d … λ 1 /d
This is a free construction as no condition is imposed on [A] (or [B]) for the moment.

Let us describe matrices (√p).P A , D A and (√p).Q A . For (√p).P A :

first line of (√p).P A : λ 0 = p-1, p real number, hence (1, λ

0 */d,…, λ 0 */d) = (1, λ 0 /d,…, λ 0 /d) = (1, (p-1) /d,…, (p-1) /d), -second line of (√p).P A : (1, λ 1 */d, λ 2 */d, λ 3 */d, …, λ d */d),
second line and second column component equals λ 1 *, -rest of matrix, that is reduced matrix noted (√p).P A ' : λ i */d with increasing circulant shifts of indices with lines (return to 1 when d is exceeded) and decreasing circulant shifts of the indices with columns (return to d when 1 is exceeded) Matrix P A ' is a right circulant matrix. For D A : It is trace matrix of d+1-uplet (μ 0 , μ 1 , μ 2 , … μ d-1 , μ d ). For (√p).Q A :

-first line = first line of (√p).P B multiplied by 1/d, except first component : (1, λ 0 /d, λ 0 /d, λ 0 /d,, …, λ 0 /d) -first column = first line of (√p).P B multiplied by 1/d, except first component : (1,1/d,1/d,1/d,…,1/d) second line and second column component equal to λ 1 /d 2 , -rest of matrix λ i /d with increasing circulant shifts of indices in columns and decreasing circulant shifts of the indices in lines. Matrix Q A ', reduction of Q A , is a right circulant matrix.

We impose now conditions and check each time induced consequences :

Let us have Q A = P A -1 , thus [P A ].[P A -1 ] = I. We have immediately, with A = B + I 1 λ 0 */d λ 0 */d … λ 0 */d λ 0 0 0 … 0 1 λ 0 /d λ 0 /d … λ 0 /d 1 λ 1 */d λ 2 */d … λ d */d 0 λ 1 0 … 0 1 λ 1 /d λ d /d … λ 2 /d B = (1/p) 1 λ d */d λ 1 */d … λ d-1 */d 0 0 λ 2 … 0 1 λ 2 /d λ 1 /d … λ 3 /d … … … … … … … … … … … … … … … 1 λ 2 */d λ 3 */d … λ 1 */d 0 0 0 … λ d 1 λ d /d λ d-1 /d … λ 1 /d Moreover : 1 λ 0 */d λ 0 */d … λ 0 */d 1 λ 0 /d λ 0 /d … λ 0 /d 1 0 0 … 0 1 λ 1 */d λ 2 */d … λ d */d 1 λ 1 /d λ d /d … λ 2 /d 0 1 0 … 0 I = (1/p) 1 λ d */d λ 1 */d … λ d-1 */d 1 λ 2 /d λ 1 /d … λ 3 /d = 0 0 1 … 0 … … … … … … … … … … … … … … … 1 λ 2 */d λ 3 */d … λ 1 */d 1 λ d /d λ d-1 /d … λ 1 /d 0 0 0 … 1
This draw along diverse identities on matrix components c (i,j) , i th line, j th column, i = 1 to d+1, j = 1 to d+1. We manipulate indices i and j as circular permutations modulo d when i > 1 or when j > 1.

Component (1,1) of I :

d c (1,1) = (1/p).(1 + Σ λ 0 */d) = (1/p).(1+p-1) = 1 I=1 This identity is self-evidently verified. Components (i+1,1) of I, i>0 : d c (i+1,1) = (1/p).(1 + Σ λ j+1-i */d) = 0
j=1 Hence, it follows the conditions :

d d Σ λ i * = -d and Σ μ i * = 0 (130) i=1 i=1 Components (1,i+1) of I, i>0 : d c (1,i+1) = (1/p).(λ 0 /d + λ 0 . Σ λ j+1-i /d 2 ) = 0 j=1 Hence, it follows : d d Σ λ i = -d and Σ μ i = 0 (131) i=1 i=1
Expressions ( 130) and ( 131) are equivalent (no contradiction to the generality of the problem). Components (i+1,i+1) of I, i>0 :

d+1-i c (i+1,i+1) = (1/p).(λ 0 /d + Σ λ j *.λ j /d 2 ) = 1 j=2-i
Hence, we get the following conditions (using also Σμ i =0, Σμ i *=0, Σ1=d for the second relation) :

P 106/390

Modulo pi abundance factors for Waring sums

d d Σ λ i *.λ i = d.(p.d-p+1) and Σ μ i *.μ i = (d-1).d.p (132) i=1 i=1
Components (i+1,j+1) of I, i≠j), i>0 and j>0 :

d c (i+1,j+1) = (1/p).(λ 0 /d + Σ λ k+1-i *.λ k+1-j /d 2 ) = 0 k=1
Hence, we get following conditions (using also Σμ i =0, Σμ i *=0, Σ1=d for the second relation) : -1 ] in following form :

d d Σ λ k+1-i *.λ k+1-j = -(p-1).d and Σ μ k+1-i *.μ k+1-j = -p.d (133) k=1 k=1 Let us write then B = [P B ].[D B ][P B
0 0 … 0 p-1 0 … … 0 1 λ 0 */d λ 0 */d … λ 0 */d λ 0 0 0 … 0 1 λ 0 /d λ 0 /d … λ 0 /d d 1 λ 1 */d λ 2 */d … λ d */d 0 λ 1 0 … 0 1 λ 1 /d λ d /d … λ 2 /d 0 = 1/p 1 λ d */d λ 1 */d … λ d-1 */d 0 0 λ 2 … 0 1 λ 2 /d λ 1 /d … λ 3 /d … U V … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … 0 1 λ 2 */d λ 3 */d … λ 1 */d 0 0 0 … λ d 1 λ d /d λ d-1 /d … λ 1 /d
Here component (1,ii+1) of matrix B is occupied by p-1 (ii = 0 case where U is reduced to void or ii = d/2 and U and V are both of dimension (d,d/2)).

We have :

0 0 … 0 p-1 0 … … 0 λ 0 λ 1 .λ 0 /d λ 2 .λ 0 /d … λ d .λ 0 /d 1 λ 0 /d λ 0 /d … λ 0 /d d λ 0 λ 1 .λ 1 */d λ 2 .λ 2 */d … λ d .λ d */d 1 λ 1 /d λ d /d … λ 2 /d 0 = 1/p λ 0 λ 1 .λ d */d λ 2 .λ 1 */d … λ d .λ d-1 */d 1 λ 2 /d λ 1 /d … λ 3 /d … U V … … … … … … … … … … … … … … … … … … … … … 0 λ 0 λ 1 .λ 2 */d λ 2 .λ 3 */d … λ d .λ 1 */d 1 λ d /d λ d-1 /d … λ 1 /d Then : Component (1,1) of B (identity verified) : d c (1,1) = (1/p).( λ 0 + λ 0 Σ λ 0 */d) = 0 i=1
Component (2,1) of B (identity verified when using relations (132)) :

d c (2,1) = (1/p).( λ 0 + Σ λ i λ i */d) = (1/p).(p-1+d.(p.d-p+1)/d = d i=1
Components (i+1,1) of B, i>1 (identity verified when using relations ( 133)) :

d c (i+1,1) = (1/p).( λ 0 + Σ λ k λ k+1-i */d) = (1/p).(p-1-d.(p-1)/d) = 0 k=1 Components (1,i+1) of B, i≠ii, i>0 : d c (1,i+1) = (1/p).(λ 0 2 /d + Σ λ 0 .λ k .λ k+1-i /d 2 ) = 0 k=1
Hence, we get the following conditions (using also Σμ i =0 for the second relation) :

d d Σ λ k .λ k+1-i = -d.(p-1) and Σ μ k .μ k+1-i = -d.p (134) k=1 k=1
Components (1,ii+1) of B :

P 107/390

Modulo pi abundance factors for Waring sums

d c (1,ii+1) = (1/p).(λ 0 2 /d + Σ λ 0 .λ k .λ k+1-ii /d 2 ) = p-1 k=1
Hence, we get following conditions (using also Σμ i =0, Σμ i *=0, Σ1=d for the second relation) :

d d Σ λ k .λ k+1-ii = d.(p.d-p+1) et Σ μ k .μ k+1-ii = (d-1).d.p (135) k=1 k=1
Components (i+1,j+1), i>0 and j>0, of B :

d c (i+1,j+1) = (1/p).(λ 0 2 /d + Σ λ k .λ k+1-i *.λ k+d+1-j /d 2 ) k=1
Hence, we get (using ( 130) to ( 135) for second identity) :

d c (i+1,j+1) = (1/p/d).((p-1) 2 +(1/d). Σ λ k .λ k+1-i *.λ k+d+1-j ) (136) k=1 If {i≠1, i≠j, j≠ii}: d c (i+1,j+1) = (1/d).(p+1+(1/p/d). Σ μ k .μ k+1-i *.μ k+1-j ) (137) k=1
If {i=1, i≠j, j≠ii} or {i≠1, i=j, j≠ii} or {i≠1, i≠j, j=ii}, that is {i=1, j≠1, j≠ii} or {i≠1, i=j, i≠ii} or {i≠1, i≠ii, j=ii}:

d c (i+1,j+1) = (1/d).(p+d-5+(1/p/d). Σ μ k .μ k+1-i *.μ k+1-j ) (138) k=1
If {i=1, i=j, j≠ii} or {i≠1, i=j, j=ii} or {i=1, i≠j, j=ii}, that is {i=j=1, i≠ii} or {i≠1, i=j=ii} or {i=1, i≠ii, j=ii}:

d c (i+1,j+1) = (1/d).(p+2.d-5+(1/p/d). Σ μ k .μ k+1-i *.μ k+1-j ) (139) k=1 If {i=1, i=j, j=ii}, that is {i=j=ii=1} : d c (i+1,j+1) = (1/d).(p+3.d-5+(1/p/d). Σ μ k .μ k+1-i *.μ k+1-j ) (140) k=1
If we impose now matrix B sums of line characteristics, it follows :

d d d Σ c (i+1,j+1) = Σ (1/p/d).((p-1) 2 +(1/d).Σ λ k .λ k+1-i *.λ k+1-j ) = p-1-d i=1 (j=1) i=1 k=1 and d d d Σ c (i+1,j+1) = Σ (1/p/d).((p-1) 2 +(1/d).Σ λ k .λ k+1-i *.λ k+1-j ) = p-1 i=1 (j>1) i=1 k=1 That is also : d d Σ Σ λ k .λ k+1-i *.λ k+1-j = -d 2 .(p.d-p+1) (141) i=1 k=1 (j=1) and d d Σ Σ λ k .λ k+1-i *.λ k+1-j = d 2 .(p-1) (142) i=1 k=1 (j>1)
In the same way, if we impose matrix B sums of columns characteristics, it follows :

P 108/390
Modulo pi abundance factors for Waring sums

d d d Σ c (i+1,j+1) = Σ (1/p/d).((p-1) 2 +(1/d).Σ λ k .λ k+1-i *.λ k+1-j ) = p-1-d j=1 (i=ii) j=1 k=1 and d d d Σ c (i+1,j+1) = Σ (1/p/d).((p-1) 2 +(1/d).Σ λ k .λ k+1-i *.λ k+1-j ) = p-1 j=1 (i≠ii) i=1 k=1 That is also : d d Σ Σ λ k .λ k+1-i *.λ k+1-j = -d 2 .(p.d-p+1) (143) i=1 k=1 (i=ii) and d d Σ Σ λ k .λ k+1-i *.λ k+1-j = d 2 .(p-1) (144) i=1 k=1 (i≠ii)
Let us consider now relations ( 133) and ( 143

). d Σ λ k-i *.λ k-j = -d.(p-1) k=1 (i≠j) d d Σ Σ λ k .λ k+1-i *.λ k+1-j = d 2 .(p-1) i=1 k=1 (j>1)
The first relation results from the second. Indeed, let us write the succession of equivalent expressions (after a valid change of indices) rising from ( 133

) : d λ m Σ λ k+1-i *.λ k+1-j = λ m .(-d).(p-1) k=1 (i≠j) d λ m+1 Σ λ k+2-i *.λ k+2-j = λ m+1 .(-d).(p-1) k=1 (i≠j) … d λ m+d-1 Σ λ k+d-i *.λ k+d-j = λ m+d-1 .(-d).(p-1) k=1 (i≠j)
By summoning the first components of the left members of above equations, we get :

d d Σ Σ λ k+m .λ k+1-i *.λ k+1-j i=1 k=1 (i≠j, m) With m = 0, it follows : d d Σ Σ λ k+m .λ k+1-i *.λ k+1-j i=1 k=1 (i≠j)
The sum of the right members, while using (131), is :

d (-d).(p-1).Σ λ k = d 2 .(p-1) k=1 Hence d Σ λ k .λ k+1-I *.λ k+1-j = d 2 .(p-1) k=1, i≠j
We can show on the same model that the set of relations ( 141), ( 142), ( 143) and ( 144) result from ( 130), ( 131), ( 132), ( 133), ( 134) and ( 135). The conditions on B (et A) matrix lines and columns sums result from the only conditions ( 130), ( 131), ( 132), ( 133), ( 134) and ( 135).

Let us now bring closer relations ( 132) and ( 135). It follows immediately that : P 109/390 Modulo pi abundance factors for Waring sums With ii ≠ 1 (=d/2+1), we bring closer relations ( 132) and ( 135) :

d d d d Σ λ k .λ k+1-ii = Σ λ i *.λ i = d.(p.d-p+1) and Σ μ k .μ k+1-ii = Σ μ i *.μ i = (d-1).d.p k=1 i=1 k=1 i=1 If ii = 1, then : d d d d Σ λ k .λ k = Σ λ i *.λ i = Σ Re(λ i 2 ) = d.(p.d-p+1) and Σ Im(λ i 2 ) = 0 k=1 i=1 i=1 i=1 and d d d d Σ μ k .μ k = Σ μ i *.μ i = Σ Re(μ i 2 ) = (d-1).d.
d d Σ λ i *.λ i = d.(p.d-p+1) and Σ μ i *.μ i = (d-1).d.p i=1 i=1 d d Σ λ i+1-m .λ i+1-m = -(p-1).d and Σ μ i+1-m .μ i+1-m = -d.p i=1 i=1
We choose m such that m = ii-d/2. Then m = 1. Hence, it follows :

d d Σ Re(λ i 2 )+Im(λ i 2 ) = d.(p.d-p+1) and Σ Re(μ i 2 )+Im(μ i 2 ) = (d-1).d.p i=1 i=1 d d Σ Re(λ i 2 )-Im(λ i 2 ) +2i.Re(λ i ).Im(λ i ) = -(p-1).d and Σ Re(μ i 2 )-Im(μ i 2 ) +2i.Re(μ i ).Im(μ i ) = -d.p i=1 i=1 Hence: d d d Σ Re(λ i 2 ) = d.(p.d/2-p+1) and Σ Im(λ i 2 ) = p.d 2 /2 and Σ Re(λ i ).Im(λ i ) = 0 I=1 i=1 i=1 d d d Σ Re(μ i 2 ) = (d/2-1).d.p and Σ Im(μ i 2 ) = d 2 .
p/2 and Σ Re(μ i ).Im(μ i ) = 0 i=1 i=1 i=1 So that : If p = 1+d mod 2d (matrices of type 2), then the set of eigenvalues of matrices A and B include imaginary numbers and Σ Re(

μ i 2 ) = (d/2-1).d.p+p 2 = p.((d/2-1).d+p), Σ Im(μ i 2 ) = d 2 .p/2, Σ Re(μ i ).Im(μ i ) = 0 for matrix A (μ 0 included) and Σ Re(λ i 2 ) = d.(p.d/2-p+1)+(p-1) 2 , Σ Im(λ i 2 ) = p.d 2 /2, Σ Re(λ i ).Im(λ i ) = 0 for matrix B (λ 0 included).
Let us study now the conditions related to properties 7. We saw that it suffice to consider two transformations, namely T2 and T4. For the T2 transformation

x' = -x+4 y' = -x+y+2+(p-1)/2 Then : d c(x,y) = (1/p).((p-1) 2 /d + Σ λ k .λ k+d+2-x *.λ k+d+2-y /d 2 ) k=1 and d c(x',y') = (1/p).((p-1) 2 /d + Σ λ k .λ k+d+2-x' *.λ k+d+2-y' /d 2 ) k=1
Hence, as d = 0 mod d, for c(x',y')=c(x,y) :

P 110/390
Modulo pi abundance factors for Waring sums

d d Σ λ k .λ k+2-x *.λ k+2-y = Σ λ k .λ k+x-2 *.λ k+x-y-(p-1)/2 k=1 k=1
That is, with the substitution k →k-x+2+(p-1)/2, at the second member:

d d Σ λ k .λ k+2-x *.λ k+2-y = Σ λ k-x+2+(p-1)/2 .λ k+(p-1)/2 *.λ k+2-y k=1 k=1
So that, rearranging the terms of the second member :

d d Σ λ k .λ k+2-x *.λ k+2-y = Σ λ k+(p-1)/2 *.λ k-x+2+(p-1)/2 . λ k+2-y (145) k=1 k=1 For transformation T4 : x' = y+(p-1)/2 y' = x+(p-1)/2 Then : d c(x,y) = (1/p).((p-1) 2 /d + Σ λ k .λ k+d+2-x *.λ k+d+2-y /d 2 ) k=1 and d c(x',y') = (1/p).((p-1) 2 /d + Σ λ k .λ k+d+2-x' *.λ k+d+2-y' /d 2 ) k=1
Hence, as d = 0 mod d, for c(x',y')=c(x,y) :

d d Σ λ k .λ k+d+2-x *.λ k+d+2-y = Σ λ k .λ k+d+2-y-(p-1)/2 *.λ k+d+2-x-(p-1)/2 k=1 k=1
So that, rearranging the terms of the second member and with the substitution -(p-1)/2 = (p-1)/2 mod d :

d d Σ λ k .λ k+d+2-x *.λ k+d+2-y = Σ λ k .λ k+d+2-x+(p-1)/2 .λ k+d+2-y+(p-1)/2 * (146) k=1 k=1
Let us consider now case (p-1)/2 = 0 mod d. We know that matrix B eigenvalues are real. Thus :

λ k+(p-1)/2 = λ k = λ k * (147)
Relations ( 145) and ( 146) are then self-evident identities by using λ k+(p-1)/2 = λ k *.

Let us consider case (p-1)/2 = d/2 mod d. Matrix B eigenvalues are eigenvalues of a matrix with integer coefficients, therefore with real coefficients. Each eigenvalue thus admits a conjugate eigenvalue. List of eigenvalues (λ 1 , λ 2 , …, λ d ) are presented with the good indice and the good rearrangement of the terms in the form (

λ 1 , λ 2 , …, λ d/2 , λ d/2+1 , λ kd/2+2 , …, λ d ) = (λ 1 , λ 2 , …, λ d/2 , λ 1 *, λ 2 *, …, λ d/2 *), that is : λ k+(p-1)/2 = λ k * (148)
Returning to relations ( 145) and ( 146), they are now self-evident identities by using relation ( 148).

Let us recapitulate all of these results :

Properties 16 (of summations of the eigenvalues)

16.1 Σ λ i = Σ λ i * = -d Σ μ i = Σ μ i * = 0 i = 1 to d 16.2 Σ λ i *.λ i = d.(p.d-p+1) Σ μ i *.μ i = (d-1).d.p i = 1 to d 16.3 Σ λ k+1-i *.λ k+1-j = -d.(p-1) Σ μ k+1-i *.μ k+1-j = -p.d k = 1 to d, i≠j 16.4 Σ λ k .λ k+1+(p-1)/2 = d.(p.d-p+1) Σ μ k .μ k+1+(p-1)/2 = (d-1).d.p k = 1 to d 16.5 Σ λ k .λ k+1-i = -d.(p-1) Σ μ k .μ k+1-i = -d.p k = 1 to d, i≠(p-1)/2 mod d 16.6 Σ Re(λ i 2 ) = p.d 2 +(p-1).(p-1-d) Σ Re(μ i 2 ) = p.d 2 +p.(p-d) p = 1 mod 2d, i = 0 to d P 111/390
Modulo pi abundance factors for Waring sums

16.7 Σ Re(λ i 2 ) = d.(p.d/2-p+1) Σ Im(λ i 2 ) = p.d 2 /2 Σ Re(λ i ).Im(λ i ) = 0 Σ Re(μ i 2 ) = (d/2-1).d.p Σ Im(μ i 2 ) = d 2 .p/2 Σ Re(μ i ).Im(μ i ) = 0 p = 1+d/2 mod 2d, i = 1 to d 16.8 λ k+(p-1)/2 = λ k * μ k+(p-1)/2 = μ k * 16.9 c B (x,y) = (1/p).((p-1) 2 /d + Σ λ k .λ k+d+2-x *.λ k+d+2-y /d 2 ) c A (x,y) = (1/p).((p-1) 2 /d + Σ μ k .λ k+d+2-x *.λ k+d+2-y /d 2 ) k = 1 to d
We would be able to establish also (not demonstrate) :

16.10 ∑ λ i 2 = (-1) (p-1)/2 .d.(p.d-p+1) ∑ μ i 2 = (-1) (p-1)/2 .(d-1)d.p i = 1 to d 16.11 ∑ λ i 3 = d.(-1+p.(-p+d(x 1 -d+2))) ∑ μ i 3 = d.p.(-1-p+d(x 1 -d+3)) i = 1 to d

Cardinal matrices properties (continuation)

Certain properties of cardinal matrices are inherited properties of square matrices in general. Let us see that. Let us have [M] a square matrix of dimension d+1 and μ i its eigenvalues, i = 0 to d. Here of course μ 0 does not have particular properties compared to the other eigenvalues. However, when [M] is a cardinal matrix of dimension d+1, the sum of each line is constant. All square matrices of this type admits that constant like an eigenvalue. Thus, for a variable of integers, one of the eigenvalues μ 0 equals p. In the same way, for a variable of prime numbers, one of the eigenvalues noted λ 0 is p-1. We consider matrix [M]-μ[I] determinant where μ is any complex number :

det(M-μ[I]) = (-1) d+1 .μ d+1 +tr([M]).μ d +...+μ 0 .μ 1 …μ d .μ 0 = (μ 0 -μ).(μ 1 -μ)…(μ d -μ) (149)
Hence, it follows :

d Σ μ i = tr[M] i = 0

Let us have [μ i ] and [P] eigenvalues and eigenvectors matrices of [M]

.

Identity [M] = [P].[μ i ].[P -1 ] involves immediately [M] n = [P].[μ i ] n .[P -1 ], hence : d Σ μ i n = tr([M] n ) (150) i = 1
Let us have then [ME] a matrix of dimension d+1 with integer components. Let us have {μ i }, i = 0 to d, eigenvalues of this matrix. The preceding expression is thus an integer for any integer n.

d Σ μ i n = tr([ME] n ) is an integer (151) i = 1
In the same way from relations (149), we deduce :

all Σ μ i .μ j …μ t is an integer (152) distinct indices, #(μ i , μ j , …, μ t ) = r
In preceding expression, the products include a constant number of terms (r chosen between 0 and d). In addition, indices permutations are excluded (for example, in the case d = 3, (1,2,3) is redundant with (2,1,3)). The term (-1) r .Σμ cei .μ cej …μ cet is the factor of μ d-r in factors decomposition ∏ (μ-μ ce1 ).(μ-μ ce2 )…(μ-μ ced ).

Returning to the trace of [M] a cardinal matrix of dimension d+1, expressed in function of the given eigenvalues sum, we have :

Properties 17 17.1 Variable x of integers tr[M] = p 17.

Variable y of prime numbers tr[M] = p-1 Foot-note :

For matrices with real eigenvalues, these properties rise self-evidently from detailed form of matrices (all x i , i = 1 to d are on the principal diagonal). However, for matrices with imaginary eigenvalues (i > 0), these properties do not rise in an obvious way of their detailed form.

Property18 (conjecture of powers)

Here is an elegant conjecture : The sum of the eigenvalues to any positive integer power of a dimension d+1cardinal matrix associated to a variable of integers is a multiple of p (at sequence p). More precisely, there exists an integer such as :

P 112/390 Modulo pi abundance factors for Waring sums d Σ μ i n = a.p 1+int((n-1)/d) (153) i = 1
Let us note that if we include i = 0 in the sum, we get another constant "a" but the conjecture remains in the same way. The conjecture rises from :

d Σ μ i d = a'.p (154) i = 1

Normalization of modulo p matrices

The abundance factors are normalized right away when we calculate them with a normalized cardinal matrix. For that, lines sum must be equal to 1 and the basic vector is not any more (1,0,0,…,0) but (p,0,0,…,0

). Thus, k u k,p 1 v k,p 0 w k,p = A p 0 … … x k,p 0 becomes k Fan(0,p) p Fan(gd -1 ,p) 0 (155) Fan(gd -2 ,p) = AN p 0 … … Fan(g 0 ,p) 0
Where [AN p ] = (1/p) [A p ] and Fan(c,p) is the normalized abundance factor of target c at sequence p. We can proceed in the same way for prime numbers variables. Lines sum, not being p, but p-1, division of [B p ] components is done by p-1. k Fan(0,p) p Fan(gd -1 ,p) 0 ( 156)

Fan(gd -2 ,p) = BN p 0 … … Fan(g 0 ,p) 0 Where [BN p ] = (1/(p-1))[B p ].
Foot-note : Matrix normalization is obtained in the same way by eigenvalues normalization :

Thus :

[AN p ] = [P A ].[μ/p].[P -1 A ] [BN p ] = [P A ].[λ/(p-1)].[P -1 A ] where : 1 [0] μ 1 /p [μ/p] = (1/p).[μ] = μ 2 /p … [0] μ d /p and 1 [0] λ 1 /(p-1) [λ/(p-1)] = (1/(p-1)).[λ] = λ 2 /(p-1) … [0] λ d /(p-1)

Condensed form per blocks

This form will next be useful in modulo p δ approach.

We can write the case general using the following definitions :

[A] = A 1 : dimension d+1 square matrix, solution of case δ = 1 for variables of integers.

[A'] = A' 1 : dimension d square matrix, resulting from [A] by removing first line and first column.

[B] = B 1 : dimension d+1 square matrix, solution of case δ = 1 for prime numbers variables.

[B'] = B' 1 : dimension d square matrix, resulting from [B] by removing first line and first column.

[μ] : dimension d+1eigenvalues square matrix of [A], trace type matrix, (with adequate order of eigenvalues, namely such as [A] [p] = [p,0…0] (d+1,1) : dimension d+1 column matrix of which all components are null except first one worth p.

We write for variables of integers :

[A] k = 1 (p-1).[J'] k [K'] [A'] [P A ] = (1/p 1/2 ). 1 (p-1).[U'] [U''] [λ'*] [P -1 A ] = (1/p 1/2 ). 1 ((p-1)/d).[U'] (1/d).[U"] (1/d 2 ).[λ'] [μ/p] k = 1 [0] k [0] [μ'/p] [P A ].[μ/p] k .[P -1 A ] = (1/p). 1 (p-1).[U'].[μ'/p] k 1 ((p-1)/d).[U'] [U''] [λ'*].[μ'/p] k (1/d).[U"] (1/d 2 ).[λ'] Fan(c,p) = [P A ].[μ/p] k .[P -1 A ][p] = 1+((p-1)/d).[U'].[μ'/p] k .[U"] [U'']+(1/d).[λ'*].[μ'/p] k .[U"] So that : Fan δ=1 (c,p) = 1+((p-1)/d).Σ(μ/p) k (157) [U'']+(1/d).[λ'*].[μ'/p] k .[U"] Σ refers to indices 1 to d (indice 0 is excluded, i.e. μ 0 = p is excluded)
We write for prime numbers variables :

[B] k = 1 (p-1).[J'] m [K'] [B'] [P B ] = (1/p 1/2 ). 1 (p-1).[U'] [U''] [λ'*] [P -1 B ] = (1/p 1/2 ). 1 ((p-1)/d).[U'] (1/d).[U"] (1/d 2 ).[λ'] [λ/(p-1)] m = 1 [0] m [0] [λ'/(p-1)] [P B ].[λ/(p-1)] m .[P -1 B ] = (1/p). 1 (p-1).[U'].[λ'/(p-1)] k 1 ((p-1)/d).[U'] [U''] [λ'*].[λ'/(p-1)] k (1/d).[U"] (1/d 2 ).[λ'] P 114/390
Modulo pi abundance factors for Waring sums

Fan(c,p)= [P B ].[λ/(p-1)] m .[P -1 A ][p] = 1+((p-1)/d).[U'].[λ'/(p-1)] m .[U"] [U'']+(1/d).[λ'*].[λ'/(p-1)] m .[U"] So that : Fan δ=1 (c,p) = 1+((p-1)/d).Σ(λ/(p-1)) m (158) [U'']+(1/d).[λ'*].[λ'/(p-1)] m .[U"]
Σ rests on indices 1 to d (indice 0 is excluded, i.e. λ 0 = p-1) These expressions will be useful at exercise 6 where we work modulo p δ .

Abundance factors construction

The diagram of construction modulo p is :

c p = 2 p = 3 … p = q p = +∞ 0 card(c=0,p=2) card(c=0,p=3) … card(c=0,p=q) … ∏ card(0,p) 1 card(c=1,p=2) card(c=1,p=3) … card(c=1,p=q) … ∏ card(1,p) 2 card(c=0,p=2) card(c=2,p=3) … card(c=2,p=q) … ∏ card(2,p) 3 card(c=1,p=2) card(c=0,p=3) … card(c=3,p=q) … ∏ card(3,p) 4 card(c=0,p=2) card(c=1,p=3) … card(c=4,p=q) … ∏ card(4,p) 5 card(c=1,p=2) card(c=2,p=3) … card(c=5,p=q) … ∏ card(5,p) 6 etc. etc. … card(c=6,p=q) … ∏ card(6,p) … … … … … … … q-1 … … … card(c=q-1,q) … ∏ card(q-1,p) q … … … etc. … ∏ card(q,p) … … … … … … c card(c mod 2,p=2) card(c mod 3,p=3) … card(c mod q,q) … ∏ card(c,p)
Of course, here, card (c,p) = Fan(c,p) is obtained with the matrix method, matrices which we will have taken care to normalize (or overlapping loops method if matrices method does not apply).

We have thus :

∞ Fan(c) = ∏ Fan(c mod p,p) (159) p = 2

Case of boundless hypervolumes

Prime numbers variables

We are first interested here in case :

y 1 n + y 2 n + … + y k n = c + y k+1 n + y k+2 n + … + y k+i n ( 160
)
To deal with Waring problem in its basic form y 1 n + y 2 n +…+ y k n = c, we have built a table of type :

card 0 card 1 card 2 … card d-1 0 g 0 .g d g 0 .g 2d … g 0 .g (p-1) g 1 .g d g 1 .g 2d … g 1 .g (p-1) g d-1 .g d g d-1 .g 2d … g d-1 .g (p-1) card = 1 0 card = d g d g 2d … g (p-1)
Cardinal matrix is here :

[B(x,y)] = [if((x,y) = (2,1), d, if((x,y) = (1,(p+1)/2 mod d), p-1, if((x,y) > (1,1), #(u,v).d / g x-2 = g u.d + g y-2 .g v.d mod p, 0)))]
Rewriting relation (160) as y 1 n + y 2 n + … + y k n -y k+1 n -y k+2 n -… -y k+i n = c, at step k+1, we draw up tables as :

card 0 card 1 card 2 … card d-1 0 g 0 .g d g 0 .g 2d … g 0 .g (p-1) g 1 .gd g 1 .g 2d … g 1 .g (p-1) g d-1 .g d g d-1 .g 2d … g d-1 .g (p-1) card = 1 0 card = d -g d -g 2d … -g (p-1)
Cardinal matrix is here : y) = (2,1),d,if((x,(p+1)/2 mod d,if((x,y) > (1,1),

[B'(x,y)] = [if((
#(u,v).d / g x-2 = -g u.d + g y-2 .g v.d mod p, 0)))]
Above, the first modification of the condition compared to B, follows from -1 = g (p-1)/2 mod p. For the second modification, we use -g i.d = -g y+1 .g j.d = 0 mod p equivalent to g i.d + g (y+(p-1)/2+1 .g j.d = 0 mod p. The last modification is self-evident.

Case study

Case p = 1 mod 2d

Then :

-g u.d = g (p-1)/2 .g u.d mod p = g r.d .g u.d mod p and x+(p-1)/2 = x mod d y+(p-1)/2 = y mod d Thanks to circulant shift u+r → u, this is equal to :

[B'(x,y)] = [if((x,y) = (2,1), d, if((x,y) = (1,(p+1)/2 mod d), p-1, if((x,y) > (1,1), #(u,v).d / g x-2 = g u.d + g y-2 .g v.d mod p, 0)))]
Hence, the matrices identity : B = B' (161) Case p = 1+d mod 2d

Let us write B as : .d mod p}] then, with p-1 choices for (u,v) : y = 2 Hence, at last, B' form :

0 0 … 0 p-1 0 … … 0 d B = 0 … U V … 0 W(x,y) = U V = [d.#(u,v) /{g x-2 = g u.d + g y-2 .g v.d }] Then for B', we have : W'(x,y) = [d.#(u,v) / {g x-2 = -g u.d + g y-2 .g v.d }] which can be written also : W'(y,x) = [d.#(u,v) / { g y-2 = -g u.d + g x-2 .g v.d }] then : W'(y,x) = [d.#(u,v) / {g x-2 = g (u-v).d + g y-2 .g -v.d }] Hence (permutation v → -v, u → u-v) : W'(y,x) = W(x,y) So that : W' = t W = t U V Characteristic equation of B' first column (except first line) is : B' =[B(x,y)] = [d.#(u) / {g x-2 = -g u.d mod p}] As : -1 = g (p-
B' = [B(x,y)] = [d.#(u,v) / {-g u.d + g y-2 .g v.d = 0 mod p}] As : B' = [B(x,y)] = [d.#(u,v) / {g u.d = g y-2 .g v
0 p-1 … … … 0 … B' = … t [U V] d … 0
Let us underline here that, even if p-1 is on the second column, t [U:V] is not symmetric, but t [V:U] (and [V:U]) is. In addition, B' is also deduced from B by permutation per block of lines 2 to 1+d/2 with lines 2+d/2 to 1+d, then permutation of columns 2 to 1+d/2 with columns 2+d/2 to 1+d. This point, self-evident for first line and column, is proven by use of T4 transform (cf. properties 7): x' = y'+(p-1)/2 and y' = x+(p-1)/2, hence x' = y' +d/2 and y' = x+d/2 what suffices to our argument.

Calculation of cardinals

Cardinals, at order m = k+i, are then obtained by :

i k u m 1 v m 0 w m = B' . B . 0 … … x m 0
Let us note that any permutation of equation y 1 n + y 2 n + … + y k n -y k+1 n -y k+2 n -… -y k+i n = c mod p members is equivalent. This means that, whatever calculation carried out with the matrix method, the result must be identical, so that matrices B and B' commutate (case B = B' being self-evident ): [B].[B'] = [B']. [B] (162)

The commutativity rises from the following points : [B] and [B']

have equal eigenvalues characteristic polynomial. Indeed, characteristic of B(λ) is -λ.det([U:V]-λ[I])+((- 1) (1+d/2) .(p-1)).d.det([F]) where [F] is matrix ([U:V]-λ[I]) deprived of column containing p-1 in first line and deprived of line containing d in first column. The characteristic polynomial of B'(λ) is -λ.det( t ([U:V]-λ[I]))+(-1).(p-1).((-1) (d/2) .d).det([G]) where [G] is matrix ( t ([U:V]-λ[I]
)) deprived of column containing p-1 in first line and deprived of line containing d in first column. Asymmetries in the characteristic polynomials calculation disappear, by elimination of quoted lines and columns, involving suggested identity. Then, we can take the same eigenvectors for these two matrices and commutativity follows :

[B] = [P B ].[λ].[P B -1 ] and [B'] = [P B ].[λ'].[P B -1 ] then [B].[B'] = [P B ].[λ].[λ'].[P B -1 ] = [P B ].[λ'].[λ].[P B -1 ] = [B'].[B]
Let us note that in [λ] and [λ'], λ 0 appears at its usual place (at trace head), the other eigenvalues observing a d/2 (mod d) shift from one to the other eigenvalues matrices, the best argument to sit this assertion being simplicity to pass from one case to the other (limited volumes towards boundless volumes).

Variables of integers

The case of variables of integers receives same remarks as previously.

For : Variables of integers :

x 1 n + x 2 n + … + x k n = c + x k+1 n + x k+2 n + … + x k+i
x 1 2 + x 2 2 + … + x k 2 = x k+1 2 + x k+2 2 + … + x k+j 2 + c mod p For p = 1 mod 4 :
We have p = 1+2d and enumerations are obtained starting from, already completely solved, case of equation x

1 2 + x 2 2 + … + x k 2 + x k+1 2 + x k+2 2 + … + x k+j 2 = c mod p : #{0} (1/p).(p k+j +(1/2).(p-1).p (k+j)/2 .(1+(-1) k+j )) #{g 0 .g 2u } = (1/p).(p k+j +(1/2).p (k+j)/2 .(-(1+(-1) k+j )+p 1/2 .(1-(-1) k+j ))) (164) #{g 1 .g 2u } (1/p).(p k+j -(1/2).p (k+j)/2 .((1+(-1) k+j )+p 1/2 .(1-(-1) k+j )))
For p = 3 mod 4 :

We have p = 1+d+2d, hence :

#{0} 1 (p-1)/2 (p-1)/2 p k p j 1 (p-1)/2 (p-1)/2 1 #{g 0 .g 2u } = (1/p) 1 (-1-i√p)/2 (-1+i√p)/2 i k p k/2 (-i) j p j/2 1 (-1+i√p)/2 (-1-i√p)/2 0 #{g 1 .g 2u } 1 (-1+i√p)/2 (-1-i√p)/2 (-i) k p k/2 (i) j p j/2 1 (-1-i√p)/2 (-1+i√p)/2 0 So that : #{0} 1 (p-1)/2 (p-1)/2 p k+j 1 #{g 0 .g 2u } = (1/p) 1 (-1-i√p)/2 (-1+i√p)/2 (-1) j (i) k+j p (k+j)/2 1 #{g 1 .g 2u } 1 (-1+i√p)/2 (-1-i√p)/2 (-1) k (i) k+j p (k+j)/2 1
Then : #{0}

(1/p).(p k+j +(1/2).(p-1).(i) k+j .p (k+j)/2 .((-1) k +(-1) j ))

#{g 0 .g 2u } = (1/p).(p k+j +(1/2).(i) k+j .p (k+j)/2 .(-(-1) k -(-1) j +i.p 1/2 .((-1) k -(-1) j ))) If k+j = 0 mod 2 : #{0}

(1/p).(p k+j +(p-1).(-1) (k-j)/2 .p (k+j)/2 ) #{g 0 .g 2u } = (1/p).(p k+j -(-1) (k-j)/2 .p (k+j)/2 ) (166)

#{g 1 .g 2u } (1/p).(p k+j -(-1) (k-j)/2 .p (k+j)/2 ) If k+j = 1 mod 2 : #{0} (1/p).(p k+j ) #{g 0 .g 2u } = (1/p).(p k+j -(-1) (k-j+1)/2 .p (k+j+1)/2 ) (167) #{g 1 .g 2u } (1/p).(p k+j +(-1) (k-j+1)/2 .p (k+j+1)/2 )
That is also gathering the two expressions :

#{0} (1/p).(p k+j +if(k+j=0 mod2,k(p-1).(-1) (k-j)/2 .p (k+j)/2 ,0) #{g 0 .g 2u } = (1/p).(p k+j -(-1) (k-j+mod(j+k,2))/2 .p (k+j+mod(j+k,2))/2 ) (168) #{g 1 .g 2u } (1/p).(p k+j -(-1) (k+j) (-1) (k-j+mod(j+k,2))/2 .p (k+j+mod(j+k,2))/2 )
We get thus, with their similarities and their differences, the following tables :

P
#{g 0 .g 2u } = (1/p).((p-1) k+j +(1/2).((p 1/2 -1) k+j+1 +(-1) k+j+1 .(p 1/2 +1) k+j+1 )) (169) #{g 1 .g 2u } (1/p).((p-1) k+j -(1/2).(p-1).((p 1/2 -1) k+j-1 +(-1) k+j-1 .(p 1/2 +1) k+j-1 ))
For p = 3 mod 4 :

#{0} 1 (p-1)/2 (p-1)/2 (p-1) k 0 0 (p-1) j 0 0 1 #{g 0 .g 2u } = (1/p) 1 (-1-i√p)/2 (-1+i√p)/2 0 (i.p 1/2 -1) k 0 0 (-1) j (i.p 1/2 +1) j 0 1 #{g 1 .g 2u } 1 (-1+i√p)/2 (-1-i√p)/2 0 0 (-1) k (i.p 1/2 +1) k 0 0 (i.p 1/2 -1) j 1 So that : #{0} 1 (p-1)/2 (p-1)/2 (p-1) k+j 0 0 1 #{g 0 .g 2u } = (1/p) 1 (-1-i√p)/2 (-1+i√p)/2 0 (-1) j .(i.p 1/2 -1) k .(i.p 1/2 +1) j 0 1 (170) #{g 1 .g 2u } 1 (-1+i√p)/2 (-1-i√p)/2 0 0 (-1) k .(i.p 1/2 +1) k .(i.p 1/2 -1) j 1
We get thus, with their similarities and their differences, the following tables :

p = 1 mod 4 k = 1, j=1 k = 2, j=1 k = 3, j=1 k = 4, j=1 #{0} 2(p-1) (p-1
p = 3 mod 4 k = 1, j=1 k = 2, j=1 k = 3, j=1 k = 4, j=1 #{0} 2(p-1) (p-1).(p-3) (p-1) 2 .(p-3) (p-1).( p 3 -4p²+9p-2) #{g 0 .g 2u } p-3 p 2 -2p+5 p 3 -4p²+5p-6 (p-1).(p 3 -4p²+5p-6) #{g 1 .g 2u } p-3 (p-1).(p-3) p 3 -4p²+9p-2 p.(p 3 -5p²+11p-15)

EXERCISE 6 : MODULO P δ ABUNDANCE FACTORS CORRECTION

This exercise may seem daunting. The reader may fly over it to retain only the essence which is that the transition from modulo p approach to modulo p δ approach proceeds of a correction which in successful outcomes (a number of variables higher than the degree of the equation) is negligible. The following developments are however essential to give orders of magnitude of this correction for abundance factors. Let us note besides that the correction modulo p δ in itself does not interest us. What we seek is the limit when δ tends towards infinite, projection which give the « right » results for asymptotic enumerations. The subject covered here is that of Waring sums with variables of integers or variables of prime numbers. It is vast in particular as the case p = 2 by itself takes as much place as the other cases p > 2. We have, according to whether we are interested in integers or prime numbers variables, the equations:

Cardinal matrices in

x 1 n + x 2 n + … +x i n + … + x k n = c modulo p δ and y 1 n + y 2 n + … +y i n + … + y k n = c modulo p δ
In [11], Waring problem induces the enumeration of the solutions of x 1 n + x 2 n + … +x i n + … + x k n = c modulo p δ where δ tends towards infinite. This enumeration is not made explicitly but a minoration of the fudge factor is given (Vinogradov minor and major arcs method). We propose here the actual calculation of fudge factor by using a "two-dimensional table method".

We will use the writing's rule d j = d[j] each time indices are difficult to distinguish in our text. Let us have g a primitive root of p for the class modulo p 2 (and thus for all classes mod p δ ). Let us have the Euler totient function : Ф(δ) = p δ-1 .(p-1). Let us have d j the greatest common divisor of n and p δ-1-j .(p-1) : d j = (n,p δ-1-j .(p-1)) = (n,Ф(δ-j)). Let us have the integer part of (δ-1)/n : δn = int((δ-1)/n). Let us have u(j) = Ф(δ-j)/d = p δ-1-j .(p-1)/d. Let us pose also v(i) ≡ {0,1,…,u(i)-1}, that we understand as v(i) successively describes all the integers ranging from 0 to u(i)-1 (or is any among these numbers).

The study of x n = c mod p δ asserts itself initially. The reader will refer to exercise 3 for modulo p δ classes' cardinals determination (cardinals that we call also multiplicities)

#{p δ } #{0} p δ-δn-1 #{p j.n .g 0 .g v(j.n).d[j.n] } #{p j.n .g 0 } d j .p j.(n-1) } #{p j.n .g 1 .g v(j.n).d[j.n] } = #{p j.n .g 1 } = 0 (1) … … … for j = 0 to δn #{p j.n .gd [j.n]-1 .g v(j.n).d[j.n] } #{p j.n .gd [j.n]-1 } 0
An important particular case is that where n does not have a factor equal to p :

p ∤ n (2) Then: d j = (n,Ф(δ-j)) = (n,p δ-1-j .(p-1)) = (n,p-1) = (n,Ф(1)) = d 0 = d (3)
Relations ( 1) simplify then :

Let us pass to the cardinal matrices.

In case δ = 1, cardinals research on the basis of the elements {0, g 0 , g 1 , …, gd -1 } earlier led to a square matrix of dimension d+1, and this owing to the fact that target g r .g w.d had equal cardinal that g r whatever w. We had :

#{p δ } #{0} p δ-δn-1 #{p j.n .g 0 .g v(j.n).d } #{p j.n .g 0 } d.p j.(n-1) } #{p j.n .g 1 .g v(j.n).d } = #{p j.n .g 1 } = 0 (4) … … … for j = 0 to δn #{p j.n .gd -1 .g v(j.n) } #{p j.n .gd -1 } 0 P 120/390
Modulo p Δ abundance factors correction .d +or(0,g y-2 .g v.d ) mod p}]+[I] where x is lines 1 to d+1 indice, y is columns 1 to d+1 indice {u, v} are integers describing [0,(p-1)

A δ=1 =[A δ=1 (x,y)] = or(#(u,v).d,#(v)) \ {or(0,g x-2 ) = g u
/d-1] 2
The element 0 constitutes a particular case in the study relating to δ = 1. For δ any natural integer, the number of particular cases increases since for x n = c mod p δ , not only 0, but all multiples of p have distinct cardinals.

We propose the two-dimensional table such as it presents itself before any explanation :

# c line card 0 card δ-1,0 card δ-1,d-1 card y1,0 card y1,d-1 card 0,0 card 0,d-1 c line (at k=k) 0 p δ-1 .g 0 .{g v(δ- 1).d[δ-1] } … p δ-1 .g d[δ -1]-1 .{g v(δ- 1).d[δ-1] } … p y1 .g 0 .{g v(y1).d[y1] } … p y1 .g d[y1]-1 .{g v(y1).d[y1] } ... p 0 .g 0 .{g v(0).d[0] } … p 0 .g d[0]-1 .{g v(0).d[0] } # c col c col (at k=1) p δ-δn-1 0 d δn .p δn.(n-1) p δn.n .g 0 p δn.n .g d[δn] … p δn.n .g u(n.δn).d[δn] … … d i1 .p i1.(n-1) p i1.n .g 0 c p i1.n .gd [i1] … p i1.n .g u(i1.n).d[i1] … … d 0 .p 0 g 0 g d[0] … g u(0)
.d [0] Second line and second column are occupied by what we will call generators of target c. The target c is, for given line and column, the sum of a generator in column c col (second column) with its multiplicity #{c col } (first column) and of a generator in line c lig (second line) with its multiplicity #{c lig } (first line). The multiplicity of c is then ∑ #{c lig }.#{c col }, the sum bearing on all c of the table having equal value.

Case p ∤ n (p>2)

The particular case, p does not divide n, gives a simplified two-dimensional table that we treat here.

# card 0 card δ-1,0 card δ-1,d-1 card y1,0 card y1,d-1 card 0,0 card 0,d-1 c (at k=k) 0 p δ-1 .g 0 .{g v(δ-1).d } … p δ-1 .g d-1 .{g v(δ-1).d } … p y1 .g 0 .{g v(y1).d } … p y1 .g d-1 .{g v(y1).d } ... p 0 .g 0 .{g v(0).d } … p 0 .g d-1 .{g v(0).d } # c (at k=1) p δ-δn-1 0 d.p δn.(n-1) p δn.n .g 0 p δn.n .g d … p δn.n .g u(n.δn).d … … d.p i1.(n-1) p i1.n .g 0 c p i1.n .g d … p i1.n .g u(j.n).d … … d.p 0 g 0 g d … g u(0).d
In the second column, we only write numbers of p i1.n .g i2.d type since the other numbers are with cardinal 0. In the second line, we make the hypothesis that p y1 .g y2 .{g v(y1 . This identity of cardinals is true at the first stage (first two columns is case k = 1). The property (equal cardinality per multiplicity for g x3.d ) is preserved when k is incremented because :

A δ (x1,x2,y1,y2) = #{(i1,i2,y3) \ or(0,p x1 .g x2 ) = or(0,p i1.n .g i2.d ) + or(0,p y1 .g y2 .g y3.d ) mod p δ (5) and A δ (x1,x2,y1,y2) = #{(i1,i2,y3) \ or(0,p x1 .g x2 .g x3.d ) = or(0,p i1.n .g (i2+x3).d ) + or(0,p y1 .g y2 .g (y3+x3).d ) mod p δ (6) are identical by circulant shifts (i2,y3) → (i2+x3,y3+x3).

It results for each quadruplet (x1, x2, y1, y2) a matrix of cardinals of dimension d (except of course for first line and column based on combinations with 0). The matrix is square of dimension δ.d+1.

We then seek the number of occurrences of target c of given value (at stage k+1) by examining all from this twodimensional table obtained sums, the goal being to get an expression like :

card' x1,x2 (c=or(0,p x1 .g x2 )) = ∑ a x1,x2,y1,y2 .card y1,y2 (7)
Here indices of "card'" are declined in lines by transposition of indices of "card" in columns. a x1,x2,y1,y2 is the result of all the contributions of the column corresponding to card y1,y2 by addition of the "good" cardinals of two-dimensional table generators. We have exactly :

card' x1,x2 (c) = ∑ (∑ #(c-x) mod p δ ).card y1,y2 (x) (8) y1,y2 x= p y1 .g y2 .{g v(y1).d } mod p δ v(y1) = 0 to u(y1)-1
This writing means that the "sum equal to c" test is carried out on all column generators x and line generators c-x. For each x = p y1 .g y2 .g v(y1).d , v(y1) integer varying from 0 to u(y1)-1, cardinal #(c-x) is registered and summation of cardinals is carried out for all the family of numbers x = p y1 .g y2 .{g v(y1).d }. The sum ∑ #(c-x) will then be the component of the cardinal matrix at (x1,x2,y1,y2). We base on the simple but fundamental result, given at the end of exercise 3 (that we called displacement mod p)  s, there exist s' and k  [[0,p δ-1 /d-1]] \ g s.d = g s'.d +k.p mod p δ hence it follows in particular :

 s, there exist k  [[0,p δ-1 /d-1]] \ g s.d = g 0 +k.p mod p δ to rewrite the two-dimensional table (all of expressions are modulo p δ and v( ) varies in its field of definition) :

# card 0 card δ--1,0 card δ--1,d-1 card y1,0 card y1,d-1 card 0,0 card 0,d-1 c (at k=k) 0 p δ-1 . (g 0 + v(δ-1).p) … p δ-1 . (gd -1 + v(δ-1).p) … p y1 . (g 0 + v(y1).p) … p y1 . (gd -1 + v(y1).p) ... p 0 . (g 0 + v(0).p) … p 0 . (gd -1 + v(0).p) # c (at k=1) p δ-δn-1 0 d.p δn.(n-1) p δn.n .(g 0 +v(n.δn).p) … … d.p i1.(n-1) p i1.n .(g 0 +v(i1.n).p) … … d.p 0 g 0 +v(0).p
This summarize as :

# card 0 card y1,y2 c (at k=k) 0 p y1 .(g y2 +v(y1).p) v(y1) = 0 to u(y1)-1 # c (at k=1) p δ-δn-1 0 d.p i1.(n-1) p i1.n .(g 0 +v(i1.n).p) v(i1.n) = 0 to u(i1.n)-
= ∑ #(c-x) = ∑ #(p x1 .(g x2 -p (y1-x1) .g y2 .g v(y1).d )) = ∑ #( p k.n .(g 0 mod p)) x= p y1 .g y2 .{g v(y1).d } v(y1) = 0 to u(y1)-1 v(y1) = 0 to u(y1)-1 v(y1) = 0 to u(y1)-1 So that : a x1,x2,y1,y2 = d.p k.(n-1) .u(y1) = d.p x1-k . p δ-y1-1 .(p-1)/d = p δ-1+x1-y1-k .(p-1) That is also : a x1=k.n,x2,y1>x1,y2 = Ф(δ+x1-y1-k) ( 10
)
This contribution to the cardinal matrix, because of the presupposed of this case, is below the trace's blocks (d dimension blocks with "centre" the matrix trace). It does not interest first column of the matrix, but begins at the second, column for which, for the concerned lines (at x1=k.n) :

a x1=k.n,x2,y1>x1,y2 (y1=δ-1) = Ф(k.(n-1)+1)
The value remains unchanged on line for d columns since the expression above does not depend on y2, then Ф(x) becomes Ф(x+1) with each d columns as long as positions of trace's blocks are not reached (condition y1 > x1).

The contribution is a minimal contribution to the here interested positions as other contributions may possibly be added by other hypothesis on (x1, x2, y1, y2) (we will see however that it is not so).

Case 2

Let us suppose : g 0 +g y2 = 0 mod p and c = p x1 .g x2 and x1 > y1 and y1 = k.n where k[ [0,δn]

] Then y2 = (p-1)/2 mod p and a x1,x2,y1,y2 = ∑ #(c-x) = ∑ #(c-x) = ∑ #(p y1 .(p (x1-y1) .g x2 +p.v(y1)+g 0 )) = ∑ #(p k.n .(g 0 +v(y2).p)) x= -p y1 .{g v(y1).d } v(y1) = 0 to u(y1)-1 x= -p y1 .(g 0 +v(y1).p) v(y1) = 0 to u(y1)-1 v(y1) = 0 to u(y1)-1 v(y2) = 0 to u(y2)-1 Thus : a x1,x2,y1,y2 = d.p k.(n-1) .u(y1) = p y1-k . p δ-y1-1 .(p-1) = p δ-k-1 .(p-1) So that : a x1>y1,x2,y1=k.n,y2=(p-1)/2 mod p = Ф(δ-k) (11)
This contribution to the cardinal matrix is, by the presupposed of this case, above the trace's blocks. On a column satisfying the condition y2 = (p-1)/2 mod p, the component remains constant from first line until a trace's block is reached (condition x1 > y1). Here the first line is actually also concerned as above calculation is applicable.

The contribution is again a minimal contribution to here interested positions (We will see however that it is not so).

p )) = ∑ #(-p x1 .g x2 .v(y1).p) = ∑ #(p x1+1 .g x2 .v(y1)) x= p y1 .g y2 .(g 0 +v(y1).p) v(y1) = 0 to u(y1)-1 v(y1) = 0 to u(y1)-1 v(y1) = 0 to u(y1)-1 v(y1) = 0 to u(y1)-1
When v(y1) varies p x1+1 .g x2 .v(y1) will describe, mod p δ , all of multiples of p x1+1 and in peculiar all multiples of p i.n such as i.n ≥ x1+1. Thus :

a x1,x2,y1,y2 = ∑ d.p i(n-1) .{v(in)} = ∑ d.p i(n-1) .u(in) = ∑ d.p i(n-1) .p δ-in-1 .(p-1)/d = ∑ p δ-i-1 .(p-1) = p δ-i i>int(x1/n) v(in) = 0 to u(i.n)-1 i>int(x1/n) i>int(x1/n) i>int(x1/n) (at i=int(x1/n)+1 = k+1) So that : a x1,x2,y1=x1,y2=x2 = p δ-1-k (12)
This contribution to cardinal matrix is located at the level of the trace (trace's blocks in fact). On this trace (trace's blocks), we have, as early as second column (x1 = 0 and k = 0), the contribution p δ-1 , contribution remaining unchanged on d.n following columns, then incremented by 1 and unchanged again on d.n following columns, etc. This contribution is a minimal contribution to the here interested positions and other hypothesis on (x1,x2,y1,y2) may intervene. We will see below under which conditions.

Case 4

Let us suppose :

c = p x1 .g x2 and x1 = y1 and x1 = k.n Then a x1,x2,y1,y2 = ∑ #(c-x) = ∑ #(p x1 .(g x2 -g y2 -g y2 .v(y1).p)) = ∑ #(p k.n .(g x2 -g y2 -v(y3).p)) x= p y1 .g y2 .(g 0 +v(y1).p) v(y1) = 0 to u(y1)-1 v(y1) = 0 to u(y1)-1 v(y3) = 0 to u(y1)-1
We then collect all the incidences of type : g x2 -g y2 -v(y3).p = g 0 .p j.n We get then : a x1,x2,y1,y2 = ∑ #(p (k+j).n .(g 0 +v(y4).p)) v(y4) = 0 to u(y1)-1 and g x2 -g y2 = g 0 mod p

This last equation induces use of the components of case δ = 1 cardinal matrix. Thus, by adopting a notation per block (that is using [ ]), we get :

[a x1,x2,y1,y2 ] = ∑ d.p i.(n-1) .{v(in)}.[A'] = ∑ d.p i.(n-1) .u(in).[A'] = ∑ p δ-i-1 .(p-1).[A'] = p δ-i .[A'] i>int(x1/n) v(in) = 0 to u(i.n)-1 i>int(x1/n) i>int(x1/n) i=int(x1/n)+1 = k+1 Hence : [a x1,x2,y1,y2 ] = p δ-1-k .[A'] (13)
This contribution to cardinal matrix is at the level of trace's blocks. Case 3 is redundant at x2 = y2 with case 4 and is thus to deduct. Thus case 4 applies for x1 = 0 mod n and case 3 at x1 ≠ 0 mod n.

From the contributions of these four cases, it rises general form from the case "p does not divide n" of cardinal matrices. The obtained components (cardinals) are minimal values. It suffices then to prove that we exhausted all possible combinations. This is done by adding each line of matrix. If the result gives p δ , obtained minimal values are also maximum values and proposed matrix is matrix satisfying to the problem. This verification does not present any particular difficulty. For that, the reader will refer on one hand to case δ = 1 to check that card line1 We represent the thus obtained below cardinal matrices in most general form and we give an example d = 2 and its various lower cases : (p,δ) = (or(1,3) mod 4, or(0,1) mod 2)).

We give directly [A δ ] eigenvalues and eigenvectors of A δ . These results can be demonstrated by simple substitution while writing [A δ ](X) = {μ δ }(X) for each column (X) of P δ . To return there, in detail, would not bring anything more here than to fill up paper. In the same way, the expression of P -1 δ is checked "self-evidently" while calculating [P -1 δ ].[P δ ] that one finds equal to [I].

Then we get

[A δ ] k = [P δ ].[μ δ ] k .[P δ -1 ] and k # (p δ ) 1 #{p δ-1 .g 0 } 0 #{p δ-1 .g 1 } 0 … ... #{p δ-1 .gd -1 } 0 ... ... #{p 1 .g 0 } = A δ 0 #{p 1 .g 1 } 0 … ... #{p 1 .gd -1 } 0 #{p 0 .g 0 } 0 #{p 0 .g 1 } 0 … ... #{p 0 .gd -1 } 0 Foot-note :
Hereafter matrices also admit a form close to "intra-unit" form (found in case δ = 1).

p δ p δ-δn-1 … Ф(δ-δn).[J'] … [0] Ф(δ-2).[J'] [0] … [0] Ф(δ-1).[J'] [0] … [0] Ф(δ).[J'] … … [0] Ф(δ-2).[J] [0] … [0] Ф(δ-1).[J] [0] … [0] Ф(δ).[J] … [0] Ф(δ-2).[J] [0] … [0] Ф(δ-1).[J] [0] … [0] Ф(δ).[J] p δn.n .[g i ] d.p δn.(n-1) .[K'] Ф((n-1).δn+1).[K] … … p δ-δn-1 .[A'] … [0] Ф(δ-2).[J] [0] … [0] Ф(δ-1).[J] [0] … [0] Ф(δ).[J] [0] [0] [0] … [0] … [0] Ф(δ-2).[J] [0] … [0] Ф(δ-1).[J] [0] … [0] Ф(δ).[J] … … … … … … [0] Ф(δ-2).[J] [0] … [0] Ф(δ-1).[J] [0] … [0] Ф(δ).[J] … [0] Ф(δ-2).[J] [0] … [0] Ф(δ-1).[J] [0] … [0] Ф(δ).[J] … … … … … … … … … … … … … … … … p 2n+1 .[g i ] [0] [0] [0] … [0] … p δ-3 .[I] Ф(δ-2).[J] [0] … [0] Ф(δ-1).[J] [0] … [0] Ф(δ).[J] p 2n .[g i ] d.p 2(n-1) .[K'] Ф(2n-1).[K] Ф(2n).[K] … Ф(δ-n.δn+2n-2).[K] … Ф(δ-3).[K] p δ-3 .[A'] [0] … [0] Ф(δ-1).[J] [0] … [0] Ф(δ).[J] … [0] [0] [0] … [0] … [0] [0] p δ-2 .[I] … [0] Ф(δ-1).[J] [0] … [0] Ф(δ).[J] p n+2 .[g i ] [0] [0] [0] … [0] … [0] [0] [0] … [0] Ф(δ-1).[J] [0] … [0] Ф(δ).[J] p n+1 .[g i ] [0] [0] [0] … [0] … [0] [0] [0] … p δ-2 .[I] Ф(δ-1).[J] [0] … [0] Ф(δ).[J] p n .[g i ] d.p n-1 .[K'] Ф(n).[K] Ф(n+1).[K] … Ф(δ-n.δn+n-1).[K] … Ф(δ-n-2).[K] Ф(δ-n-1).[K] Ф(δ-n).[K] … Ф(δ-2).[K] p δ-2 .[A'] [0] … [0] Ф(δ).[J] … [0] [0] [0] … [0] … [0] [0] [0] … [0] [0] p δ-1 .[I] … [0] Ф(δ).[J] p 2 .[g i ] [0] [0] [0] … [0] … [0] [0] [0] … [0] [0] [0] … [0] Ф(δ).[J] p 1 .[g i ] [0] [0] [0] … [0] … [0] [0] [0] … [0] [0] [0] … p δ-1 .[I] Ф(δ).[J] p 0 .[g i ] d.[K'] Ф(1).[K] Ф(2).[K] … Ф(δ-n.δn).[K] … Ф(δ-2n-1).[K] Ф(δ-2n).[K] Ф(δ-2n+1).[K] … Ф(δ-n-1).[K] Ф(δ-n).[K] Ф(δ-n+1).[K] … Ф(δ-1).[K] p δ-1 .[A']
Foot note : In first column, out of matrix, [g i ] represents column vector {g 0 , g 1 , g 2 , …, gd -1 }. For the record Ф(t) = p t-1 .(p-1).

P 126/390

Modulo p Δ abundance factors correction Eigenvalues of A δ , followed by eigenvectors matrix P δ (general case) (λ i = μ i -1) (matrix P is built easily while starting on left and bottom) (k varies 1 to d) (first column of this table does not belong to matrix)

{μ δ } p δ p δ-1 [μ'] p δ-1 [I] … p δ-1 [I] p δ-2 [μ'] p δ-2 [I] … p δ-2 [I] p δ-3 [μ'] p δ-3 [I] … p δ-3 [I] p δ-δn-1 [μ'] p δ-δn-1 [I] … if(δ = 1 mod n, p δ-δn-1 [μ'] , p δ-δn-1 [I] p δ [1] (p-1).[U'] (p-1).[U'] … (p-1).[U'] (p-1).[U'] (p-1).[U'] … (p-1).[U'] (p-1).[U'] (p-1).[U'] … (p-1).[U'] (p-1).[U'] (p-1).[U'] … (p-1).[U'] … … … … … … … … … … … … … … … … [λ'*] [U"] (p-1).[U] (p-1).[U] … (p-1).[U] (p-1).[U] (p-1).[U] … (p-1).[U] (p-1).[U] (p-1).[U] … (p-1).[U] (p-1).[U] (p-1).[U] … [0] [U"] (p-1).[U] (p-1).[U] (p-1).[U] (p-1).[U] (p-1).[U] … (p-1).[U] (p-1).[U] (p-1).[U] … (p-1).[U] (p-1).[U] [λ'*] … … p δn.n .[g i ] [U"] (p-1).[U] (p-1).[U] … (p-1).[U] (p-1).[U] (p-1).[U] … (p-1).[U] (p-1).[U] (p-1).[U] … (p-1).[U] [λ'*] [0] … [0] [U"] (p-1).[U] (p-1).[U] … (p-1).[U] (p-1).[U] (p-1).[U] … (p-1).[U] (p-1).[U] (p-1).[U] … [λ'*] [0] [0] … [0] [U"] (p-1).[U] (p-1).[U] … (p-1).[U] (p-1).[U] (p-1).[U] … (p-1).[U] (p-1).[U] (p-1).[U] … [0] [0] [0] … [0] … … … … … … … … … … … … … … … … … … p 2n+1 .[g i ] [U"] (p-1).[U] (p-1).[U] … (p-1).[U] (p-1).[U] (p-1).[U] … (p-1).[U] (p-1).[U] [λ'*] … [0] [0] [0] … [0] p 2n .[g i ] [U"] (p-

1).[U] (p-1).[U] … (p-1).[U] (p-1).[U] (p-1).[U] … (p-1).[U]

[λ'*] [0] … [0] [0] [0] … [0] p 2n-1 .[g i ] [U"] (p-1).[U] (p-1).[U] … (p-1).[U] (p-1).[U] (p-1).[U] … [λ'*] [0] [0] … [0] [0] [0] … [0] … [U"] (p-1).[U] (p-1).[U] … (p-1).[U] (p-1).[U] (p-1).[U] … [0] [0] [0] … [0] [0] [0] … [0] p n+2 .[g i ] [U"] (p-1).[U] (p-1).[U] … (p-1).[U] (p-1).[U] (p-1).[U] … [0] [0] [0] … [0] [0] [0] … [0] p n+1 .[g i ] [U"] (p-1).[U] (p-1).[U] … (p-1).[U] (p-1).[U] [λ'*] … [0] [0] [0] … [0] [0] [0] … [0] p n .[g i ] [U"] (p-1).[U] (p-1).[U] … (p-1).[U] [λ'*] [0] … [0] [0] [0] … [0] [0] [0] … [0] … [U"] (p-1).[U] (p-1).[U] … [λ'*] [0] [0] … [0] [0] [0] … [0] [0] [0] … [0] p 2 .[g i ] [U"] (p-1).[U] (p-1).[U] … [0] [0] … [0] [0] [0] … [0] [0] [0] … [0] p 1 .[g i ] [U"] (p-1).[U] [λ'*] … [0] [0] … [0] [0] [0] … [0] [0] [0] … [0] p 0 .[g i ] [U"] [λ'*] [0] … [0] [0] … [0] [0] … [0] [0] [0] … [0] Foot note : [λ'] = [ t λ
).[U] (p-1)/(d 2 .p n-1 ).[U] … 1/(d 2 p).[λ' ] [0] … … … … … … … … … … … … … … 1/(d.p δ-n+1 ).[U"] (p-1)/(d 2 .p δ-n+1 ).[U] (p-1)/(d 2 .p δ-n ).[U] … (p-1)/(d 2 .p n+3 ).[U] (p-1)/(d 2 .p n+2 ).[U] (p-1)/(d 2 .p n+1 ).[U] … (p-1)/(d 2 .p 3 ).[U] (p-1)/(d 2 .p 2 ).[U] 1/(d 2 p).[λ' ] … [0] [0] 1/(d.p δ-n ).[U"]] (p-1)/(d 2 .p δ-n ).[U] (p-1)/(d 2 .p δ-n-1 ).[U] … (p-1)/(d 2 .p n+2 ).[U] (p-1)/(d 2 .p n+1 ).[U] (p-1)/(d 2 .p n ).[U] … (p-1)/(d 2 .p 2 ).[U] 1/(d 2 p).[ λ'] [0] … [0] [0] 1/(d.p δ-n-1 ).[U"] (p-1)/(d 2 .p δ-n-1 ).[U] (p-1)/(d 2 .p δ-n-2 ).[U] … (p-1)/(d 2 .p n ).[U] (p-1)/(d 2 .p n-1 ).[U] (p-1)/(d 2 .p n-1 ).[U] … 1/(d 2 p).[λ' ] [0] [0] … [0] [0] … … … … … … … … … … … … … … 1/(d.p δ-2n+1 ).[U"] (p-1)/(d 2 .p δ-2n+1 ).[U] (p-1)/(d 2 .p δ-2n ).[U] … (p-1)/(d 2 .p 3 ).[U] (p-1)/(d 2 .p 2 ).[U] 1/(d 2 p).[λ' ] … [0] [0] [0] … [0] [0] 1/(d.p δ-2n ).[U"] (p-1)/(d 2 .p δ-2n ).[U] (p-1)/(d 2 .p δ-2n-1 ).[U] … (p-1)/(d 2 .p 2 ).[U] 1/(d 2 p).[ λ'] [0] … [0] [0] [0] … [0] [0] 1/(d.p δ-2n-1 ).[U"] (p-1)/(d 2 p δ-2n-1 ).[U] (p-1)/(d 2 p δ-2n-2 ).[U] … 1/(d 2 p).[λ' ] [0] [0] … [0] [0] [0] … [0] [0] … … … … … … … … … … … … … … … 1/(d.p 4 ).[U''] (p-1)/(d 2 .p 4 ).[U] (p-1)/(d 2 .p 3 ).[U] … [0] [0] [0] … [0] [0] [0] … [0] [0] p 2 .[g i ] 1/(d.p 3 ).[U"] (p-)/(d 2 .p 3 ).[U] (p-)/(d 2 .p 2 ).[U] … [0] [0] [0] … [0] [0] [0] … [0] [0] p 1 .[g i ] 1/(d.p 2 ).[U"] (p-1)/(d 2 .p 2 ).[U] 1/(d 2 p).[λ' ] … [0] [0] [0] … [0] [0] [0] … [0] [0] p 0 .[g i ] 1/(d.p).[U"] 1/(d 2 p).[λ' ] [0] … [0] [0] [0] … [0] [0] [0] … [0] [0] P 128/390 Modulo p Δ abundance factors correction Case of hypersphere p ∤ n, d = 2, p = 1 mod 4, δ = 2t Matrix A δ p t 0 0 p t (p-1) 0 0 0 p t+1 (p-1) 0 0 0 p t+2 (p-1) 0 0 0 … 0 0 p 2t-1 (p-1) 0 p t 0 p t (p-1) 0 0 0 p t+1 (p-1) 0 0 0 p t+2 (p-1) 0 0 0 … 0 0 p 2t-1 (p-1) 0 0 p t p t (p-1) 0 0 0 p t+1 (p-1) 0 0 0 p t+2 (p-1) 0 0 0 … 0 0 p 2t-1 (p-1) 2p t-1 p t-1 (p-1) p t-1 (p-1) p t (p-3)/2 p t (p-1)/2 0 0 p t+1 (p-1) 0 0 0 p t+2 (p-1) 0 0 0 … 0 0 p 2t-1 (p-1) 0 0 0 p t (p-1)/2 p t (p+1)/2 0 0 p t+1 (p-1) 0 0 0 p t+2 (p-1) 0 0 0 … 0 0 p 2t-1 (p-1) 0 0 0 0 0 p t+1 0 p t+1 (p-1) 0 0 0 p t+2 (p-1) 0 0 0 … 0 0 p 2t-1 (p-1) 0 0 0 0 0 0 p t+1 p t+1 (p-1) 0 0 0 p t+2 (p-1) 0 0 0 … 0 0 p 2t-1 (p-1) 2p t-2 p t-2 (p-1) p t-2 (p-1) p t-1 (p-1) p t-1 (p-1) p t (p-1) p t (p-1) p t+1 (p-3)/2 p t+1 (p-1)/2 0 0 p t+2 (p-1) 0 0 0 … 0 0 p 2t-1 (p-1) 0 0 0 0 0 0 0 p t+1 (p-1)/2 p t+1 (p+1)/2 0 0 p t+2 (p-1) 0 0 0 … 0 0 p 2t-1 (p-1) 0 0 0 0 0 0 0 0 0 p t+2 0 p t+2 (p-1) 0 0 0 … 0 0 p 2t-1 (p-1) 0 0 0 0 0 0 0 0 0 0 p t+2 p t+2 (p-1) 0 0 0 … 0 0 p 2t-1 (p-1) 2p t-3 p t-3 (p-1) p t-3 (p-1) p t-2 (p-1) p t-2 (p-1) p t-1 (p-1) p t-1 (p-1) p t (p-1) p t (p-1) p t+1 (p-1) p t+1 (p-1) p t+2 (p-3)/2 p t+2 (p-1)/2 0 0 … 0 0 p 2t-1 (p-1) 0 0 0 0 0 0 0 0 0 0 0 p t+2 (p-1)/2 p t+2 (p+1)/2 0 0 … 0 0 p 2t-1 (p-1) 0 0 0 0 0 0 0 0 0 0 0 0 0 p t+2 0 … 0 0 p 2t-1 (p-1) 0 0 0 0 0 0 0 0 0 0 0 0 0 0 p t+3 … 0 0 p 2t-1 (p-1) … … … … … … … … … … … … … … … … … …
2p p(p-1) p(p-1) p 2 (p-1) p 2 (p-1) p 3 p-1) p 3 (p-1) p 4 (p-1) p 4 (p-1) p 5 (p-1) p 5 (p-1) p 6 (p-1) p 6 (p-1) p 7 (p-1) P 7 (p-1) … 0 0 p 2t-1 (p-1)

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 p 2t-1 (p-1) 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 … p 2t-1 0 p 2t-1 (p-1) 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 … 0 p 2t-1 p 2t-1 (p-1) 2 p-1 p-1 p(p-1) p(p-1) p 2 (p-1) p 2 (p-1) p 3 (p-1) p 3 (p-1) p 4 (p-1) p 4 (p-1) p 5 (p-1) p 5 (p-1) p 6 (p-1) p 6 (p-1) … (p-1)p 2t-2 (p-1)p 2t-2 p 2t-1 (p-3)/2 p 2t-1 (p-1)/2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 … 0 0 p 2t-1 (p-1)/2 p 2t-1 (p+1)/2 P 129/390
Modulo p Δ abundance factors correction Eigenvalues of A δ , followed by corresponding eigenvectors matrix P and inverse matrix P -1 (case of hypersphere p ∤ n, d = 2, p = 1 mod 4, δ = 2t)

p 2t -p 2t-1 √p p 2t-1 √p p 2t-1 p 2t-1 -p 2t-2 √p p 2t-2 √p … p t+1 p t+1 -p t √p p t √p p t p t 1 p-1 p-1 p-1 p-1 p-1 p-1 … p-1 p-1 p-1 p-1 p-1 p-1 1 p-1 p-1 p-1 p-1 p-1 p-1 … p-1 p-1 p-1 p-1 -1-√p -1+√p 1 p-1 p-1 p-1 p-1 p-1 p-1 … p-1 p-1 p-1 p-1 -1+√p -1-√p 1 p-1 p-1 p-1 p-1 p-1 p-1 … p-1 p-1 -1-√p -1+√p 0 1 p-1 p-1 p-1 p-1 p-1 p-1 … p-1 p-1 -1+√p -1-√p 0 1 p-1 p-1 p-1 p-1 p-1 p-1 … -1-√p -1+√p 0 0 0 1 p-1 p-1 p-1 p-1 p-1 p-1 … -1+√p -1-√p 0 0 0 1 p-1 p-1 p-1 p-1 p-1 p-1 … 0 0 0 0 0 1 p-1 p-1 p-1 p-1 p-1 p-1 … 0 0 0 0 0 … … … … … … … … … … … … 1 p-1 p-1 p-1 p-1 -1-√p -1+√p … 0 0 0 0 0 1 p-1 p-1 p-1 p-1 -1+√p -1-√p 0 0 0 0 0 1 p-1 p-1 -1-√p -1+√p 0 0 … 0 0 0 0 0 1 p-1 p-1 -1+√p -1-√p 0 0 … 0 0 0 0 0 1 -1-√p -1+√p 0 0 0 0 … 0 0 0 0 0 1 -1+√p -1-√p 0 0 0 0 … 0 0 0 0 0 1/p 2t (p-1)/2p 2t (p-1)/2p 2t (p-1)/2p 2t-1 (p-1)/2p 2t-1 (p-1)/2p 2t-2 (p-1)/2p 2t-2 … (p-1)/2p 3 (p-1)/2p 3 (p-1)/2p 2 (p-1)/2p 2 (p-1)/2p
(p-1)/2p 1/2p 2t (p-1)/4p 2t (p-1)/4p 2t (p-1)/4p 2t-1 (p-1)/4p 2t-1 (p-1)/4p 2t-2 (p-1)/4p 2t-2 … (p-1)/4p 3 (p-1)/4p 3 (p-1)/4p 2 (p-1)/4p 2 (-1-√p)/4p (-1+√p)/4p 1/2p 2t (p-1)/4p 2t (p-1)/4p 2t (p-1)/4p 2t-1 (p-1)/4p 2t-1 (p-1)/4p 2t-2 (p-1)/4p 2t-2 … (p-1)/4p 3 (p-1)/4p 3 (p-1)/4p 2 (p-1)/4p 2 (-1+√p)/4p (-1-√p)/4p 1/2p 2t-1 (p-1)/4p 2t-1 (p-1)/4p 2t-1 (p-1)/4p 2t-2 (p-1)/4p 2t-2 (p-1)/4p 2t-3 (p-1)/4p 2t-3 … (p-1)/4p 2 (p-1)/4p 2 (-1-√p)/4p (-1+√p)/4p 0 1/2p 2t-1 (p-1)/4p 2t-1 (p-1)/4p 2t-1 (p-1)/4p 2t-2 (p-1)/4p 2t-2 (p-1)/4p 2t-3 (p-1)/4p 2t-3 … (p-1)/4p 2 (p-1)/4p 2 (-1+√p)/4p (-1-√p)/4p 0 1/2p 2t-2 (p-1)/4p 2t-2 (p-1)/4p 2t-2 (p-1)/4p 2t-3 (p-1)/4p 2t-3 (p-1)/4p 2t-4 (p-1)/4p 2t-4 … (-1-√p)/4p (-1+√p)/4p

0 0 0 1/2p 2t-2 (p-1)/4p 2t-2 (p-1)/4p 2t-2 (p-1)/4p 2t-3 (p-1)/4p 2t-3 (p-1)/4p 2t-4 (p-1)/4p 2t-4 … (-1+√p)/4p (-1-√p)/4p 0 0 0 … … … … … … … … … … … … 1/2p 3 (p-1)/4p 3 (p-1)/4p 3 (p-1)/4p 2 (p-1)/4p 2 (-1-√p)/4p (-1+√p)/4p … 0 0 0 0 0 1/2p 3 (p-1)/4p 3 (p-1)/4p 3 (p-1)/4p 2 (p-1)/4p 2 (-1+√p)/4p (-1-√p)/4p 0 0 0 0 0 1/2p 2 (p-1)/4p 2 (p-1)/4p 2 (-1-√p)/4p (-1+√p)/4p 0 0 … 0 0 0 0 0 1/2p 2 (p-1)/4p 2 (p-1)/4p 2 (-1+√p)/4p (-1-√p)/4p 0 0 … 0 0 0 0 0 1/2p (-1-√p)/4p (-1+√p)/4p 0 0 0 0 … 0 0 0 0 0 1/2p (-1+√p)/4p (-1-√p)/4p 0 0 0 0 … 0 0 0 0
0 0 0 0 0 p t+1 0 0 p t+1 (p-1) 0 0 0 p t+2 (p-1) … 0 0 p 2t-1 (p-1) 0 0 0 0 0 0 p t+1 0 p t+1 (p-1) 0 0 0 p t+2 (p-1) … 0 0 p 2t-1 (p-1) 2p t-2 p t-2 (p-1) p t-2 (p-1) p t-1 (p-1) p t-1 (p-1) p t (p-1) p t (p-1) p t+1 (p-1)/2 p t+1 (p-3)/2 0 0 0 p t+2 (p-1) … 0 0 p 2t-1 (p-1) 0 0 0 0 0 0 0 p t+1 (p+1)/2 p t+1 (p-1)/2 0 0 0 p t+2 (p-1) … 0 0 p 2t-1 (p-1) 0 0 0 0 0 0 0 0 0 p t+2 0 0 p t+2 (p-1) … 0 0 p 2t-1 (p-1) 0 0 0 0 0 0 0 0 0 0 p t+2 0 p t+2 (p-1) … 0 0 p 2t-1 (p-1) 2p t-3 p t-3 (p-1) p t-3 (p-1) p t-2 (p-1) p t-2 (p-1) p t-1 (p-1) p t-1 (p-1) p t (p-1) p t (p-1) p t+1 (p-1) p t+1 (p-1) p t+2 (p-1)/2 p t+2 (p-3)/2 … 0 0 p 2t-1 (p-1) 0 0 0 0 0 0 0 0 0 0 0 p t+2 (p+1)/2 p t+2 (p-1)/2 … 0 0 p 2t-1 (p-1) 0 0 0 0 0 0 0 0 0 0 0 0 0 … 0 0 p 2t-1 (p-1) 0 0 0 0 0 0 0 0 0 0 0 0 0 … 0 0 p 2t-1 (p-1) … … … … … … … … … … … … … … … … … …
2p p(p-1) p(p-1) p 2 (p-1) p 2 (p-1) p 3 p-1) p 3 (p-1) p 4 (p-1) p 4 (p-1) p 5 (p-1) p 5 (p-1) p 6 (p-1)

p 6 (p-1) … 0 0 p 2t-1 (p-1) 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 p 2t-1 (p-1) 0 0 0 0 0 0 0 0 0 0 0 0 0 … p 2t-1 0 p 2t-1 (p-1) 0 0 0 0 0 0 0 0 0 0 0 0 0 … 0 p 2t-1 p 2t-1 (p-1) 2 p-1 p-1 p(p-1) p(p-1) p 2 (p-1) p 2 (p-1) p 3 (p-1) p 3 (p-1) p 4 (p-1) p 4 (p-1) p 5 (p-1) p 5 (p-1) … (p-1)p 2t-2 (p-1)p 2t-2 p 2t-1 (p-1)/2 p 2t-1 (p-3)/2 0 0 0 0 0 0 0 0 0 0 0 0 0 … 0 0 p 2t-1 (p+1)/2 p 2t-1 (p-
1 p-1 p-1 p-1 p-1 p-1 p-1 … p-1 p-1 p-1 p-1 p-1 p-1 p-1 p-1 1 p-1 p-1 p-1 p-1 p-1 p-1 … p-1 p-1 p-1 p-1 p-1 p-1 (-1-i√p) (-1+i√p) 1 p-1 p-1 p-1 p-1 p-1 p-1 … p-1 p-1 p-1 p-1 p-1 p-1 (-1+i√p) (-1-i√p) 1 p-1 p-1 p-1 p-1 p-1 p-1 … p-1 p-1 p-1 p-1 (-1-i√p) (-1+i√p) 0 1 p-1 p-1 p-1 p-1 p-1 p-1 … p-1 p-1 p-1 p-1 (-1+i√p) (-1-i√p) 0 1 p-1 p-1 p-1 p-1 p-1 p-1 … p-1 p-1 (-1-i√p) (-1+i√p) 0 0 0 1 p-1 p-1 p-1 p-1 p-1 p-1 … p-1 p-1 (-1+i√p) (-1-i√p) 0 0 0 1 p-1 p-1 p-1 p-1 p-1 p-1 … (-1-i√p) (-1+i√p) 0 0 0 0 0 1 p-1 p-1 p-1 p-1 p-1 p-1 … (-1+i√p) (-1-i√p) 0 0 0 0 0 … … … … … … … … … … … … … … 1 p-1 p-1 p-1 p-1 (-1-i√p) (-1+i√p) … 0 0 0 0 0 0 0 1 p-1 p-1 p-1 p-1 (-1+i√p) (-1-i√p) 0 0 0 0 0 0 0 1 p-1 p-1 (-1-i√p) (-1+i√p) 0 0 … 0 0 0 0 0 0 0 1 p-1 p-1 (-1+i√p) (-1-i√p) 0 0 … 0 0 0 0 0 0 0 1 (-1-i√p) (-1+i√p) 0 0 0 0 … 0 0 0 0 0 0 0 1 (-1+i√p) (-1-i√p) 0 0 0 0 … 0 0 0 0 0 0 0 1/p 2t (p-1)/2p 2t (p-1)/2p 2t (p-1)/2p 2t-1 (p-1)/2p 2t-1 (p-1)/2p 2t-2 (p-1)/2p 2t-2 … (p-1)/2p 4 (p-1)/2p 4 (p-1)/2p 3 (p-1)/2p 3 (p-1)/2p 2 (p-1)/2p 2 (p-1)/2p
(p-1)/2p 1/2p 2t (p-1)/4p 2t (p-1)/4p 2t (p-1)/4p 2t-1 (p-1)/4p 2t-1 (p-1)/4p 2t-2 (p-1)/4p 2t-2 … (p-1)/4p 4 (p-1)/4p 4 (p-1)/4p 3 (p-1)/4p 3 (p-1)/4p 2 (p-1)/4p 2 (-1+i√p)/4p (-1-i√p)/4p 1/2p 2t (p-1)/4p 2t (p-1)/4p 2t (p-1)/4p 2t-1 (p-1)/4p 2t-1 (p-1)/4p 2t-2 (p-1)/4p 2t-2 … (p-1)/4p 4 (p-1)/4p 4 (p-1)/4p 3 (p-1)/4p 3 (p-1)/4p 2 (p-1)/4p 2 (-1-i√p)/4p (-1+i√p)/4p 1/2p 2t-1 (p-1)/4p 2t-1 (p-1)/4p 2t-1 (p-1)/4p 2t-2 (p-1)/4p 2t-2 (p-1)/4p 2t-3 (p-1)/4p 2t-3 … (p-1)/4p 3 (p-1)/4p 3 (p-1)/4p 2 (p-1)/4p 2 (-1+i√p)/4p (-1-i√p)/4p 0 1/2p 2t-1 (p-1)/4p 2t-1 (p-1)/4p 2t-1 (p-1)/4p 2t-2 (p-1)/4p 2t-2 (p-1)/4p 2t-3 (p-1)/4p 2t-3 … (p-1)/4p 3 (p-1)/4p 3 (p-1)/4p 2 (p-1)/4p 2 (-1-i√p)/4p (-1+i√p)/4p 0 1/2p 2t-2 (p-1)/4p 2t-2 (p-1)/4p 2t-2 (p-1)/4p 2t-3 (p-1)/4p 2t-3 (p-1)/4p 2t-4 (p-1)/4p 2t-4 … (p-1)/4p 2 (p-1)/4p 2 (-1+i√p)/4p (-

1-i√p)/4p 0 0 0 1/2p 2t-2 (p-1)/4p 2t-2 (p-1)/4p 2t-2 (p-1)/4p 2t-3 (p-1)/4p 2t-3 (p-1)/4p 2t-4 (p-1)/4p 2t-4 … (p-1)/4p 2 (p-1)/4p 2 (-1-i√p)/4p (-1+i√p)/4p 0 0 0 1/2p 2t-3 (p-1)/4p 2t-3 (p-1)/4p 2t-3 (p-1)/4p 2t-4 (p-1)/4p 2t-4 (p-1)/4p 2t-5 (p-1)/4p 2t-5 … (-1+i√p)/4p (-1-i√p)/4p 0 0 0 0 0 1/2p 2t-3 (p-1)/4p 2t-3 (p-1)/4p 2t-3 (p-1)/4p 2t-4 (p-1)/4p 2t-4 (p-1)/4p 2t-5 (p-1)/4p 2t-5 … (-1-i√p)/4p (-1+i√p)/4p 0 0 0 0 0 … … … … … … … … … … … … … … 1/2p 3 (p-1)/4p 3 (p-1)/4p 3 (p-1)/4p 2 (p-1)/4p 2 (-1+i√p)/4p (-1-i√p)/4p … 0 0 0 0 0 0 0 1/2p 3 (p-1)/4p 3 (p-1)/4p 3 (p-1)/4p 2 (p-1)/4p 2 (-1-i√p)/4p (-1+i√p)/4p 0 0 0 0 0 0 0 1/2p 2 (p-1)/4p 2 (p-1)/4p 2 (-1+i√p)/4p (-1-i√p)/4p 0 0 … 0 0 0 0 0 0 0 1/2p 2 (p-1)/4p 2 (p-1)/4p 2 (-1-i√p)/4p (-1+i√p)/4p 0 0 … 0 0 0 0 0 0 0 1/2p (-1+i√p)/4p (-1-i√p)/4p 0 0 0 0 … 0 0 0 0 0 0 0 1/2p (-1-i√p)/4p (-1+i√p)/4p 0 0 0 0 … 0 0 0 0 0 0
0 0 0 0 p t+1 p t+1 (p-1) 0 0 0 p t+2 (p-1) 0 0 0 p t+3 (p-1) 0 … 0 0 p 2t-p-1) 2p t-1 p t-1 (p-1) p t-1 (p-1) p t (p-1) p t (p-1) p t+1 (p-3)/2 p t+1 (p-1)/2 0 0 p t+2 (p-1) 0 0 0 p t+3 (p-1) 0 … 0 0 p 2t-p-1) 0 0 0 0 0 p t+1 (p-1)/2 p t+1 (p+1)/2 0 0 p t+2 (p-1) 0 0 0 p t+3 (p-1) 0 … 0 0 p 2t (p-1) 0 0 0 0 0 0 0 p t+2 0 p t+2 (p-1) 0 0 0 p t+3 (p-1) 0 … 0 0 p 2t (p-1) 0 0 0 0 0 0 0 0 p t+2 p t+2 (p-1) 0 0 0 p t+3 (p-1) 0 … 0 0 p 2t (p-1) 2p t-2 p t-2 (p-1) p t-2 (p-1) p t-1 (p-1) p t-1 (p-1) p t (p-1) p t (p-1) p t+1 (p-1) p t+1 (p-1) p t+2 (p-3)/2 p t+2 (p-1)/2 0 0 p t+3 (p-1) 0 … 0 0 p 2t (p-1) 0 0 0 0 0 0 0 0 0 p t+2 (p-1)/2 p t+2 (p+1)/2 0 0 p t+3 (p-1) 0 … 0 0 p 2t (p-1) 0 0 0 0 0 0 0 0 0 0 0 p t+2 0 p t+3 (p-1) 0 … 0 0 p 2t (p-1) 0 0 0 0 0 0 0 0 0 0 0 0 p t+3 p t+3 (p-1) 0 … 0 0 p 2t (p-1)
2p t-3 p t-3 (p-1) p t-3 (p-1) p t-2 (p-1) p t-2 (p-1) p t-1 (p-1) p t-1 (p-1) p t (p-1) p t (p-1) p t+1

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 … 0 0 p 2t (p-1) … … … … … … … … … … … … … … … … … … 2p p(p-
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 p 2t (p-1) 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 … p 2t 0 p 2t (p-1) 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 … 0 p 2t p 2t (p-1) 2 p-1 p-1 p(p-
p-1 p-1 p-1 p-1 p-1 p-1 … p-1 p-1 p-1 p-1 p-1 p-1 p-1 p-1 1 p-1 p-1 p-1 p-1 p-1 p-1 … p-1 p-1 p-1 p-1 p-1 p-1 -1-√p -1+√p 1 p-1 p-1 p-1 p-1 p-1 p-1 … p-1 p-1 p-1 p-1 p-1 p-1 -1+√p -1-√p 1 p-1 p-1 p-1 p-1 p-1 p-1 … p-1 p-1 p-1 p-1 -2 0 0 1 p-1 p-1 p-1 p-1 p-1 p-1 … p-1 p-1 p-1 p-1 0 -2 0 1 p-1 p-1 p-1 p-1 p-1 p-1 … p-1 p-1 -1-√p -1+√p 0 0 0 1 p-1 p-1 p-1 p-1 p-1 p-1 … p-1 p-1 -1+√p -1-√p 0 0 0 1 p-1 p-1 p-1 p-1 p-1 p-1 … -2 0 0 0 0 0 0 1 p-1 p-1 p-1 p-1 p-1 p-1 … 0 -2 0 0 0 0 0 … … … … … … … … … … … … … … 1 p-1 p-1 p-1 p-1 -1-√p -1+√p … 0 0 0 0 0 0 0 1 p-1 p-1 p-1 p-1 -1+√p -1-√p 0 0 0 0 0 0 0 1 p-1 p-1 -2 0 0 0 … 0 0 0 0 0 0 0 1 p-1 p-1 0 -2 0 0 … 0 0 0 0 0 0 0 1 -1-√p -1+√p 0 0 0 0 … 0 0 0 0 0 0 0 1 -1+√p -1-√p 0 0 0 0 … 0 0 0 0 0 0
)/4p 0 1/2p 2t-1 (p-1)/4p 2t-1 (p-1)/4p 2t-1 (p-1)/4p 2t-2 (p-1)/4p 2t-2 (p-1)/4p 2t-3 (p-1)/4p 2t-3 … (p-1)/4p 2 (p-1)/4p 2 (-1-√p)/4p (-1+√p)/4p 0 0 0 1/2p 2t-1 (p-1)/4p 2t-1 (p-1)/4p 2t-1 (p-1)/4p 2t-2 (p-1)/4p 2t-2 (p-1)/4p 2t-3 (p-1)/4p 2t-3 … (p-1)/4p 2 (p-1)/4p 2 (-1+√p)/4p (-1-√p)/4p 0 0 0 1/2p 2t-2 (p-1)/4p 2t-2 (p-1)/4p 2t-2 (p-1)/4p 2t-3 (p-1)/4p 2t-3 (p-1)/4p 2t-4 (p-1)/4p 2t-4 … -(p+1)/4p (p-1)/4p 0 0 0 0 0 1/2p 2t-2 (p-1)/4p 2t-2 (p-1)/4p 2t-2 (p-1)/4p 2t-3 (p-1)/4p 2t-3 (p-1)/4p 2t-4 (p-1)/4p 2t-4 … (p-1)/4p -(p+1)/4p 0 0 0 0 0 … … … … … … … … … … … … … … 1/2p 3 (p-1)/4p 3 (p-1)/4p 3 (p-1)/4p 2 (p-1)/4p 2 (-1-√p)/4p (-1+√p)/4p … 0 0 0 0 0 0 0 1/2p 3 (p-1)/4p 3 (p-1)/4p 3 (p-1)/4p 2 (p-1)/4p 2 (-1+√p)/4p (-1-√p)/4p 0 0 0 0 0 0 0 1/2p 2 (p-1)/4p 2 (p-1)/4p 2 -(p+1)/4p (p-1)/4p 0 0 … 0 0 0 0 0 0 0 1/2p 2 (p-1)/4p 2 (p-1)/4p 2 (p-1)/4p -(p+1)/4p 0 0 … 0 0 0 0 0 0 0 1/2p (-1-√p)/4p (-1+√p)/4p 0 0 0 0 … 0 0 0 0 0 0 0 1/2p (-1+√p)/4p (-1-√p)/4p 0 0 0 0 … 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 p t+2 (p+1)/2 p t+2 (p-1)/2 0 0 0 p t+3 (p-1) … 0 0 p 2t (p-1) 0 0 0 0 0 0 0 0 0 0 0 p t+2 0 0 p t+3 (p-1) … 0 0 p 2t (p-1) 0 0 0 0 0 0 0 0 0 0 0 0 p t+3 0 p t+3 (p-1) … 0 0 p 2t (p-1)
2p t-3 p t-3 (p-1) p t-3 (p-1) p t-2 (p-1) p t-2 (p-1) p t-1 (p-1) p t-1 (p-1) p t (p-1) p t (p-1) p t+1

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 … 0 0 p 2t (p-1) … … … … … … … … … … … … … … … … … … 2p p(p-
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 p 2t (p-1) 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 … p 2t 0 p 2t (p-1) 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 … 0 p 2t 0 p 2t (p-1) 2 p-1 p-1 p(p-
p-1 p-1 p-1 p-1 p-1 p-1 … p-1 p-1 p-1 p-1 p-1 p-1 p-1 p-1 1 p-1 p-1 p-1 p-1 p-1 p-1 … p-1 p-1 p-1 p-1 p-1 p-1 (-1-i√p) (-1+i√p) 1 p-1 p-1 p-1 p-1 p-1 p-1 … p-1 p-1 p-1 p-1 p-1 p-1 (-1+i√p) (-1-i√p) 1 p-1 p-1 p-1 p-1 p-1 p-1 … p-1 p-1 p-1 p-1 -2 0 0 1 p-1 p-1 p-1 p-1 p-1 p-1 … p-1 p-1 p-1 p-1 0 -2 0 1 p-1 p-1 p-1 p-1 p-1 p-1 … p-1 p-1 (-1-i√p) (-1+i√p) 0 0 0 1 p-1 p-1 p-1 p-1 p-1 p-1 … p-1 p-1 (-1+i√p) (-1-i√p) 0 0 0 1 p-1 p-1 p-1 p-1 p-1 p-1 … -2 0 0 0 0 0 0 1 p-1 p-1 p-1 p-1 p-1 p-1 … 0 -2 0 0 0 0 0 … … … … … … … … … … … … … … 1 p-1 p-1 p-1 p-1 (-1-i√p) (-1+i√p) … 0 0 0 0 0 0 0 1 p-1 p-1 p-1 p-1 (-1+i√p) (-1-i√p) 0 0 0 0 0 0 0 1 p-1 p-1 -2 0 0 0 … 0 0 0 0 0 0 0 1 p-1 p-1 0 -2 0 0 … 0 0 0 0 0 0 0 1 (-1-i√p) (-1+i√p) 0 0 0 0 … 0 0 0 0 0 0 0 1 (-1+i√p) (-1-i√p) 0 0 0 0 … 0 0 0 0 0 0
)/4p 0 1/2p 2t-1 (p-1)/4p 2t-1 (p-1)/4p 2t-1 (p-1)/4p 2t-2 (p-1)/4p 2t-2 (p-1)/4p 2t-3 (p-1)/4p 2t-3 … (p-1)/4p 2 (p-1)/4p 2 (-1+i√p)/4p (-1-i√p)/4p 0 0 0 1/2p 2t-1 (p-1)/4p 2t-1 (p-1)/4p 2t-1 (p-1)/4p 2t-2 (p-1)/4p 2t-2 (p-1)/4p 2t-3 (p-1)/4p 2t-3 … (p-1)/4p 2 (p-1)/4p 2 (-1-i√p)/4p (-1+i√p)/4p 0 0 0 1/2p 2t-2 (p-1)/4p 2t-2 (p-1)/4p 2t-2 (p-1)/4p 2t-3 (p-1)/4p 2t-3 (p-1)/4p 2t-4 (p-1)/4p 2t-4 … -(p+1)/4p (p-1)/4p 0 0 0 0 0 1/2p 2t-2 (p-1)/4p 2t-2 (p-1)/4p 2t-2 (p-1)/4p 2t-3 (p-1)/4p 2t-3 (p-1)/4p 2t-4 (p-1)/4p 2t-4 … (p-1)/4p -(p+1)/4p 0 0 0 0 0 … … … … … … … … … … … … … … 1/2p 3 (p-1)/4p 3 (p-1)/4p 3 (p-1)/4p 2 (p-1)/4p 2 (-1+i√p)/4p (-1-i√p)/4p … 0 0 0 0 0 0 0 1/2p 3 (p-1)/4p 3 (p-1)/4p 3 (p-1)/4p 2 (p-1)/4p 2 (-1-i√p)/4p (-1+i√p)/4p 0 0 0 0 0 0 0 1/2p 2 (p-1)/4p 2 (p-1)/4p 2 -(p+1)/4p (p-1)/4p 0 0 … 0 0 0 0 0 0 0 1/2p 2 (p-1)/4p 2 (p-1)/4p 2 (p-1)/4p -(p+1)/4p 0 0 … 0 0 0 0 0 0 0 1/2p (-1+i√p)/4p (-1-i√p)/4p 0 0 0 0 … 0 0 0 0 0 0 0 1/2p (-1-i√p)/4p (-1+i√p)/4p 0 0 0 0 … 0 0 0 0 0 0 0

Important remark :

According to the eigenvalues summation properties (properties 16 of exercise 5), we have in accordance with the preceding definitions of However, when more general than monomial z n diophantine expressions are used and so when the eigenvalues structure is more complex, one will return to the [λ'] and [λ'*] decomposition.

Case p \ n

If p divides n, the equation to start with will be identical, that is card' x1,x2 (c=or(0,p x1 .g x2 )) = ∑ a x1,x2,y1,y2 .card y1,y2 (relation ).

However, the matrix, translation of this writing, becomes more complicated. We attend to a "dispersal" of the components which were aligned by columns above trace blocks and by lines under trace blocks in former case p ∤ n. We give some samples of matrices next. The arrows materialize dispersal of virtual initial positions (when p ∤ n).

Example 1 : p= 3, n = 3, δ = 3 :g= 2, p δ = 27, g , 32, 24, 18, 76, 57, 74, 118, 26, 82, 124, 93, 101, 107, 49, 68, 51, 7, 99, 43 5 20 25 50 2, 64, 48, 36, 27, 114, 23, 111, 52, 39, 123, 61, 77, 89, 98, 11, 102, 14, 73, 86 5 5 5 5 5 25 25 25 4, 3, 96, 72, 54, 103, 46, 97, 104, 78, 121, 122, 29, 53, 71, 22, 79, 28, 21, 47 5 5 5 5 5 75 8, 6, 67, 19, 108, 81, 92, 69, 83, 31, 117, 119, 58, 106, 17, 44, 33, 56, 42, 94 5 5 5 5 5 50 25 25 16, 12, 9, 38, 91, 37, 59, 13, 41, 62, 109, 113, 116, 87, 34, 88, 66, 112, 84 This case seems most complex. A priori, it seems impossible to make full turn of possible configurations of "contracted" matrices (infinity of case). We will admit that the expressions obtained thereafter when p\n are, a priori, quite the same as those obtained when p ∤ n as case p \ n is wrapped between two cases p ∤ n. This intuitive argument is not worth proof of course. If we want to be more rigorous, we will return to general case : card'(i) = ∑ #(i-j mod n).card(j) j = 0 to n-1 which can be written as a right circulant matrix whose eigenvalues and eigenvectors are given at exercise (4).

Case p even (p=2) 1.2.1. Generalities

Concerning only one value of p, it would be pleasant to spend little time to this case relying simply on previous results of exercise 4 concerning circulant matrices. However, we propose an alternative study in accordance with latter remarks. Let us note first, that there are no primitive roots modulo 2 δ (cf. [1]). To generate odd numbers family between 0 and 2 δ -1, it is necessary to resort to two generators which are 5 and -5.

What was written in the odd case : p i .g r = p k .g u.d + p i .g s .g v.d mod p δ becomes ().2 i .5 r = ().2 k .5 u.d + ().2 i .5 s .5 v.d mod 2 δ (15) This alternative of the signs does not threaten the principles of calculation used in the case of odd sequences. Cardinal matrices equation (15) decomposes no more in one but in eight cases. By addition, the identity rule of the cardinals #{(x 1 , x 2 , …, x k ) / x 1 n +x 2 n +…+x k n = c mod 2  } (for variables of integers, respectively #{(y 1 , y 2 , …, y k ) / y 1 n +y 2 n +…+y k n = c mod 2  } for variables of prime numbers) of sets of targets is trivially retained. The structure of these sets is however more complex than in odd sequences cases where we only distinguish sets like g k .p r.n (the other targets, except 0, having null cardinals). This structure was established at exercise 3 : 1) .(#{1}) other targets 0

Variable of integers c #{c} 0 or 2 δn.n 2 δ-δn-1 2 r..n (1+2.(#{1}).k) 2 r.(n-
Variable of prime numbers (deduced from preceding table

) c #{c} 1+2.(#{1}).k #{1} other targets 0
The difference is in term #{1}, which in odd sequences cases is restricted to 1, but here (p = 2) is so only when n is odd. This single point has large consequences. Without anticipating on the remainder of the article, it is as convenient not to contract cardinal matrices (in opposite way to odd sequences). This means to go back and replacing p = 2 in the equations developed at exercise 4 causing the dissatisfaction due to eigenvalues in sum type forms (thus difficult manipulations).

Hence, we will develop the alternative building of "cardinal" matrices without contractions. The name of cardinal matrices is abusive. The reader must retain that, if this alternatives do interest us for Waring sums cases, the corresponding change of base matrices (that is. environment matrices while anticipating latter work) are no more unique for different monomials (and thus not compatible without precaution with the concept of environment). When variables of different exponents are associated (starting with x 2 +x 3 ), the reader will refer to the tools that we propose at exercise 11, tools based on (right) circulant matrices for which the change of base matrix is unique for a given dimension.

The method, which interest us underneath, is based on the following table (where Φ(i)=p i-1 .(p-1)=2 i-1 ) :

P 139/390 Modulo p Δ abundance factors correction Current cardinals #c'1 #c'2 #c'3 … #c'2^ Target at stage k c'1 c'2 c'3 … c'2^ Cardinals stage 1 Targets (generators) stage 1 #{u / u n = 0 mod 2  } c1 = 0 j → #{u / u n = 5 0 .2 -1 mod 2  } c2 = 5 0 .2 -1 #{u / u n = 5 0 .2 -2 mod 2  } c3 = 5 0 .2 -2 #{u / u n = -5 0 .2 -2 mod 2  } c4 = -5 0 .2 -2 #{u / u n = 5 1 .2 -3 mod 2  } c5 = 5 1 .2 -3 i #{u / u n = 5 0 .2 -3 mod 2  } c6 = 5 0 .2 -3 #{u / u n = -5 1 .2 -3 mod p  } c7 = -5 1 .2 -3 #{u / u n = -5 0 .2 -3 mod p  } c8 = -5 0 .2 -3 ↓ … #{u / u n = 5 (Φ(j)/2-1) .2 δ-j mod 2  } 5 (Φ(j)/2-1) .2 δ-j c(i,j) … #{u / u n = 5 0 .2 δ-j mod 2  } 5 0 .2 δ-j #{u / u n = -5 (Φ(j)/2-1) .2 δ-j mod 2  } -5 (Φ(j)/2-1) .2 δ-j … #{u / u n = -5 0 .2 δ-j mod 2  } -5 0 .2 δ-j … #{u / u n = 5 (Φ(δ)/2-1) .2 0 mod 2  } 5 (Φ(δ)/2-1) .2 0 … #{u / u n = 5 0 .2 0 mod 2  } 5 0 .2 0 #{u / u n = -5 (Φ(δ)/2-1) .2 0 mod 2  } -5 (Φ(δ)/2-1) .2 0 … #{u / u n = -5 0 .2 0 mod 2  } c2  = -5 0 .2 0

Writing convention 1

In above table, we define the order in which are given line and column generating elements of targets c(i,j). This has an identical repercussion on generating vectors of the cardinal matrix.

Let us note (i,j) the intersection of line i and column j of characteristic matrices mod 2 δ . Let us have c i the generator at position i and c' j the generator at position j.

By construction, we have c' i = c i .

The set of numbers 0 to 2 δ -1 figure in generators within j axis (and within i axis also). It follows the immediate consequence :

Property 1 « of components permutations »

Each column is a permutation of elements of the first column, the permutations not depending on n.

The condition allowing characteristic matrix construction is equality of targets c(i,j) and c(i',j'), that is, by subtraction, for two positions (i,j) and (i',j') in matrix :

if c' j' -c i' = c' j -c i mod 2 δ => #(i',j') = #(i,j) (16)
Let us consider the first column (c' 1 =0) :

#(i,1) = #{u / u n = c i = c i -c' 1 mod 2 δ } and c' j' -c i' = c' 1 -c i mod 2 δ => #(i',j') = #(i,1) that is c' j' -c i' = -c i mod 2 δ => #(i',j') = #{u / u n = c i mod 2 δ } hence #(i',j') = #{u / u n = c i' -c' j' mod 2 δ } and then (with c' i = c i ) #(i,j) = #{u / u n = c i -c j mod 2 δ } (17)
We use this result to index matrix components.

Property 2 « periodicity of cardinals for 2 δ distant generators »

Modulo 2 δ distant generators (subtraction of value of generators c i according to lines on one hand and of value of generators c j according to columns on the other hand equal 0 modulo 2 δ ) correspond to equal cardinals, the value of the literally distinct cardinals (possibly not distinct numerically) being given by the first column of cardinal matrix (equivalent notation c[i] and c i :

u c[i] = #(i,1) = #{u / u n = c i mod 2 δ } (18)
Writing convention 2 : Reference sequence

The order, in which first column indices do appear, is called the reference sequence. We get :

u c = #{u / u n = c mod 2 δ } (19)
This notation is particularly practical. We use this sequence to index cardinal matrices eigenvectors. For simplicity, indices are handled modulo 2 δ without being indicated like such in the text. Thus, the reader will not be astonished by negative values in intermediate calculations concerning aforementioned indices.

Let us give some examples to illustrate at the same time the rule (concerning reference sequence) and properties 1 and 2 :

=1 =2 =3 0 1 0 2 1 -1 0 4 2 -2 5 1
-5 -1 0 u0 u1 0 u0 u2 u3 u1 0 u0 u4 u6 u2 u3 u7 u5 u1 1 u1 u0 2 u2 u0 u1 u3 4 u4 u0 u2 u6 u7 u3 u1 u5 1 u1 u3 u0 u2 2 u2 u6 u0 u4 u5 u1 u7 u3 -1 u3 u1 u2 u0 -2 u6 u2 u4 u0 u1 u5 u3 u7 5 u5 u1 u3 u7 u0 u4 u2 u6 1 u1 u5 u7 u3 u4 u0 u6 u2 -5 u3 u7 u1 u5 u6 u2 u0 u4 -1 u7 u3 u5 u1 u2 u6 u4 u0 u4 u12 u0 u8 u10 u2 u14 u6 u7 u11 u15 u3 u1 u13 u9 u5 -4 u12 u4 u8 u0 u2 u10 u6 u14 u15 u3 u7 u11 u9 u5 u1 u13 10 u10 u2 u6 u14 u0 u8 u4 u12 u13 u1 u5 u9 u7 u3 u15 u11 2 u2 u10 u14 u6 u8 u0 u12 u4 u5 u9 u13 u1 u15 u11 u7 u3 -10 u6 u14 u2 u10 u12 u4 u0 u8 u9 u13 u1 u5 u3 u15 u11 u7 -2 u14 u6 u10 u2 u4 u12 u8 u0 u1 u5 u9 u13 u11 u7 u3 u15 13 u13 u5 u9 u1 u3 u11 u7 u15 u0 u4 u8 u12 u10 u6 u2 u14 9 u9 u1 u5 u13 u15 u7 u3 u11 u12 u0 u4 u8 u6 u2 u14 u10 5 u5 u13 u1 u9 u11 u3 u15 u7 u8 u12 u0 u4 u2 u14 u10 u6 1 u1 u9 u13 u5 u7 u15 u11 u3 u4 u8 u12 u0 u14 u10 u6 u2 -13 u3 u11 u15 u7 u9 u1 u13 u5 u6 u10 u14 u2 u0 u12 u8 u4 -9 u7 u15 u3 u11 u13 u5 u1 u9 u10 u14 u2 u6 u4 u0 u12 u8 -5 u11 u3 u7 u15 u1 u9 u5 u13 u14 u2 u6 u10 u8 u4 u0 u12 -1 u15 u7 u11 u3 u5 u13 u9 u1 u2 u6 u10 u14 u12 u8 u4 u0

Hence results, after resolution of first column : n=2 The empty boxes are occupied by 0 (masked for better reading).

=1 =2 =3 0 1 0 2 1 -1 0 4 2 -2 5 1 -5 -1 0 1 1 0 2 2 0 2 2 4 1 1 1 2 2 2 4 2 2 4 1 2 2 2 2 2 4 -1 2 2 -2 2
The case n = 2 shows some complexity even with small values of .

n = 3 =1 =2 =3 0 1 0 2 1 -1 0 4 2 -2 5 -5 -1 0 1 1 0 2 1 1 0 4 1 1 1 1 1 1 2 2 1 1 4 4 1 1 1 1 1 1 2 2 4 1 1 1 -1 1 1 2 -2 4 1 1 1 5 1 1 1 1 4 1 1 1 1 1 -5 1 1 1 1 4 -1 1 1 1 1 4 =4 0 8 4 -4 10 2 -10 -2 13 9 5 1 -13 -9 -5 -1 0 4 4 1 1 1 1 1 1 1 8 4 4 1 1 1 1 1 1 1 4 4 4 1 1 1 1 1 1 1 -4 4 4 1 1 1 1 1 1 1 10 4 4 1 1 1 1 1 1 1 2 4 4 1 1 1 1 1 1 1 -10 4 4
For n = 3, the basic diagram, very simple until  = 3, worsens to become of a great complexity (see below).

For n odd, #(i,1) = #{u / u n = c i mod 2 δ } = 1 if c i odd (according to exercise 3), so that we get blocks of size 2 δ-1 with components 1 on both sides of trace, characteristic blocks of this case.

For n even, the diagram of distribution is more heterogeneous.

The reader will note that we deal here with magic squares (except for diagonals however) with a peculiar method of generation of these squares.

Preservation of "symmetries" by matrices exponentiation

We can index matrix components u(i,j) as c i -c j , this indices being implicitly handled modulo δ . Then, let us consider the result of the product of two matrices at components coordinates (i,j) and (i',j') :

#(i,j) = Σ u ci-ck .u ck-cj #(i',j') = Σ u ci'-ck' .u ck'-cj' = Σ u ci+ci'-ci-ck' .u ck'-cj'+cj-cj = Σ u ci+ci'-ci-(-cj'+cj)-ck .u ck-cj = Σ u ci+(ci'-cj')-(ci-cj)-ck .u ck-cj
That is also : if c j' -c j' = c i -c j mod 2 δ then #(i',j') = Σ u ci-ck .u ck-cj = #(i,j) Hence the property :

Cardinal matrix components literal equalities C δ are transmitted to C δ n .

Foot-note: This property also applies to C δ -1 and C δ -n . (We will not demonstrate these points for they are of little interest in the next course).

Property 3 « eigenvalues types »

Matrices A δ,n eigenvalues are real numbers when n is odd. Matrices A δ,n eigenvalues are imaginary numbers when n is even, except for the two eigenvalues λ = 0 and λ = 2 δ .

n odd   λ[A δ, n ], λ[A δ, n ]   n even   λ[A δ, n ], [A δ, n ]   except λ = 0 and λ = 2 δ
We will demonstrate later this result by explicit evaluation of matrices [A δ, n ] eigenvalues.

Property 4 : External dichotomy rule

Let us have λ an eigenvalue of cardinal matrix A δ,n . Then, with the same multiplicity, 2λ is eigenvalue of the cardinal matrix A δ+1,n .



λ[A δ+1,n ] \ λ[A δ+1,n ] = 2.λ[A δ,n ]
We easily conceive that this property is of interest to get eigenvalues. Half of matrix A δ,n eigenvalues are immediately deduced from matrix A δ-1,n eigenvalues by elementary by 2 multiplication. The second half however remains to be determined. Using this property, which we later prove by explicit evaluation of the eigenvalues, we get the following table :

#(λ) 2 0 2 0 2 1 2 2 2 3 2 4 … δ = 0 2 0 .λ 0 δ = 1 2 1 .λ 0 2 0 .λ 1 δ = 2 2 2 .λ 0 2 1 .λ 1 2 0 .{λ 2 } δ = 3 2 3 .λ 0 2 2 .λ 1 2 1 .{λ 2 } 2 0 .{λ 3 } δ = 4 2 4 .λ 0 2 3 .λ 1 2 2 .{λ 2 } 2 1 .{λ 3 } 2 0 .{λ 4 } δ = 5 2 5 .λ 0 2 4 .λ 1 2 3 .{λ 2 } 2 2 .{λ 3 } 2 1
.{λ 4 } 2 0 .{λ 5 } … (λ 0 = 1, λ 1 = 0, {λ i } are eigenvalues families depending on the studied cases, n≠2)

Property 5 : Interior dichotomy rule

Let us have δn = int((δ-1)/n). Let us have q integer, δ-δn+1 < q < δ-1. Let us have λ q = 2 q .ε cardinal matrix A δ,n eigenvalue of multiplicity m q . Then λ q ' = 2 q-1 .ε is eigenvalue of cardinal matrix A δ,n of multiplicity 2 n .m q . By iteration, we get : {{ λ q = 2 q .ε, #(λ q ) = m q }  { λ q ' = 2 q-1 .ε, #(λ q ') = 2 n .m q }}, q = δ-2 to δ-δn Let us give, in addition (by direct research for example), the values taken by ε for a sample of A δ,n matrices :

n = 2 n = 3 n = 4 n = 5 n = 6 ε 21 = 1+i ε 22 = 1-i ε 21 = 1 ε 22 = 1 ε 21 = 1+i ε 22 = 1-i ε 21 = 1 ε 22 = 1 ε 21 = 1+i ε 22 = 1-i ε 41 = (√2)/2.(1+i) ε 42 = (√2)/2.(1-i) ε 43 = (√2)/2.(-1+i) ε 44 = (√2)/2.(-1-i) ε 41 = 1 to ε 44 = 1 ε 41 = 1+(√2)/2.(1+i) ε 42 = 1+(√2)/2.(1-i) ε 43 = 1+(√2)/2.(-1+i) ε 44 = 1+(√2)/2.(-1-i) ε 41 = 1 to ε 44 = 1 ε 41 = 1+(√2)/2.(1+i) ε 42 = 1+(√2)/2.(1-i) ε 43 = 1+(√2)/2.(-1+i) ε 44 = 1+(√2)/2.(-1-i) ε 81 = 0 to ε 88 = 0 ε 81 = 1+(√(2+√2))/2+(√(2-√2))/2.i ε 82 = 1+(√(2+√2))/2-(√(2-√2))/2.i ε 83 = 1+(√(2-√2))/2+(√(2+√2))/2.i ε 84 = 1+(√(2-√2))/2-(√(2+√2))/2.i ε 85 = 1-(√(2+√2))/2+(√(2-√2))/2.i ε 86 = 1-(√(2+√2))/2-(√(2-√2))/2.i ε 87 = 1-(√(2-√2))/2+(√(2+√2))/2.i ε 88 = 1-(√(2-√2))/2-(√(2+√2))/2.i ε 81 = 1 to ε 88 = 1 ε 81 = 1 to ε 88 = 1 ε 16 1 = 0 to ε 16 16 = 0 ε 16 1 = 1 to ε 16 16 = 1 ε 16 1 = 1 to ε 16 16 = 1 ε 32 1 = 0 to ε 32 32 = 0 ε 32 1 = 1 to ε 32 32 = 1 ε 64 1 = 0 to ε 64 64 = 0
The sum of each line (and column) components of a cardinal matrix are doubled while passing from a matrix of dimension 2 δ to dimension 2 δ+1 . However, this condition is not sufficient to explain interior and external dichotomies properties as it is easy to build counter-examples.

There again, we will demonstrate the property by explicit evaluation of matrices [A δ,n ] eigenvalues.

Eigenvalues and eigenvectors evaluation 1.2.8.1. Generalities

Cardinal matrices eigenvalues and eigenvectors evaluation is done simultaneously. For given δ and n, there is an infinity of matrices of eigenvectors. It suffice however to give explicitly one of the matrices to solve the problem. We start with an example. We start from matrix of dimension 2 5 general form. 0123456781 -13 -9 -21 -17 -29 -25 -5 -1 0 0 u0 u16 u24 u8 u12 u28 u20 u4 u6 u14 u22 u30 u26 u18 u10 u2 u19 u23 u11 u15 u3 u7 u27 u31 u13 u9 u21 u17 u29 u25 u5 u1 x0 16 u16 u0 u8 u24 u28 u12 u4 u20 u22 u30 u6 u14 u10 u2 u26 u18 u3 u7 u27 u31 u19 u23 u11 u15 u29 u25 u5 u1 u13 u9 u21 u17 x16 8 u8 u24 u0 u16 u20 u4 u28 u12 u14 u22 u30 u6 u2 u26 u18 u10 u27 u31 u19 u23 u11 u15 u3 u7 u21 u17 u29 u25 u5 u1 u13 u9 x8 -8 u24 u8 u16 u0 u4 u20 u12 u28 u30 u6 u14 u22 u18 u10 u2 u26 u11 u15 u3 u7 u27 u31 u19 u23 u5 u1 u13 u9 u21 u17 u29 u25 x24 20 u20 u4 u12 u28 u0 u16 u8 u24 u26 u2 u10 u18 u14 u6 u30 u22 u7 u11 u31 u3 u23 u27 u15 u19 u1 u29 u9 u5 u17 u13 u25 u21 x20 4 u4 u20 u28 u12 u16 u0 u24 u8 u10 u18 u26 u2 u30 u22 u14 u6 u23 u27 u15 u19 u7 u11 u31 u3 u17 u13 u25 u21 u1 u29 u9 u5 x4 -20 u12 u28 u4 u20 u24 u8 u0 u16 u18 u26 u2 u10 u6 u30 u22 u14 u31 u3 u23 u27 u15 u19 u7 u11 u25 u21 u1 u29 u9 u5 u17 u13 x12 -4 u28 u12 u20 u4 u8 u24 u16 u0 u2 u10 u18 u26 u22 u14 u6 u30 u15 u19 u7 u11 u31 u3 u23 u27 u9 u5 u17 u13 u25 u21 u1 u29 x28 26 u26 u10 u18 u2 u6 u22 u14 u30 u0 u8 u16 u24 u20 u12 u4 u28 u13 u17 u5 u9 u29 u1 u21 u25 u7 u3 u15 u11 u23 u19 u31 u27 x26 18 u18 u2 u10 u26 u30 u14 u6 u22 u24 u0 u8 u16 u12 u4 u28 u20 u5 u9 u29 u1 u21 u25 u13 u17 u31 u27 u7 u3 u15 u11 u23 u19 x18 10 u10 u26 u2 u18 u22 u6 u30 u14 u16 u24 u0 u8 u4 u28 u20 u12 u29 u1 u21 u25 u13 u17 u5 u9 u23 u19 u31 u27 u7 u3 u15 u11 x10 2 u2 u18 u26 u10 u14 u30 u22 u6 u8 u16 u24 u0 u28 u20 u12 u4 u21 u25 u13 u17 u5 u9 u29 u1 u15 u11 u23 u19 u31 u27 u7 u3 x2 -26 u6 u22 u30 u14 u18 u2 u26 u10 u12 u20 u28 u4 u0 u24 u16 u8 u25 u29 u17 u21 u9 u13 u1 u5 u19 u15 u27 u23 u3 u31 u11 u7 x6 -18 u14 u30 u6 u22 u26 u10 u2 u18 u20 u28 u4 u12 u8 u0 u24 u16 u1 u5 u25 u29 u17 u21 u9 u13 u27 u23 u3 u31 u11 u7 u19 u15 x14 -10 u22 u6 u14 u30 u2 u18 u10 u26 u28 u4 u12 u20 u16 u8 u0 u24 u9 u13 u1 u5 u25 u29 u17 u21 u3 u31 u11 u7 u19 u15 u27 u23 x22 -2 u30 u14 u22 u6 u10 u26 u18 u2 u4 u12 u20 u28 u24 u16 u8 u0 u17 u21 u9 u13 u1 u5 u25 u29 u11 u7 u19 u15 u27 u23 u3 u31 x30 13 u13 u29 u5 u21 u25 u9 u1 u17 u19 u27 u3 u11 u7 u31 u23 u15 u0 u4 u24 u28 u16 u20 u8 u12 u26 u22 u2 u30 u10 u6 u18 u14 x13 9 u9 u25 u1 u17 u21 u5 u29 u13 u15 u23 u31 u7 u3 u27 u19 u11 u28 u0 u20 u24 u12 u16 u4 u8 u22 u18 u30 u26 u6 u2 u14 u10 x9 21 u21 u5 u13 u29 u1 u17 u9 u25 u27 u3 u11 u19 u15 u7 u31 u23 u8 u12 u0 u4 u24 u28 u16 u20 u2 u30 u10 u6 u18 u14 u26 u22 x21 17 u17 u1 u9 u25 u29 u13 u5 u21 u23 u31 u7 u15 u11 u3 u27 u19 u4 u8 u28 u0 u20 u24 u12 u16 u30 u26 u6 u2 u14 u10 u22 u18 x17 29 u29 u13 u21 u5 u9 u25 u17 u1 u3 u11 u19 u27 u23 u15 u7 u31 u16 u20 u8 u12 u0 u4 u24 u28 u10 u6 u18 u14 u26 u22 u2 u30 x29 25 u25 u9 u17 u1 u5 u21 u13 u29 u31 u7 u15 u23 u19 u11 u3 u27 u12 u16 u4 u8 u28 u0 u20 u24 u6 u2 u14 u10 u22 u18 u30 u26 x25 5 u5 u21 u29 u13 u17 u1 u25 u9 u11 u19 u27 u3 u31 u23 u15 u7 u24 u28 u16 u20 u8 u12 u0 u4 u18 u14 u26 u22 u2 u30 u10 u6 x5 1 u1 u17 u25 u9 u13 u29 u21 u5 u7 u15 u23 u31 u27 u19 u11 u3 u20 u24 u12 u16 u4 u8 u28 u0 u14 u10 u22 u18 u30 u26 u6 u2 x1 -13 u19 u3 u11 u27 u31 u15 u7 u23 u25 u1 u9 u17 u13 u5 u29 u21 u6 u10 u30 u2 u22 u26 u14 u18 u0 u28 u8 u4 u16 u12 u24 u20 x19 -9 u23 u7 u15 u31 u3 u19 u11 u27 u29 u5 u13 u21 u17 u9 u1 u25 u10 u14 u2 u6 u26 u30 u18 u22 u4 u0 u12 u8 u20 u16 u28 u24 x23 -21 u11 u27 u3 u19 u23 u7 u31 u15 u17 u25 u1 u9 u5 u29 u21 u13 u30 u2 u22 u26 u14 u18 u6 u10 u24 u20 u0 u28 u8 u4 u16 u12 x11 -17 u15 u31 u7 u23 u27 u11 u3 u19 u21 u29 u5 u13 u9 u1 u25 u17 u2 u6 u26 u30 u18 u22 u10 u14 u28 u24 u4 u0 u12 u8 u20 u16 x15 -29 u3 u19 u27 u11 u15 u31 u23 u7 u9 u17 u25 u1 u29 u21 u13 u5 u22 u26 u14 u18 u6 u10 u30 u2 u16 u12 u24 u20 u0 u28 u8 u4 x3 -25 u7 u23 u31 u15 u19 u3 u27 u11 u13 u21 u29 u5 u1 u25 u17 u9 u26 u30 u18 u22 u10 u14 u2 u6 u20 u16 u28 u24 u4 u0 u12 u8 x7 -5 u27 u11 u19 u3 u7 u23 u15 u31 u1 u9 u17 u25 u21 u13 u5 u29 u14 u18 u6 u10 u30 u2 u22 u26 u8 u4 u16 u12 u24 u20 u0 u28 x27 -1 u31 u15 u23 u7 u11 u27 u19 u3 u5 u13 u21 u29 u25 u17 u9 u1 u18 u22 u10 u14 u2 u6 u26 u30 u12 u8 u20 u16 u28 u24 u4 u0 x31

For this matrix, as for the lower or higher dimensions, indices difference between two lines is constant mod 2 δ for "in column facing" (1 st and 2 nd lines with difference 16, 1 st and 3 rd lines with difference 8…). The same remark is true for a reasoning lines/columns instead of columns/lines. Let us have (x 0 , x 16 , x 8 , x 24 , x 20 , x 4 , x 12 , x 28 , x 26 , x 18 , x 10 , x 2 , x 6 , x 14 , x 22 , x 30 , x 13 , x 9 , x 21 , x 17 , x 29 , x 25 , x 5 , x 1 , x 19 , x 23 , x 11 , x 15 , x 3 , x 7 , x 27 , x 31 ) a matrix eigenvector, indices of x being those of above matrix first column.

Basic equation

We can then write (we use ^ for exponents) :

2  -1 λ.x i = ∑ u i+j .x 2^-j i = 0 to 2  -1 j = 0
The values of u i are deduced from The enumerations of u c = #{x n = c mod 2  } = #{c} given at exercise 3 and pointed out below :

c #{c} Definition domain 0 or 2 δn.n 2 δ-δn-1 δn = int((δ-1)/n) 2 r..n (1+2.(#{1}).k) 2 r.(n-1) .(#{1}) k = 0, 1, …, 2 δ-1-r..n /(#{1})-1 r = 0 to δn-1 #{1} = #{x / x n = 1 mod 2 δ , x = 0,1, …, 2 δ -1} = min(2 m+1 , 2 δ-1 )
where m is either multiplicity of factor 2 in n even, or m = -1 if n is odd From : P 144/390 Modulo p Δ abundance factors correction 2  -1 λ.x i = ∑ u i+j .x 2^-j and i+j = c j = 0 we deduce :

2  -1 2  -1 λ.x i = ∑ u c .x i+2^-c = ∑ u c .x i-c c = 0 c = 0 then : δn-1 2 δ-1-r.n /(#{1})-1 λ.x i = (∑ ∑ 2 r.(n-1) .#{1}.x i-((2^(r..n)).(1+2. #{1}.k)) ) + 2 δ-δn-1 .(x i-2^(n.n) +x i ) (20) r = 0 k = 0
Numerical example Before continuing, let us examine with attention a numerical example. Matrix modulo 2 [START_REF] Dieudonné | Abrégé d'histoire des mathématiques[END_REF] for n = 2 is presented below : We deduce immediately : Eigenvectors matrix and eigenvalues are obtained by the simple routine that we develop here and applicable in the general case, case which we will later reconsider. The heart of the problem interesting number 2, it is natural to proceed by a dichotomic approach, here and further. This routine rests on eigenvalues components indices 2 k differences. Hypothesis validation is done, a posteriori, by simply observing that no contradictions (or conditions losing generality) is induced.

Stage 1 (1 th vector) Let us suppose :

x i+1 = x i , i = 0 to 30 It results one and only distinct equation : 32x 0 = λ.x 0 Hence, we deduce : λ = 32

Choice of x 0 is open, but we do not leave it so, for reasons which will be explained in the general case :

x 0 = 2 -/2 = 1/(4√2) Stage 2 (2 nd vector) Let us suppose :

x i+1 = -x i , i = 0 to 30 Hence, we deduce equation : λ.x 0 = 0 Thus : λ = 0 Choice of x 0 is not left to chance :

x 0 = 2 -/2 = 1/(4√2) Stage 3 (3 rd and 4 th vectors) We suppose first :

x i+2 = -x i , i = 0 to 29 Hence, we deduce 2 equations : 16x 0 -16x 1 = λ.x 0 16x 0 +16x 1 = λ.x 1 then :

(16-λ)/16 = -α = x 1 /x 0 = -x 0 /x 1 hence:

α 2 = -1 thus : α = ±i λ = 16.(1+α) = 16.(1±i) x 1 = -α.x 0
Choice of x 0 is not left to chance : x 0 = 2 -/2 = 2 -5/2 = 1/(4√2) Stage 4 (5 th to 8 th vectors) We suppose first :

x i+4 = -x i , i = 0 to 27 Hence, we deduce 2 2 equations :

-16x 3 = λ.x 0 16x 1 = λ.x 2 16x 0 = λ.x 1 16x 2 = λ.x 3 then :

λ/16 = α = -x 3 /x 0 = x 1 /x 2 = x 0 /x 1 = x 2 /x 3 hence:

α 4 = -1 α 2 = ±i thus : α = √2/2.(±1±i) λ = 16α = 8√2.(±1±i) x 3 = -α.x 0 x 2 = α.x 3 = -α 2 .x 0 x 1 = α.x 2 = -α 3 .x 0
Choice of x 0 is not left to chance :

x 0 = 2 -/2 = -5/2 = 1/(4√2) Stage 5 (9 th to 16 th vectors) We suppose first :

x i+8 = -x i , i = 0 to 15 Hence, we deduce 2 3 equations : 8x 0 -8x 4 = λ.x 0 8x 2 -8x 6 = λ.x 2 8x 5 +8x 1 = λ.x 5 8x 3 -8x 7 = λ.x 3 8x 0 +8x 4 = λ.x 4 8x 2 +8x 6 = λ.x 6 -8x 5 +8x 1 = λ.x 1 8x 3 +8x 7 = λ. Choice of x 0 , x 1 ,x 2 and x 3 is open with the proviso of forming a set of free eigenvectors.

Our choice is based on unitary matrix with a coefficient of adjustment (√4 follows from 4 possible choices for x 0 ) : λ/8 = α = -x 12 /x 0 = x 0 /x 4 = x 4 /x 8 = x 8 /x 12 = -x 13 /x 1 = x 1 /x 5 = x 5 /x 9 = x 9 /x 13 = -x 14 /x 2 = x 2 /x 6 = x 6 /x 10 = x 10 /x 14 = -x 15 /x 3 = x 3 /x 7 = x 7 /x 11 = x 11 /x 15 and :

coeff = (√4).2 -/2 = 2 -3/2 = 1/(2√2) choice 1 2 3 4 x 0 1/(2√2) 0 0 0 x 1 0 1/(2√2) 0 0 x 2 0 0 1/(2√2) 0 x 3 0 0 0 1/(2√2)
α 4 = -1 α 2 = ±i thus : α = √2/2.(±1±i) λ = 8α = 4√2.(±1±i)
x i+12 = -α.x i x i+8 = α.x i+12 = -α 2 .x i x i+4 = α.x i+8 = -α 3 .x i where i = 0 to 3

Choice of x 0 , x 1 ,x 2 and x 3 is open with the proviso of forming free eigenvectors.

Our choice is based on unitary matrix with a coefficient of adjustment (√4 follows from 4 possible choices for x 0 ) :

coeff = (√4).2 -/2 = 2 -3/2 = 1/(2√2) choice 1 2 3 4 x 0 1/(2√2) 0 0 0 x 1 0 1/(2√2) 0 0 x 2 0 0 1/(2√2) 0 x 3 0 0 0 1/(2√2)
We deduce cardinal matrix of modulo 2 5 case. We have put eigenvalues above each corresponding eigenvectors.

-x0 -x0 -x0 -x0 -x0 -x0 -x0 -x0 -x8 -x8 -x8 -x8 -x8 -x8 -x8 -x8 -x8 -x8 -x8 -x8 -x8 -x8 -x8 -x8 x0 x0 x0 x0 -x0 -x0 -x0 -x0 x4 x4 x4 x4 x4 x4 x4 x4 -x4 -x4 -x4 -x4 -x4 -x4 -x4 -x4 -x4 -x4 -x4 -x4 -x4 -x4 -x4 -x4 x0 x0 x0 x0 -x0 -x0 -x0 -x0 x4 x4 x4 x4 x4 x4 x4 x4 x4 x4 x4 x4 x4 x4 x4 x4 x4 x4 x4 x4 x4 x4 x4 x4 x0 x0 x0 x0 -x0 -x0 -x0 -x0 -x4 -x4 -x4 -x4 -x4 -x4 -x4 -x4 x12 x12 x12 x12 x12 x12 x12 x12 x12 x12 x12 x12 x12 x12 x12 x12 x0 x0 x0 x0 -x0 -x0 -x0 -x0 -x4 -x4 -x4 -x4 -x4 -x4 -x4 -x4 -x12 -x12 -x12 -x12 -x12 -x12 -x12 -x12 -x12 -x12 -x12 -x12 -x12 -x12 -x12 -x12 x0 x0 -x0 -x0 x2 x2 x2 x2 -x2 -x2 -x2 -x2 -x2 -x2 -x2 -x2 -x10 -x10 -x10 -x10 -x10 -x10 -x10 -x10 -x10 -x10 -x10 -x10 -x10 -x10 -x10 -x10 x0 x0 -x0 -x0 x2 x2 x2 x2 x2 x2 x2 x2 x2 x2 x2 x2 -x2 -x2 -x2 -x2 -x2 -x2 -x2 -x2 -x2 -x2 -x2 -x2 -x2 -x2 -x2 -x2 x0
-x3 -x3 -x3 -x3 -x3 -x3 -x3 -x3 -x3 -x3 -x3 -x3 -x3 -x3 -x3 -x3 x0 -x0 -x1 -x1 -x3 -x3 -x3 -x3 x7 x7 x7 x7 x7 x7 x7 x7 -x7 -x7 -x7 -x7 -x7 -x7 -x7 -x7 -x7 -x7 -x7 -x7 -x7 -x7 -x7 -x7 x0 -x0 -x1 -x1 x3 x3 x3 x3 -x3 -x3 -x3 -x3 -x3 -x3 -x3 -x3
-x0 -x1 -x1 x3 x3 x3 x3 -x3 -x3 -x3 -x3 -x3 -x3 -x3 -x3 -x11 -x11 -x11 -x11 -x11 -x11 -x11 -x11 -x11 -x11 -x11 -x11 -x11 -x11 -x11 -x11 x0 -x0 -x1 -x1 -x3 -x3 -x3 -x3 -x7 -x7 -x7 -x7 -x7 -x7 -x7 -x7 -x15 -x15 -x15 -x15 -x15 -x15 -x15 -x15 -x15 -x15 -x15 -x15 -x15 -x15 -x15 -x15
Foot-note: x 0 varies from one column to another (according to the value of the generator given above each column).

Thus we get finally for the eigenvalues and the unitary matrix P of corresponding eigenvectors : (this matrix is to be divided by 4√2 so that P. t P* = I thanks to above choices of x 0 )

λ=32 λ=0 λ=16(1-i) λ=16(1+i) λ=8√2.(1+i) λ=8√2.(1-i) λ=8√2.(-1+i) λ=8√2.(-1-i) λ=8.(1+i) λ=8.(1-i) λ=8.(1+i) λ=8.(1-i) λ=8.(1+i) λ=8.(1-i) λ=8.(1+i) λ=8.(1-i) λ=4√2.(1+i) λ=4√2.(1-i) λ=4√2.(-1+i) λ=4√2.(-1-i) λ=4√2.(1+i) λ=4√2.(1-i) λ=4√2.(-1+i) λ=4√2.(-1-i) λ=4√2.(1+i) λ=4√2.(1-i) λ=4√2.(-1+i) λ=4√2.(-1-i) λ=4√2.(1+i) λ=4√2.(1-i) λ=4√2.(-1+i) λ=4√2.(-1-i) 1 1 1 1 1 1 1 1 2 2 0 0 0 0 0 0 2 2 2 2 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 2 2 0 0 0 0 0 0 -2 -2 -2 -2 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 -2 -2 0 0 0 0 0 0 -2i 2i 2i -2i 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 -2 -2 0 0 0 0 0 0 2i -2i -2i 2i 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 -1 -1 -1 -1 2i -2i 0 0 0 0 0 0 √2.(-1+i) √2. (-1-i)/2 √2.(1+i)/2 √2.(1-i)/2 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 -1 -1 -1 -1 2i -2i 0 0 0 0 0 0 √2.(1-i)/2 √2.(1+i)/2 √2.(-1-i)/2 √2.(-1+i)/2 0 0 0 0 0 0 0 0 0 0 0 P 148/390
Modulo p Δ abundance factors correction

1 1 1 1 -1 -1 -1 -1 -2i 2i 0 0 0 0 0 0 √2.(-1-i)/2 √2.(-1+i)/2 √2.(1-i)/2 √2.(1+i)/2 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 -1 -1 -1 -1 -2i 2i 0 0 0 0 0 0 √2.(1+i)/2 √2.(1-i)/2 √2.(-1+i)/2 √2.(-1-i)/2 0 0 0 0 0 0 0 0 0 0 0 0 1 1 -1 -1 -i i i -i 0 0 0 0 -2 -2 0 0 0 0 0 0 0 0 0 0 2i -2i -2i 2i 0 0 0 0 1 1 -1 -1 -i i i -i 0 0 0 0 2 2 0 0 0 0 0 0 0 0 0 0 -2 -2 -2 -2 0 0 0 0 1 1 -1 -1 -i i i -i 0 0 0 0 -2 -2 0 0 0 0 0 0 0 0 0 0 -2i 2i 2i -2i 0 0 0 0 1 1 -1 -1 -i i i -i 0 0 0 0 2 2 0 0 0 0 0 0 0 0 0 0 2 2 2 2 0 0 0 0 1 1 -1 -1 i -i -i i 0 0 0 0 2i -2i 0 0 0 0 0 0 0 0 0 0 √2.(1-i) √2.(1+i) √2.(-1-i) √2.(-1+i) 0 0 0 0 1 1 -1 -1 i -i -i i 0 0 0 0 -2i 2i 0 0 0 0 0 0 0 0 0 0 √2.(-1-i) √2.(-1+i) √2.(1-i) √2.(1+i) 0 0 0 0 1 1 -1 -1 i -i -i i 0 0 0 0 2i -2i 0 0 0 0 0 0 0 0 0 0 √2.(-1+i) √2.(-1-i) √2.(1+i) √2.(1-i) 0 0 0 0 1/(4√2) 1 1 -1 -1 i -i -i i 0 0 0 0 -2i 2i 0 0 0 0 0 0 0 0 0 0 √2.(1+i) √2.(1-i) √2.(-1+i) √2.(-1-i) 0 0 0 0 1 -1 i -i (-1+i)/√2 (-1-i)/√2 (1+i)/√2 (1-i)/√2 0 0 -2i 2i 0 0 0 0 0 0 0 0 √2.(-1-i) √2.(-1+i) √2.(1-i) √2.(1+i) 0 0 0 0 0 0 0 0 1 -1 i -i (1-i)/√2 (1+i)/√2 (-1-i)/√2 (-1+i)/√2 0 0 -2 -2 0 0 0 0 0 0 0 0 -2i 2i 2i -2i/ 0 0 0 0 0 0 0 0 1 -1 i -i (-1+i)/√2 (-1-i)/√2 (1+i)/√2 (1-i)/√2 0 0 2i -2i 0 0 0 0 0 0 0 0 √2.(-1+i) √2.(-1-i) √2.(1+i) √2.(1-i) 0 0 0 0 0 0 0 0 1 -1 i -i (1-i)/√2 (1+i)/√2 (-1-i)/√2 (-1+i)/√2 0 0 2 2 0 0 0 0 0 0 0 0 -2 -2 -2 -2 0 0 0 0 0 0 0 0 1 -1 i -i (-1+i)/√2 (-1-i)/√2 (1+i)/√2 (1-i)/√2 0 0 -2i 2i 0 0 0 0 0 0 0 0 √2.(1+i) √2.(1-i) √2.(-1+i) √2.(-1-i) 0 0 0 0 0 0 0 0 1 -1 i -i (1-i)/√2 (1+i)/√2 (-1-i)/√2 (-1+i)/√2 0 0 -2 -2 0 0 0 0 0 0 0 0 2i -2i -2i 2i 0 0 0 0 0 0 0 0 1 -1 i -i (-1+i)/√2 (-1-i)/√2 (1+i)/√2 (1-i)/√2 0 0 2i -2i 0 0 0 0 0 0 0 0 √2.(1-i) √2.(1+i) √2.(-1-i) √2.(-1+i) 0 0 0 0 0 0 0 0 1 -1 i -i (1-i)/√2 (1+i)/√2 (-1-i)/√2 (-1+i)/√2 0 0 2 2 0 0 0 0 0 0 0 0 2 2 2 2 0 0 0 0 0 0 0 0 1 -1 -i i (-1-i)/√2 (-1+i)/√2 (1-i)/√2 (1+i)/√2 0 0 0 0 0 0 2 2 0 0 0 0 0 0 0 0 0 0 0 0 -2 -2 -2 -2 1 -1 -i i (1+i)/√2 (1-i)/√2 (-1+i)/√2 (-1-i)/√2 0 0 0 0 0 0 2i -2i 0 0 0 0 0 0 0 0 0 0 0 0 √2.(-1+i) √2.(-1-i) √2.(1+i) √2.(1-i) 1 -1 -i i (-1-i)/√2 (-1+i)/√2 (1-i)/√2 (1+i)/√2 0 0 0 0 0 0 -2 -2 0 0 0 0 0 0 0 0 0 0 0 0 -2i 2i 2i -2i 1 -1 -i i (1+i)/√2 (1-i)/√2 (-1+i)/√2 (-1-i)/√2 0 0 0 0 0 0 -2i 2i 0 0 0 0 0 0 0 0 0 0 0 0 √2.(-1-i) √2.(-1+i) √2.(1-i) √2.(1+i) 1 -1 -i i (-1-i)/√2 (-1+i)/√2 (1-i)/√2 (1+i)/√2 0 0 0 0 0 0 2 2 0 0 0 0 0 0 0 0 0 0 0 0 2 2 2 2 1 -1 -i i (1+i)/√2 (1-i)/√2 (-1+i)/√2 (-1-i)/√2 0 0 0 0 0 0 2i -2i 0 0 0 0 0 0 0 0 0 0 0 0 √2.(1-i) √2.(1+i) √2.(-1-i) √2.(-1+i) 1 -1 -i i (-1-i)/√2 (-1+i)/√2 (1-i)/√2 (1+i)/√2 0 0 0 0 0 0 -2 -2 0 0 0 0 0 0 0 0 0 0 0 0 2i -2i -2i 2i 1 -1 -i i (1+i)/√2 (1-i)/√2 (-1+i)/√2 (-1-i)/√2 0 0 0 0 0 0 -2i 2i 0 0 0 0 0 0 0 0 0 0 0 0 √2.(1+i) √2.(1-i) √2.(-1+i) √2.(-1-i) 1.2.8.3. Case n = 2
More generally for n = 2 and an order 2 δ matrix, we get (with #|1}=2 2 for δ≥3) :

δn-1 2 δ-3-2r -1 λ.x i = ∑ ∑ 2 r+2 .x i-((2^(2r)).( 1+8k
)) +2 δ-δn-1 .(x i-2^(2.n) +x i ) i = 0 to 2  -1 r = 0 k = 0 For x i+1 = x i , it is self-evident that : λ = 2 δ With x 0 = 2 -δ/2 ,

we get

x i = 2 -δ/2 for any i. In the same way for x i+1 = -x i , it come easily : λ.x i = 0.x i Thus : λ = 0 With x 0 = 2 -δ/2 , we get x 2i = 2 -δ/2 , x 2i+1 = -2 -δ/2 for any i. Let us suppose, for some t, 1≤t≤δ-1, that :

x i+2^t = -x i Then for given i :

2 δ-3 -1 2 δ-3-2.(r-1) -1 2 δ-3-2.r -1 2 δ-3-2.(r+1) -1 λ.x i = ∑ 2 2 .x i-1-(2^3).k + … +(. ∑ 2 r+1 .x i-2^(2.(r-1))-2^(2.(r-1)+3).k + ∑ 2 r+2 .x i-2^(2.r)-2^(2.r+3).k + ∑ 2 r+3 .x i-2^(2.(r+1))-2^(2.(r+1)+3). k ) k = 0 k = 0 k = 0 k = 0 +2 δ-3-r .x i +…+2 δ-δn+1 .x i +2 δ-δn .x i +2 δ-δn-1 .x i +2 δ-δn-1 .
x i If moreover t = 2r, then :

x i+2^(2r) = -x i x i+2^(2r+1) = x i and with signs alternative changes of x j :

if z≤2r, ∑ k x i+(2^z).k = 0 (equivalent to ∑ k x i-(2^z).k = 0) then : 2 δ-1-2r -1 2 δ-3-2r -1 2 δ-5-2r -1 λ.x i = 0+…+0+( ∑ 2 r+1 .x i-2^(2r-2) -∑ 2 r+2 .x i +∑ 2 r+3 .x i )+(2 δ-3-r +…+2 δ-δn+1 +2 δ-δn +2 δ-δn-1 +2 δ-δn-1 ).x i k = 0 k = 0 k = 0 Thus : λ.x i = 2 δ-r .x i-2^(2r-2) -2 δ-1-r .x i +2 δ-2-r .x i +(2 δ-2-r ).x i = 2 δ-r .x i-2^(2r-2) hence: λ/2 δ-t/2 = x i-2^(t-2) /x i = (x i /x i+1.2^(t-2) = x i+1.2^(t-2) /x i+2.2^(t-2) = x i+2.2^(t-2) /x i+3.2^(t-2) = x i+3.2^(t-2) /x i+4.2^(t-2) ) = x i+3.2^(t-2) /(-x i ) Then (λ/2 δ-t/2 ) 4 = -1 (λ/2 δ-t/2 ) 2 = ±i and : if t even α = (√2/2).(±1±i) λ = 2 δ-t/2 .α x i+2^(t-2) = α -1 .x i
Choice of x 0 to x 2^(t-2)-1 is open. The most practical one (see further), derived from canonical form, is : (X) j = 2 (t-δ)/2 .[I] j j = 1 to 2 (t-2) This writing means simply that column vector x = [x 0 , x 1 , x 2 , …, x 2^(t-2)-1 ] identifies with j th column of unitary matrix order 2 (t-2) and corrective factor 2 -δ/2 .

If now t = 2r+1, then :

x i+2^(2r+1) = -x i x i+2^(2r+2) = x i if z≤2r+1, ∑ k x i-(2^z).k = 0 Thus : 2 δ-3 -1 2 δ-3-2(r-1) -1 2 δ-3-2r -1 2 δ-3-2(r+1) -1 λ.x i = ∑ 2 2 .x i-1-(2^3).k + … +( ∑ 2 r+1 .x i-2^(2(r-1))-2^(2(r-1)+3).k + ∑ 2 r+2 .x i-2^(2r)-2^(2r+3).k + ∑ 2 r+3 .x i-2^(2(r+1))-2^(2(r+1)+3).k ) k = 0 k = 0 k = 0 k = 0 +(2 δ-3-r +…+2 δ-δn+1 +2 δ-δn +2 δ-δn-1 +2 δ-δn-1
).x i is written also :

2 δ-3-2.r -1 2 δ-5-2.r -1 λ.x i = 0+…+(0+ ∑ 2 r+2 .x i-2^(2r) +∑ 2 r+3 .x i )+(2 δ-3-r +…+2 δ-δn+1 +2 δ-δn +2 δ-δn-1 +2 δ-δn-1 ).x i k = 0 k = 0 Thus : λ.x i = 2 δ-1-r .x i-2^(2r) +2 δ-2-r .x i +2 δ-2-r .x i = 2 δ-(t+1)/2 .(x i-2^(t-1) +x i ) hence: (λ-2 δ-(t+1)/2 )/2 δ-(t+1)/2 = x i-2^(t-1) /x i = (x i /x i+1.2^(t-1) = x i+1.2^(t-1) /x i+2.2^(t-1) ) = x i+2^(t-1) /(-x i ) then : ((λ-2 δ-(t+1)/2 )/2 δ-(t+1)/2 ) 2 = -1 and : if t odd α = ±i λ = 2 δ-(t+1)/2 .(1+α)
x i+2^(t-1) = α -1 .x i Choice of x 0 to x 2^(t-1)-1 is free. The most practical form, with above writing rules, remains :

(X) j = 2 (t-δ)/2 .[I] j j = 1 to 2 (t-1) Foot-note :

In any rigour, it is necessary to show that by dichotomy the eigenvectors set, we get here, is an independent system of vectors. Dichotomy creates a methodical shift between the non-null values of components of each vector what contributes to this point. We will further reconsider this point by another approach (P unitary matrix).

Case n = 2 is summarized by extrapolation in the following table (relations 21) :

Recurrent elements ε (except the first two) Eigenvalues λ # of eigenvectors Eigenvectors (any x i whose indices are not indicated are equal to 0)

Entry ε 0 = 1 2 δ 2 0 x i = 2 -δ/2 ε 2^0 1 = 0 0 2 0 x i+2^0 = -x i x 0 = 2 -δ/2 Loop 1 (possibly partly) ε 2^1 1 = 1+i ε 2^1 2 = 1-i 2 δ-1 .ε(α 1 ) 2 1 x i+2^1 = -x i or(x i+2^0 = (i) -1 .x i , x i+2^0 = (-i) -1 .x i ) x 0 = 2 (1-δ)/2 ε 2^2 1 = √2/2.(1+i) ε 2^2 2 = √2/2.(1-i) ε 2^2 3 = √2/2.(-1+i) ε 2^2 4 = √2/2.(-1-i) 2 δ-1 .ε(α 2 ) 2 2 x i+2^2 = -x i or(x i+2^0 = (α 2 ) -1 .x i , x i+2^0 = (-α 2 ) -1 .x i , x i+2^0 = (α 2 3 ) -1 .x i , x i+2^0 = (-α 2 3 ) -1 .x i ) x 0 = 2 (2-δ)/2
Loop 2 (possibly partly)

ε 2^3 1 = 1+α 3 ε 2^3 2 = 1-α 3 … ε 2^3 2^3-1 = 1+α 3 2^3-1 ε 2^3 2^3 = 1-α 3 2^3-1 2 δ-2 .ε(α 3 ) 2 3 x i+2^3 = -x i or(x i+2^0 = (α 3 ) -1 .x i , x i+2^0 = (-α 3 ) -1 .x i ) (X) j = 2 (3-δ)/2 .[I] j j = 1 to 2 3 ε 2^4 1 = 1+α 4 ε 2^4 2 = 1-α 4 … ε 2^4 2^4-1 = 1+α 4 2^4-1 ε 2^4 2^4 = 1-α 4 2^4-1 2 δ-2 .ε(α 4 ) 2 4
x i+2^4 = -x i or(x i+2^0 = (α 4 ) -1 .x i , x i+2^0 = (-α 4 ) -1 .x i , x i+2^0 = (α 4

3 ) -1 .x i , x i+2^0 = (-α 43

) -1 .x i ) (X) j = 2 (4-δ)/2 .[I] j j = 1 to 2 4 … … … Loop δn+1 (possibly partly)
pending value of δ 2 δ-δn-1 .ε 2 δ-1 pending value of δ

Case n > 2

Let us return first to the general case. We had :

δn-1 2 δ-1-r.n /(#{1})-1 λ.x i = (∑ ∑ 2 r.(n-1) .#{1}.x i-((2^(r..n)).(1+2. #{1}.k)) ) + 2 δ-δn-1 .(x i-2^(n.n) +x i ) i = 0 to 2  -1 r = 0 k = 0
We start with x i+1 = x i (this first relation making somewhat exception to following ones), then x i+2 = -x i , then x i+2^t = -x i with t = 2, 3, etc. until exhaustion of values to be sought.

For x i+1 = x i , case is self-evident : λ.x i = 2 δ .x i So that, choosing x 0 = 2 -δ/2 , λ = 2 δ x i = 2 -δ/2 , i = 0 to 2 δ -1
In the same way for x i+1 = -x i , it come easily : λ.x i = 0.x i Thus : λ = 0 With x 0 = 2 -δ/2 , we get :

x 2i = 2 -δ/2 , x 2i+1 = -2 -δ/2 for any i Let us suppose then :

x i+2^t = -x i Then, using #{1} = 2 m+1 for δ≥3,

δn-1 2 δ-m-2-r.n -1 λ.x i =. ( ∑ ∑ 2 r.(n-1)+m+1 .x i-2^(r.n)-(2^(r.n+m+2)).k ) + 2 δ-δn-1 .(x i-2^(n.n) +x i ) r = 0 k = 0 then : 2 δ-m-2 -1 2 δ-m-2-r.n+n -1 2 δ-m-2-r.n -1 λ.x i = ∑ 2 m+1 .x i-1-(2^(m+2)).k +…+( ∑ 2 (r-1).(n-1)+m+1 .x i-2^(r.n-n)-(2^(r.n-n+m+2)).k )+ ∑ 2 r.(n-1)+m+1 .x i-2^(r.n)-(2^(r.n+m+2)).k + k = 0 k = 0 k = 0 2 δ-m-2-r.n-n -1 ∑ 2 (r+1).(n-1)+m+1 .x i-2^(r.n+n)-(2^(r.n+n+m+2)).k )+2 δ-3-r .x i + … +2 δ-δn+1 .x i +2 δ-δn .x i +2 δ-δn-1 .x i +2 δ-δn-1 .x i k = 0
Let us consider successive indices of x j sums between brackets, that is :

i-2^(r.n)-(2^(r.n+m+2)).k i-2^(r.n+n)-(2^(r.n+n+m+2)).k … Let us forget i and signs -: 2^(r.n)+(2^(r.n+m+2)).k 2^(r.n+n)+(2^(r.n+n+m+2)).k
Then, let us view coefficients of variable k in their order of appearance, that is : Foot-note:

2 (r.n) 2 (r.n+m+2)
For n = 2, sign of y is negative. There is an indices values overlapping which constitutes an exception to the other calculations that we will carry out below. However this case was already completely solved and it is not useful to return to it.

For other values of n (n > 2), we have y ≥ 0.

We consider three cases then :

Case 1 : t = r.n x i+2^(r.n) = -x i x i+2^(r.n+1) = x i if z ≤ r.n, ∑ k x i-(2^z).k = 0 hence : 2 δ-m-2-r.n -1 2 δ-m-2-r.n-n -1 λ.x i = 0+…+0+(0-∑ 2 r.(n-1)+m+1 .x i )+ ∑ 2 (r+1).(n-1)+m+1 .x i )+(2 δ-3-r +…+2 δ-δn+1 +2 δ-δn +2 δ-δn-1 +2 δ-δn-1 ).x i k = 0 k = 0 then : λ.x i = ((-2 δ-1-r +2 δ-2-r )+2 δ-2+r ).x i = 0 so that : λ = 0
Choice of x 0 to x 2^r.n-1 is arbitrary and chosen form is written :

(X) j = 2 (t-δ)/2 .[I] j j = 1 to 2 r.n Case 2 : t = r.n+s, 0 < s < m+2

x i+2^(r.n+s) = -x i x i+2^(r.n+s+1) = x i if z≤r.n+s, ∑ k x i-(2^z).k = 0 Then, r.n = t-s, hence :

if z≤t, ∑ k x i-(2^z).k = 0 Then : 2 δ-m-2 -1 2 δ-m-2-t+s+n -1 2 δ-m-2-t+s -1 λ.x i = ∑ 2 m+1 .x i-1-(2^(m+2)).k +…+( ∑ 2 t-s-r-n+m+2 .x i-2^(t-s-n)-(2^(t-s-n+m+2)).k + ∑ 2 t-s-r+m+1 .x i-2^(t-s)-(2^(t-s+m+2)).k + k = 0 k = 0 k = 0 2 δ-m-2-t+s-n -1 ∑ 2 t-s-r+n+m .x i-2^(t-s+n)-(2^(t-s+n+m+2)).k )+2 δ-3-r .x i +…+2 δ-δn+1 .x i +2 δ-δn .x i +2 δ-δn-1 .x i +2 δ-δn-1 .x i k = 0 If 0 < s < m+2, then t < t-s+m+2 < t+m+2 and : 2 δ-m-2-t+s+n -1 ∑ 2 t-s-r-n+m+2 .x i-2^(t-s-n)-(2^(t-s-n+m+2)).k ) = x i -x i +x i -x i +…+x i -x i = 0 k = 0 2 δ-m-2-t+s -1 ∑ 2 t-s-r+m+1 .x i-2^(t-s)-(2^(t-s+m+2)).k = 2 δ-1-r .x i-2^(t-s)) k = 0 2 δ-m-2-t+s-n -1 ∑ 2 t-s-r+n+m .x i-2^(t-s+n)-(2^(t-s+n+m+2)).k = 2 δ-2-r .x i k = 0 Hence : λ.x i = 0+2 δ-1-(t-s)/n .x i-2^(t-s) +2 δ-1-(t-s)/n .x i Then : (λ-2 δ-1-(t-s)/n )/2 δ-1-(t-s)/n = x i-2^(t-s) /x i
As x i+2^t = -x i , we deduce :

((λ-2 δ-1-(t-s)/n )/2 δ-1-(t-s)/n ) 2^s = -1 Let us pose for 2 s th root of -1 : α = 2^s √-1 Then λ = 2 δ-1-(t-s)/n .(1+α) = 2 δ-1-r .(1+α) = 2 δ-1-r .ε x i+2^(t-s) = α -1 .x i (X) j = 2 (t-δ)/2 .[I] j j = 1 to 2 2^(t-s)
Let us note that case 2 is impossible for n odd (as m = -1 and 0 < s < m+2 = 1 has no solution s).

Case 3 : t = r.n+s+m+2, 0 ≤ s < n-m-2 x i+2^(r.n+s+m+2) = -x i x i+2^(r.n+s+m+3) = x i if z≤r.n+s+m+2, ∑ k x i-(2^z).k = 0 Then, r.n = t-s-m-2, hence : if z≤t, ∑ k x i-(2^z).k = 0 thus : 2 δ-m-2 -1 2 δ-t+s+n -1 2 δ-t+s -1 λ.x i = ∑ 2 m+1 .x i-1-(2^(m+2)).k +…+( ∑ 2 t-s-r-n-2 .x i-2^(t-s-n-m-2)-(2^(t-s-n)).k + ∑ 2 t-s-r-1 .x i-2^(t-s-m-2)-(2^(t-s)).k + k = 0 k = 0 k = 0 2 δ-t+s-n -1 ∑ 2 t-s-r+n-2 .x i-2^(t-s+n-m-2)-(2^(t-s+n)).k )+2 δ-3-r .x i +…+2 δ-δn+1 .x i +2 δ-δn .x i +2 δ-δn-1 .x i +2 δ-δn-1 .x i k = 0
If 0≤s<n-m-2, then t-s≤t and t<t+n-s-m-2 :

2 δ-t+s+n -1 ∑ 2 t-s-r-n-2 .x i-2^(t-s-n-m-2)-(2^(t-s-n)).k = x i -x i +x i -x i +…+x i -x i = 0 k = 0 2 δ-t+s -1 ∑ 2 t-s-r-1 .x i-2^(t-s-m-2)-2^(t-s).k = x i -x i +x i -x i +…+x i -x i = 0 k = 0 but : 2 δ-t+s-n -1 ∑ 2 t-s-r+n-2 . x i-2^(t-s+n)-2^(t-s+n+m+2).k = 2 δ-2-r .x i k = 0 Hence : λ.x i = 2 δ-1-r .x i = 2 δ-1-(t-s-m-2)/n .x i Then : λ = 2 δ-1-(t-s-m-2)/n = 2 δ-1-r .(1+α) = 2 δ-1-r .ε (X) j = 2 (t-δ)/2 .[I] j j = 1 to 2 2^t
Before drawing up all the results, we will seek values corresponding to 2^s √-1.

Calculation of 2 k th root of unit.

The calculation of these roots is essential to our purpose. Let us note α s a « generative » root of 2^s √-1.

We get the full picture of cases (writing convention for m is still the same) (relations 22) :

n = 2 n even (n>2) multiplicity of ε ij n even n odd (m=-1) multiplicity of ε ij n odd entry ε 0 = 1 ε 0 = 1 2 0 ε 0 = 1 2 0 ε 11 = 0 ε 11 = 0 2 0 ε 11 = 0 2 0 loop ε 21 = 1+i ε 22 = 1-i ε 21 = 1+α 1 ε 22 = 1-α 1 2 0 ε 21 = 1 ε 22 = 1 2 1
Case 2

ε 41 = (1/2).(√2)(1+i) ε 42 = (1/2).(√2)(1-i) ε 43 = (1/2).(√2)(-1+i) ε 44 = (1/2).(√2)(-1-i) ε 41 = 1+α 2 ε 42 = 1-α 2 ε 43 = 1+α 2 3 ε 44 = 1-α 2 3 2 0 ε 41 = 1 to ε 44 = 1 2 2 … … ε 2^(m+1) 1 = 1+α m+1 ε 2^(m+1) 2 = 1-α m+1 ε 2^(m+1) 3 = 1+α m+1 3 ε 2^(m+1) 4 = 1-α m+1 3 ε 2^(m+1) 5 = 1+α m+1 5 ε 2^(m+1) 6 = 1-α m+1 5 … ε 2^(m+1) 2^(m+1)-1 = 1+α m+1 2^(m+1)- 1 ε 2^(m+1) 2^(m+1) = 1-α m+1 2^(m+1)-1 2 0 ε 2^(m+1) 1 = 1 to ε 2^(m+1) 2^(m+1) = 1 2 m+1 ε 2^(m+2) 1 = 1 to ε 2^(m+2) 2^(m+2) = 1 2 m+2 ε 2^(m+2) 1 = 1 to ε 2^(m+2) 2^(m+2) = 1 2 m+2 Case 3 … … ε 2^(n-1) 1 = 1 to ε 2^(n-1) 2^(n-1) = 1 2 n-1 ε 2^(n-1) 1 = 1 to ε 2^(n-1) 2^(n-1) = 1 2 n-1 ε 2^n 1 = 0 to ε 2^n 2^n = 0 2 n ε 2^n 1 = 0 to ε 2^n 2^n = 0 2 n Case 1
We should not be astonished that neither case n = 2 is distinct from odd cases, nor is an exception to the other even numbers : 2 is definitely a peculiar number.

Let us note that with odd n (since m = -1), expressions using ε = ε (α) are reduced to an empty set.

Let us note also that for n = 4, m = 2 there is no solution x of m+2 ≤ x ≤n-1 and thus in this case there is no solution of type ε = 1 (except in the entry).

We can now easily check (and demonstrate) external and interior dichotomy properties.

P 154/390

Modulo p Δ abundance factors correction Indeed, passage of loop i to loop i+1 correspond to interior dichotomy property. In addition, to a recurrent element ε of loop i corresponds eigenvalue 2 δ-i .ε for a matrix of order δ (dimension 2 δ ) and eigenvalue 2 δ+1-i .ε for a matrix of order δ+1 (dimension 2 δ+1 ), thus external dichotomy property.

Illustrations

Let us give two illustrations to the results obtained for n = 3 and n = 8 :

Case n = 3 (here δ = 5)

Cardinal matrix is : 012345678

1 x8 -8 4 4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 x24 20 8 4 4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 x20 4 8 4 4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 x4 -20 4 4 8 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 x12 -4 4 4 8 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 x28 26 8 4 4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 x26 18 4 8 4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 x18 10 4 8 4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 x10 2 4 4 8 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 x2 -26 8 4 4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 x6 -18 4 8 4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 x14 -10 4 8 4 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 x22 -2 4 4 8 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 x30 13 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 8 4 4 x13 9 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 8 4 4 x9 21 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 8 4 x21 17 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 8 4 x17 29 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 8 4 x29 25 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 8 4 x25 5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 4 8 x5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 4 8 x1 -13 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 8 4 4 x19 -9 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 8 4 4 x23 -21 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 8 4 x11 -17 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 8 4 x15 -29 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 8 4 x3 -25 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 8 4 x7 -5 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 4 8 x27 -1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 4 8 x31
We get :

entry loop 1 loop 2 (partly) ε = 1 ε 1 = 0 ε 2 1 = ε 2 2 = 1 ε 4 1 to ε 4 4 = 1 ε 8 1 to ε 8 8 = 0 ε 16 1 to ε 16 16 = 1 λ = 2 δ .1 = 32 λ 1 = 0 λ 2 1 to λ 2 2 = 2 δ-1 .1 = 16 λ 4 1 to λ 4 4 = 2 δ-1 .1 = 16 λ 8 1 to λ 8 8 = 2 δ-1 .0 = 0 λ 16 1 to λ 16 16 = 2 δ-2 .1 = 8
Eigenvalues and eigenvectors are ( here P is multiplied by = 2 δ/2 = 2 We have P = P* and thus P -1 = t P*. P is an unitary matrix.

2 2√2 4 1 √2 2 2√2 -4 1 √2 2 -2√2 4 1 √2 2 -2√2 -4 1 √2 -2 2√2 4 1 √2 -2 2√2 -4 1 √2 -2 -2√2 4 1 √2 -2 -2√2 -4 1 -√2 2 2√2 4 1 -√2 2 -2√2 4 1 -√2 2 2√2 -4 1 -√2 2 -2√2 -4 1 -√2 -2 2√2 4 1 -√2 -2 -2√2 4 1 -√2 -2 2√2 -4 1 -√2 -2 -2√2 -4 -1 √2 2 2√2 4 -1 √2 -2 2√2 4 -1 √2 2 -2√2 4 -1 √2 -2 -2√2 4 -1 √2 2 2√2 -4 -1 √2 -2 2√2 -4 -1 √2 2 -2√2 -4 -1 √2 -2 -2√2 -4 -1 -√2 2 2√2 4 -1 -√2 -2 2√2 4 -1 -√2 2 -2√2 4 -1 -√2 -2 -2√2 4 -1 -√2 2 2√2 -4 -1 -√2 -2 2√2 -4 -1 -√2 2 -2√2 -4 -1 -√2 -2 -2√2 -4 1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 √2 √2 √2 √2 √2 √2 √2 √2 -√2 -√2 -√2 -√2 -√2 -√2 -√2 -√2 √2 √2 √2 √2 √2 √2 √2 √2 -√2 -√2 -√2 -√2 -√2 -√2 -√2 -√2 2 2 2 -2 -2 -2 -2 2 2 2 2 -2 -2 -2 -2 2 -2 2 -2 2 -2 2 -2 2 -2 2 -2 2 -2 2 -2 2√2 2√2 -2√2 -2√2 2√2 2√2 -2√2 -2√2 2√2 -2√2 2√2 -2√2 2√2 -2√2 2√2 -2√2 2√2 -2√2 2√2 -2√2 2√2 -2√2 2√2 -2√2 2√2 -2√2 2√2 -2√2 2√2 -2√2 2√2 -2√2 -4 4
Case n = 8

Let us give, to clarify the purpose, an illustration with case n = 8 for a matrix of dimension 2 δ (m = 3). The choice of an even number is taken intentionally since more complex, giving thus a richer illustration (relations 23

Entry ε 0 = 1 2 δ 2 0 x i = 2 -δ/2 ε 2^0 1 = 0 0 2 0 x i+2^0 = -x i x 0 = 2 -δ/2 Loop 1 (possibly partly) ε 2^1 1 = 1+α 1 ε 2^1 2 = 1-α 1 2 δ-1 .ε(α 1 ) 2 1 x i+2^1 = -x i or(x i+2^0 = (α 1 ) -1 .x i , x i+2^0 = (-α 1 ) -1 .x i ) x 0 = 2 (1-δ)/2 Case 2 ε 2^2 1 = 1+α 2 ε 2^2 2 = 1-α 2 ε 2^2 3 = 1+α 2 3 ε 2^2 4 = 1-α 2 3 2 δ-1 .ε(α 2 ) 2 2 x i+2^2 = -x i or(x i+2^0 = (α 2 ) -1 .x i , x i+2^0 = (-α 2 ) -1 .x i , x i+2^0 = (α 2 3 ) -1 .x i , x i+2^0 = (-α 2 3 ) -1 .x i ) x 0 = 2 (2-δ)/2 ε 2^3 1 = 1+α 3 ε 2^3 2 = 1-α 3 … ε 2^3 2^3-1 = 1+α 3 2^3-1 ε 2^3 2^3 = 1-α 3 2^3-1 2 δ-1 .ε(α 3 ) 2 3 x i+2^3 = -x i or(x i+2^0 = (α 3 ) -1 .x i , x i+2^0 = (-α 3 ) -1 .x i , … x i+2^0 = (α 3 2^3-1 ) -1 .x i , x i+2^0 = (-α 3 2^3-1 ) -1 .x i ) x 0 = 2 (3-δ)/2 ε 2^4 1 = 1+α 4 ε 2^4 2 = 1-α 4 … ε 2^4 2^4-1 = 1+α 4 2^4-1 ε 2^4 2^4 = 1-α 4 2^4-1 2 δ-1 .ε(α 4 ) 2 4 x i+2^4 = -x i or(x i+2^0 = (α 4 ) -1 .x i , x i+2^0 = (-α 4 ) -1 .x i , … x i+2^0 = (α 4 2^4-1 ) -1 .x i , x i+2^0 = (-α 4 2^4-1 ) -1 .x i ) x 0 = 2 (4-δ)/2 ε 2^5 1 = 1 to ε 2^5 2^5 = 1 2 δ-1 .ε (ε=1) 2 5 x i+2^5 = -x i (X) j = 2 (5-δ)/2 .[I] j j = 1 to 2 5 Case 3 ε 2^6 1 = 1 to ε 2^6 2^6 = 1 2 δ-1 .ε (ε=1) 2 6 x i+2^6 = -x i (X) j = 2 (6-δ)/2 .[I] j j = 1 to 2 6 ε 2^7 1 = 1 to ε 2^7 2^7 = 1 2 δ-1 .ε (ε=1) 2 7 x i+2^7 = -x i (X) j = 2 (7-δ)/2 .[I] j j = 1 to 2 7 ε 2^8 1 = 0 to ε 2^8 2^8 = 0 2 δ-1 .ε (ε=1) 2 8 x i+2^8 = -x i (X) j = 2 (8-δ)/2 .[I] j j = 1 to 2 8 Case 1 Loop 2 (possibly partly) ε 2^9 1 = 1+α 9 ε 2^9 2 = 1-α 9 … ε 2^9 2^9-1 = 1+α 9 2^9-1 ε 2^9 2^9 = 1-α 9 2^9-1 2 δ-2 .ε(α 9 ) 2 9 x i+2^9 = -x i or(x i+2^0 = (α 9 ) -1 .x i , x i+2^0 = (-α 9 ) -1 .x i ) (X) j = 2 (8-δ)/2 .[I] j j = 1 to 2 8 Case 2 ε 2^10 1 = 1+α 10 ε 2^10 2 = 1-α 10 … ε 2^10 2^10-1 = 1+α 10 2^10-1 ε 2^10 2^10 = 1-α 10 2^10-1 2 δ-2 .ε(α 10 ) 2 10 x i+2^10 = -x i or(x i+2^0 = (α 10 ) -1 .x i , x i+2^0 = (-α 10 ) -1 .x i , x i+2^0 = (α 10 3 ) -1 .x i , x i+2^0 = (-α 10 3 ) -1 .x i ) (X) j = 2 (8-δ)/2 .[I] j j = 1 to 2 8 ε 2^11 1 = 1+α 11 ε 2^11 2 = 1-α 11 … ε 2^11 2^11-1 = 1+α 11 2^11-1 ε 2^11 2^11 = 1-α 11 2^11-1 2 δ-2 .ε(α 11 ) 2 11 x i+2^11 = -x i or(x i+2^0 = α 11 -1 .x i , x i+2^0 = (-α 11 ) -1 .x i , … x i+2^0 = (α 11 2^3-1 ) -1 .x i , x i+2^0 = (-α 11 2^3-1 ) -1 .x i ) (X) j = 2 (8-δ)/2 .[I] j j = 1 to 2 8 P 157/390 Modulo p Δ abundance factors correction ε 2^12 1 = 1+α 12 ε 2^12 2 = 1-α 12 … ε 2^12 2^12-1 = 1+α 12 2^12-1 ε 2^12 2^12 = 1-α 12 2^12-1 2 δ-2 .ε(α 12 ) 2 12 x i+2^12 = -x i or(x i+2^0 = α 12 -1 .x i , x i+2^0 = (-α 12 ) -1 .x i , … x i+2^0 = (α 12 2^4-1 ) -1 .x i , x i+2^0 = (-α 12 2^4-1 ) -1 .x i ) (X) j = 2 (8-δ)/2 .[I] j j = 1 to 2 8 ε 2^13 1 = 1 to ε 2^13 2^13 = 1 2 δ-2 .ε (ε=1) 2 13
x i+2^13 = -x i (X) j = 2 (13-δ)/2 .[I] j j = 1 to 2 [START_REF] Nikolaï | Calcul différentiel and intégral[END_REF] Case 3

ε 2^14 1 = 1 to ε 2^14 2^14 = 1 2 δ-2 .ε (ε=1) 2 14
x i+2^14 = -x i (X) j = 2 (14-δ)/2 .[I] j j = 1 to 2 14

ε 2^15 1 = 1 to ε 2^15 2^15 = 1 2 δ-2 .ε (ε=1) 2 15 x i+2^15 = -x i (X) j = 2 (15-δ)/2 .[I] j j = 1 to 2 15 ε 2^16 1 = 0 to ε 2^16 2^16 = 0 2 δ-2 .ε (ε=1) 2 16
x i+2^16 = -x i (X) j = 2 (16-δ)/2 .[I] j j = 1 to 2 16 For this property, we study the eigenvectors matrix we proposed higher. Let us have P(i,j) the component at line i and column j of P. We must show that :

2 δ 2 δ  (i,j) c [[1, … , 2 δ ]]x[[1, … , 2 δ ]] ∑ P(i,k).P(j,k)* = 0 and k = 1 i ≠ j ∑ P(i,k).P(i,k)* = 1 k = 1
This involves close examination of P(i,j) terms.

For n odd, through given examples, the trace terms equal :

δ-1 (2 δ/2 ) 2 +∑ (2 -δ/2+i/2 ) 2 = (2 -δ ).(1+2 0 +2 1 +…+2 δ-1 ) = 1 i = 0 The other terms equal : k-1 (2 δ/2 ) 2 + ( ∑ (2 -δ/2+i/2 ) 2 ) -(2 -δ/2+k/2 ) 2 = 0 i = 0
For n even, calculation principle is the same one (calculation per block), but indices handling is undoubtedly more complex (and is not presented here).

Cardinal matrices in Waring prime numbers variables enumeration 2.1. Case p odd (p>2)

We study enumeration of the solutions of :

y 1 n + y 2 n + … +y i n + … + y m n = c modulo p δ
Here variable y i does not describe entire class {0, 1, …, p δ -1}, but only this class numbers which are not multiples of p (including 0). That is :

{1, 2, …, p-1, p+1, …, 2p-1, 2p+1, … i.p-1, i.p+1, …, p δ -1}
In this case, matrices B δ construction is much simpler than in case of integers' variables enumeration : First column of matrix B δ is given by the column vector (0, …,0, d, 0, …, 0), where exactly d-1 zeros finish this list. Matrix construction obeys the same rules as those used in the case of integers' variables. We thus get general case of matrix B δ (first column of the table does not belong to the matrix) :

P 158/390

Modulo p Δ abundance factors correction

p δ 0 0 0 … 0 Ф(δ).[J'] [0] [0] [0] … [0] Ф(δ).[J] [0] [0] [0] [0] Ф(δ).[J] … [0] [0] [0] … [0] Ф(δ).[J] p 2 .g i [0] [0] [0] … [0] Ф(δ).[J] p 1 .g i [0] [0] [0] … [0] Ф(δ).[J] p 0 .g i d.[K'] Ф(1).[K] Ф(2).[K] … Ф(δ-1).[K] p δ-1 .[B'] Definitions of [K'], [K], [I],
[J] are those which were given higher in this exercise. [B'] is the reduced matrix of [B], i.e. [B] without first line and first column.

The eigenvectors matrix P δ is the same as in case of integers' variables (see page 126).

The eigenvalues {λ δ } of [B δ ] are given by : p δ-1 .(p-1), p δ-1 {λ'}, 0, …, 0, p δ-2 {λ'}, 0, …, 0, p δ-3 {λ'}, 0, … 0, … Between p δ-i {λ'} and p δ-i-1 {λ'}, we count each time d-1 zeros.

It is easy to check, by developing expressions that :

[B δ ] = [P δ ].[λ δ ].[P δ -1 ] We have then : [B δ ] m = [P δ ].[λ δ ] m .[P δ -1 ]
2.2. Case p even (p=2)

Position of the problem

In this case, multiples of 2 are eliminated in cardinal matrices method. Hence, for #(i,1) = #{u / u n = c i mod 2 δ }, we will have #(i,1) = 0 for i = 1 to 2 δ-1 . The first half of cardinal matrix first column includes only null components. By construction, dividing the matrix into four parts of equal size 2 δ-1 by 2 δ-1 , the two square parts which include matrix trace are with null components. The values of u i are deduced from enumerations of u c = #{y n = c mod 2  } = #{c}. Here y being a prime numbers variable, we get :

c #{c} Definition domain 1+2.(#{1}).k #{1} k = 0, 1, …, 2 δ-1 /(#{1})-1 #{1} = #{y / y n = 1 mod 2 δ , y = 1, 3, 5, …, 2 δ -1} = min(2 m+1 , 2 δ-1 )
where m is either multiplicity of factor 2 if n is even, or m = -1 if n is odd We set out then using : P 159/390 Modulo p Δ abundance factors correction 2  -1 λ.x i = ∑ u i+j .x 2^-j and i+j = c j = 0 so that :

2  -1 2  -1 λ.x i = ∑ u c .x i+2^-c = ∑ u c .x i-c c = 0 c = 0 then : 2 δ-1 /(#{1})-1 λ.x i = ∑ #{1}.x i-(1+2. #{1}.k) i = 0 to 2  -1 k = 0 Thus, using #{1} = 2 m+1 for δ≥3, 2 δ-m-2 -1 λ.x i = ∑ 2 m+1 .x i-(1+(2^(m+2)).k) i = 0 to 2  -1 k = 0 So that : λ.x i = 2 m+1 .(x i-1 +x i-1-2^(m+2) +x i-1-2.2^(m+2) +…+x i-1-(2^(δ-m-2)-1).(2^(m+2)) ) i = 0 to 2  -1
We build then a satisfying choice of eigenvectors by tests and errors (which are not developed here) while starting with x i+1 = x i , then x i+2 = -x i , then x i+2^t = -x i with t = 2, 3, etc. until exhaustion of values to be sought. For x i+1 = x i , we have :

λ.x i = 2 δ-1 .x i So that, choosing x 0 = 2 -δ/2 , λ = 2 δ-1 x i = 2 -δ/2 , i = 0 to 2 δ -1
In the same way for x i+1 = -x i , we get immediately :

λ.x i = -2 δ-1 .x i So that : λ = -2 δ-1 With x 0 = 2 -δ/2 , it follows :

x 2i = 2 -δ/2 , x 2i+1 = -2 -δ/2 for any i Let us suppose then :

x i+2^t = -x i Expression (x i-1 +x i-1-2^(m+2) +x i-1-2.2^(m+2) +…+x i-1-(2^(δ-m-2)-1).(2^(m+2))
) contains an even quantity of equal terms in absolute value and alternate signs. Thus : λ = 0

The eigenvectors being built here with the same conditions as in integers' variables case, the eigenvectors matrix is supplemented with the same components, and are forming then an unitary matrix. The eigenvectors matrices obtained in the case of integers' variables are "reusable" in the case of prime numbers variables.

The summary table for λ i j = 2 δ-1 .ε i j is very simple :

entry ε 0 = 1 ε 11 = 1 loop ε i j = 0 (i>1)

Case n odd

When c is odd, then #{c} = 1. Thus, all B 2, matrices are as follows :

0 even [0] [1] … odd … [1] [0] … Matrices B 2,
m are equal to :

P 160/390 Modulo p Δ abundance factors correction even … [(1+(-1) m ).2 (-1).(m-1)-1 ] [(1+(-1) m-1 ).2 (-1).(m-1)-1 ] … odd … [(1+(-1) m-1
).2 (-1).(m-1)-1 ] [(1+(-1) m ).2 (-1).(m-1)-1 ] … It is the simplicity which prevails here. Eigenvalues and eigenvectors research is useless for evaluation of B 2, m .

Case n even

Even case n the case being more complex, we suppose that it interpolates the preceding case.

Normalization of matrices modulo p δ

Case p > 2 (odd)

Modulo p δ process is in any point similar to case mod p. Instead of making evaluation of abundance factors with matrices A δ (or B δ pending on studied case) and then to carry out normalization, we may normalize matrices so as to get normalized abundance factors right away. We gave implementing rules of normalization at exercise 2 : Stage 1 -for a variable of integers (type x), cardinal must be divided by p δ at sequence p, -for a variable of prime numbers (type y), cardinal must be divided by p δ- Implicitly, that means that matrices AN δ and BN δ are commutative. This rises immediately owing to the fact that their eigenvectors matrices (and inverse matrices) are identical (for given p and δ) : P A,δ = P B,δ and P -1 A,δ = P -1 B,δ

Case p = 2 (even)

Cardinal matrices do not have similar form and dimension as for p > 2. However, the same constraint remains : the sum following each line and column is equal to p δ , that is here 2 δ .

Step 1 -for a variable of integers, cardinal must be divided by 2 δ , -for a variable of prime numbers, cardinal must be divided by 2 δ-1 , Step 2

-restore sum to 2 δ by multiplication of factors with 2 δ .

Thus :

[AN δ ] = 1/2 δ .[A δ ] k Fan{c=2 δ ,2,δ} 2 δ Fan{2 δ-1 .5 0 ,2,δ} = AN δ 0 … 0 Fan{-2 0 .5 0 ,2,δ} 0
In the same way :

[BN δ ] = (1/(2 δ-1 )
). [B δ ] and for m prime numbers variables : m Fan{c=2 δ ,2,δ}

2 δ Fan{2 δ-1 .5 0 ,2,δ} = BN δ 0 … 0 Fan{-2 0 .5 0 ,2,δ} 0
We can also carry out mix of variables of integers and prime numbers in matrices products (while maintaining the remark on commutative matrices). -if n is odd, the sign (-) of x k+j n can be ignored, -if n is even, the sign is taken into account as follows.

The cardinal matrices, for n even and indices greater than k, are written by using above developed rules with some corrections as follow : The general form of the matrices results from the study of the integers' variables enumeration (done higher in this exercise). Thus, generic elements are not mentioned below.

Matrix C δ (for variables of indices greater than k)

p δ p δ-δn-1 … … [0] Ф(δ-1).[J' 0 ] [0] … [0] Ф(δ).[J' 0 ] … … [0] Ф(δ-1).[J 0 ] [0] … [0] Ф(δ).[J 0 ] … [0] Ф(δ-1).[J 0 ] [0] … [0] Ф(δ).[J 0 ] p δn.n .g i d.p δn.(n-1) .[K' d/2 ] Ф((n-1).δn+1).[K d/2 ] … … … [0] Ф(δ-1).[J 0 ] [0] … [0] Ф(δ).[J 0 ] [0] [0] [0] … … [0] Ф(δ-1).[J 0 ] [0] … [0] Ф(δ).[J 0 ] … … … … … [0] Ф(δ-1).[J 0 ] [0] … [0] Ф(δ).[J 0 ] … [0] Ф(δ-1).[J 0 ] [0] … [0] Ф(δ).[J 0 ] … … … … … … … … … … … p 2n+1 .g i [0] [0] [0] … … [0] Ф(δ-1).[J 0 ] [0] … [0] Ф(δ).[J 0 ] p 2n .g i d.p 2(n-1) .[K' d/2 ] Ф(2n-1).[K d/2 ] Ф(2n).[K d/2 ] … … [0] Ф(δ-1).[J 0 ] [0] … [0] Ф(δ).[J 0 ] … [0] [0] [0] … … [0] Ф(δ-1).[J 0 ] [0] … [0] Ф(δ).[J 0 ] p n+2 .g i [0] [0] [0] … … [0] Ф(δ-1).[J 0 ] [0] … [0] Ф(δ).[J 0 ] p n+1 .g i [0] [0] [0] … … p δ-2 .[I] Ф(δ-1).[J 0 ] [0] … [0] Ф(δ).[J 0 ] p n .g i d.p n-1 .[K' d/2 ] Ф(n).[K d/2 ] Ф(n+1).[K d/2 ] … … Ф(δ-2).[K d/2 ] p δ-2 . t [A'] [0] … [0] Ф(δ).[J 0 ] … [0] [0] [0] … … [0] [0] p δ-1 .[I] … [0] Ф(δ).[J 0 ] p 2 .g i [0] [0] [0] … … [0] [0] [0] … [0] Ф(δ).[J 0 ] p 1 .g i [0] [0] [0] … … [0] [0] [0] … p δ-1 .[I] Ф(δ).[J 0 ] p 0 .g i d.[K' d/2 ] Ф(1).[K d/2 ] Ф(2).[K d/2 ] … … Ф(δ-n-1).[K d/2 ] Ф(δ-n).[K d/2 ] Ф(δ-n+1).[K d/2 ] … Ф(δ-1).[K d/2 ] p δ-1 . t [A']
The eigenvalues of C δ are given by :

{μ δ } p δ p δ-1 {μ k d/2 } p δ-1 … p δ-1 … p δ-δn-1 {μ k d/2 } p δ-δn-1 … p δ-δn-1
The eigenvectors matrix [P δ ] is the same as in general case. We get then :

[C δ ] i = [P δ ].[μ δ ] i .[P δ -1 ]
The expression of normalized cardinal matrix is :

[AN δ ] k .[CN δ ] i

Case p odd (p>2) and p\n

This case is not studied here.

Case p = 2

This case is not studied here.

Cardinal matrices in Waring prime numbers enumeration, boundless volumes, logarithmic mesh

We examine case p >

This paragraph is following the precedent example :

-if n is odd, there is no evolution compared to the problem with m+i variables -if n is even, evolutions below result

The cardinal matrices, for even n and indices greater than m, are written by using again above developed rules with some corrections :

t [B'] = B' 1 : square matrix of dimension d, transposed of matrix [B'] previously defined.

Matrix D δ (for variables with indices greater than m)

p δ p δ-δn-1 … … [0] Ф(δ-1).[J' 0 ] [0] … [0] Ф(δ).[J' 0 ] … … [0] Ф(δ-1).[J 0 ] [0] … [0] Ф(δ).[J 0 ] … [0] Ф(δ-1).[J 0 ] [0] … [0] Ф(δ).[J 0 ] p δn.n .g i d.p δn.(n-1) .[K' d/2 ] Ф((n-1).δn+1).[K d/2 ] … … … [0] Ф(δ-1).[J 0 ] [0] … [0] Ф(δ).[J 0 ] [0] [0] [0] … … [0] Ф(δ-1).[J 0 ] [0] … [0] Ф(δ).[J 0 ] … … … … … [0] Ф(δ-1).[J 0 ] [0] … [0] Ф(δ).[J 0 ] … [0] Ф(δ-1).[J 0 ] [0] … [0] Ф(δ).[J 0 ] … … … … … … … … … … … p 2n+1 .g i [0] [0] [0] … … [0] Ф(δ-1).[J 0 ] [0] … [0] Ф(δ).[J 0 ] p 2n .g i d.p 2(n-1) .[K' d/2 ] Ф(2n-1).[K d/2 ] Ф(2n).[K d/2 ] … … [0] Ф(δ-1).[J 0 ] [0] … [0] Ф(δ).[J 0 ] … [0] [0] [0] … … [0] Ф(δ-1).[J 0 ] [0] … [0] Ф(δ).[J 0 ] p n+2 .g i [0] [0] [0] … … [0] Ф(δ-1).[J 0 ] [0] … [0] Ф(δ).[J 0 ] p n+1 .g i [0] [0] [0] … … p δ-2 .[I] Ф(δ-1).[J 0 ] [0] … [0] Ф(δ).[J 0 ] p n .g i d.p n-1 .[K' d/2 ] Ф(n).[K d/2 ] Ф(n+1).[K d/2 ] … … Ф(δ-2).[K d/2 ] p δ-2 . t [A'] [0] … [0] Ф(δ).[J 0 ] … [0] [0] [0] … … [0] [0] p δ-1 .[I] … [0] Ф(δ).[J 0 ] p 2 .g i [0] [0] [0] … … [0] [0] [0] … [0] Ф(δ).[J 0 ] p 1 .g i [0] [0] [0] … … [0] [0] [0] … p δ-1 .[I] Ф(δ).[J 0 ] p 0 .g i d.[K' d/2 ] Ф(1).[K d/2 ] Ф(2).[K d/2 ] … … Ф(δ-n-1).[K d/2 ] Ф(δ-n).[K d/2 ] Ф(δ-n+1).[K d/2 ] … Ф(δ-1).[K d/2 ] p δ-1 . t [B']
Eigenvalues of D δ (P δ and P δ -1 remain unchanged)

{λ δ } p δ p δ-1 {λ k d/2 } 0 … 0 … p δ-δn-1 {λ k d/2 } 0 … 0
We have then :

[D δ ] j = [P δ ].[λ δ ] j .[P δ -1 ]
The normalized cardinal matrix has following expression :

[BN δ ] k .[DN δ ] j
Cases (p>2, p\n) and p =2 are not studied here.

Case of hyperplanes, mixed unit and logarithmic meshes

The problems with a mix of variables are treated by simple multiplication of matrix using A δ , B δ , C δ and D δ with exponents adapted to the numbers of respective variables. Normalizations are made with afore-suggested routine.

Abundance factors for hyperplanes, unit mesh

Let us have :

x 1 + x 2 + … + x k = c (27)
There is no lower cases p > 2 and p = 2. They are worked out together.

Here x i are positive integers.

It is case d = 1.

We studied in preceding exercise matrix A δ=1,d=1 and found μ 0 = p, μ 1 = 0, hence we can deduce eigenvalues of A δ,d=1 : μ 0 = p → μ 0,δ = p δ μ 1 = 0 → μ 1,δ = 0.p δ-1 , μ 2,δ = 0.p δ-2 , …, μ δ,δ = 0.p 0

We have then (p>2 and p=2) deduced from general model of A δ,d :

P 164/390 Modulo p Δ abundance factors correction Matrix A δ,d=1 1 p-1 p(p-1) p 2 (p-1) … p δ-2 (p-1) p δ-1 (p-1) 1 p-1 p(p-1) p 2 (p-1) … p δ-2 (p-1) p δ-1 (p-1) … … … … … … … 1 p-1 p(p-1) p 2 (p-1) … p δ-2 (p-1) p δ-1 (p-1)
Eigenvalues of A δ,d=1 , followed by eigenvectors matrix P δ,d=1

p δ 0 0 0 … 0 0 1 p-1 p-1 p-1 … p-1 p-1 1 p-1 p-1 p-1 … p-1 -1 1 p-1 p-1 p-1 … -1 0 … … … … … … 1 p-1 p-1 -1 … 0 0 1 p-1 -1 0 … 0 0 1 -1 0 0 … 0 0 Inverse matrix P -1 δ,d=1 1/p δ (p-1)/p δ (p-1)/p δ-1 (p-1)/p δ-2 … (p-1)/p 2 (p-1)/p 1/p δ (p-1)/p δ (p-1)/p δ-1 (p-1)/p δ-2 … (p-1)/p 2 -1/p 1/p δ-1 (p-1)/p δ-1 (p-1)/p δ-2 (p-1)/p δ-3 … -1/p 0 1/p δ-2 (p-1)/p δ-2 (p-1)/p δ-3 (p-1)/p δ-4 … 0 0 1/p δ-3 (p-1)/p δ-3 (p-1)/p δ-4 (p-1)/p δ-5 … 0 0 … … … … … … … 1/p 3 (p-1)/p 3 (p-1)/p 2 -1/p … 0 0 1/p 2 (p-1)/p 2 -1/p 0 … 0 0 1/p -1/p 0 0 … 0 0
Let us note the first line and second line identity except for last terms.

It is easy to verify that we get here normalized abundance factors all equal to 1, awaited result identical to case δ=1.

Abundance factors for hyperplanes, logarithmic mesh

y 1 + y 2 + … + y m = c ( 28
)
There is no lower cases p > 2 and p = 2. They are worked out together.

Here y i are positive prime numbers.

As previously, we can anticipate cardinal matrices B δ,d=1 eigenvalues. For matrix B δ=1,d=1 , we had μ 0 = p-1, μ 1 = -1, hence :

μ 0 = p-1 → μ 0,δ = p δ-1 .(p-1) μ 1 = -1 → μ 1,δ = (-1).p δ-1 μ 2,δ = 0, …, μ δ,δ = 0
We have :

Matrix B δ,d=1 0 0 0 0 … 0 p δ-1 (p-1) 0 0 0 0 … 0 p δ-1 (p-1) … … … … … … … 0 0 0 0 0 p δ-1 (p-1) 1 p-1 p(p-1) p 2 (p-1) … p δ-2 (p-1) p δ-1 (p-2)
Eigenvalues of B δ,d=1 , followed by eigenvectors matrix P δ,d=1

P 165/390

Modulo p Δ abundance factors correction

p δ-1 .(p-1) -p δ-1 0 0 … 0 0 1 p-1 p-1 p-1 … p-1 p-1 1 p-1 p-1 p-1 … p-1 -1 1 p-1 p-1 p-1 … -1 0 … … … … … … 1 p-1 p-1 -1 … 0 0 1 p-1 -1 0 … 0 0 1 -1 0 0 … 0 0 Inverse matrix P -1 δ,d=1 1/p δ (p-1)/p δ (p-1)/p δ-1 (p-1)/p δ-2 … (p-1)/p 2 (p-1)/p 1/p δ (p-1)/p δ (p-1)/p δ-1 (p-1)/p δ-2 … (p-1)/p 2 -1/p 1/p δ-1 (p-1)/p δ-1 (p-1)/p δ-2 (p-1)/p δ-3 … -1/p 0 1/p δ-2 (p-1)/p δ-2 (p-1)/p δ-3 (p-1)/p δ-4 … 0 0 1/p δ-3 (p-1)/p δ-3 (p-1)/p δ-4 (p-1)/p δ-5 … 0 0 … … … … … … … 1/p 3 (p-1)/p 3 (p-1)/p 2 -1/p … 0 0 1/p 2 (p-1)/p 2 -1/p 0 … 0 0 1/p -1/p 0 0 … 0 0
Then we use :

[B δ ] m = [P δ ].[λ δ ] m .[P δ -1 ] [BN δ ] = (1/(p δ-1 .(p-1))).[B δ ]
to write normalized abundance factors (here [p δ ] is column vector (p δ , 0, 0, …, 0)) :

[BN δ ] m [p δ ] = (1/(p δ-1 .(p-1))) m .[P δ ].[λ δ ] m .[P δ -1 ].[p δ ]
All calculations done, we get, for this expression, the column vector :

(1-(-1) m-1 /(p-1) m-1 ) … (1-(-1) m-1 /(p-1) m-1 ) (1-(-1) m /(p-1) m )
All lines of this vector are identical except the last one. The first line relates to targets c such as c = k.p δ , the second such as c = k.p δ-1 , etc., the last such as c = k.p 0 (exclusive character of writing is understood here). Hence :

Fan(c,p,δ) = П (1- (-1) m ) П (1- (-1) m-1 ) (29) (p i -1) m (p i -1) m-1 p ∤ c p ∖ c
This result is in any point identical to that obtained in case δ = 1.

We get the remarkable equality :

Fan(c,p,δ)= Fan(c,p,1) (30)
This confirm mathematical literature expressions for :

-m = 2 : (Goldbach) -m = 3 : Vinogradov -etc. There is no signs incidence on the abundance factors in these cases (n = 1 odd).

Abundance factors for hyperplanes, logarithmic mesh, with negative signs

This confirm mathematical literature expressions for :

P 166/390

Modulo p Δ abundance factors correction -m = 2 : De Polignac, twin prime numbers, and so on.

-etc.

Abundance factors of Waring, unit mesh

We study : Cases where n is even are complex in their evaluation because of imaginary components. Thus, we work first on cases where n is odd with aim of extrapolating results.

x 1 n + x 2 n + … +x i n + … + x k n = c modulo p δ (

Case n = 3

We can draw up a table with the relative contributions of eigenvalues and of eigenvectors matrices P and P -1 , that is :

P -1 2 0 2 0 2 1/2 2 1 2 3/2 2 2 2 5/2 2 3 2 7/2 2 4 2 9/2 … λ k 2 k.δ 0 2 k.(δ-1) 2 k.(δ-1) 0 2 k.(δ-2) 2 k.(δ-2) 0 2 k.(δ-3) 2 k.(δ-3) 0 … P c\2 0 and c∤2 1 1 -1 c\2 1 and c∤2 2 1 1 -1.2 1/2 c\2 2 and c∤2 3 1 1 1.2 1/2 -1.2 1 c\2 3 and c∤2 4 1 1 1.2 1/2 1.2 1 -1.2 3/2 c\2 4 and c∤2 5 1 1 1.2 1/2 1.2 1 1.2 3/2 -1.2 2 c\2 5 and c∤2 6 1 1 1.2 1/2 1.2 1 1.2 3/2 1.2 2 -1.2 3/2 c\2 6 and c∤2 7 1 1 1.2 1/2 1.2 1 1.2 3/2 1.2 2 1.2 3/2 -1.2 3 c\2 7 and c∤2 8 1 1 1.2 1/2 1.2 1 1.2 3/2 1.2 2 1.2 3/2 1.2 3 -1.2 7/2 c\2 8 and c∤2 9 1 1 1.2 1/2 1.2 1 1.2 3/2 1.2 2 1.2 3/2 1.2 3 .2 7/2 -1.2 4 c\2 9 and c∤2 10 1 1 1.2 1/2 1.2 1 1.2 3/2 1.2 2 1.2 3/2 1.2 3 .2 7/2 .2 4 -1. 2 9/2 … … … … … … … … … … … … …
By readjusting eigenvectors by adapted multiplicative constants, P and P -1 contribute also as follows (general form case) : 4) … P c\2 0 and c∤2 1 1 -1

P -1 2 0 2 0 2 1 2 2 2 3 2 4 2 5 2 6 2 7 2 8 2 9 2 10 2 11 … λ k 2 k.δ 0 2 k.(δ-1) 2 k.(δ-1) 0 2 k.(δ-2) 2 k.(δ-2) 0 2 k.(δ-3) 2 k.(δ-3) 0 2 k.(δ-4) 2 k.(δ-
c\2 1 and c∤2 2 1 1 -1 c\2 2 and c∤2 3 1 1 1 -1 c\2 3 and c∤2 4 1 1 1 1 -1 c\2 4 and c∤2 5 1 1 1 1 1 -1 c\2 5 and c∤2 6 1 1 -1 … … … … … … … … … … … … … … …
Let us pose it = int(i/3).

We get for line c\2 i and c∤2 i+1 : 1) .2 k.(δ-it) +r So that : P 167/390 Modulo p Δ abundance factors correction it # c = 2 k.δ .(1+3/4.∑ 2 (3-k).j )+r/2 k.δ ) j = 1 or # c = 2 k.δ .(1+(6/2 k ).if(k=3,it,((2 (3-k).it -1)/(2 (3-k) -1)))+r/2 k.δ )

# c = 2 k.δ +(2 1 +2 2 ).2 k.(δ-1) +(2 1 +2 2 ).2 3 .2 k.(δ-2) +(2 1 +2 2 ).2 6 .2 k.(δ-3) +…+(2 1 +2 2 ).2 3.(it-
Three cases are to be considered for last term of this arithmetic expression :

if i = 0 mod 3 r = 0 if i = 1 mod 3 r = (-2 i ).2 k.(δ-it-1) = (-2 i+1-i mod 3 ).2 k.(δ-it-1) if i = 2 mod 3 r = (2 i-1 -2 i ).2 k.(δ-it-1) = (-2 i+1-i mod 3 ).2 k.(δ-it-1)
That gives normalized factors for line c\2 i and c∤2 i+1 :

Fan(c,2,n=3) = 1+(3/2 k-1
).if(k=3,it,((2 (3-k).it -1)/(2 (3-k) -1)))+if(i=0mod3,0,if(i=1mod3,-2 i .2 i-k.(it+1) ,-2 i-1-k.(it+1) ))

Case n odd

The results for n = 3 can acquire general character. The eigenvalues are built on modulo n basis (instead of mod 3) as follows :

multiplicity 2 0 2 0 2 1 2 n-1 2 n 2 n+1 2 2n-1 … 2 (i-1).n 2 (i-1).n+1 … 2 i.n-1 … λ 2 δ 0 2 (δ-1) … 2 (δ-1) 0 2 (δ-2) … 2 (δ-2) … 0 2 (δ-i) … 2 (δ-i) … 1 zero and 2 n -2 values 2 (δ-1)
2 n zeros and 2 2..n -2 n+1 values 2 (δ-1) …

2 (i-1).n zeros and 2 i..n -2 i..n-n+1 values 2 (δ-i)

…

We can draw up a table with the relative contributions of eigenvalues and of eigenvectors matrices P and P -1 :

P -1 2 0 2 0 2 1 2 n-1 2 n 2 n+1 2 2n-1 2 2n 2 2n+1 … 2 δ-3 2 δ-2 2 δ-1 λ k 2 k.δ 0 2 k.(δ-1) … 2 k.(δ-1) 0 2 k.(δ-2) … 2 k.(δ-2) 0 2 k.(δ-3) … … … P c\2 0 and c∤2 1 1 -1 c\2 1 and c∤2 2 1 1 -1 … … … … … c\2 n-1 and c∤2 n 1 1 1 … -1 c\2 n and c∤2 n+1 1 1 1 … 1 -1 c\2 n+1 and c∤2 n+2 1 1 1 … 1 1 -1 … … … … … … … … … c\2 2n-1 and c∤2 2n 1 1 1 1 1 1 -1 c\2 2n and c∤2 2n+1 1 1 1 1 1 1 1 -1 c\2 2n+1 and c∤2 2n+2 1 1 1 1 1 1 1 1 -1 … … … … … … … … … … … c\2 δ-2 and c∤2 δ-1
[P A(δ) ] was written at the line corresponding to [c]=[p i.n+j .g r .g u.d ] (with δ>>i.n+j) :

[U"] (p-1).[U] (p-1).[U] … (p-1).[U] (p-1).[U] (p-1).[U]

…. … [λ'*] [0] … [0]
Here expression [λ'*] is on column i.n+j+2.

Concerning [P -1 A(δ) ].[p δ ],
it is simply the first column of matrix [P -1 A(δ) ] multiplied by p δ , so that we get column vector :

[1] 1/d.[U"] p/d.[U"] p 2 /d.[U"] … … … … … … … p δ-2 /d.[U"] p δ-1 /d.[U"]
Then let us put in correspondence these three vectors in the order of matrix multiplication :

1 2 3 … n+1 … (i-1).n+2 (i-1).n+3 … (i-1).n+n+1 i.n+2 i.n+3 … i.n+j+1 i.n+j+2 i.n+j+3 1 0 1 … n-1 … 0 1 … n-1 0 1 … j-1 j j+1 [U"] (p-1).[U] (p-1).[U] … (p-1).[U] … (p-1).[U] (p-1).[U] … (p-1).[U] if(j=0,[λ'*], (p-1).[U]) if(j=0,[0], (p-1).[U]) … if(j=0,[0], (p-1).[U]) [λ'*] [0] … 1 [μ'/p] k [I/p] k … [I/p] k … [μ'/p i ] k [I/p i ] k … [I/p i ] k [μ'/p i+1 ] k [I/p i+1 ] k … [I/p i+1 ] k [I/p i+1 ] k [I/p i+1 ] k 1 1/d.[U"] p/d.[U"] p n-1 /d.[U"] … p (i-1).n /d.[U"] p (i-1).n+1 /d.[U"] … p (i-1).n+n-1 /d.[U"] p i.n /d.[U"] p i.n+1 /d.[U"] … p i..n+j-1 /d.[U"] p i..n+j /d.[U"] p i.n+j+1 /d.[U"]
We have of course j ≤ n-1.

Matrices multiplication [P δ ].[λ δ /(p δ )] k leads, for the line corresponding to target p i..n+j .g r .g u.d , to the line vector : ).[I/p] k +p 1.n .[μ'/p 2 ] k +(p n+1 +…+p 2n-1 ).[I/p 2 ] k +… +p (i-1).n .[μ'/p i ] k +(p (i-1).n+1 +…+p (i-1).n+n-1 ).[I/p i ] k ) +if(j≤1, [0], (p i.n+1 +…+p i.n+j-1 ).[I/p (i+1) ] k )

([U''], (p-1).[U].[μ'/p] k ,(p-1).[U].[I/p] k , …, (p-1).[U].[I/p] k , 1+(n-1) blocks if i ≥ 1 (p-1).[U].[μ'/p 2 ] k ,(p-1).[U].[I/p 2 ] k , …, (p-1).[U].[I/p 2 ] k , 1+(n-1) blocks if i ≥ 2 … (p-1).[U].[μ'/p i ] k ,(p-1).[U].[I/p i ] k , …, (p-1).[U].[I/p i ] k , 1+(n-1) blocks (last current term) if(j=0,[ λ'*],(p-1).[U]).[μ'/p (i+1) ] k , (p-1).[U].[I/p (i+1) ] k , …, (p-1).[U].[I/p (i+1) ] k , 1+(j-1) blocks (j-1 blocs if j ≥ 2) if(j=0,[0],[λ'*].[μ'/p (i+1) ] k , 1 block if j ≠ 0 ,[0]…[0]) Complement null
).

[U''] +if(j=0, (p i.n /d)[λ'*].[μ'/p i+1 ] k .[U''], (p i.n+j /d).[λ'*].[I/p (i+1) ] k .[U'']) +if(j=0,[0], [U].(p-1).[μ'/p (i+1) ] k .[U"].p i..n /d)
Some terms are rearranged to give :

[U''] +[U].((p-1)/d).( +if(i=0, [0], (1+p 1.(n-k) +…+p (i-1).(n-k) ).([μ'/p] k +(p n+1 +…+p 2n-1 ).[I/p] k ) ) +if(j≤1, [0], p i.(n-k)+1 .((p j-1 -1)/(p-1)).[I/p] k ) ).[U''] +if(j=0, (p i.(n-k) /d)[λ'*].[μ'/p] k .[U''], (p i.(n-k)+j-k /d).[λ'*].[U'']) + p i..(n-k) .((p-1)/d).if(j=0, [0], [U].[μ'/p] k .[U"])
We distinguish case n = k using the equality 1+p 1.(n-k) +…+p (i-1).(n-k) = if(n = k, i, (p i.(n-k) -1)/(p n-k -1)) in order to write :

[U''] +[U].(1/d).( +if(i=0, [0], if(n = k, i, (p i.(n-k) -1)/(p n-k -1)).((p-1).[μ'/p] k +(p n -p).[I/p] k )) +if(j=0, [0], p i.(n-k)+1-k .(p j-1 -1)) P 170/390 Modulo p Δ abundance factors correction ).[U''] +(1/d).p i.(n-k) .if(j=0, [λ'*].[μ'/p] k .[U''], p j-k .[λ'*].[U'']+(p-1)[U].[μ'/p] k .[U"])
Condition on j has being made uniform as for j = 1 we have p j-1 -1 = 0. Then, we use the equalities

[U][U''] = d.[U''], [λ'].[U''] = [λ'*].[U''] = -d.[U''] and [U].[μ'/p] k = ∑(μ'/p) k .
[I], with sum bearing on d terms, μ 0 = p being rejected, to get :

(1 +(1/d).if(i=0, [0], if(n = k, i, (p i.(n-k) -1)/(p n-k -1

)).((p-1).∑(μ'/p) k .[I]+(p n -p).[I/p

] k ) +(1/d).p i.(n-k) .(p j-k -p 1-k ).if(j=0, [0], [I]) +p i.(n-k) .if(j=0, [λ'*].[μ'/p] k /d, -p j-k .[I]+((p-1)/d).∑(μ'/p) k .[I]) ).[U'']
Hence, while gathering conditions on j and making some adaptations on conditions on i :

[fan δ=∞ (c = p i.n+j .g r .g

u.d )] = (1 +(1/d) . if(n = k, i, (p i.(n-k) -1)/(p n-k -1)) . if(i=0, [0], (p-1).∑(μ'/p) k .[I]+(p n -p).[I/p] k ) +(1/d) . p i.(n-k) . if(j=0, [λ'*].[μ'/p] k , (p-1).∑(μ'/p) k .[I] +p.((1-d).p j-1 -1).[I/p] k ) ).[U'']
As we already mentioned above, this expression concerns matrix blocks of d lines. The conditional value for j = 0 is the only one to be differentiated. The remainder of expression has equal value for the other d-1 lines in question.

We draw from that the following cases ([1] = [U''] is a column vector whose components are 1) :

n° Conditions

[fan δ=∞ (c,p)] (if j = 0) or fan δ=∞ (c,p) (if j ≠ 0) 1 i = 0, j = 0 [1]+(1/d).[λ'*].[μ'/p] k .[1] 2 i = 0, j > 1 i = 0, j =1 (1+p j-k .(1-d-p -j+1 )+((p-1)/d).∑ [μ'/p] k ).[1] (1-d.p 1-k +((p-1)/d).∑ [μ'/p] k ).[1]
k +(1-p 1-k )))).[1] +[λ'*].[μ'/p] k ).[1] 6 i ≠ 0, j > 0, n = k (1+(1/d).((i+1).(p-1).∑(μ'/p) k +i-(i+1).p 1-k +(1-d).p j-k )).[1] 7 n << k ≈ [1]

Consequences

Let us return to per blocks condensed form of δ = 1 case (cf. relation (157) at exercise 5) :

Fan(c,p,δ=1) = 1+((p-1)/d).Σ(μ'/p) k [U'']+(1/d).[λ'*].[μ'/p] k .[U'']
The immediate result is that we get the same abundance factor, at sequence p, with δ = 1 and with any δ (and thus δ = +∞) for any target c not-multiple of p (case i = j = 0) : p∤c => Fan(c,p,δ→∞) = Fan(c,p,δ=1)

Otherwise, a correction intervenes, correction which depends on multiplicity of factor p in c. The correction is ratio (relation (34)) : fac lines r (c,p,δ→∞) = (1+(1/d) . if(n = k, i, (p i.(n-k) -1)/(p n-k -1)) . if(i=0, [0], (p-1).∑(μ'/p) k .

[I]+(p n -p).[I/p] k ) +(1/d) . p i.(n-k) . if(j=0, [λ'*].[μ'/p] k , (p-1).∑(μ'/p) k .[I] +p.((1-d).p j-1 -1).[I/p] k )).[U''] lines r / ([U'']+(1/d).[λ'*].[μ'/p] k .[U'']) lines r
We also notice the importance of the difference n-k. As μ i < p (i > 0), finite sum Σ(μ'/p) k tends towards 0 when k increases and we have, as well for i = 0 as for i ≠ 0, for k >> n, fan δ=∞,k>>n (c,p) ≈ 1

This point is, of course, foreseeable without any calculation.

Otherwise, the ratio moves away from 1 by several decades when k < n.

fan δ=∞,k<<n (c,p) >> 1 (36)

For n = k, we get a decade like ratio.

Capital notice

The simplest, but longest basic method, for abundance factors evaluation is use of exercise 2 algorithm. Pending on δ choice, relative values of abundance factors are not fixed. This stabilizes only when the dimension of the matrix δn.n+1 is equal or higher then i.n+j+2, that is when : δn.n > in+j Abundance factors of some compound target (not prime) can thus be difficult to calculate with precision with this simplest means.

Case n = 2

We can calculate the factor in an explicit literal form in this case. We get :

If p = 1 mod 4 μ 1A = -√p, μ 2A = √p, ∑(μ'/p) k = (1/p k/2 ).(1+(-1) k ) [λ'*] = -1-√p -1+√p -1+√p -1-√p [μ'/p] k = 1 (-1) k 0 p k/2 0 1 So that : [λ'*].[μ'/p] k .[U''] = 1 -1-√p -1+√p (-1) k 0 1 p k/2 -1+√p -1-√p 0 1 1 Thus : [λ'*].[μ'/p] k .[U''] =
1 if(k= 0 mod 2,-1,√p) p k/2 if(k= 0 mod 2,-1,-√p) Then :

[fan(c,p,δ=1)] = 1+if(k= 0 mod 2, -p -k/2 , p (1-k)/2 ) 1+if(k= 0 mod 2, -p -k/2 , -p (1-k)/2 )

If p = 3 mod 4 μ 1A = i.√p μ 2A = -i.√p, ∑(μ'/p) k = (i/p 1/2 ) k .(1+(-1) k ) [λ'*] = -1-i.√p -1+i.√p -1+i.√p -1-i.√p [μ'/p] k = (i) k 1 0 p k/2 0 (-1) k So that : [λ'*].[μ'/p] k .[U''] = (i) k -1-i.√p -1+i.√p 1 0 1 p k/2 -1+i.√p -1-i.√p 0 (-1) k 1 Thus : [λ'*].[μ'/p] k .[U''] = 2 if(k= 0 mod 2,-(-1) k/2 ,-(-1) (k+1)/2 .√p) p k/2 if(k= 0 mod 2,-(-1) k/2 ,(-1) (k+1)/2 .√p) Then : [fan(c,p,δ=1)] = 1+ if(k= 0 mod 2, -(-p) -k/2 , (-p) (1-k)/2 ) 1+ if(k= 0 mod 2, -(-p) -k/2 , -(-p) (1-k)/2 )
The abundance factor is obtained by substitution in the literal expression presented earlier.

P 172/390

Modulo p Δ abundance factors correction

[fan δ=∞ (c,p, δ→∞)] = [1] +((p -i.(k-2) /d).if(j=0, [ λ '*].[μ'/p] k .[1], (p j-k .(1-d-p -j+1 )+(p-1).∑ [μ'/p] k ).[1]) +if(i=0,[0],p -(k-2) .((p-1)/d).( (if(k=2, i-1, (1-p -(i-1).(k-2) )/(1-p -(k-2) ))).∑ [μ'/p] k .[1]+((1-p -1 ).if(n=k, i, (1-p i.(2-k) )/(1-p -(k-2) ))).[1]))

Case c = 0

First line evaluation (correspondent to p δ ) is similar except that we have no conditional term and that expression does not concern blocks of matrices but single components. The table to be considered is :

1 2 3 … n+1 … i.n+2 i.n+3 … i.n+j+1 … 1 0 1 … n-1 … 0 1 … n-1 … [1] (p-1).[U'] (p-1).[U'] … (p-1).[U'] … (p-1).[U'] (p-1).[U'] … (p-1).[U'] … 1 [μ'/p] k [I/p] k … [I/p] k … [μ'/p i+1 ] k [I/p i+1 ] k … [I/p i+1 ] k … 1 1/d.[U"] p/d.[U"] p n-1 /d.[U"] … p i.n /d.[U"] p i.n+1 /d.[U"] … p i..n+n-1 /d.[U"] …
The type c = 0 mod p δ target is taken into account by the preceding case when using a matrix of dimension greater than δn+1. Thus, the preceding table interests indeed only target c =0.

Capital notice

Unlike in case c ≠ 0, above table can be used ad infinitum. The calculation, by data processing, of abundance factors by standard algorithm (cf. exercise 2) will be always approximate for target 0, because ratios between target 0 and targets p i vary when δ increases as new appearances of targets p i arise (higher degrees i).

Let us calculate actual limit of the abundance factor in Waring sums case. 1.n .[μ'/p 2 ] k +(p n+1 +…+p 2n-1 ).[I/p 2 ] k +… +p i.n .[μ'/p i+1 ] k +(p i.n+1 +…+p i.n+n-1 ).[I/p i+1 ] k +..

1 +[U'].((p-1)/d).( p 0.n .[μ'/p] k +(p 1 +…+p n-1 ).[I/p] k +p

. ).[U'']

It follows after rearrangements of terms :

1

+[U'].((p-1)/d).( (1+p (n-k) +...+p i.(n-k) ).[μ'/p] k +(p 1 +…+p n-1 ).(1+p n-k +…+p i.(n-k) ).[I/p] k +... ).[U''] So that : 1 +[U'].((p-1)/d).( (1+p (n-k) +...+p i.(n-k) ).([μ'/p] k +(p 1 +…+p n-1 ).[I/p] k ) +... ).[U''] That is : 1 +[U'].((p-1)/d).( +if(n=k, (i+1), (1-p (i+1).(n-k) )/(1-p n-k )).([μ'/p] k +p n-k-1 .((1-p -n+1 )/(1-p -1

)).[I])

).

[U'']

Let us recall that [U'], respectively [U''], is a line, respectively column, matrix with d components all equal to 1. Hence the expression : 1 +((p-1)/d).( +if(n=k, (i+1), (1-p (i+1).(n-k) )/(1-p n-k )).(∑ [μ'/p] k +d.p n-k-1 .((1-p -n+1 )/(1-p -1 )))

)

Target c = 0 abundance factor is obtained while tending i towards infinite P 173/390

Modulo p Δ abundance factors correction fan δ=∞ (0) = 1 + ((p-1)/d) . if (n=k, (i+1), (1-p (i+1).(n-k) )/(1-p n-k )) . (∑ (μ'/p) k +d.p n-k . (1-p -n+1 )) i→∞ Hence the summary table :   n° Conditions fan δ=∞ (c,p)

1 n > k ∞ 2 n = k ∞ 3 n < k 1+((p-1)/d).(1/(1-p -(k-n) )).(∑ (μ'/p) k +d.p -(k-n) .(1-p -(n-1) ))
When k >> n, the last term tends towards 1+((p-1)/d).∑ (μ'/p) k ) which is the expression of case δ =1. The ratio is given by : fac(0,p,δ→∞) = 1+(p-1).

(

/(1-p -(k-n) )-1).∑(μ'/p) k /d + (1/(1-p -(k-n) )).(1-p -(n-1) )).p -(k-n) 1+((p-1)/d).Σ(μ'/p) k 1
which we can write somewhat more clearly :

( 1 -1).∑(μ'/p) k + d.(1-1/p (n-1) ) fac(0,p,δ→∞) = 1+ (p-1) . 1-1/p (k-n) p (k-n) .(1-1/p (k-n) ) (37) d 1+((p-1)/d).Σ(μ'/p) k
This ratio, as we said earlier, tends towards 1 when k >> n (the more likely as p is greater) : fac δ=∞,k>>n (0,p) ≈ 1 (38)

Once again, this last result is quite practical.

Abundance factors of Waring, logarithmic mesh

The object of the study is : After normalization we get :

even … [(1+(-1) m )/2] [(1+(-1) m-1 )/2] … odd … [(1+(-1) m-1 )/2] [(1+(-1) m )/2] … So that : For m odd : 0 even [0] [1] … odd … [1] [0] … For m even : 0 even [1] [0] … odd … [0] [1] … P 174/390 Modulo p Δ abundance factors correction
The result is identical to case δ = 1

Case p > 2

We have a complete eigenvalues and eigenvectors expression in this general case. We use

[B δ ] m = [P δ ].[λ δ ] m .[P δ -1 ] [BN δ ] = (1/(p δ-1 .(p-1))).[B δ ]
to write normalized abundance factors (here [p δ ] is column vector (p δ , 0, 0, …, 0)) :

Fan(c,p,δ) = [BN δ ] m [p δ ] = [P δ ].[(λ δ /(p δ-1 .(p-1))] m .[P δ -1 ].[p δ ]
We calculate this expression using dimension d blocks of matrices, except for first line and first column which receive a special line with column treatment.

Case c ≠ 0

Let us have c = p i.n+j .g r (thus r-1 = n° of line in a block of B δ matrix of dimension d).

We must replace in the preceding expressions [μ'] by [λ'] and [I] by [0] (and k by m) getting following table :

1 2 3 … n+1 … (i-1).n+2 (i-1).n+3 … (i-1).n+n+1 i.n+2 i.n+3 … i.n+j+1 i.n+j+2 i.n+j+3 1 0 1 … n-1 … 0 1 … n-1 0 1 … j-1 j j+1 [U"] (p-

1).[U] (p-1).[U] … (p-1).[U] … (p-1).[U] (p-1).[U] … (p-1).[U] if(j=0, [ λ '*], (p-1).[U] (p-1).[U] … (p-1).[U]

[λ'*] [0] … 1 [λ'/p] m [0] … [0] … [λ'/p i ] m [0] … [0] [λ'/p i+1 ] m [0] … [0] [0] [0] 1 1/d.[U"] p/d.[U"] p n-1 /d.[U"] … p (i-1).n /d.[U"] p (i-1).n+1 /d.[U"] … p (i-1).n+n-1 /d.[U"] p i.n /d.[U"] p i.n+1 /d.[U"] … p i..n+j-1 /d.[U"] p i..n+j /d.[U"] p i.n+j+1 /d.[U"]
By multiplication, it follows :

[U''] +if(j=0, (p i.(n-m) /d)[ λ '*].[λ'/p] m , p i.(n-m) .((p-1)/d).[U].[λ'/p] m ).[U''] +[U].((p-1)/d).( +if(i=0, [0], [λ'/p] m +p n /p m .[λ'/p] m +...+p (i-1).n /p (i-1).m .[λ'/p] m
).[U''] Then :

[

fan δ=∞ (c)] = [U''] +(p i.(n-m) /d).if(j=0, [ λ '*].[λ'/p] m .[U''], (p-1).∑ [λ'/p] m .[U''] +if(i=0, [0], p (n-m) .((p-1)/d).(if(n=m, i, (1-p i.(n-m) )/(1-p n-m )).∑ [λ'/p] m ).[U'']
As we already mentioned above, this expression concerns blocks of d lines. Only the conditional value for j = 0 is differentiated. Thus, we get the following cases (with [U''] = [1]) :

n° Conditions

[fan δ=∞ (c,p)] (if j = 0) or fan δ=∞ (c,p) (if j ≠ 0) 1 i = 0, j = 0 [1]+(1/d).[ λ '*].[λ'/p] m .[1] 2 i = 0, j >0 1+(p-1).∑ [λ'/p] m )/d 3 i ≠ 0, j = 0, n ≠ m [1]+(p i.(n-m) /d).[ λ '*].[λ'/p] m .[1] +p (n-m) .((p-1)/d).((1-p i.(n-m) )/(1-p n-m )).∑ [λ'/p] m ).[1] 4 i ≠ 0, j > 0, n ≠ m 1+p n-m .((p-1)/d).∑ [λ'/p] m .( p (i-1).(n-m) +((1-p i.(n-m) )/(1-p n-m ))) 5 i ≠ 0, j = 0, n = m [1]+(p n-m . /d).( p (i-1).(n-m) .[ λ '*].[λ'/p] m .[1]+i.(p-1).∑ [λ'/p] m .[1]) 6 i ≠ 0, j > 0, n = m 1+((p-1)/d).p (n-m) .(i+p (i-1).(n-m) ).∑ [λ'/p] m 7 n << m ≈ [1]

Consequences

Let us return to per blocks condensed form in δ = 1 case (cf. relation ( 158 Again, the immediate result is that we get the same abundance factor, at sequence p, with δ = 1 and with any δ (and thus δ = +∞) for any target c not-multiple of p (case i = j = 0) : p∤c => Fan(c,p,δ→∞) = Fan(c,p,δ=1) (40)

Otherwise, a correction intervenes, correction which depends on multiplicity of factor p in c. The correction is ratio (relation ( 41)) :

fac lines r (c,p,δ→∞) = ([U''] +(p i.(n-m) /d).if(j=0, [ λ '*].[λ'/p] m ).[U''], (p-1).∑ [λ'/p] m .[U''] +if(i=0, [0], p (n-m) .((p-1)/d).(if(n=m, i, (1-p i.(n-m) )/(1-p n-m )).∑ [λ'/p] m ).[U'']))) lines r /([U'']+(1/d).[ λ '*].[μ'/p] k .[U'']) lines r
We notice again the importance of difference n-k. As μ i < p (i > 0), finite sum Σ(μ'/p) k tends towards 0 when k increases and we have, as well for i = 0 as for i ≠ 0, for k >> n, fan δ=∞,k>>n (c,p) ≈ 1 (42)

But the ratio moves away from 1 by several decades when k < n :

fan δ=∞,k<<n (c,p) >> 1 (43)

For n = k, we get a decade like ratio. (1-p (i+1).(n-m) ) (44) (1-p (n-m) ) i → +∞ Hence the following cases : n° Conditions fan δ=∞ (0,p)

1 n ≥ m ∞ 2 n < m 1+((p-1)/d).∑ [λ'/p] m . 1 (1-1/p (m-n) )
The reader can compare this last result to the case fan δ=1 (0,p) = 1+((p-1)/d).Σ(μ'/p) k . When m >> n, the normalized abundance factor modulo p tends towards the normalized abundance factor modulo p δ value.

Case n even

The remark and the conclusion are the same one as above.

Correction modulo p δ

Study modulo p  is carried out primarily to deduce abundance factors behaviour when  tends towards +∞. We wish to compared normalized enumerations modulo p  with normalized enumerations modulo p (when  tends towards +∞). Let us have target c for which we study this correction. Let us have prime factors decomposition : c = П p i j In general case, we build this table :

P 176/390 Modulo p Δ abundance factors correction p = 2 p = 3 … p = q p = +∞ infinite product c Fan(c,2) Fan(c,3) … Fan(c,q) … ∏ Fan(c)
The incidence (even if that can appear self-evident ) concerns only prime factors of c in the sought enumeration. Thus : ∞ ∞ ∞ Fan δ=∞ (c) = ∏ Fan δ=∞ (c,p) = Fan δ=∞ (c,2). ∏ Fan δ=∞ (c,p) = Fac δ→∞ (c,2).Fan δ=1 (c,2). ∏ Fac δ→∞ (c,p).Fan δ=1 (c,p) p=2 p=3 p=3

These terms are not difficult to find with good precision.

Fundamental results of correction modulo p δ study

The modulo p approach is the base of abundance factors evaluation (modulo p δ ). The factor of correction relates only to the divisors of target c, divisors which number is generally reduced (and finite in any case). If c is a prime number, only one calculation of correction is necessary.

When k increases, abundance factors will progressively tend towards 1. Otherwise, from now on, we are better equipped for treating cases like : n < k ≤ 2 n +1 and even like n ≥ k

Our point of view on the last case is that divergence between modulo p and modulo p ∞ factors also translates the amplitude of variations (undulations) around the asymptotic limit, translation which remains to be invented.

We will return at exercise 11 to modulo p δ resolution essential to most general cases.

EXERCISE 7 : ELEMENTARY OPERATORS

We seek matrix like operators associated with elementary operations for abundance factors evaluation. These operators are subjected to a precise framework because even an operation such as addition does not result in general case in a product of matrices (when overlapping variables prevail).

Addition of a linear prime numbers variable

We give the abundance factors of an expression R (x, …) = y + c starting from already found abundance factors of R (x, …) = c. Here (x, …) are indifferently integers or prime numbers variables, but y is a prime numbers variable. Incidentally, we pass thus from a non-asymptotic (c is fixed by advance) to an asymptotic problem. We reason initially on only one variable x of integers . For each sequence p i , the following table is set up :

+ \ - y 1 2 … p i -1 0 x R(x) 0 R(0) R(0)-1 R(0)-2 … R(0)-(p i -1) R(0) 1 R(1) R(1)-1 R(1)-2 … R(1)-(p i -1) R(1) 2 R(2) R(2)-1 R(2)-2 … R(2)-(p i -1) R(2) … … … … … … … p i -1 R(p i -1) R(p i -1)-1 R(p i -1)-2 … R(p i -1)-(p i -1) R(p i -1)
It is then a question of counting in this table the number of occurrences of target c modulo p i . If we add to the table (framed part) the column represented on his right, we get the same number of occurrences for targets c = 0 to p i -1, that is p i for each c. We get enumeration by cutting off cases resulting from right column (out of table). Thus at sequence p i , target c cardinal equals :

#c = p i -#{n / R(n) = c mod p i , n = 0 to p i -1}
With k integers' variables and m prime numbers variables, that is p k .(p-1) m occurrences for all targets (modulo p), we get, by the same arguments as previously, target c cardinal :

#c = p k .(p-1) m -#{(x 1 , …, x k ,y 1 , …, y m ) / R(x 1 , …, x k ,y 1 , …, y m ) = c mod p}
In matrix writing, if we have for R(x 1 , …, x k ,y 1 , …, y m ) = c mod p :

#{0} 1 #{g 0 .g u.cm } 0 #{g 1 .g u.cm } = [M] 0 … … #{g cm-1 .g u.cm } 0 it follows for R(x 1 , …, x k ,y 1 , …, y m ) = y+c mod p : #{0} 1 1 #{g 0 .g u.cm } 1 0 #{g 1 .g u.cm } = p k .(p-1) m . 1 -[M]. 0 (1) … … … #{g cm-1 .g u.cm } 1 0
We give at exercise 9 relation (31) a multiplicative expression by a matrix operator of the preceding relation.

Addition of a linear integers variable

We examine evolution of enumeration when crossing over from expression R (x1,…) = c to expression R (x1,…) = x+c.

Here x is an integers variable. By posing the problem in term of two-dimensional tables, we see immediately that the additional variable does not affect normalized abundance factor. This steadiness is used implicitly at exercise 5 relation (5) and in particular for Goldbach problem y 1 +y 2 = c+n (and y 1 +y 2 = c+2n) which has identical normalized abundance factors as problem y 1 +y 2 = c.

Matrix of permutation

Elementary permutation matrix

We will call elementary permutation matrix (of dimension d+1), the matrix derived from identity matrix [I] which contains one and only one non-null component equal to 1 per column and per line.

0 1 0 … … … 0 0 … 0 1 0 … 0 1 0 … … … … 0 [MP] = 0 0 1 0 … … 0 (2) 0 … … … 0 … 1 … … … … … … … 0 … … 0 1 … 0
Properties of elementary permutation matrices

[MP i ] = t [MP d-i ] [MP i ].[MP j ] = [MP k ] (3) [MP i ] -1 = [MP d-i ] = t [MP i ]
Indice i represents here line or column position of a given component. The product of two matrices of permutation is still a matrix of permutation.

Linear permutation matrix

We call linear permutation matrix [MOA i ] (from French designation), the elementary permutation matrix corresponding to a shift of i column of the trace identity matrix [I] except for first line and first column. The dimension 1+d of current matrix is implied.

1 0 … … 0 0 0 … … 0 1 0 … … … 0 … … 0 1 … … [MOA i ] = … … … 0 … … 0 … 0 (4) … 0 … … 0 … … … 1 1 0 … … 0 … … 0 0 … … … … … … … … … 1 0 … … … … 0 … … 0 1 0 … … 0
Properties of linear permutation matrices

[MOA i ] = t [MOA d-i ] [MOA i+j ] = [MOA i ].[MOA j ] (5) [MOA i ] -1 = [MOA d-i ] = t [MOA i ]
We justify the choice of name below.

Multiplication by an integer coefficient in Waring sums

General case (p ≠ 2)

We are interested in the evaluation of abundance factors for following equations :

a 1 x 1 n + a 2 x 2 n + … + a k x k n = c mod p or a 1 y 1 n + a 2 y 2 n + … + a m y m n = c mod p
Let us have g a primitive root of p and d = (n, p-1). We start with following remarks. If a divides p, then a = 0 mod p and a.x n = 0 mod p. The abundance factors are then all null except for c = 0 with #{0} = p (or p-1 in the case a.y n = 0 mod p). If a does not divide p, then there is i (i < d) and j such as a = g i .g j.d , thus a.x n = g i .g j.d .x n mod p. This enables us then to proceed to the usual method of two-dimensional tables with an, below illustrated, adequate correction. Let us have p = 13, n = 9 and g= 2. card 0 card 1 card 2 card 3 0 g 0d .g 0 g 1d .g 0 g 2d .g 0 g 3d .g 0 g 0d .g 0 g 1d .g 0 g 2d .g 0 g 3d .g 0 g 0d .g 0 g 1d .g 0 g 2d .g 0 g 3d .g 0 This table was that of a preceding example ((exercise 5).

Let us observe its evolution when we change the column generator : card 0 card 1 card 2 card 3 0 g 0d .g 0 g 1d .g 0 g 2d .g 0 g 3d .g 0 g 0d .g 0 g 1d .g 0 g 2d .g 0 g 3d .g 0 g 0d .g 0 g 1d .g 0 g 2d .g 0 g 3d .g 0 g 0d .g 0 g 1d .g 0 g 2d .g 0 g 3d .g 0 g 0d .g 0 g 1d .g 0 g 2d .g 0 g 3d .g 0 g 0d .g 0 g 1d .g 0 g 2d .g 0 g 3d .g 0 With the same matrix, we may calculate the abundance factors with the proviso of using a circulant shift of cardinals. More generally, we form the following table :   card card 1 card 2 … card d-1 card 0 g 0 .g 0d g 0 .g 1d … g 0 .g pi-1-d g 1 .g 0d g 1 .g 1d … g 1 .g pi-1-d gd -1 .g 0d gd -1 .g 1d … gd -1 .g pi-1-d 1 g i .0 d g i .g 0d g i .g 1d … g i .g ((pi-1)/d-1).d For the first line, the incidence of a = g i .g j.d is null (a.0 = 0). For the other lines, if c = g u.d + g j .g v.d , then g i .c = g i .g u.d + g i .g j .g v.d , which means exactly preceding example result. Let us have thus for an integers' variable (respectively a prime numbers' variable), u varying in the usual field of definition 0 to (p-1)/d-1 :

card' 0 card 0 card'{g i .g u.d } card {g i .g u.d } card'{g i+1 .g u.d } = [A] (or [B]) card {g i+1 .g u.d } … … card'{g i+d-1 .g u.d } card {g i+d-1 .g u.d }
In addition, by a trace vector of [I] shift of i columns, except for the first component, we have

card 0 1 0 … … 0 card 0 card {g i .g u.d } 0 0 1 card {g 0 .g u.d } card {g i+1 .g u.d } 0 1 card {g 1 .g u.d } card {g i+2 .g u.d } = … 0 … card {g 2 .g u.d } card {g i+3 .g u.d } … 0 1 card {g 3 .g u.d } … … … card {g i+d-1 .g u.d } 0 1 0 card {gd -1 .g u.d }
In the same way, by a shift of -i columns, we have

P 180/390 Elementary operators card' 0 1 0 … … 0 card' 0 card' {g 0 .g u.d } 0 0 1 card' {g i .g u.d } card' {g 1 .g u.d } 0 card' {g i+1 .g u.d } card' {g 2 .g u.d } = … 1 0 card' {g i+2 .g u.d } card' {g 3 .g u.d } … 1 0 card' {g i+3 .g u.d } … … … … card' {gd -1 .g u.d } 0 1 0 card' {g i+d-1 .g u.d } Hence: card' 0 1 0 … … 0 1 0 … … 0 card 0 card' {g 0 .g u.d } 0 0 1 0 0 1 card {g 0 .g u.d } card' {g 1 .g u.d } 0 0 1 card {g 1 .g u.d } card' {g 2 .g u.d } = … 1 0 [A] (or [B]) … 0 … card {g 2 .g u.d } (6) card' {g 3 .g u.d } … 1 0 … 0 1 card {g 3 .g u.d } … … … … … card' {gd -1 .g u.d } 0 1 0 0 1 0 card {gd -1 .g u.d }
The product of the three matrices gives the required transformation.

1 0 … … 0 1 0 … … 0 0 0 1 0 0 1 0 0 1 [A'] (or [B']) = … 1 0 [A] (or [B]) … 0 … … 1 0 … 0 1 … … … 0 1 0 0 1 0
For a 1 x 1 n + a 2 x 2 n + … + a k x k n = c mod p, respectively a 1 y 1 n + a 2 y 2 n + … + a m y m n = c mod p, we determine abundance factors by matrices products :

[A'] = [A 1 '].[A 2 ']…[A k '] and [B'] = [B 1 '].[B 2 ']…[B m ']
Let us notice that these matrices are commutative. In addition, combinations of integers and prime numbers variables are welcome.

Case d = 2 (p ≠ 2)

Initial data

We have here only three alternatives for abundance factors evaluation at sequence p, either a = 0 mod p, either a = g 0 .g 2u , or a = g 1 .g 2u . If a = 0 mod p, the resolution is immediate. If a = g 0 .g 2u , the cardinal matrix is that of case a = 1, therefore unchanged. If a = g 1 .g 2u , we have a priori a logical incidence on target c abundance factors which is a simple permutation between the previous values for c = g 0 .g 2u and c = g 1 .g 2u . We can verify this as follows :

1 0 0 x

Determinants

The determinants for eigenvalues research are the same ones as in the initial cases :

Det(M1-μI) = -μ 3 +p.μ 2 +p.μ-p 2 = -(μ-p)(μ 2 -p) = -(μ-p)(μ 2 +√p)(μ-√p) Det(M3-μI) = -μ 3 +p.μ 2 -p.μ+p 2 = -(μ-p)(μ 2 +p) = -(μ-p)(μ 2 +i√p)(μ-i√p)

Eigenvalues and eigenvectors

For p = 1 mod 4

Eigenvalues of M1 are μ 0M1 p μ 1M1 = √p μ 2M1
-√p and are associated to eigenvectors matrix :

1 (p-1)/ 2 (p-1)/2 P A = (1/p 1/2 ). 1 (-1-√p)/2 (-1+√p)/2 1 (-1+√p)/2 (-1-√p)/2
and inverse matrix P -1 A = P A Thus :

#{0} 1 (p-1)/ 2 (p-1)/2 p k 0 0 1 (p-1)/ 2 (p-1)/2 #{g 0 .g 2u } = (1/p) 1 (-1-√p)/2 (-1+√p)/2 0 p k/2 0 1 (-1-√p)/2 (-1+√p)/2 #{g 1 .g 2u } 1 (-1+√p)/2 (-1-√p)/2 0 0 (-1) k .p k/2 1 (-1+√p)/2 (-1-√p)/2
All calculations done, we get : #{0}

(1/p).(p k +(1/2).(p-1).p k/2 .(1+(-1) k ))

#{g 0 .g 2u } = (1/p).(p k +(1/2).p k/2 .(-(1+(-1) k )+p 1/2 .(-1+(-1) k )))

#{g 1 .g 2u } (1/p).(p k -(1/2).p k/2 .((1+(-1) k )+p 1/2 .(-1+(-1) k )))

Here #{g 0 .g 2u } and #{g 1 .g 2u } are actually the initial literal values of #{g 1 .g 2u } and #{g 0 .g 2u } as the reader will be able to check it while referring to exercise 5.

For p = 3 mod 4

Eigenvalues of M3 are :

μ 0M3 p μ 1M3 = -i.√p μ 2M3
i.√p and are associated to eigenvectors matrix : k/2 (1+(-1) k ) #{g 0 .g 2u } = (1/p).(p k -(1/2).p k/2 .((-1) k/2 (1+(-1) k )-p 1/2 .(-1) (k+1)/2 (1+(-1) k+1 ))

(8) #{g 1 .g 2u } (1/p).(p k -(1/2).p k/2 .((-1) k/2 .(1+(-1) k ))+p 1/2 .(-1) (k+1)/2 .(1+(-1) k+1 ))
Here, again, #{g 0 .g 2u } and #{g 1 .g 2u } are initial literal values of #{g 1 .g 2u } and #{g 0 .g 2u }.

P 182/390 Elementary operators Prime numbers variables cardinal matrices are treated in the same way and we observe the same permutation between #{g 0 .g 2u } and #{g 1 .g 2u }.

Foot-note : We find a fundamental result which is the possibility of choosing (as other choices exist of course) the same eigenvectors matrices for x n and for a.x n . It is a general point as we will still see it in this article (matrix of Fermat Catalan, quadratic matrices…).

Translations

Elementary translations

Translated polynomials

Let us consider the case of Waring sums on the basis of polynomial of type (x-c) n , c a constant integer. We want to get abundance factors in the case of an equation :

a 1 .(x 1 -c 1 ) n + a 2 .(x 2 -c 2 ) n + … + a k .(x k -c k ) n = c mod p.
The passage of x n to (x-c) n corresponds again to a translation of the set {0,1,… p-1} for each variable x i who requires again simply the recourse to a permutation matrix. We will take into account the two shifts r i and s i such as (with d = (n, p-1)) :

Equations Permutation matrices a i = g ri .gd .u mod p [MOA ri ] (9) c i = g si .g 2d.u mod p and permutation 0 with a residue [MP] Then, if initially we had [M], we get :

[M'] = [MOA -ri ].[M].[MOA ri ].[MP]
This operation is repeated for each member of the equation.

Equations of second degree

Let us have an expression u.x 2 +v.x+w. Its discriminant is Δ = v 2 -4.u.w. Then we have : u.x 2 +v.x+w = u.(x+v/2u) 2 -Δ/(4u)

This form is immediately usable for permutation matrices.

EXERCISE 8 : HARDY-LITTLEWOOD ABUNDANCE FACTORS

We enumerate, under the title of Hardy-Littlewood problems, the solutions of integers or prime numbers variables polynomial expressions :

P 1 (x 1 ) + P 2 (x 2 ) + … + P k (x k ) + Q 1 (y 1 ) + Q 2 (y 2 ) + … + Q m (y m ) = y + c.
The justification of abundance factors modulo p evaluation (and not modulo p ∞ ) is deferred to exercise 11.

Prime numbers generation by a polynomial

P(x) -y = c (1)
Here it is about a simple application of exercise 7 concerning the addition of a prime numbers variable : Thus, at sequence p i , target c cardinal is :

p i -#{(x) / P(x) = c mod p i , x = 0 to p i -1}
Then, normalized abundance factor for target c will be an infinite product, taking account of the number of integers' variables (that is 1) and of prime numbers variables (that is 1), implying multiplication by (1/p i 1 ).(1/(p i -1) 1 ).p i :

Fan(c) = ∞ П ( p i -#{ (x) / P(x) = c mod p i , x = 0 to p i -1} ) p i -1 p i then Fan(c) = ∞ П (1+ 1 (1 -#{ (x) / P(x) = c mod p i , x = 0 to p i -1})) (2) p i -1 p i
Foot-note: The result is identical with P(x)+y = c.

Prime numbers generation by several polynomials

P 1 (x 1 ) + P 2 (x 2 ) + … + P k (x k ) -y = c (3)
Here x i are integers' variables and y a prime numbers variable. The case is very close to the precedent one. We get, by adding to the last column on the right for the prime numbers variable, a table with k+1 dimensions including the same number of occurrences for targets from c = 0 to c = p i -1, that is p i k+1 /p i occurrences for each value of c. We get the enumeration by cutting off cases resulting from the column on the right (out of table). Thus at sequence p i , target c cardinal is : p i k -#{(x 1 , ..., x i , ..., x k ) / P 1 (x 1 ) + … + P i (x i ) + … + P k (x k ) = c mod p i , x i = 0 to p i -1, i = 1 to k} Then, normalized abundance factor for target c will be an infinite product, taking account of the number of integers' variables (that is k) and of prime numbers variables (that is 1), implying multiplication by (1/p i k ).(1/(p i -1) 1 ).p i :

Fan(c) = ∞ П ( p i k -#{(x 1 , ..., x i , ..., x k ) / P 1 (x 1 ) + … + P i (x i ) + … + P k (x k ) = c mod p i , x i = 0 to p i -1, i = 1 to k} ) p i k-1 (p i -1) p i then Fan(c) = ∞ k П (1+ 1 (1- #{(x 1 , ..., x i , ..., x k ) / ∑ P i (x i ) = c mod p i , x i = 0 to p i -1, i = 1 to k} )) (4) (p i -1) p i k-1 p i

Prime numbers generation by a monomial with unit coefficient

x n -y = c ( 5)

From preceding case we deduce immediately :

P 184/390

Hardy-Littlewood abundance factors

Fan(c) = ∞ П (1- 1 (1-#{(x) / x n = c mod p i , x = 0 to p i -1})) (p i -1) p i
To go further, we lean on the resolution of equation x n = c mod p i than we led at exercise 3. We know that this equation has 1 solution for x = 0 and d i = (n,p i -1) solutions when x satisfies Euler criterion (generalized to n > 2). Hence the result :

Fan(c) = ∞ П (1+ 1 (1-if(p i \c, 1, if(c (pi-1)/di =1 mod p i , d i , 0)) )) (6) (p i -1) p i

Prime numbers generation by a monomial

a.x n -y = c (7) The abundance factor

Fan(c) = ∞ П (1+ 1 (1-#{(x) / a.x n = c mod p i , x = 0 to p i -1})) (p i -1) p i equals Fan(c) = ∞ П (1+ 1 (1-if(et(p i \c, p i \a), p i , if(et(p i \c, p i ∤a), 0, if(c (pi-1)/di =1 mod p i , d i , 0))) )) (p i -1) p i that is also Fan(c) = ∞ П (1+ 1 (1-if(et(p i \c, p i \a), p i , if(c (pi-1)/di =1 mod p i , d i , 0)) )) (8) (p i -1) p i

Hardy-Littlewood problem

We examine the case of a second degree polynomial whose we seek to enumerate prime numbers production : ax²+bx-c = y Starting from already mentioned results, we have :

Fan(c) = ∞ П (1+ 1 (1 -#{ (x) / ax²+bx = c mod p i , x = 0 to p i -1})) (9) p i -1 p i
A polynomial of degree n (also while reasoning in congruence terms) has to more than n solutions. If x is a solution of a second degree equation mod p i , then p i -x is also solution. Thus, this type of equation has either 0 or 2 distinct solutions (except case p i =2). The problem is solved completely using quadratic residues theory (the reader will refer for example to [9]). Thus, using Legendre symbol and second degree polynomial discriminant Δ = b²+4ac, we get :

Prime numbers generation by Waring sum

Let us find following diophantine equation abundance factors :

x 1 n + x 2 n + … + x k n = p + c (13)
Here, x i are integers' variables and p a prime numbers variable. Immediate application of exercise 7 leads to the following abundance factors at sequence p i (relations 13) : c 0 gd.g 0 g 2d .g 0 … g (pi-1) .g 0 gd.g 1 g 2d .g 1 … g (pi-1) .g 1 gd.gd -1 g 2d .gd -1 … g (pi-1) .gd

-1 # p i k -u k p i k -v k p i k -w k … p i k -x k where k u k 1 v k 0 w k = A δ=1 0 … … x k 0
7. Prime numbers generation by squares sum

x 1 2 + x 2 2 + … + x k 2 = p + c (14)
The first degree being of no interest, we tackle the problem of squares here. We know that the enumeration can be carried out completely using the equations established at exercise 5. Thus :

For p i = 2 : We get #{0} = #{1} = 2 k -2 k-1 , hence #{0} = 2 k-1 (15) #{1} 2 k-1 For p i = 1 mod 4 : #{0} p i k -(1/p i ).(p i k +(1/2).(p i -1).p i k/2 .(1+(-1) k )) #{g 0 .g 2u } = p i k -(1/p i ).(p i k +(1/2).p i k/2 .(-(1+(-1) k )+p i 1/2 .(1-(-1) k ))) (16) #{g 1 .g 2u } p i k -(1/p i ).(p i k -(1/2).p i k/2 .((1+(-1) k )+p i 1/2 .(1-(-1) k )))
For p i = 3 mod 4 :

#{0} p i k -(1/p i ).(p i k +(1/2).(p i -1).p i k/2 .if(mod(k,2)=0,2.(-1) k/2 ,0)) #{g 0 .g 2u } = p i k -(1/p i ).(p i k -(1/2).p i k/2 .(if(mod(k,2)=0,2.(-1) k/2 ,0)+p i 1/2 .if(mod(k+1,2)=0,2.(-1) (k+1)/2 ,0))) (17) #{g 1 .g 2u } p i k -(1/p i ).(p i k -(1/2).p i k/2 .(if(mod(k,2)=0,2.(-1) k/2 ,0)-p i 1/2 .if(mod(k+1,2)=0,2.(-1) (k+1)/2 ,0)))
Hence as examples :

For p i = 1 mod 4 k = 1 k = 2 k = 3 k = 4 #{0} p i -1 (p i -1) 2 p i 3 -p i 2 p i (p i 3 -p i 2 -p i +1) #{g 0 .g 2u } p i -2 p i 2 -p i +1 p i 3 -p i 2 -p i p i (p i 3 -p i 2 +1) #{g 1 .g 2u } p i p i 2 -p i +1 p i 3 -p i 2 +p i p i (p i 3 -p i 2 +1
)

For p i = 3 mod 4 k = 1 k = 2 k = 3 k = 4 #{0} p i -1 p i 2 -1 p i 3 -p i 2 p i (p i 3 -p i 2 -p i +1) #{g 0 .g 2u } p i -2 p i 2 -p i -1 p i 3 -p i 2 +p i p i (p i 3 -p i 2 +1) #{g 1 .g 2u } p i p i 2 -p i -1 p i 3 -p i 2 -p i p i (p i 3 -p i 2 +1) P 186/390
Hardy-Littlewood abundance factors

Prime numbers generation by a prime numbers powers sum

We study diophantine equation y 1 n + y 2 n + … + y m n = p + c (18) like previously. We get immediately abundance factors at sequence p i (relations 19) : c 0 gd.g 0 g 2d .g 0 … g (pi-1) .g 0 gd.g 1 g 2d .g 1 … g (pi-1) .g 1 gd.gd -1 g 2d .gd -1 … g (pi-1) .gd

-1 # (p i -1) m -u m (p i -1) m -v m (p i -1) m -w m … (p i -1) m -x m where m u m 1 v m 0 w m = B δ=1 0 … … x m 0
9. Prime numbers generation by prime numbers squares sum

y 1 2 + y 2 2 + … + y m 2 = p + c ( 20
)
The first degree having already been studied (variable p has same representative as -p, relation ( 5) of exercise 5 gives normalized factors expression of y 1 + y 2 + … + y m = p + c by replacing m by m+1), we tackle the problem of squares here.

We know that the enumeration can be carried out completely using the equations established at exercise 5. Thus, we have : or : #{0} (p i -1) m -(1/p i ).((p i -1) m +(p i -1).(p i +1) m/2 .cos(k.arcos (-1/(p i +1) 1/2 ))) #{g 0 .g 2u } = (p i -1) m -(1/p i ).((p i -1) m +(p i +1) (m+1)/2 .cos((m-1).arcos (-1/(p i +1) 1/2 ))) (24) #{g 1 .g 2u } (p i -1) m -(1/p i ).((p i -1) m +(p i +1) (m+1)/2 .cos((m+1).arcos (-1/(p i +1) 1/2 )))

For p i = 2 : #{0} = (1/2).(1-(-1) m ) (21) #{1} (1/2).(1+(-1) m ) For p i = 1 mod 4 : #{0} (p i -1) m -(1/p i ).((p i -1) m +(1/2).(p i -1).((p i 1/2 -1) m +(-1) m .(p i 1/2 +1) m )) #{g 0 .g 2u } = (p i -1) m -(1/p i ).((p i -1) m +(1/2).((p i 1/2 -1) m+1 +(-1) m+1 .(p i 1/2 +1) m+1 )) (22)
Hence as examples :

For p i = 1 mod 4 m = 1 m = 2 m = 3 m = 4 #{0} p i -1 (p i -1).(p i -3) (p i -1).(p i ²-3p i +6) (p i -1).(p i 3 -4p i ²+5p i -10) #{g 0 .g 2u } p i -3 p i ²-3p i +6 p i 3 -4p i ²+5p i -10 p i 4 -5p i 3 +10p i ²-5p i +15 #{g 1 .g 2u } p i -1 (p i -1).(p i -2) (p i -1).(p i ²-3p i +4) (p i -1).(p i 3 -4p i ²+6p i -7) For p i = 3 mod 4 m = 1 m = 2 m = 3 m = 4 #{0} p i -1 (p i -1)² p i .(p i -1).(p i -3) (p i -1).(p i 3 -4p i ²+5p i +2) #{g 0 .g 2u } p i -3 p i ²-3p i +4 (p i -1).(p i ²-3p i +4) p i 4 -5p i 3 +10p i ²-13p i +3 #{g 1 .g 2u } p i -1 p i .(p i -3) p i 3 -4p i ²+5p i +2 p i 4 -5p i 3 +10p i ²-5p i -5

EXERCISE 9 : FERMAT-CATALAN ABUNDANCE FACTORS

Sometimes a construction finds an ideal framework which gives him all its glare. It is the case of cardinal matrices method which also shows all its flexibility here. It makes it possible to consider innumerable situations where the reader will be able to exert his talents. We explore this vast subject, certain, sometimes secondary, sometimes essential, aspects remaining at the stage of conjectures. The developments are led modulo p (and not modulo p ∞ ) in order to simplify things. This approach is not systematically appropriate what we will specify at exercise 11. We leave to the reader the care to imagine additional methods and matrices to be set up in this case (on the basis of what was developed at exercise 6 giving all clues).

General framework

We are interested, without strict limit to it, to diophantine equations like :

x 1 (a1) + x 2 (a2) +…+ x k (ak) + y 1 (b1) + y 2 (b2) +…+ y m (bm) = c (1)
where (a 1 ), (a 2 ), …, (a k ) respectively (b 1 ), (b 2 ), …, (b m ) are any positive integers and x i and respectively y i integers and prime numbers variables. Generalization to linear expressions and other avatars of the basic expression is possible thanks to exercise 7.

The general framework rests on the existence of a cardinal matrix for each monomial, these monomials being separate. For each sequence p, we seek, by the usual two-dimensional table method, these cardinal matrices modulo p corresponding, pending on variables types, to monomials x i or y j from suggested diophantine equation. We note them [A] i,cm or [B] j,cm , as sequence p can be implied, indice cm being defined in the following paragraph. When i (or j) is equal to cm, we deal with cardinal matrices of the environment. They are fundamental because all the other matrices result from then very simply as we will see further. We will call the matrices, when i is a true divisor of cm (≠1 and ≠cm), Fermat-Catalan cardinal matrices.

Concept of environment

Let us have p a given sequence.

Let us have cm the lowest common multiple (lcm) of following numbers :

d i = (p-
We will call cm the environment.

Primitives equations

We seek the contribution of monomial z i (ai) with d =(p-1,a i ) in environment cm. As in the case studied at exercise 5, the two-dimensional table method leads immediately to equations using the primitives of p. We get still here matrices whose first line and the first column are to be distinguished from the other components. The difference between this case and that of exercise 5 resides quite simply on the exponentiation of one of the terms (g u.d instead of g u.cm ).

First line of matrix

The equation is :

c(1,y) = #{(u,v) \ 0 = g u.d + g y-2 .g v.cm mod p} (3)

First column of matrix

The equation is : c(x,1) = #{(u,v) \ g x-2 = g u.d mod p} (4)

Block of the reduced matrix Matrix components c(x,y) are located respectively with lines (x) and columns (y). For reduced matrix, co-ordinates x and y start at 2. Thus : c(x,y) = d . #{(u,v) \ g x-2 = g u.d + g y-2 .g v.cm mod p} (5) The definition domain of (u,v) is u = 0 to (p-1)/d-1, v = 0 to (p-1)/cm-1.

Let us note that : Reduced matrix characteristic equation is c(x,y) = d.#{(u,v) \ g x-2 = g u.d + g y-2 .g v.cm mod p} = d.#{(u,v) \ gd.g x+d-2 = gd.(g u.d + g y-2 .g v.cm ) mod p} = d.#{(u,v) \ g x+d-2 = g u.d+d + g y-2+d .g v.cm mod p} which is the required principal diagonal property mod d.

The addition of {0} in two-dimensional table involves, as in standard cardinal matrix case, the following relations:

[A(i)] = [B(i)] if i ≠ 1 [A(1)] = [B(1)] + [I] P 189/390
Fermat-Catalan abundance factors

Property of components average

Let us have [A] d,cm the Fermat-Catalan cardinal matrix of monomial x d in environment cm and let us have [A] cm,cm the standard cardinal matrix of monomial x cm in same environment for integers variable, [B] d,cm and [B] cm,cm being our matrices in prime numbers variable cases. The components of the first matrices ( [A] d,cm and [B] d,cm ) result from the seconds ( [B] cm,cm and [B] cm,cm ) by an average operation : c [A](d,cm) (i,j) = (d/cm).∑ c [A](cm,cm) (i+n.d,j+n.d) ( 9) n = 0 to cm/d-1 i>1, j>1 c [A](d,cm) (1,j) = (d/cm).∑ c [A](cm,cm) (1,j+n.d) (10) n = 0 to cm/d-1 j>1 c [A](d,cm) (i,1) = (d/cm).∑ c [A](cm,cm) (i+n.d,1) ( 11) n = 0 to cm/d-1 i>1 c [B](d,cm) (i,j) = (d/cm).∑ c [B](cm,cm) (i+n.d,j+n.d) ( 12) n = 0 to cm/d-1 i>1, j>1 c [B](d,cm) (1,j) = (d/cm).∑ c [B](cm,cm) (1,j+n.d) ( 13) n = 0 to cm/d-1 j>1 c [B](d,cm) (i,1) = (d/cm).∑ c [B](cm,cm) (i+n.d,1) ( 14) n = 0 to cm/d-1 i>1

Proof

For first line and first column components, the relations were proved by direct calculation in the preceding paragraph. For dimension d reduced matrices blocks, we must compare : d.#{(u,v) / g x-2 = g u.d + g y-2 .g v.cm mod p} and d.#{(u',v') / g x-2 = g u'.cm + g y-2 .g v'.cm mod p} Using the cardinals with positions in the diagonals distant of d steps, and with couple (x,y) given in advance, we get the following table :   Position in reduced matrix # (x,y) d.#{(u,v) / g x-2 = g u.cm + g y-2 .g v.cm mod p} (x+d mod cm,y+d mod cm) d.#{(u,v) / g x+d-2 = g u.cm + g y+d-2 .g v.cm mod p} (x+d mod cm,y+d mod cm) d.#{(u,v) / g x+2d-2 = g u.cm + g y+2d-2 .g v.cm mod p} … … (x+(cm/d-1).d mod cm,y+(cm/d-1).d mod cm) d.#{(u,v) / g x+(cm/d-1).d-2 = g u.cm + g y+(cm/d-1).d-2 .g v.cm mod p}



Position in reduced matrix # (x,y) d.#{(u,v) / g x-2 = g (u.cm/d).d + g y-2 .g v.cm mod p} (x+d mod cm,y+d mod cm) d.#{(u,v) / g x-2 = g (u.cm/d-1)..d + g y-2 .g v.cm mod p} (x+2d mod cm,y+2d mod cm) d.#{(u,v) / g x-2 = g (u.cm/d-2).d + g y-2 .g v.cm mod p} … … (x+(cm/d-1).d mod cm,y+(cm/d-1).d mod cm) d.#{(u,v) / g x-2 = g (u.cm/d-cm/d+1).d + g y-2 .g v.cm mod p}

The median terms of right-hand side column, including u, thus describe the set {g u'.d } what involves immediately the property (12).

Property of components splitting

Let us note temporarily the Fermat-Catalan cardinal matrices of monomial x d in environment d (standard cardinal matrix) and environment cm respectively [A (d,d) ] and [A (d,cm) ]. The first square matrix is of dimension d+1 and second is of dimension cm+1. let us have c A(d,d) (i,j) and c A(d,cm) (i,j) the components of these matrices. We get :

c A(d,d) (i,j) = ∑ c A(d,cm) (i,j+n.d) (15) n = 0 to cm/d-1 i>1, j>1 c A(d,d) (2,1) = c A(d,cm) (2,1) (16)
This result is translated immediately in same terms for prime numbers variables matrices. c B(d,d)

(i,j) = ∑ c B(d,cm) (i,j+n.d) (17) n = 0 to cm/d-1 i>1, j>1 c B(d,d) (2,1) = c B(d,cm) (2,1) (18)
Proof Targets set {g 0 .g u.d , g 1 .g u.d , …, gd -1 .g u.d } is split in d sets {g 0.d .g u.cm , g 1.d .g u.cm , …, g (cm/d-1).d .g u.cm }, {g 1 .g 0.d .g u.cm , g 1 .g 1.d .g u.cm , …, g 1 .g (cm/d-1).d .g u.cm }, …, {gd -1 .g 0.d .g u.cm , gd -1 .g 1.d .g u.cm , …, gd -1 . g (cm/d-1).d .g u.cm } with their respective cardinals what corresponds to the stated property.

Notice

We must notice two differences between the properties which we have just described. The property of the average on one hand interests matrices of equal dimension and sweeps diagonals. The property of splitting on the other hand distinguishes matrices of different dimensions and applies according to lines.

Numerical example

Environment cm = 6, p = 97 The reader will be able to verify on these examples average and splitting properties. Other alternatives on the same topic can be proposed. Thus, while considering this time, matrices A (d,d) and A (cm,cm) , it follows :

c A(d,d) (i,j) = ∑ ∑ c A(cm,cm) (i+m.d,j+n.d) n = 0 to cm/d-1 m = 0 to cm/d-1 i>1, j>1

Diagonalisation of Fermat Catalan cardinal matrices

Proposition Fermat-Catalan cardinal matrices (that is [A] i,cm for integers' variable x i and [B] j,cm for prime numbers' variables y j ) are diagonalisable. For a given environment cm, we can use eigenvectors matrix [P B ] of standard cardinal matrix common with any Fermat-Catalan cardinal matrices.

P 191/390

Fermat-Catalan abundance factors

Rule of writing

We adopt notations λ i,d,cm and respectively μ i,d,cm for Fermat-Catalan cardinal matrices eigenvalues of a monomial x d (integers variable) and respectively y d (prime numbers variable) if qualification of environment cm is necessary. If not, we will use λ i,d and μ i,d . For a given environment cm, we can imply certain indices and simplify notations with μ i,cm,cm = μ i . Sequence p where calculation is carried out is generally implied even if literal results are function of p. We note [DA] i,cm and [DB] i,cm the eigenvalues matrices of [A] i,cm and [B] i,cm respectively.

General expressions

We have for a given sequence p : [A] i,cm = [P B ] i,cm .

[DA] i,cm .[P B

-1

] i,cm [B] ] j,cm and (19) [A] cm,cm = [P B ] cm,cm .[DA] cm,cm .[P B -1 ] cm,cm [B] cm,cm = [P B ] cm,cm .[DB] cm,cm .[P B -1 ] cm,cm where:

[P B ] i,cm = [P B ] j,cm = [P B ] cm,cm = [P B ]
(20) and (relations 21)

μ 0 0 0 … 0 0 μ 1 … 0 [DA] cm,cm = 0 0 μ 2 … 0 … … … … … 0 0 0 … μ cm λ 0 0 0 … 0 0 λ 1 … 0 [DB] cm,cm = 0 0 λ 2 … 0 … … … … … 0 0 0 … λ cm 1 λ 0 /cm λ 0 /cm … λ 0 /cm 1 λ 1 /cm λ 2 /cm … λ cm /cm [P B ] = (1/√p) 1 λ cm /cm λ 1 /cm … λ cm-1 /cm … … … … … 1 λ 2 /cm λ 3 /cm … λ 1 /cm 1 λ 0 */cm λ 0 */cm … λ 0 */cm 1 λ 1 */cm λ cm */cm … λ 2 */cm [P B -1 ] = (1/√p) 1 λ 2 */cm λ 1 */cm … λ 3 */cm … … … … … 1 λ cm */cm λ cm-1 */cm … λ 1 */cm
Then, cardinal matrix [E] eq corresponding to expression x 1 (a1) + x 2 (a2) +…+ x k (ak) + y 1 (b1) + y 2 (b2) +…+ y m (bm) = c, at sequence p, is given by a product of matrices (relations 22) :

[E] eq = [A] eq . [B] Here the matrices eigenvalues starting from μ 1,(ai) , respectively μ 1,(bj) , repeat cm/(ai) and respectively cm/(bj) times :

(μ 0 , μ 1,(ai) , μ 2,(ai) , μ 3,(ai) , …, μ cm,(ai) ) = (μ 0 , μ 1,(ai) , μ 2,(ai) , …, μ (ai),(ai) , μ 1,(ai) , μ 2,(ai) ,…μ (ai),(ai) , …, μ 1,(ai) , μ 2,(ai) , …, μ (ai),(ai) ) and (λ 0 , λ 1,(bj) , λ 2,(bj) , λ 3,(bj) , …, λ cm,(bj) ) = (λ 0 , λ 1,(bj) , λ 2,(bj) , …, λ (bj),(bj) , λ 1,(bj) , λ 2,(bj) ,… λ (bj),(bj) , …, λ 1,(bj) , λ 2,(bj) , …, λ (bj),(bj) )

Eigenvalues {μ 0 , μ 1,(ai) , μ 2,(ai) , …, μ (ai),(ai) } are standard cardinal matrix eigenvalues of monomial x (ai) , i.e. within environment (ai). In the same way {λ 0 , λ 1,(bj) , λ 2,(bj) , …, λ (bj),(bj) } are standard cardinal matrix eigenvalues of monomial x (bj) within environment (bj). Thus "complete" Fermat-Catalan cardinal matrix corresponding to sums (or differences with matrices holding account of this) of separate variables is expressed in the form of a multiplication of "elementary" Fermat-Catalan cardinal matrices. Eigenvalues of the elementary matrices are exclusive to the "initial" environment (ai) or (bj) while eigenvectors are common and result from global environment cm. Thus, we are led to multiply eigenvalues diagonal matrices of single monomials, which are elementary operations. We will call the common matrix [P B ] cm,cm the matrix of the environment. (or environment matrix)

Property of eigenvalues average

We have : Let us take an example : cm =6. We start with standard cardinal matrix which we present facing its eigenvectors matrix and its eigenvalues (noted E0 stage). We carry out successively a set of permutations pending on columns and lines as follows :

μ i,d = (d/cm).∑ μ i+n.d,cm (24)
(E0) 1 p 0 0 0 0 0 1 λ 0 /cm λ 0 /cm λ 0 /cm λ 0 /cm λ 0 /cm λ 0 /cm μ 0 cm a 11 a 12 a 13 a 14 a 15 a 16 1 λ 1 /cm λ 2 /cm λ 3 /cm λ 4 /cm λ 5 /cm λ 6 /cm μ 1 0 a 21 a 22 a 23 a 24 a 25 a 26 1 λ 6 /cm λ 1 /cm λ 2 /cm λ 3 /cm λ 4 /cm λ 5 /cm μ 2 0 a 31 a 32 a 33 a 34 a 35 a 36 1 λ 5 /cm λ 6 /cm λ 1 /cm λ 2 /cm λ 3 /cm λ 4 /cm μ 3 0 a 41 a 42 a 43 a 44 a 45 a 46 1 λ 4 /cm λ 5 /cm λ 6 /cm λ 1 /cm λ 2 /cm λ 3 /cm μ 4 0 a 51 a 52 a 53 a 54 a 55 a 56 1 λ 3 /cm λ 4 /cm λ 5 /cm λ 6 /cm λ 1 /cm λ 2 /cm μ 5 0 a 61 a 62 a 63 a 64 a 65 a 66 1 λ 2 /cm λ 3 /cm λ 4 /cm λ 5 /cm λ 6 /cm λ 1 /cm μ 6 1 0 0 p 0 0 0 1 λ 0 /cm λ 0 /cm λ 0 /cm λ 0 /cm λ 0 /cm λ 0 /cm μ 0 cm a 13 a 14 a 11 a 12 a 15 a 16 1 λ 3 /cm λ 4 /cm λ 1 /cm λ 2 /cm λ 5 /cm λ 6 /cm μ 1 0 a 23 a 24 a 21 a 22 a 25 a 26 1 λ 2 /cm λ 3 /cm λ 6 /cm λ 1 /cm λ 4 /cm λ 5 /cm μ 2 0 a 33 a 34 a 31 a 32 a 35 a 36 1 λ 1 /cm λ 2 /cm λ 5 /cm λ 6 /cm λ 3 /cm λ 4 /cm μ 3 0 a 43 a 44 a 41 a 42 a 45 a 46 1 λ 6 /cm λ 1 /cm λ 4 /cm λ 5 /cm λ 2 /cm λ 3 /cm μ 4 0 a 53 a 54 a 51 a 52 a 55 a 56 1 λ 5 /cm λ 6 /cm λ 3 /cm λ 4 /cm λ 1 /cm λ 2 /cm μ 5 0 a 63 a 64 a 61 a 62 a 65 a 66 1 λ 4 /cm λ 5 /cm λ 2 /cm λ 3 /cm λ 6 /cm λ 1 /cm μ 6 1 0 0 p 0 0 0 1 λ 0 /cm λ 0 /cm λ 0 /cm λ 0 /cm λ 0 /cm λ 0 /cm μ 0 0 a 33 a 34 a 31 a 32 a 35 a 36 1 λ 1 /cm λ 2 /cm λ 5 /cm λ 6 /cm λ 3 /cm λ 4 /cm μ 3 0 a 43 a 44 a 41 a 42 a 45 a 46 1 λ 6 /cm λ 1 /cm λ 4 /cm λ 5 /cm λ 2 /cm λ 3 /cm μ 4 cm a 13 a 14 a 11 a 12 a 15 a 16 1 λ 3 /cm λ 4 /cm λ 1 /cm λ 2 /cm λ 5 /cm λ 6 /cm μ 1 0 a 23 a 24 a 21 a 22 a 25 a 26 1 λ 2 /cm λ 3 /cm λ 6 /cm λ 1 /cm λ 4 /cm λ 5 /cm μ 2 0 a 53 a 54 a 51 a 52 a 55 a 56 1 λ 5 /cm λ 6 /cm λ 3 /cm λ 4 /cm λ 1 /cm λ 2 /cm μ 5 0 a 63 a 64 a 61 a 62 a 65 a 66 1 λ 4 /cm λ 5 /cm λ 2 /cm λ 3 /cm λ 6 /cm λ 1 /cm μ 6

P 193/390

Fermat-Catalan abundance factors 1 0 0 0 0 p 0 1 λ 0 /cm λ 0 /cm λ 0 /cm λ 0 /cm λ 0 /cm λ 0 /cm μ 0 a 33 a 34 a 35 a 36 a 31 a 32 1 λ 1 /cm λ 2 /cm λ 3 /cm λ 4 /cm λ 5 /cm λ 6 /cm μ 0 a 43 a 44 a 45 a 46 a 41 a 42 1 λ 6 /cm λ 1 /cm λ 2 /cm λ 3 /cm λ 4 /cm λ 5 /cm μ cm a 13 a 14 a 15 a 16 a 11 a 12 1 λ 3 /cm λ 4 /cm λ 5 /cm λ 6 /cm λ 1 /cm λ 2 /cm μ 0 a 23 a 24 a 25 a 26 a 21 a 22 1 λ 2 /cm λ 3 /cm λ 4 /cm λ 5 /cm λ 6 /cm λ 1 /cm μ 0 a 53 a 54 a 55 a 56 a 51 a 52 1 λ 5 /cm λ 6 /cm λ 1 /cm λ 2 /cm λ 3 /cm λ 4 /cm μ 0 a 63 a 64 a 65 a 66 a 61 a 62 1 λ 4 /cm λ 5 /cm λ 6 /cm λ 1 /cm λ 2 /cm λ 3 /cm μ (E1) 1 0 0 0 0 p 0 1 λ 0 /cm λ 0 /cm λ 0 /cm λ 0 /cm λ 0 /cm λ 0 /cm μ 0 a 33 a 34 a 35 a 36 a 31 a 32 1 λ 1 /cm λ 2 /cm λ 3 /cm λ 4 /cm λ 5 /cm λ 6 /cm μ 0 a 43 a 44 a 45 a 46 a 41 a 42 1 λ 6 /cm λ 1 /cm λ 2 /cm λ 3 /cm λ 4 /cm λ 5 /cm μ 0 a 53 a 54 a 55 a 56 a 51 a 52 1 λ 5 /cm λ 6 /cm λ 1 /cm λ 2 /cm λ 3 /cm λ 4 /cm μ 0 a 63 a 64 a 65 a 66 a 61 a 62 1 λ 4 /cm λ 5 /cm λ 6 /cm λ 1 /cm λ 2 /cm λ 3 /cm μ cm a 13 a 14 a 15 a 16 a 11 a 12 1 λ 3 /cm λ 4 /cm λ 5 /cm λ 6 /cm λ 1 /cm λ 2 /cm μ 0 a 23 a 24 a 25 a 26 a 21 a 22 1 λ 2 /cm λ 3 /cm λ 4 /cm λ 5 /cm λ 6 /cm λ 1 /cm μ 1 0 0 0 0 p 0 1 λ 0 /cm λ 0 /cm λ 0 /cm λ 0 /cm λ 0 /cm λ 0 /cm μ 0 a 35 a 36 a 33 a 34 a 31 a 32 1 λ 3 /cm λ 4 /cm λ 1 /cm λ 2 /cm λ 5 /cm λ 6 /cm μ 0 a 45 a 46 a 43 a 44 a 41 a 42 1 λ 2 /cm λ 3 /cm λ 6 /cm λ 1 /cm λ 4 /cm λ 5 /cm μ 0 a 55 a 56 a 53 a 54 a 51 a 52 1 λ 1 /cm λ 2 /cm λ 5 /cm λ 6 /cm λ 3 /cm λ 4 /cm μ 0 a 65 a 66 a 63 a 64 a 61 a 62 1 λ 6 /cm λ 1 /cm λ 4 /cm λ 5 /cm λ 2 /cm λ 3 /cm μ cm a 15 a 16 a 13 a 14 a 11 a 12 1 λ 5 /cm λ 6 /cm λ 3 /cm λ 4 /cm λ 1 /cm λ 2 /cm μ 0 a 25 a 26 a 23 a 24 a 21 a 22 1 λ 4 /cm λ 5 /cm λ 2 /cm λ 3 /cm λ 6 /cm λ 1 /cm μ 1 0 0 0 0 p 0 1 λ 0 /cm λ 0 /cm λ 0 /cm λ 0 /cm λ 0 /cm λ 0 /cm μ 0 a 55 a 56 a 53 a 54 a 51 a 52 1 λ 1 /cm λ 2 /cm λ 5 /cm λ 6 /cm λ 3 /cm λ 4 /cm μ 0 a 65 a 66 a 63 a 64 a 61 a 62 1 λ 6 /cm λ 1 /cm λ 4 /cm λ 5 /cm λ 2 /cm λ 3 /cm μ 0 a 35 a 36 a 33 a 34 a 31 a 32 1 λ 3 /cm λ 4 /cm λ 1 /cm λ 2 /cm λ 5 /cm λ 6 /cm μ 0 a 45 a 46 a 43 a 44 a 41 a 42 1 λ 2 /cm λ 3 /cm λ 6 /cm λ 1 /cm λ 4 /cm λ 5 /cm μ cm a 15 a 16 a 13 a 14 a 11 a 12 1 λ 5 /cm λ 6 /cm λ 3 /cm λ 4 /cm λ 1 /cm λ 2 /cm μ 0 a 25 a 26 a 23 a 24 a 21 a 22 1 λ 4 /cm λ 5 /cm λ 2 /cm λ 3 /cm λ 6 /cm λ 1 /cm μ 1 0 0 p 0 0 0 1 λ 0 /cm λ 0 /cm λ 0 /cm λ 0 /cm λ 0 /cm λ 0 /cm μ 0 a 55 a 56 a 51 a 52 a 53 a 54 1 λ 1 /cm λ 2 /cm λ 3 /cm λ 4 /cm λ 5 /cm λ 6 /cm μ 0 a 65 a 66 a 61 a 62 a 63 a 64 1 λ 6 /cm λ 1 /cm λ 2 /cm λ 3 /cm λ 4 /cm λ 5 /cm μ 0 a 35 a 36 a 31 a 32 a 33 a 34 1 λ 3 /cm λ 4 /cm λ 5 /cm λ 6 /cm λ 1 /cm λ 2 /cm μ 0 a 45 a 46 a 41 a 42 a 43 a 44 1 λ 2 /cm λ 3 /cm λ 4 /cm λ 5 /cm λ 6 /cm λ 1 /cm μ cm a 15 a 16 a 11 a 12 a 13 a 14 1 λ 5 /cm λ 6 /cm λ 1 /cm λ 2 /cm λ 3 /cm λ 4 /cm μ 0 a 25 a 26 a 21 a 22 a 23 a 24 1 λ 4 /cm λ 5 /cm λ 6 /cm λ 1 /cm λ 2 /cm λ 3 /cm μ (E2) 1 0 0 p 0 0 0 1 λ 0 /cm λ 0 /cm λ 0 /cm λ 0 /cm λ 0 /cm λ 0 /cm μ 0 a 55 a 56 a 51 a 52 a 53 a 54 1 λ 1 /cm λ 2 /cm λ 3 /cm λ 4 /cm λ 5 /cm λ 6 /cm μ 0 a 65 a 66 a 61 a 62 a 63 a 64 1 λ 6 /cm λ 1 /cm λ 2 /cm λ 3 /cm λ 4 /cm λ 5 /cm μ cm a 15 a 16 a 11 a 12 a 13 a 14 1 λ 5 /cm λ 6 /cm λ 1 /cm λ 2 /cm λ 3 /cm λ 4 /cm μ 0 a 25 a 26 a 21 a 22 a 23 a 24 1 λ 4 /cm λ 5 /cm λ 6 /cm λ 1 /cm λ 2 /cm λ 3 /cm μ 0 a 35 a 36 a 31 a 32 a 33 a 34 1 λ 3 /cm λ 4 /cm λ 5 /cm λ 6 /cm λ 1 /cm λ 2 /cm μ 0 a 45 a 46 a 41 a 42 a 43 a 44 1 λ 2 /cm λ 3 /cm λ 4 /cm λ 5 /cm λ 6 /cm λ 1 /cm μ In (E0), (E1) and (E2) steps, the change of basis matrices [P] are identical. We then take the average of matrices [A] at these steps which according to components average property is the required Fermat-Catalan cardinal matrix. In same time, we check as well that Fermat-Catalan cardinal matrix eigenvalues are steps (E0), (E1) and (E2) eigenvalues averages. These considerations are generalized for any divisor of cm and thus for the whole set of Fermat-Catalan cardinal matrices of any environment cm.

Literal resolution of Fermat-Catalan enumerations

Here the average property is essential. The knowledge of "standard" cardinal matrix permits to evaluate without difficulty any Fermat-Catalan cardinal matrices in a given environment.

Conversely, the splitting property may be useful to discover by groping the standard matrix.

Eigenvalues of the matrix, in an adapted order, are roughly : Then the eigenvalues are given in the order of the principal diagonal while taking first μ 0 in the following way : 193, 58.723704, 54.48156, -53.31949, 17.898724, -50.92077, 57.09326, -5.542397, -88.18578, -7.964582, 2.3828543, -24.33113, 39.684041 With n dimensions, we can build a table with n axis and we follow co-ordinates (i, i,…, i).

Orthogonality of eigenvalues in disjoined environments

We call disjoined environments two environments d 1 and d 2 where (d 1 ,d 2 ) = 1. Let us have the product d = d 1 .d 2 . Let us pose μ cm = (μ 1,cm , μ 2,cm , …, μ cm,cm ) the ordered vector of the Fermat-Catalan cardinal matrix eigenvalues that we get by the principal diagonal method (except μ 0 ). Then for any m ranging between 1and d, we can write the scalar products :

<μ d1 .μ d2 > d,m = ∑ μ i,d1 .μ i+m,d2 = 0 (27)
This has interesting consequences, we use thereafter (often in an implied way).

Let us note that the sum is extended to indices i ranging between 1 and d by repeating the indices modulo d 1 or modulo d 2 as many time as necessary.

Proof

Terms μ i,d1 .μ i+m,d2 and μ i,d1 .μ i+m',d2 (m ≠ m') are identical except for a circulant shift. We summon ∑∑ μ i,d1 .μ i+m,d2 on indices i and m. The sum is null as ∑μ i = 0.

Matrix of degree 1

Here, we are primarily interested by the following problem. Let us have a diophantine equation of Fermat-Catalan type : R(x i ,y j ) = x 1 (a1) +x 2 (a2) +…+x k (ak) +y 1

+…+ y m (bm) = c (28)

For this equation, we suppose to have already got cardinal matrix [M] allowing us to calculate the abundance factors. We attend to pass then to abundance factors evaluation when a prime numbers variable is added on the right member of this equation :

x 1 (a1) +x 2 (a2) +…+x k (ak) +y 1 (b1) +y 2 (b2) +…+y m (bm) = y+c ( 29
)
We wish an expression in the form of a matrix [Y] with the rule of a multiplicative operator.

Elementary approach

We use initially the two-dimensional tables' method. Let us have cm lowest common multiple such as definite higher.

[M] is of dimension cm+1.

The equation ( 28) includes k integers' variables and m prime numbers variables, which is in all p k .(p-1) m occurrences for the targets (modulo p). We have then for the cardinal of a target c :

p k .(p-1) m -#{(x 1 , …, x k ,y 1 , …, y m ) / R(x 1 , …, x k ,y 1 , …, y m ) = c mod p}
In matrix writing, we have for the relation (28) :

P 196/390 Fermat-Catalan abundance factors #{0} 1 m 0 #{g 0 .g u.cm } 0 m 1 #{g 1 .g u.cm } = [M] 0 = m 2 … … … #{g cm-1 .g u.cm } 0 m cm Then we have : #{0} 1 m 0 #{g 0 .g u.cm } 1 m 1 #{g 1 .g u.cm } = p k .(p-1) m . 1 -m 2 … … … #{g cm-1 .g u.cm } 1 m cm
We consider the matrices :

[N] = 1 (p-1)/cm (p-1)/cm … (p-1)/cm (30) 1 (p-1)/cm (p-1)/cm … (p-1)/cm … … … … … 1 (p-1)/cm (p-1)/cm … (p-1)/cm [N]-[I] = 0 (p-1)/cm (p-1)/cm … (p-1)/cm (31) 1 (p-1)/cm-1 (p-1)/cm … (p-1)/cm … … … … … 1
(p-1)/cm (p-1)/cm … (p-1)/cm-1 Then we have :

1 1 m 0 0 1 m 1 ([N]-[I]).[M]. 0 = m 0 +(p-1)/cm.(m 1 +m 2 +…+m cm ) 1 -m 2 … … … 0 1 m cm Matrix [N]-[I]
is appropriate for our research provided that :

m 0 +(p-1)/cm.(m 1 +m 2 +…+m cm ) = p k .(p-1) m (32)
However, when k = 1 and m = 0, we have m 0 = 1, m 1 = cm and the other m i are null. When k = 0 and m = 1, we have m 0 = 0, m 1 = cm and the other m i are null. We verify the initial condition. We proceed then by recurrence. On the basis of m 0 +(p-1)/cm.(m 1 +m 2 +…+m (p-1)/cm ) = p k .(p-1) m , we evaluate m 0 '+(p-1)/cm.(m 1 '+m 2 '+…+m (p-1)/cm '). Here, values m i ' result from m i thanks to a cardinal matrix with a ij components whose each line sums are p-1 in the case of a prime numbers variable. When p = 1+2i.cm, matrix general configuration is :

0 p-1 0 … 0 cm [B] = 0 U … 0
We thus have, the sums relating to indices 1 to (p-1)/cm, m 0 ' = (p-1).m 1 , m 1 ' = cm.m 0 +∑a 1j .m j , m i ' = ∑m j .∑a ij . It results that m 0 '+(p-1)/cm.(m 1 '+m 2 '+…+m (p-1)/cm ') = (p-1).m 1 +(p-1)/cm.(cm.m 0 +∑a 1j .m j +∑m j .∑a ij ) = (p-1).m 0 +(p-1)/cm.(cm.m 1 +∑a 1j .m j +∑m j .∑a ij ). As a ij = a ji it follows m 0 '+(p-1)/cm.(m 1 '+m 2 '+…+m (p-1)/cm ') = (p-1).m 0 +(p-1)/cm.(∑m j .∑a ji ), indices of the last sum bearing on 0 to (p-1)/cm, sum which equals thus p-1. Then m 0 '+(p-1)/cm.(m 1 '+m 2 '+…+m (p-1)/cm ') = (p-1).(m 0 +(p-1)/cm.(m 1 +m 2 +…+m (p-1)/cm )) = p k .(p-1) m+1 what is the sought result. We could proceed with the same approach for the cases where p = 1+cm+2i.cm. For an integers variable, we get m 0 '+(p-1)/cm.(m 1 '+m 2 '+…+m (p-1)/cm ') = p.(m 0 +(p-1)/cm.(m 1 + m 2 +…+m (p-1)/cm )) = p k+1 .(p-1) m .

Efficient approach

Let us note still that the average property, applied to case d = 1, led to the same matrices [N] (for x) and [N]-[I] (for y), results which are of course awaited.

Eigenvalues and eigenvectors

Matrix [N] eigenvalues are p and 0, the eigenvalue 0 being of multiplicity p-1. Matrix [N-I] eigenvalues of are p-1 and -1, the eigenvalue -1 being of multiplicity p-1. The two matrices share a common intra-unitary matrix of eigenvectors.

Environment cm = 2. Hua matrices

We illustrate this case for the Luogeng Hua conjecture concerning the conditions of existence of an integer decomposition into a prime and a square prime, that is :

y 1 +y 2 2 = c
We examine here only the asymptotic behaviour, meaning the solutions' enumeration when c tends towards infinity. The representative of variable -p is identical to that of variable p. We thus have an analogue problem, which can be solved for any c, with : y 2 2 = y 1 +c

We have two principal cases from which we deduce the environment and two lower cases as we will see further :

p (p-1,d) variable y 1 (p-1,d) variable y 2 2 cm lower cases 2 1 1 1 ≠ 2 1 2 2 p = 1 mod 4 p = 3 mod 4
We go on by simple multiplication of the matrices, using the already obtained results in environments 1 and 2.

Var x or -x Var y or -y

#{0} = 1 p-1 #{0} = [M0] = 0 p-1 #{g u } 1 p-1 #{g u } 1 p
For p = 2 2 #{0} = 0 1 1 = 1 #{1} 1 0 0 0 For p ≠2 #{0} 1 0 p-1 #{g 2u } = [M0].or([M1],[M2]) = 0 = [M0] 2 = p-3 #{g.g 2u } 0 0 p-1

It is easy to verify that [M0] commutate with [M1] or [M2].

In addition, there is no need for distinction between the lower case p = 1 mod 4 and 3 mod 4. This distinction will arise only when more variables intervene in the equation.

Environment cm = 4. Matrices of Friedlander and Iwaniec

Friedlander and Iwaniec showed in 1996 the infinity of the prime numbers of x 1 4 +x 2 2 type. We give here the asymptotic evaluation of this result and others (c is an integer) :

x

#{0} = 1 1 1 = 2 #{1} 1 1 0 2 For p = 3 mod 4
We recover the cardinal matrix corresponding to d = 2 which it suffice to carry squared and to multiply by (1,0,0)

getting abundance factors ([M1] = [M2]) : 2 #{0} 1 0 p-1 1 1 0 p-1 1 1 #{g 2u } = [M1].[M2] = 2 (p-1)/2 (p-3)/2 0 = 2 (p-1)/2 (p-3)/2 2 = p+1 #{g.g 2u } 0 (p+1)/2 (p-1)/2 0 0 (p+1)/2 (p-1)/2 0 p+1
For p = 1 mod 8 We must consider two matrices. The first is that corresponding to the monomial x 4 which we obtained in exercise 5 : Foot-note : The reader will be caution here because of the confusion of notations between the variables (x 1 and x 2 ) and the components of the matrices (x 1 , x 2 , x 3 , etc.)

1 p-1 0 0 0 4 x 1 -3 x 2 x 3 x 4 [M1] = 0 x 2 x 4 +1
The second matrix is that corresponding to the monomial x 2 when cm = 4. All made calculations, we have : For the abundance factors evaluation, we have then :

P 199/390

Fermat-Catalan abundance factors

#{0} 1 1 #{g 4u } 0 0 #{g.g 4u } = [M1].[M2]. 0 = [M2].[M1]. 0 #{g 2 .g 4u } 0 0 #{g 3 .g 4u } 0 0
The simplest thing is to retain the first equality, that is also :

#{0} 1 p-1 0 0 0 1 #{g 4u } 4 x 1 -3 x 2 x 3 x 4 2 #{g.g 4u } = 0 x 2 x 4 +1 x 5 x 5 0 #{g 2 .g 4u } 0 x 3 x 5 x 3 +1 x 5 2 #{g 3 .g 4u } 0 x 4 x 5 x 5 x 2 +1 0 Thus : #{0} 1 2p-1 2p-1 #{g 4u } 0 2(x 1 +x 3 -1) p-1+2.(-1) (β+1)/2 .β #{g.g 4u } = [M1].[M2]. 0 = 2(x 2 +x 5 ) = p-1+4α.if(x 4 >x 2 ,-1,1) #{g 2 .g 4u } 0 2(2x 3 +1) p-1-2.(-1) (β+1)/2 .β #{g 3 .g 4u } 0 2(x 4 +x 5 ) p-1-4α.if(x 4 >x 2 ,-1,1)
For p = 5 mod 8

We must consider two matrices. The first is that corresponding to the monomial x 4 which we obtained in exercise 5 :

1 0 0 p-1 0 4 x 3 +1 x 5 x 3 x 5 [M1] = 0 x 4 x 5 +1 x 5 x 2 0 x 1 x 2 x 3 +1 x 4 0 x 2 x 4 x 5 x 5 +1 where x 1 (p+1)/4+(-1) (β+1)/2 .(3/2).β x 2 (p+1)/4+2α.if(x 4 >x 2 ,-1,1)-(-1) (β+1)/2 .(1/2).β x 3 = (p-7)/4-(-1) (β+1)/2 .(1/2).β x 4 (p+1)/4-2α.if(x 4 >x 2 ,-1,1)-(-1) (β+1)/2 .(1/2).β x 5
(p-3)/4+(-1) (β+1)/2 .(1/2).β with again the decomposition into integers of p : p = (2α) 2 +β 2

The second matrix is that corresponding to the monomial x 2 when cm = 4. All made calculations, we have : For the abundance factors evaluation, we have then :

#{0} 1 1 #{g 4u } 0 0 #{g.g 4u } = [M1].[M2]. 0 = [M2].[M1]. 0 #{g 2 .g 4u } 0 0 #{g 3 .g 4u } 0 0
The simplest thing is again to retain the first equality, that is also :

P 200/390 Fermat-Catalan abundance factors #{0} 1 0 0 p-1 0 1 #{g 4u } 4 x 3 +1 x 5 x 3 x 5 2 #{g.g 4u } = 0 x 4 x 5 +1 x 5 x 2 0 #{g 2 .g 4u } 0 x 1 x 2 x 3 +1 x 4 2 #{g 3 .g 4u } 0 x 2 x 4 x 5 x 5 +1 0 It follows then : #{0} 1 2p-1 2p-1 #{g 4u } 0 2(2x 3 +3) p-1-2.(-1) (β+1)/2 .β #{g.g 4u } = [M1].[M2]. 0 = 2(x 4 +x 5 ) = p-1-4α.if(x 4 >x 2 ,-1,1) #{g 2 .g 4u } 0 2(x 1 +x 3 +1) p-1+2.(-1) (β+1)/2 .β #{g 3 .g 4u } 0 2(x 2 +x 5 ) p-1+4α.if(x 4 >x 2 ,-1,1)
The passage of expression x 1 4 +x 2 2 = c to expression x 1 4 +x 2 2 = y+c is made according to the now standard process of prime numbers variable addition. For two variables of integers in the left member, we must withdraw from p 2 the values obtained for x 1 4 +x 2 2 = c. As follows :

For p = 2 #{c} = 2 For p = 3 mod 4 #{0} (p-1)(p+1) #{g 2u } = p 2 -p-1 #{g.g 2u } p 2 -p-1 For p = 1 mod 8 #{0} (p-1) 2 #{g 4u } p 2 -p+1-2.(-1) (β+1)/2 .β #{g.g 4u } = p 2 -p+1-4α.if(x 4 >x 2 ,-1,1) #{g 2 .g 4u } p 2 -p+1+2.(-1) (β+1)/2 .β #{g 3 .g 4u } p 2 -p+1+4α.if(x 4 >x 2 ,-1,1) For p = 5 mod 8 #{0} (p-1) 2 #{g 4u } p 2 -p+1+2.(-1) (β+1)/2 .β #{g.g 4u } = p 2 -p+1+4α.if(x 4 >x 2 ,-1,1) #{g 2 .g 4u } p 2 -p+1-2.(-1) (β+1)/2 .β #{g 3 .g 4u } p 2 -p+1-4α.if(x 4 >x 2 ,-1,1)
Let us recall that the condition between x 2 and x 4 (x 4 >x 2 or x 2 >x 4 ) depends simply on the choice of g and that α and β result from the decomposition into positive integers of p.

We can produce with these relations the abundance factors of more general diophantine equations :

x We have immediately : and also -(p-7)/6+2y 3 = (p+2)/3-(x 1 +x 3 +x 5 )/3 (p-1)/3-y 3 = (p-1)/12+(x 1 +x 3 +x 5 )/6

For p = 2 #{c} = 2 i+j-1 For p = 1 mod 4 #{0} p i+j .(p-1) k+m 1 #{g 4u } p i+j .(p-1) k+m 0 #{g.g 4u } = p i+j .(p-1) k+m -[M1] i .[M2] j .([M1]-[I]) k .([M2]-[I]) m . 0 #{g 2 .g 4u } p i+j .(p-
The equality y 4 = (p-1)/6 remains to be proved. From symmetry properties of g x-2 = g 2u + g y-2 .g 6v mod p, we deduce the components c(5,2) = 2{#{g 3 = g 2u + g 0 .g 6v mod p} = 2{#{g 5 = g 2u + g 2 .g 6v mod p} = c(7,4) = 2{#{g 1 = g 2u + g 4 .g 6v mod p} = c (3,6). In the same way, we have immediately c(2,5) = c(4,7) = c (6,3). As the reduced matrix is symmetric, the whole set of these components are equal. The sum of this set is p-1 (sum on a diagonal), hence a cardinal (p-1)/6 for these components.

Finally, let us work on [M2] using splitting property of the components:

1 (p-1)/2 0 0 (p-1)/2 0 0 3 z 1 +1 z 2 z 3 x 1(3) -3-z 1 x 2(3) -z 2 x 3(3) -z 3 0 z 2 z 4 +1 (p-1)/2-x 3(3) -z 2 +z 3 x 2(3) -z 2 x 3(3) -z 4 (p-1)/2-x 2(3) +z 2 -z 3 [M2] = 0 z 3 (p-1)/2-x 3(3) -z 2 +z 3 (p-5)/2-z 1 -z 4 x 3(3) -z 3 (p-1)/2-x 2(3) +z 2 -z 3 -(p-7)/2+x 2(3) +z 1 +z 4 3 x 1(3) -3-z 1 x 2(3) -z 2 x 3(3) -z 3 z 1 +1 z 2 z 3 0 x 2(3) -z 2 x 3(3) -z 4 (p-1)/2-x 2(3) +z 2 -z 3 z 2 z 4 +1 (p-1)/2-x 3(3) -z 2 +z 3 0 x 3(3) -z 3 (p-1)/2-x 2(3) +z 2 -z 3 -(p-7)/2+x 2(3) +z 1 +z 4 z3 (p-1)/2-x 3(3) -z 2 +z 3 (p-5)/2-z 1 -z 4
Values x 1(3) , x 2(3) , x 3(3) are those of the cardinal matrix of dimension 3, values totally clarified at exercise 5. Matrix [M2] is deduce from dimension 3 blocks repetitions along diagonals (of reduced matrix) on one hand and splitting property of standard dimension 3 cardinal matrix components on the other hand.

Using the method of averages, we get also : We proposed until now literal expressions of the matrices of dimension 6+1 starting from considerations of symmetries and sums. Partial results are obtained. We show here what can be done while using eigenvalues and eigenvectors of Fermat-Catalan cardinal matrices. This example can be useful in the study of an environment different from 6. We initially observe a set of conjectures (using notation conventions μ i,6,6 = μ i ) :

z 1 +1 = (x 1 +x 4 )/2-2 z 2 = (x 2 +x 8 )/2 z 3 = (p-1)/4+(x 3 -x 4 -x 8 )/2 z 4 = (p-
μ 1 -μ 2 +μ 3 -μ 4 +μ 5 -μ 6 = -6√p μ 1 2 -μ 2 2 +μ 3 2 -μ 4 2 +μ 5
2 -μ 6 2 = 24.z.√p μ 1 .μ 2 -μ 2 .μ 3 +μ 3 .μ 4 -μ 4 .μ 5 +μ 5 .μ 6 -μ 6 .μ 1 = -36.y.√p μ 1 .μ 3 -μ 2 .μ 4 +μ 3 .μ 5 -μ 4 .μ 6 +μ 5 .μ 1 -μ 6 .μ 2 = -12.z.√p μ 1 .μ 4 -μ 2 .μ 5 +μ 3 .μ 6 -μ 4 .μ 1 +μ 5 .μ 2 -μ 6 .μ 3 = 0 as well as the following results which rise from the preceding ones : μ 1 .μ 5 -μ 2 .μ 6 +μ 3 .μ 1 -μ 4 .μ 2 +μ 5 .μ 3 -μ 6 .μ 4 = 12z.√p μ 1 .μ 6 -μ 2 .μ 1 +μ 3 .μ 2 -μ 4 .μ 3 +μ 5 .μ 4 -μ 6 .μ 5 = 36y.√p Parameters y and z are integer functions of p. We observe that : p = 3y 2 +z 2 (33) y = 0 mod 2 z = 1 mod 6

We find here a peculiar Euler decomposition for 1 mod 12 prime numbers. Parameter y is even here (p = 1 mod 12) what is not any more the case further (p = 7 mod 12). The decomposition is unique (when y and z have given signs), proposition which is not demonstrated here. Let us note that y and z are not mandatory of positive signs. The property of the averages of the eigenvalues leads to : where we had the choice of expressions : θ = if(x 3 >x 2 ,-1,1).arcos ((3x 1 -p-1)/(2.√p)) mod 2π, θ = 2π/3+if(x 1 >x 3 ,-1,1).arcos ((3x 2 -p+2)/(2.√p)) mod 2π, θ = -2π/3-if(x 1 >x 2 ,-1,1).arcos ((3x 3 -p+2)/(2.√p)) mod 2π.

We have a 2 +a.b+b 2 = b 2 +b.c+c 2 = c 2 +c.a+a 2 = -(a.b+b.c+c.a) = (a 2 +b 2 +c 2 )/2 = 3, a.b.c = 2f/√p. By construction, we also have (a 3 +b 3 +c 3 ).p 1/2 = 3x 1(3) -p-1, (a 2 .b+b 2 .c+c 2 .a).p 1/2 = 3x 2(3) -p+2, (a 2 .c+b 2 .a+c 2 .b).p 1/2 = 3x 3(3) -p+2.

After substitution of μ 1 +μ 4 = a√p, μ 2 +μ 5 = b√p, μ 3 +μ 6 = c√p in μ 1 -μ 2 +μ 3 -μ 4 +μ 5 -μ 6 = -6√p, μ 1 .μ 2 -μ 2 .μ 3 +μ 3 .μ 4 -μ 4 .μ 5 +μ 5 .μ 6μ 6 .μ 1 = -36y.√p, it follows all calculations done : P 204/390 Fermat-Catalan abundance factors μ 1 = (a-1).p 1/2 -(b-c).y+a.z (34) μ 2 = (b+1).p 1/2 +(c-a).y-b.z μ 3 = (c-1).p 1/2 -(a-b).y+c.z μ 4 = (a+1).p 1/2 +(b-c).y-a.z μ 5 = (b-1).p 1/2 -(c-a).y+b.z μ 6 = (c+1).p 1/2 +(a-b).y-c.z

It is here about a solution among other permutations. We know that the number of choices to build the cardinal matrix is φ(d=6) = 2 according to the remarks made for the case general (exercise 5 page 57). This results here in the choice of the sign of y subject to that of g (primitive of p) for the two acceptable permutations {g 0 , g 1 , g 2 , g 3 , g 4 , g 5 }.g 6u and {g 0 , g 5 , g 4 , g 3 , g 2 , g 1 }.g 6u . In addition, the sign of z is defined by observing that z is always 1 mod 6. Its sign is thus given by the expression : sign(z) = (-1) 1+abs(z) mod 3 (35)

The literal expression of the whole set of Fermat-Catalan cardinal matrices in environment 6 rise from the preceding eigenvalues. We also use

μ i = (a i +(-1) i ).p 1/2 +(-1) i .b i λ i = (a i +(-1) i ).p 1/2 +(-1) i .b i -1 λ i .λ j = (a i .b j +(-1) j .a i +(-1) i .a j +(-1) i+j ).p+((-1) i+j .b i .b j -(-1) i .b i -(-1) j .b j +1) +p 1/2 .((-1) j .a i .b j +(-1) i .a j .b i +(-1) i+j .(b i +b j )-a i -a j -(-1) i -(-1) j )
We get for sums bearing on indices i = 1 to 6 and j a given integer :

∑ a i = 0 ∑ b i = 0 ∑ a i .b i = -12.z ∑ a i .b i+1 = 6.(3y+z) ∑ a i .b i+2 = -6.(3y-z) ∑ (-1) i .a i .b i+j = 0 ∑ b i .b i = 12.p ∑ b i .b i+1 = -6.p ∑ b i .b i+2 = -6.p So that :
(1/p).∑ λ i .p 1/2 .λ i / 36 = (2z+1)/3 (1/p).∑ λ i .p 1/2 .λ i+1 / 36 = (z-1)/3 (1/p).∑ λ i .p 1/2 .λ i+2 / 36 = y

Then matrix [M1] presents, all made calculations, in the form (g is selected such as y 2 -y 6 is sign of y) : For the evaluation of this expression, we need the expressions drawn from exercise 5 relation 71 :

∑ a i 3 = 6f/p 1/2 ∑ a i 2 .a i+1 = 6f 3 /p 1/2 ∑ a i 2 .a i+2 = 6f 2 /p 1/2 where f = f 1 = (3x 3(3) -p-1)/2 f 2 = (3x 2(3) -p+2)/2 f 3 = (3x 3(3) -p+2)/2 Let us pose also : Δx 1(3) = x 1(3) -(p-1)/3 Δz 1 = z 1 -(p-1)/6 Δx 2(3) = x 2(3) -(p-1)/3 Δz 2 = z 2 -(p-1)/6 Δx 3(3) = x 3(3) -(p-1)/3 Δz 3 = z 3 -(p-1)/6
.a i+2 . Thus, we only need three numerical results to evaluate the components below (the choice of g is made such as x 2 -x 6 is sign of y) :

Numerical examples

p gmin y z x1 x2 x3 y2 y3 z1 z2 z3 z4 μ1 (3,3) μ2 (3,3) μ3 (3,3) μ1 (6,6) μ2 (6,6) μ3 (6,6) μ4 (6,6) μ5 (6,6) μ6(6,6) 13 2 2 1 3 3 6 2 0 0 3 3 0 1,42325 0,50524 -1, 9285 7,8168 -3,255 -7,982 2,4464 -10,65 11, ,85516 -0,2804 -1,5747 3,2078 -21,33 -23,26 28,493 -5,582 18,465 97 5 -4 7 39 27 30 18 20 12 15 15 24 1,99209 -0,8421 -1,15 22,484 17,91 -11,47 16,755 -40,56 -5,

(x3-x4-x8)/2 0 x2(3)-z2 (p-1)/2-x4-x8 (p-1)/4- (x3-x4+x5)/2 (x2+x8)/2 (p+1)/2- (x1+x2+x4+x5)/2 -(p-1)/4+ (x1+x3+2x4+x5)/2 0 3(p-1)/4-(x1+x2+x3+ 2x4+x5+x8)/2 (p-1)/4-(x3-x4+x5)/2 x8 (p-1)/4+(x3-x4-x8)/2 (p-1)/2-x3(3)-z2+z3 (x2+x5)/2+1 1 (p-1)/2 0 0 (p-1)/2 0 0 3 z 1 +1 z 2 z 3 x 1(3) -3-z 1 x 2(3) -z 2 x 3(3) -z 3 0 z 2 z 4 +1 (p-1)/2-x 3(3) - z 2 +z 3 x 2(3) -z 2 x 3(3) -z 4 (p-1)/2-x 2(3) +z 2 - z 3 [M2] = 0 z 3 (p-1)/2-x 3(3) - z 2 +z 3 (p-5)/2-z 1 -z 4 x 3(3) -z 3 (p-1)/2- x 2(3) +z 2 -z 3 -(p-7)/2 +x 2(3) +z 1 +z 4 3 x 1(3) -3-z 1 x 2(3) -z 2 x 3(3) -z 3 z 1 +1 z 2 z 3 0 x 2(3) -z 2 x 3(3) -z 4 (p-1)/2-x 2(3) +z 2 - z 3 z 2 z 4 +1 (p-1)/2-x 3(3) - z2+z 3 0 x 3(3) -z 3 (p-1)/2- x 2(3) +z 2 -z 3 -(p-7)/2 +x 2(3) +z 1 +z 4 z 3 (p-1)/2-x 3(3) - z 2 +z 3 (p-5)/2-z 1 -z 4
The form of this last matrix is identical to case p = The sign of z is determined by the fact that z is 2 mod 6 (conjecture), so that : z = abs(z).(-1) abs(z) mod 3 (38)

We find again the expressions a, b and c drawn from environment n = 3 case.

The case p = 7 mod 12 allows to understand in a finer way the contributions of environment cm = 2 and cm = 3 that the case p = 1 mod 12 did. Let us recall that for p = 3 mod 4 : We can bring this pleasant expression closer to the case p = 1 mod 12 by noting that it includes it.

μ 1,3 = a.p 1/2 μ 2,3 = b.p 1/2 μ 3,3 = c.p 1/2 μ 1,2 = i.p 1/2 μ 2,2 = -i.p 1/2 Hence: μ 1 = μ 1,3 +μ 1,2 +((μ 2,3 -μ 3,3 ).y-μ 1,3 .z).μ 1,2 /p μ 2 = μ 2,3 +μ 2,2 +((μ 3,3 -μ 1,3 ).y-μ 2,3 .z).μ 2,2 /p μ 3 = μ 3,3 +μ 1,2 +((μ 1,3 -μ 2,3 ).y-μ 3,3 .z).μ 1,2 /p μ 4 = μ 1,3 +μ 2,2 +((μ 2,3 -μ 3,3 ).y-μ 1,3 .z)).μ 2,2 /p μ 5 = μ 2,3 +μ 1,2 +((μ 3,3 -μ 1,3 ).y-μ 2,3 .z).μ 1,2 /p μ 6 = μ 3,3 +μ 2,2 +((μ 1,3 -μ 2,3 ).y-μ 3,3 .z)).μ 2,2 /p So that : μ 1 = μ 1,3 + μ 1,2 +(1/p). -z 0 -y 0 y 0 μ 1,3 .μ 1,2 μ 2 μ 2,3 μ 2,2 0 -z 0 -y 0 y μ 2,3 .μ 2,2 μ 3 μ 3,
We then get for matrix The reader will be able to find thanks to the method of the averages the forms of the matrices [M1] and [M2].

Foot-note : Because of the problem angle of attack with eigenvalues, we obtained literal expressions for Fermat Catalan matrices in environment cm = 6 by exploiting p = 3y 2 +z 2 . Another approach starting from p = t 1 2 +t 1 .t 2 +t 2 2 is also possible. We further defer it in this exercise.

For abundance factors evaluation, we have then :

P 208/390 Fermat-Catalan abundance factors #{0} 1 #{g 6u } 0 #{g.g 6u } 0 #{g 2 .g 6u } = [M1].[M2].[M3]. 0 #{g 3 .g 6u } 0 #{g 4 .g 6u } 0 #{g 5 .g 6u } 0
The passage of expression x 1 2 +x 2 4 +x 3 6 = c to x 1 2 +x 2 4 +x 3 6 = y+c is made according to from now on familiar process. For two variables of integers in the left member, we must withdraw the values obtained for x 1 2 +x 2 4 +x 3 6 = c to p 3 .

Abundance factors in environment cm = 6

We can produce using above relations the abundance factors of a more general diophantine equation :

x We have immediately :

For p = 2 #{c} = 2 i+j+k-1
For p = 1 mod 4

#{0} p i+j+k .(p-1) r+s+t 1 #{g 6u } p i+j+k .(p-1) r+s+t 0 #{g.g 6u } = p i+j+k .(p-1) r+s+t 0 #{g 2 .g 6u } p i+j+k .(p-1) r+s+t -[M1] i .[M2] j .[M3] k .([M1]-[I]) r .([M2]-[I]) s .([M3]-[I]) t 0 #{g 3 .g 6u } p i+j+k .(p-1) r+s+t 0 #{g 4 .g 6u } p i+j+k .(p-1) r+s+t 0 #{g 5 .g 6u } p i+j+k .(p-1) r+s+t 0 For p = = 3 mod 4 #{0} p i+j+k .(p-1) r+s+t 1 #{g 2u } = p i+j+k .(p-1) r+s+t -[M1] i .[M2] j .[M3] k .([M1]-[I]) r .([M2]-[I]) s .([M3]-[I]) t 0 #{g.g 2u } p i+j+k .(p-1) r+s+t 0

Equation of decomposition into relative integers (Z)

We get the abundance factors thanks to an Euler product build at successive sequences p. During the process of evaluation of these factors, we had recourse each time to a decomposition of p into integer numbers. In n = 2 case, we use Euler decomposition p = y 2 +z 2 . For n = 4, it acts of a similar case while adding constraints of congruence. For n = 3, decomposition y 2 +y.z+z 2 is more subtle and case n = 6 made us discover another alternative p = 3y 2 +z 2 (which can be considered in environment n = 3). Constraints related to congruencies appear in these last two cases.

In fact, we note the omnipresence of this type of expressions when we evaluate abundance factors literal expressions. We will call equation of decomposition into relative integers at sequence p (in shorter equation of decomposition), the aforementioned decomposition.

The approach touches with the intimacy of numbers. The node of the problem is to find these equations of decomposition.

We will see that it concerns in fact, not a single equation, but of a system of equations which can be expressed in matrix form. Before that, we work on the range of problems relating to environment 2d which admits a homogeneous particular approach.

Environment 2d

We place ourselves in the case (d,2) = 1. We note μ i,(d) the eigenvalues of the matrix of environment d and μ i the eigenvalues of the matrix of environment 2d. Of course, we have μ 0,(d) = μ 0 = p. We are brought to use circulant shifts of indice i of μ i . In these permutations from 1 to 2d, 0 are excluded and we will thus pass implicitly from 2d to 1 (and not to 0) in indices notations. We suppose that the eigenvalues μ i,(d) are known at each sequence p in environment d. We pose :

a i = μ i,(d) /p 1/2
We observe a conjecture which immediately erases all of environment 2d hidden ways, recalling that μ 0 is excluded in these sums : 2d ∑ (-1) i-1 .μ i .μ i+k = s k (39) i = 1 2.d.√p Indeed, we observe that s k are integers and moreover, indices k ranging 0 to 2d-1 :

2d-1 p = 1 ∑ s k 2 (40) d.(d-1) k = 0
We also observe, within the same conjectural framework, constraints on the values modulo 2d of s k (in the case of real eigenvalues) : s 0 = -1+d mod 2d s 2k = -1 mod 2d s 2k+1 = 0 mod 2d (41)

Moreover

s k = (-1) k .s 2d-k (42)
This decomposition of p is the key to get literal expressions of eigenvalues associated with environment 2d when literal expressions of eigenvalues associated with environment d are known. Indeed, let us suppose that eigenvalues associated with environment 2d, except μ 0 which one knows equal to p (a 0 is excluded from notations), are like : k = d μ j = (a j +(-1) j ).p 1/2 +(-1) j .∑ r k .a j+k-1 (43) k = 1The coefficients r k are supposed relative integers and they form, when the preceding expression is put in matrix form, a right circulant matrix [r d ].

Let us form the sum of products of two factors μ j with alternative signs : j = 2d j = 2d ∑ -(-1) j .μ j .μ j+m = ∑ -(-1) j .((a j +(-1) j ).p 1/2 +(-1) j .∑r k .a j+k-1 ).((a j+m +(-1) j+m ).p 1/2 +(-1) j+m .∑r k .a j+k+m-1 ) j = 1 j = 1

As, with indices varying from 1 to d, ∑ a j = 0 with indices varying from 1 to 2d, because of signs alternation, for a sum bearing on j (k and m given indices) ∑ (-1) j .a j+k .a j+m = 0 it follows that we are left with terms factors of p 1/2 , j = 2d j = 2d ∑ -(-1) j .μ j .μ j+m = ∑ -(-1) j .((a j +(-1) j ).(-1) j+m .∑r k .a j+k+m-1 +(-1) j .∑r k .a j+k .(a j+m-1 +(-1) j+m )).p 1/2 j = 1 j = 1 being still simplified for the same reasons of signs alternation j = 2d j = 2d ∑ -(-1) j .μ j .μ j+m = (∑ a j .(-1) m .∑ r k .a j+k+m-1 +∑ r k .a j+k .a j+m-1 ).p 1/2 j = 1 j = 1

In the second member, we clarify the sum with implicit indice k : j = 2d k = d j = 2d ∑ -(-1) j .μ j .μ j+m = p 1/2 .∑ r k ∑ (-1) m .a j .a j+k+m-1 +a j+k .a j+m-1 j = 1 k = 1 j = 1

We then use properties 16.

s 1 = -1+d -1 -1 -1 -1 -1 … -1 -1 -1 -1 -1 r 1 s 2 0 d/2 0 0 0 0 … 0 0 0 0 -d/2 r 2 s 3 -1 -1 -1+d/2 -1 -1 -1 … -1 -1 -1 -1+d/2 -1 r 3 s 4 0 0 0 d/2 0 0 … 0 0 -d/2 0 0 r 4 s 5 -1 -1 -1 -1 -1+d/2 -1 … -1 -1+d/2 -1 -1 -1 r 5 … … … … … … … … … … … … … … s d-3 0 0 0 0 -d/2 0 … 0 d/2 0 0 0 r d-3 s d-2 -1 -1 -1 -1+d/2 -1 -1 … -1 -1 -1+d/2 -1 -1 r d-2 s d-1 0 0 -d/2 0 0 0 … 0 0 0 d/2 0 r d-1 s d -1 -1+d/2 -1 -1 -1 -1 … -1 -1 -1 -1 -1+d/2 r d
The first line of the table is special. At semi-table, the elements comprising d/2 do cross.

As each line of matrix is with null sum, the determinant is null. The hypothesis that the eigenvalues μ j are of supposed form (relation 43) thus implies that the integers s k are interdependent :

s 1 = -2.(s 3 +s 5 +s 7 +…+s d ) (44)
Expressions of r k function of s k are obtained by a simple process. We can, indeed, write : We can write this using a matrix of a remarkable simplicity :

s 1 = d.
r 1 -r 1 = 1/d 0 0 0 0 0 … 0 0 0 0 s 1 r 2 -r 1 -1 1 0 0 0 … 0 0 0 1 s 2 r 3 -r 1 -1 0 1 0 0 … 0 0 -1 0 s 3 r 4 -r 1 -1 0 0 1 0 … 0 1 0 0 s 4 r 5 -r 1 -1 0 0 0 1 … -1 0 0 0 s 5 … … … … … … … … … … … … r d-3 -r 1 -1 0 0 0 1 … 1 0 0 0 s d-3 r d-2 -r 1 -1 0 0 -1 0 … 0 1 0 0 s d-2 r d-1 -r 1 -1 0 1 0 0 … 0 0 1 0 s d-1 r d -r 1 -1 -1 0 0 0 … 0 0 0 1 s d
The choice of r 1 being arbitrary, we can adopt the following one quite simply :

r 1 = 0 (45) P 211/390
Fermat-Catalan abundance factors

The decomposition of p = d.(d-1) ∑s k 2 allows to get the integers s k which then give the values of r k (which are also integers, then : k = d μ j = (a j +(-1) j ).p 1/2 +(-1) j .∑ r k .a j+k-1 k = 1 So that : k = d μ j = a j .p 1/2 +(-1) j .p 1/2 +∑ r j-k+1 .a k .p 1/2 .(-1) j .p 1/2 /p k = 1 Then in vector notations : k = d (μ j ) = (μ j mod d,(d) )+(μ j mod 2,( 2 )/3 In addition : p = 3y 2 +z 2 = (2z) 2 /6+((3y) 2 +z 2 )/3 Hence the possible choice :

s 0 = -2z s 1 = 3y s 2 = z Then : r 1 0 0 0 -2z 0 r 2 -r 1 = 1/3 -1 1 1 3y = y+z r 3 -r 1 -1 -1 1 z -y+z
Thus, by carrying out a translation of -z on the whole r i set, it follows r 1 = -z, r 2 = y, r 3 = -y. So that :

(-z) (y) (-y) r = (-y) (-z) (y) (y) (-y) (-z)
Then, by respecting the alternation of the signs at each line for matrix [r] :

μ 1 a -1 (-z) (y) (-y) a μ 2 = p 1/2 . b + p 1/2 . 1 + -(-y) -(-z) -(y) b μ 3 c -1 (y) (-y) (-z) c
What is the awaited result : k = 3 (μ j ) = (μ j mod 3,( 3) )+(μ j mod 2,( 2) )+∑ [r k ].(μ j mod 3,(3) .μ j mod 2,( 2) )/p k = 1

Inversely, expression of s i , function of r i , is of course :

s 1 -1+3 -1 -1 -z -2z s 2 = 0 3/2 -3/2 y = 3y s 3 -1 -1+3/2 -1+3/2 -y z Numerical example 2d = 30
We show with an example how to pass from environment 15 cardinal matrix eigenvalues (for a prime numbers variable) whose eigenvalues are 11.14787519, -24.7906862, -36.6228965, -6.61213164, 23.11108437, 0.56822968, -8.8566794, -54.2734895, -6.6933044, 51.55940736, -37.1364493, 43.55842799, 21.21869406, 26.3000555, -2.47813723 to environment 30 cardinal matrix eigenvalues. We seek first the thirty relative integers s k :

14, 30, -1, 0, 29, 30, -1, 0, -31, 0, -31, 30, 29, 0, -1, 0, -1, 0, 29, -30, -31, 0, -31, 0, -1, -30, 29, 0, -1, -30

P 212/390
Fermat-Catalan abundance factors

First half of this list, list which answers actually to p = 1/(d.(d-1)).∑ s k 2 , is used to constitute a column vector.

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 14 1 -1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 30 -1 -1 0 1 0 0 0 0 0 0 0 0 0 0 -1 0 -1 1 -1 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 -1 -1 0 0 0 1 0 0 0 0 0 0 -1 0 0 0 29 -1 -1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 30 -1 -1 0 0 0 0 0 1 0 0 -1 0 0 0 0 0 -1 -3 =1/15 -1 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 -3 -1 0 0 0 0 0 0 -1 1 0 0 0 0 0 0 -31 -1 -1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 -5 -1 0 0 0 0 -1 0 0 0 0 1 0 0 0 0 -31 3 -1 0 0 0 1 0 0 0 0 0 0 1 0 0 0 30 1 -1 0 0 -1 0 0 0 0 0 0 0 0 1 0 0 29 -1 -1 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 -3 -1 -1 0 0 0 0 0 0 0 0 0 0 0 0 1 -1
This column vector is the generator of the change of basis matrix of environment 15 matrix eigenvalues to environment 30 matrix eigenvalues Δμ 30 variations :

56,393716 0 1 -1 1 -1 -1 -1 -3 -3 -1 -5 3 1 -1 -3 11,14787519 -25,68434 -3 0 1 -1 1 -1 -1 -1 -3 -3 -1 -5 3 1 -1 -24,7906862 -1,614042 -1 -3 0 1 -1 1 -1 -1 -1 -3 -3 -1 -5 3 1 -36,6228965 -1,346757 1 -1 -3 0 1 -1 1 -1 -1 -1 -3 -3 -1 -5 3 -6,61213164 -25,05768 3 1 -1 -3 0 1 -1 1 -1 -1 -1 -3 -3 -1 -5 23,11108437 -49,77964 -5 3 1 -1 -3 0 1 -1 1 -1 -1 -1 -3 -3 -1 0,56822968 -15,82769 -1 -5 3 1 -1 -3 0 1 -1 1 -1 -1 -1 -3 -3 -8,8566794 3,1632938 = 1/ (61) 1/2 -3 -1 -5 3 1 -1 -3 0 1 -1 1 -1 -1 -1 -3 -54,2734895 56,259954 -3 -3 -1 -5 3 1 -1 -3 0 1 -1 1 -1 -1 -1 -6,6933044 6,1525426 -1 -3 -3 -1 -5 3 1 -1 -3 0 1 -1 1 -1 -1 51,55940736 -7,722526 -1 -1 -3 -3 -1 -5 3 1 -1 -3 0 1 -1 1 -1 -37,1364493 -9,313315 -1 -1 -1 -3 -3 -1 -5 3 1 -1 -3 0 1 -1 1 43,55842799 32,659471 1 -1 -1 -1 -3 -3 -1 -5 3 1 -1 -3 0 1 -1 21,21869406 13,397762 -1 1 -1 -1 -1 -3 -3 -1 -5 3 1 -1 -3 0 1 26,3000555 -31,68074 1 -1 1 -1 -1 -1 -3 -3 -1 -5 3 1 -1 -3 0 -2,47813723
We find then the sought eigenvalues by summation μ 30 = Δμ -31,68074439 -37,43559072 11,14787519 7,810249676 -56,39371559 -58,28527377 -24,79068623 -7,810249676 -25,68433786 -27,19860458 -36,62289649 7,810249676 1,614042231 -15,76913809 -6,612131636 -7, Let us note that on the basis of eigenvalues associated with environment 15 in the good order, we find eigenvalues of environment 30 in the good order. 958 0,257 -1,57 0,051 -0,36 1,905 -1,12 -1,69 1,359 1,741 -1,79 1,989 1,502 0,657 -1,43 -1,28 -1,98 -0,94 -0,56 1,833 -1,93 0,459 -2 -1,87 1,63 -0,75 0,848 -0,15 87 1,75 0,15 1,03 1,2 1,958 0,57 0,36 1,69 1,359 1,79 1,989 1,502 0,56 1,93 0,459 0,87 1,75 0,15 1,03 1,2 1,958 0,57 0,36 1,69 1,359 1,79 1,989 1,502 0,56 1,93 0,87 1,75 0,15 1,03 1,2 1,958 0,57 0,36 1,69 1,359 1,79 1,989 1,502 0,56 1,833 1 ,833 -1,93 0,459 -2 -1,87 1,63 -0,75 0,848 -0,15 1,03 1,2 1,958 0,257 -1,57 0,051 -0,36 1,905 -1,12 -1,69 1,359 1,741 -1,79 1,989 1,502 0,657 -1,43 -1,28 -1,98 -0,94 -0,56 -0,56 1,833 -1,93 0,459 -2 -1,87 1,63 -0,75 0,848 -0,15 1,03 1,2 1,958 0,257 -1,57 0,051 -0,36 1,905 -1,12 -1,69 1,359 1,741 -1,79 1,989 1,502 0,657 -1,43 -1,28 -1,98 -0,94 -0,94 -0,56 1,833 -1,93 0,459 -2 -1,87 1,63 -0,75 0,848 -0,15 1,03 1,2 1,958 0,257 -1,57 0,051 -0,36 1,905 -1,12 -1,69 1,359 1,741 -1,79 1,989 1,502 0,657 -1,43 -1,28 -1,98 -1,98 -0,94 -0,56 1,833 -1,93 0,459 -2 -1,87 1,63 -0,75 0,848 -0,15 1,03 1,2 1,958 0,257 -1,57 0,051 -0,36 1,905 -1,12 -1,69 1,359 1,741 -1,79 1,989 1,502 0,657 -1,43 -1,28 -1,28 -1,98 -0,94 -0,56 1,833 -1,93 0,459 -2 -1,87 1,63 -0,75 0,848 -0,15 1,03 1,2 1,958 0,257 -1,57 0,051 -0,36 1,905 -1,12 -1,69 1,359 1,741 -1,79 1,989 1,502 0,657 -1,43 -1,43 -1,28 -1,98 -0,94 -0,56 1,833 -1,93 0,459 -2 -1,87 1,63 -0,75 0,848 -0,15 1,03 1,2 1,958 0,257 -1,57 0,051 -0,36 1,905 -1,12 -1,69 1,359 1,741 -1,79 1,989 1,502 0,657 0,657 -1,43 -1,28 -1,98 -0,94 -0,56 1,833 -1,93 0,459 -2 -1,87 1,63 -0,75 0,848 -0,15 1,03 1,2 1,958 0,257 -1,57 0,051 -0,36 1,905 -1,12 -1,69 1,359 1,741 -1,79 1,989 1,502 1,502 0,657 -1,43 -1,28 -1,98 -0,94 -0,56 1,833 -1,93 0,459 -2 -1,87 1,63 -0,75 0,848 -0,15 1,03 1,2 1,958 0,257 -1,57 0,051 -0,36 1,905 -1,12 -1,69 1,359 1,741 -1,79 1,989 1,989 1,502 0,657 -1,43 -1,28 -1,98 -0,94 -0,56 1,833 -1,93 0,459 -2 -1,87 1,63 -0,75 0,848 -0,15 1,03 1,2 1,958 0,257 -1,57 0,051 -0,36 1,905 -1,12 -1,69 1,359 1,741 -1,79 -1,79 1,989 1,502 0,657 -1,43 -1,28 -1,98 -0,94 -0,56 1,833 -1,93 0,459 -2 -1,87 1,63 -0,75 0,848 -0,15 1,03 1,2 1,958 0,257 -1,57 0,051 -0,36 1,905 -1,12 -1,69 1,359 1,741 1,741 -1,79 1,989 1,502 0,657 -1,43 -1,28 -1,98 -0,94 -0,56 1,833 -1,93 0,459 -2 -1,87 1,63 -0,75 0,848 -0,15 1,03 1,2 1,958 0,257 -1,57 0,051 -0,36 1,905 -1,12 -1,69 1,359 1,359 1,741 -1,79 1,989 1,502 0,657 -1,43 -1,28 -1,98 -0,94 -0,56 1,833 -1,93 0,459 -2 -1,87 1,63 -0,75 0,848 -0,15 1,03 1,2 1,958 0,257 -1,57 0,051 -0,36 1,905 -1,12 -1,69 -1,69 1,359 1,741 -1,79 1,989 1,502 0,657 -1,43 -1,28 -1,98 -0,94 -0,56 1,833 -1,93 0,459 -2 -1,87 1,63 -0,75 0,848 -0,15 1,03 1,2 1,958 0,257 -1,57 0,051 -0,36 1,905 -1,12 -1,12 -1,69 1,359 1,741 -1,79 1,989 1,502 0,657 -1,43 -1,28 -1,98 -0,94 -0,56 1,833 -1,93 0,459 -2 -1,87 1,63 -0,75 0,848 -0,15 1,03 1,2 1,958 0,257 -1,57 0,051 -0,36 1,905 1,905 -1,12 -1,69 1,359 1,741 -1,79 1,989 1,502 0,657 -1,43 -1,28 -1,98 -0,94 -0,56 1,833 -1,93 0,459 -2 -1,87 1,63 -0,75 0,848 -0,15 1, 03 1,2 1,958 0,57 0,36 1,69 1,359 1,79 1,989 1,502 0,56 1,93 0,87 1,75 0,15 1,03 1,2 1,958 0,57 0,051 0,36 1,69 1,359 1,79 1,989 1,502 0,56 1,93 0,87 1,75 0,15 1,03 1,2 1,958 0,57 0,36 1,69 1,359 1,79 1,989 1,502 0,56 1,93 0,87 1,75 0,15 1,03 1,2 1,958 0,257 0,57 0,36 1,69 1,359 1,79 1,989 1,502 0,56 1,93 0,87 1,75 0,15 1,03 1,2 1, 958 1,2 1,15 0,75 1,93 1,43 0,657 1,502 1,79 1,741 1,12 1,36 0,57 0,257 0,257 1,958 1,2 1,15 0,75 1,93 1,43 0,657 1,502 1,79 1,741 1,12 1,36 0,57 0,257 1,958 1,2 1,15 0,75 1,93 1,43 0,657 1,502 1,79 1,741 1,12 1,36 0,051 0,57 0,257 1,958 1,2 1,15 0,75 1,93 1,43 0,657 1,502 1,79 1,741 1,12 1,36 0,57 0,257 1,958 1,2 1,15 0,75 1,93 1,43 0,657 1,502 1,79 1,741 1,12 1,905 1,36 0,57 0,257 1,958 1,2 1,15 0,75 1,93 1,43 0,657 1,502 1,79 1,741 1,12 1,36 0,57 0,257 1,958 1,2 1,15 0,75 1,93 1,43 0,657 1,502 1,79 1,741 1,12 1,36 0,57 0,257 1,958 1,2 1,15 0,75 1,93 1,43 0,657 1,502 1,79 1,741 1,359 1,12 1,36 0,57 0,257 1,958 1,2 1,15 0,75 1,93 1,43 0,657 1,502 1,79 1,741 1,741 1,12 1,36 0,57 0,257 1,958 1,2 1,15 0,75 1,93 1,43 0,657 1,502 1,79 1,741 1,12 1,36 0,57 0,257 1,958 1,2 1,15 0,75 1,93 1,43 0,657 1,502 1,989 1,79 1,741 1,12 1,36 0,57 0,257 1,958 1,2 1,15 0,75 1,93 1,43 0,657 1,502 1,502 1,79 1,741 1,12 1,36 0,57 0,257 1,958 1,2 1,15 0,75 1,93 1,43 0,657 0,657 1,502 1,79 1,741 1,12 1,36 0,57 0,257 1,958 1,2 1,15 0,75 1,93 1,43 0,657 1,502 1,79 1,741 1,12 1,36 0,57 0,257 1,958 1,2 1,15 0,75 1,93 1,43 0,657 1,502 1,79 1,741 1,12 1,36 0,57 0,257 1,958 1,2 1,15 0,75 1,93 1,43 0,657 1,502 1,79 1,741 1,12 1,36 0,57 0,257 1,958 1,2 1,15 0,75 1,93 1,43 0,657 1,502 1,79 1,741 1,12 1,36 0,57 0,257 1,958 1,2 1,15 0,75 1,93 1,43 0,657 1,502 1,79 1,741 1,12 1,36 0,57 0,257 1,958 1,2 1,15 0,75 1,93 1,833 1,43 0,657 1,502 1,79 1,741 1,12 1,36 0,57 0,257 1,958 1,2 1,15 0,75 1,93 1,43 0,657 1,502 1,79 1,741 1,12 1,36 0,57 0,257 1,958 1,2 1,15 0,75 1,459 0,93 1,43 0,657 1,502 1,79 1,741 1,12 1,36 0,57 0,257 1,958 1,2 1,15 0,75 1,-2 0,459 -1,93 1,833 -0,56 -0,94 -1,98 -1,28 -1,43 0,657 1,502 1,989 -1,79 1,741 1,359 -1,69 -1,12 1,905 -0,36 0,051 -1,57 0,257 1,958 1,2 1,03 -0,15 0,848 -0,75 1,63 -1,87 -1,87 -2 0,459 -1, 93 1,833 -0,56 -0,94 -1,98 -1,28 -1, 43 0,657 1,502 1,79 1,741 1,12 1,36 0,57 0,257 1,958 1,2 1,15 0,75 1,63 1,93 1,43 0,657 1,502 1,79 1,741 1,12 1,36 0,57 0,257 1,958 1,2 1,15 0,75 1,93 1,43 0,657 1,502 1,79 1,741 1,12 1,36 0,57 0,257 1,958 1,2 1,15 0,848 0,75 1,93 1,43 0,657 1,502 1,79 1,741 1,12 1,36 0,57 0,257 1,958 1,2 1,15 0,75 1,93 1,43 0,657 1,502 1,79 1,741 1,12 1,36 0,57 0,257 1,958 1,2 1,03 1,15 0,75 1,93 1,43 0,657 1,502 1,79 1,741 1,12 1,36 0,57 0,257 1,958 1,2 1,2 1,03 -0, 15 0,848 -0,75 1,63 -1,87 -2 0,459 -1,93 1,833 -0,56 -0,94 -1,98 -1,28 -1,43 0,657 1,502 1,989 -1,79 1,741 1,359 -1,69 -1,12 1,905 -0,36 In this expression, the coefficients r k are integers (or a right circulant matrix with integer coefficients when using matrix notations).

Matrix P 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1,
Example 1 : d = 15 = 3.5

The example is applied at sequence p = 61. The eigenvalues μ i,( 5) are roughly -8.4734, 3.30369, -23.226, 4.33154, 24.0641. The eigenvalues μ i,(3) are 13.6914, -13.358, -0.3335. We verify that : 11,1479 -8,4734 13,6914 5,92989 -24,791 3,30369 -13,358 -14,736 -36,623 -23,226 -0,3335 -13,063 -6,6121 4,33154 13,6914 -24,635 23,1111 24,0641 -13,358 12,4049 0,56823 -8,4734 -0,3335 9,37521 -8,8567 3,30369 13,6914 -25,852 -54,273 = -23,226 + -13,358 + -17,69 -6,6933 4,33154 -0,3335 -10,691 51,5594 24,0641 13,6914 13,8039 -37,136 -8,4734 -13,358 -15,305 43,5584 3,30369 -0,3335 40,5883 21,2187 -23,226 13,6914 30,7532 26,3001 4,33154 -13,358 35,3264 -2,4781 24,0641 -0,3335 -26,209 where: -8,4734 -13,358 -0,01806 3,30369 -0,3335 -5,21305 -23,226 13,6914 -0,94853 4,33154 -13,358 -0,13158 24,0641 -0,3335 We verify well, in this particular case, the existence of a right circulant matrix allowing to go from the eigenvalues of the initial environments to the eigenvalues of the product environment.

5,92989 -2 -3 1 -2 0 3 -3 -2 -1 3 0 2 0 2 -1,90186 -14,736 2 -2 -3 1 -2 0 3 -3 -2 -1 3 0 2 0 -0,72345 -13,063 2 2 -2 -3 1 -2 0 3 -3 -2 -1 3 0 2 0,126995 -24,635 0 2 2 -2 -3 1 -2 0 3 -3 -2 -1 3 0 0,972213 12,4049 2 0 2 2 -2 -3 1 -2 0 3 -3 -2 -1 3 -5,26961 9,37521 0 2 0 2 2 -2 -3 1 -2 0 3 -3 -2 -1 0,046331 -25,852 3 0 2 0 2 2 -2 -3 1 -2 0 3 -3 -2 -1 0,741512 -17,69 = -1 3 0 2 0 2 2 -2 -3 1 -2 0 3 -3 -2 5,086052 -10,691 -2 -1 3 0 2 0 2 2 -2 -3 1 -2 0 3 -3 -0,02368 13,8039 -3 -2 -1 3 0 2 0 2 2 -2 -3 1 -2 0 5,401185 -15,305 3 -3 -2 -1 3 0 2 0 2 2 -2 -3 1 -2 1,855532 40,5883 0 3 -3 -2 -1 3 0 2 0 2 2 -2 -3 1 -2 -0,01806 30,7532 -2 0 3 -3 -2 -1 3 0 2 0 2 2 -2 -3 -5,21305 35,3264 1 -2 0 3 -3 -2 -1 3 0 2 0 2 2 -2 -3 -0,94853 -26,209 -3 1 -2 0 3 -3 -2 -1 3 0 2 0 2 2 -2 -0,
P 215/390

Fermat-Catalan abundance factors

Example 2 : d = 12

The example is applied at sequence p = 193. We already examined this example by constituting the "brick" of eigenvalues averages as follows:

58, 723704 2,3828543 -5,542397 17,898724 18,365721 -50,92077 54,48156 -24,33113 -88,185776 -27,23903 -7,964582 57,09326 -53,31949 39,684041 8,8733076 -0,053882 37,985892 -27,73101 -10,201004 We verify then the right circulant matrix existence with :

Comment

The proof of the existence of a matrix (r d ) with integer components, using the general expression of cardinal matrices eigenvalues (see exercise (5) relation 72), remains an open question. For as much, we develop thereafter a conjectural method to characterize them.

Infinity of integer solutions

We have ∑μ i .μ j = ∑μ i mod d1 .μ j mod d2 , then ∑ r k .μ i mod d1 .μ j mod d2 = 0 owing to the fact that ∑μ . Moreover according to Bachet/Bézout theorem, there is (u,v) such as u.d 1 +v.d 2 = 1. Thus, there is at least a variation of the first line of [r d ']-[r d ] equal to the unit. From there, we can build a solution such that the components sum of a line of [r d ] is null. By the same modulo arguments, there is an infinity of solutions such that the sum of a line is null. In the two preceding examples, we gave such matrices.

The fact of the infinity of solutions allows to find easily one solution (on a computer). However, it would be more useful to have a solution starting from a systematic method. To arrive there, we will have recourse to three matrix entities : environment founding matrices, constructive of sequence matrices and increment of sequence matrices. They are presented in this order for reasons of methodical logic but their discovery was made in an exactly opposite order.

Elementary environment founding matrix

Let us consider the column vector (t d1 ) with d 1 components (d 1 -1, -1, -1,… -1) where d 1 = p 1 r1 . Here d 1 is power of only one prime number. We build the elementary environment founding matrix [t d1 ] starting from this column vector.

d 1 -1 -1 … -1 -1 d 1 -1 … -1 … … … … -1 -1 … d 1 -1

Compound environment founding matrix

Let us consider the vector with d i components (d i -1, -1, -1,…, -1). We repeat, in this order, this succession of numbers d/d i times to form a vector v i with d components. The components of same row of the vectors v i , i = 1 to k, are multiplied to form the column vector (t d ).

Let us have then respectively the right [t d ] and left [t' d ] circulant matrices of dimension d whose first column thus presents as follows:

(-1) k .(d 1 -1).(d 2 -1)…(d k -1) in position 1 (-1) k-1 .(d 2 -1)…(d k -1) in position 1 mod d/d 1 … (-1) k-1 .(d 1 -1)…(d k-1 -1) in position 1 mod d/d k … -(d 1 -1) in position 1 mod d 1 … -(d k -1) in position 1 mod d k 1 otherwise
We call [t' d ] the (left) environment founding d matrix.

It is an invariant of the environment as this matrix does not use the concept of sequence (p).

Properties

[t d ] is a similitude and [t' d ] is in some way also. Indeed, the reader will be able to easily verify (initially for d of type p i ) that (relations 48) :

[t d ].[t d ] = d.[t d ] [t' d ].[t' d ] = d.[t d ] and [t' d ].[t d ] = [t d ].[t' d ] = d.[t' d ]
The sum of each line and column is equal to zero : ∑ (t i,d ) = 0 (49) We also have :

∑ (t i,d ) 2 = d.(d 1 -1).(d 2 -1)…(d k -1) (50)
P 217/390

Fermat-Catalan abundance factors

The environment founding matrix is not invertible (null determinant) because of property (49). Moreover, the sum of the whole set of n positions spaced components is also null when n is a divisor of d :

∑ (t i,d ) = 0) (51) i = c mod n n \ d Example d = 12 We have d 1 = 4, d 2 = 3, d 1 -1 = 3, d 2 -1 = 2. Then : [t' 12 ] = 6 1 1 -2 -3 1 -2 1 -3 -2 1 1 1 1 -2 -3 1 -2 1 -3 -2 1 1 6 1 -2 -3 1 -2 1 -3 -2 1 1 6 1 -2 -3 1 -2 1 -3 -2 1 1 6 1 1 -3 1 -2 1 -3 -2 1 1 6 1 1 -2 1 -2 1 -3 -2 1 1 6 1 1 -2 -3 -2 1 -3 -2 1 1 6 1 1 -2 -3 1 1 -3 -2 1 1 6 1 1 -2 -3 1 -2 -3 -2 1 1 6 1 1 -2 -3 1 -2 1 -2 1 1 6 1 1 -2 -3 1 -2 1 -3 1 1 6 1 1 -2 -3 1 -2 1 -3 -2 1 6 1 1 -2 -3 1 -2 1 -3 -2 1
Foot-note

Owing to the fact that environment matrix is circulant, we have the stability of the product by circulant shift of the first column. Thus for example (shift of a notch of the first column) :

1 1 -2 -3 1 -2 1 -3 -2 1 1 6 1 1 -2 -3 1 -2 1 -3 -2 1 1 6 6 1 1 -2 -3 1 -2 1 -3 -2 1 1 1 -2 -3 1 -2 1 -3 -2 1 1 6 1 1 -2 -3 1 -2 1 -3 -2 1 1 6 1 1 1 -2 -3 1 -2 1 -3 -2 1 1 6 -2 -3 1 -2 1 -3 -2 1 1 6 1 1 -2 -3 1 -2 1 -3 -2 1 1 6 1 1 1 -2 -3 1 -2 1 -3 -2 1 1 6 1 -3 1 -2 1 -3 -2 1 1 6 1 1 -2 -3 1 -2 1 -3 -2 1 1 6 1 1 -2 -2 -3 1 -2 1 -3 -2 1 1 6 1 1 1 -2 1 -3 -2 1 1 6 1 1 -2 -3 1 -2 1 -3 -2 1 1 6 1 1 -2 -3 -3 1 -2 1 -3 -2 1 1 6 1 1 -2 -2 1 -3 -2 1 1 6 1 1 -2 -3 1 -2 1 -3 -2 1 1 6 1 1 -2 -3 1 1 -2 1 -3 -2 1 1 6 1 1 -2 -3 1 -3 -2 1 1 6 1 1 -2 -3 1 -2 . 1 -3 -2 1 1 6 1 1 -2 -3 1 -2 = 12. -2 1 -3 -2 1 1 6 1 1 -2 -3 1 -3 -2 1 1 6 1 1 -2 -3 1 -2 1 -3 -2 1 1 6 1 1 -2 -3 1 -2 1 1 -3 -2 1 1 6 1 1 -2 -3 1 -2 -2 1 1 6 1 1 -2 -3 1 -2 1 -3 -2 1 1 6 1 1 -2 -3 1 -2 1 -3 -3 -2 1 1 6 1 1 -2 -3 1 -2 1 1 1 6 1 1 -2 -3 1 -2 1 -3 -2 1 1 6 1 1 -2 -3 1 -2 1 -3 -2 -2 1 1 6 1 1 -2 -3 1 -2 1 -3 1 6 1 1 -2 -3 1 -2 1 -3 -2 1 1 6 1 1 -2 -3 1 -2 1 -3 -2 1 1 1 6 1 1 -2 -3 1 -2 1 -3 -2 6 1 1 -2 -3 1 -2 1 -3 -2 1 1 6 1 1 -2 -3 1 -2 1 -3 -2 1 1 1 6 1 1 -2 -3 1 -2 1 -3 -2 1

Uniqueness of the environment founding matrices

The environment founding matrix is unique by construction.

Other We illustrate this result, which is shown without difficulty, with the example of the assembly of the environment 3 and 4 founding matrices:

3 -1 -1 -1 3 -1 -1 -1 3 -1 -1 -1 . 2 -1 -1 2 -1 -1 2 -1 -1 2 -1 -1 =[0] -1 -1 -1 3 -1 -1 -1 3 -1 -1 -1 3 -1 -1 2 -1 -1 2 -1 -1 2 -1 -1 2 -1 -1 3 -1 -1 -1 3 -1 -1 -1 3 -1 -1 2 -1 -1 2 -1 -1 2 -1 -1 2 -1 -1 3 -1 -1 -1 3 -1 -1 -1 3 -1 -1 2 -1 -1 2 -1 -1 2 -1 -1 2 -1 -1 3 -1 -1 -1 3 -1 -1 -1 3 -1 -1 -1 -1 -1 2 -1 -1 2 -1 -1 2 -1 -1 2 -1 -1 -1 3 -1 -1 -1 3 -1 -1 -1 3 -1 2 -1 -1 2 -1 -1 2 -1 -1 2 -1 -1 -1 3 -1 -1 -1 3 -1 -1 -1 3 -1 2 -1 -1 2 -1 -1 2 -1 -1 2 -1 -1 -1 3 -1 -1 -1 3 -1 -1 -1 3 -1 -1 -1 -1 2 -1 -1 2 -1 -1 2 -1 -1 2 3 -1 -1 -1 3 -1 -1 -1 3 -1 -1 -1 -1 2 -1 -1 2 -1 -1 2 -1 -1 2 -1 -1 -1 -1 3 -1 -1 -1 3 -1 -1 -1 3 2 -1 -1 2 -1 -1 2 -1 -1 2 -1 -1 -1 -1 3 -1 -1 -1 3 -1 -1 -1 3 -1 -1 -1 2 -1 -1 2 -1 -1 2 -1 -1 2 -1 3 -1 -1 -1 3 -1 -1 -1 3 -1 -1 -1 2 -1 -1 2 -1 -1 2 -1 -1 2 -1

Eigen-elements of environment founding matrices

Eigenvalues and eigenvectors are easy to get from the general case of circulant matrices eigen-elements. Matrix d ,…,μ d,d ), μ d1 = (μ 1,d1 , μ 2,d1 , …, μ d1,d1 ) and μ d2 = (μ 1,d2 , μ 2,d2 , …, μ d2,d2 ) the ordered vectors of Fermat-Catalan cardinal matrices (or standard cardinal matrices) eigenvalues obtained by the method of the principal diagonal or by incrementing of k in μ k,d = 1+d.∑ r e (-2πi/p).g^(k-1+r.d) .

We place ourselves in the case p = 1 mod 2d. Let us pose :

Δμ i,d = μ i,d -μ i,d1 -μ i,d2 (53)
From relations (16.2) and following ones in exercise 5, we draw immediately :

(if(m = 1 mod d,d-1,-1)) = (∑ μ i,d .μ i+m-1,d ) p.d
The sums are bearing here respectively on i = 0 to (p-1)/d 1 -1, i = 0 to (p-1)/d 2 -1 and i = 0 to (p-1)/d-1.

So that :

(t m,d1 ) = (∑ μ i,d1 .μ i+m-1,d1 ) (54) p.d 1 and (t m,d2 ) = (∑ μ i,d2 .μ i+m-1,d2 ) (55) p.d 2 but again (if(m = 1 mod d,d-1,-1)) = (∑ μ i,d .μ i+m-1,d ) (56) p.d
Then, with sum bearing on i = 1 to d, it follows : where matrix [Δμ i,d ] is a left circulant matrix based on the d components column vector (Δμ i,d ).

(t m,d1 ) = (∑ μ i,d1 .μ i+m-1,d )
From relations ( 54) and ( 55), the sum bearing on indices i = 1 to d, it follows also immediately :

(t m,d ) = (∑ μ i,d1 .μ i+m-1,d1 .μ i,d2 .μ i+m-1,d2 ) (62) p 2 .d
This last relation can be rewritten in matrix form :

[t d ] = [μ i,d1 .μ i,d2 ] 2 p 2 .d
It is a left circulant matrix built starting from d components column vector (μ i,d1 .μ i,d2 ). This can be written using relation ( 61 -306,3 199,269 7,68209 -28,04 141,626 66,7662 63,4706 -12,659 -115,15 -8,9521 -71,153 40,6999 -26,478 -57,814 102,351 -516,95 -243,7 -231,67 46,208 420,304 32,6762 259,715 -148,56 96,6477 211,028 -28,04 2611 1230,89 1170,14 -233,39 -2122,9 -165,04 -1311,8 750,338 -488,15 -1065,9 141,626 -516,95 580,275 551,632 -110,02 -1000,8 -77,804 -618,4 353,728 -230,12 -502,47 66,7662 -243,7 1230,89 524,404 -104,59 -951,37 -73,964 -587,87 336,268 -218,77 -477,67 63,4706 -231,67 1170,14 551,632 20,8614 189,753 14,7522 117,253 -67,069 43,6332 95,2721 -12,659 46,208 -233,39 -110,02 -104,59 1725,97 134,184 1066,52 -610,06 396,883 866,585 -115,15 420,304 -2122,9 -1000,8 -951,37 189,753 10,4321 82,9157 -47,428 30,8554 67,3719 -8,9521 32,6762 -165,04 -77,804 -73 6.97.12 1.97.12 1.97.12 -2.97.12 -3.97.12 1.97.12 -2.97.12 1.97.12 -3.97.12 -2.97.12 1.97.12 1.97.12

When p = 1+d mod 2d, some evolutions are to be taken into account in the preceding indices.

Constructive of sequence matrix

Definition

Let us have a left circulant matrix [s' d ] of dimension d with positive integer components such as:

[s' d ].[s' d ] = p.d.[t d ] (64)
The matrix [s' d ] will be called a "constructive of sequence matrix" if, moreover, we dictate two additional constraints :

p = 1 mod d (s' d ) = -(t' d ) mod d (65) (66)
The concept of sequence intervenes here thanks to prime factor p. In turn, we can also use :

Minor identities

These identities are given only as a bonus. We observe :

[s' d ].[s d ] = p.d.[t' d ] [s' d ].[t d ] = [s d ].[t' d ] = d.[s' d ] [s' d ].[t' d ] = [s d ].[t d ] = d.[s d ] [t d ].[s' d ] = d.[s' d ] [t d ] .[s d ] = d.[s d ] (68)
Important foot-note Preceding left and right circulant matrices do not commutate.

[

s d ].[s' d ] ≠ [s' d ].[s d ] (69)
We also observe on numerical examples that : This point obliges to a permanent vigilance when the literal problems are to be resolved and to retain the left constructive matrix as reference tool (on the left of a product of matrices).

[s d ].[s d ] ≠ p.d.[t d ] [s d ].[s' d ] ≠ p.d.[t' d ] and [t' d ] .[s' d ] ≠ d.[s d ] [t' d ] .[s d ] ≠ d.[s' d ] (70)

Polynomial of decomposition and matrix of decomposition

We proposed in the study of case 2d the equality :

p = 1 . ∑ (s i,d ) 2 d.(d 1 -1).(d 2 -1)
The concept of polynomial of decomposition is an incomplete concept because it offers many solutions which all are not suitable. The whole set of constraints is seized only by the matrix equality :

[s' d ].[s' d ] = p.d.[t d ]

Non-inversion

The constructive of sequence matrix is not invertible. Indeed, the square of the determinant of [s' d ] equals p.d times the determinant of [t d ], which is null. Hence the result.

Existence of the constructive of sequence matrices

Existence of a left circulant matrix results immediately from the relation ( 63) except that we seek a matrix with integer coefficients here what is not the case in (63).

Case d = 2 n

It is easy to check that there is no constructive of sequence matrix for d = 2 (the literal development led to the fact that a prime number has several factors what is absurd). In the same way, we could not highlight of such matrix for d = 2 n , which leads us to think that there is none. This is the Achilles' heel of the method which we develop thereafter since that prevents from a complete generalization.

Case d = ∏ p k i

When p k is an odd prime number, the equations brought into play are :

∑ s i,d 2 = d.(d-1).p i = 1 to d ∑ s i,d .s i+j,d = -d.p (j constant ≠ 0) i = 1 to d (71)
Let us reconsider properties 16.i of exercise 4 (cf. page 110) :

Σ μ i,d *.μ i,d = (d-1).d.p Σ μ k+1-I,d *.μ k+1-j,d = -d.p Σ μ k,d .μ k+1+(p-1)/2,d = (d-1).d.p Σ μ k,d .μ k+1-I,d = -d.p
In spite of second members identical to those of equations ( 71), these relations do not allow, a priori, to find the desired integer quantifications. We adopt the following alternatives :

If p = 1 mod 2d, d odd 2d s k,2d = ∑ (-1) i-1 .μ i,2d .μ i+k,2d (72)
If p = 1+d mod 2d, d odd d s k,d = 2. ∑ (-1) i-1 .Re(μ i,d ).Im(μ i+k,d ) (74) i = 1 d.√p (73)
The first expression is proposed higher (relation 39 of this exercise). The second is another conjecture. Let us show then how the set of constructive of sequence matrices is implemented. We consider the decomposition into positive integers at sequence p. p = a 2 +a.b+b 2

This decomposition is unique except exchange of a and b. Taking in account possible combinations modulo 6 of a and b, this exchange is no more to be considered : Family (a mod6, b mod 6) 1 (4,3) or (0,5)

We can then rewrite p (among other ways) as :

p = ((a+2b) 2 +(2a+b) 2 +(a-b) 2 +(-(a+2b)) 2 +(-(2a+b)) 2 +(-(a-b)) 2 )/12 p = ((a+2b).(2a+b)+(2a+b).(a-b)+(a-b).(-(a+2b))+(-(a+2b)).(-(2a+b))+(-(2a+b)).(-(a-b))+(-(a-b)).(a+2b))/6
Then first column of the constructive of sequence matrices can be given, among others possibilities, by the following rule (and the following table) :

' 6 ) 6 1 a+2b 2a+b a-b -(a+2b) -(2a+b) -(a-b) 2 -(a-b) a+2b 2a+b a-b -(a+2b) -(2a+b) 3 -(2a+b) -(a-b) a+2b 2a+b a-b -(a+2b) 4 -(a+2b) -(2a+b) -(a-b) a+2b 2a+b a-b 5 a-b -(a+2b) -(2a+b) -(a-b) a+2b 2a+b 6 2a+b a-b - Family (s' 6 ) 1 (s' 6 ) 2 (s' 6 ) 3 (s' 6 ) 4 (s' 6 ) 5 (s
-2 -3 1 -2 1 -3 -2 1 1
We carry out the sums ∑ t i mod 6 , i varying modulo 6. Then :

(1/2).∑ t 1 mod 6 (1/2).∑ t 2 mod 6 (1/2).∑ t 3 mod 6 (1/2).∑ t 4 mod 6 (1/2).∑ t 5 mod 6 (1/2).∑ t 6 mod 6 All p (6-2)/2 = 2 (1+1)/2 = 1 (1-3)/2 = -1 (-2-2)/2 = -2 (-3+1)/2 = -1 (1+1)/2 = 1
It is indeed the first column of environment 6 founding matrix.

The property is spread to the constructive of sequence matrix. For a given sequence, when we consider the constructive matrix of dimension d and the constructive matrix of dimension d 1 , the components of the second one are obtained by simple average modulo d/d 1 of the components of the first one. This property is employed of course in a reverse way, i.e. we seek the dimension d matrix elements from that of dimension d 1 by the study of the induced variations.

Example: Research of literal solutions in environment d = 12

Case p = 1+2k.d

The

-b+2u -(a+2b)+4v -(2a+b)-2u -(a-b)+2v a+2b-4u 2a+b-2v a-b-2u -(a+2b)-4v -(2a+b)+2u -(a-b)-2v -(a-b)+4u a+2b+2v 2a+b+2u a-b+4v -(a+2b)-2u -(2a+b)+2v -(a-b)-4u a+2b-2v 2a+b-2u a-b-4v -(a+2b)+2u -(2a+b)-2v -(2a+b)+4u -(a-b)+2v a+2b+2u 2a+b+4v a-b-2u -(a+2b)+2v -(2a+b)-4u -(a-b)-2v a+2b-2u 2a+b-4v a-b+2u -(a+2b)-2v -(a+2b)+4u -(2a+b)+2v -(a-b)+2u a+2b+4v 2a+b-2u a-b+2v -(a+2b)-4u -(2a+b)-2v -(a-b)-2u a+2b-4v 2a+b+2u a-b-2v a-b+4u -(a+2b)+2v -(2a+b)+2u -(a-b)+4v a+2b-2u 2a+b+2v a-b-4u -(a+2b)-2v -(2a+b)-2u -(a-b)-4v a+2b+2u 2a+b-2v 2a+b+4u a-b+2v -(a+2b)+2u -(2a+b)+4v -(a-b)-2u a+2b+2v 2a+b-4u a-b-2v -(a+2b)-2u -(2a+b)-4v -(a-b)+2u a+2b-2v
For (s 12 ) components squares sum, expressions in a.u, a.v, b.u and b.v are eliminated. There remain thus the squares resulting from environment d = 6 and the squares resulting from variations. Thus :

∑ (s i,12 ) 2 = 72p = 24p+(4u) 2 +(2v) 2 +(2u) 2 +(4v) 2 +(-2u) 2 +(2v) 2 +(-4u) 2 +(-2v) 2 +(-2u) 2 +(-4v) 2 +(2u) 2 +(-2v) 2
Then : p = u 2 +v 2

This shows actually the coherence of the hypothesis and the simplicity of the assemblies.

P 224/390

Fermat-Catalan abundance factors Indeed, let us summarize the situation. The preceding decomposition interests sequences 1 mod 4.

Constructive of sequence matrices components for environment 12 = 3.4 are generated by a lineage p = 1 mod 3 (either p = 1 mod 6) and a lineage p = 1 mod 4. Hence two generating decompositions : p = 1 mod 4 p = u 2 +v 2 p = 1 mod 6 p = a 2 +a.b+b 2 => p = 1 mod 12 p = u 2 +v 2 et p = a 2 +a.b+b 2 We laid down higher the rules for (a,b). For (u,v), these rules are (so that the conjecture may hold) : v = 0 mod 3 sign(u) = (-1) abs(u) mod 3 sign(v) = ? (expression to be found) Example: Research of literal solutions in environment d = 24

We can continue the reverse process the basis of d = 12. However, in spite of promising beginnings, the prospect singularly complicates for d = 24 as p is broken up not only according to the couples of variables (u,v) and (a,b), but also a system of eight integers' variables : a 1 a 2 a 3 a 4 a 5 a 6 a 7 a 8 a 9 = -a 1 +a 5 a 10 = -a 2 +a 6 a 11 = -a 3 +a 7 a 12 = -a 4 +a 8 a 13 = -a 1 a 14 = -a 2 a 15 = -a 3 a 16 = -a 4 a 17 = -a 5 a 18 = -a 6 a 19 = -a 7 a 20 = -a 8 a 21 = -a 9 a 22 = -a 10 a 23 = -a 11 a 24 = -a 12

This handicap is recurrent in the applications.

Splitting property

We wish to deduce by a simple method the constructive of sequence matrices for d = 3 k+1 from that for d =3 k when (s' d ) = -(t' d ) mod d. Unfortunately, an elementary procedure does not seem to exist (such as that relating to the "false matrices" developed below). We will be content with only some properties as initial ways of resolution. We have with indice i mod 3 k : s' i,3^k = (s' i,3^(k+1) +s' i+3^k,3^(k+1) +s' i+2.3^k,3^(k+1) )/3

Let us take again the values of the table and examine other ratios :

(p- The generalization of these ratios to the case d = p i k+1 is elementary. Their exploitation is indeed more difficult.

Enumeration of the solutions

The number of solutions for constructive of sequence matrix is a multiple of d.φ(d), redundancies included. Indeed, the circulant shift of a solution is also solution (hence factor d). In addition, there exist φ(d) solutions that we get bearing on the first column vector by pointing these components with a step relative prime with d (see below).

Method of the step relative prime with the environment

Numbers {1, 2,4,7,8,11,13,14} are The other solutions of this family are found by carrying out fifteen circulant shifts of this table.

Redundancies

According to the values of the solutions, redundancies appear, decreasing in fact the actual number of solutions. The solution with step 7 merges with that with step 2, step 11 with 1, step 13 with 8, the step 14 with 4.

Variations with the sequence

We give the solutions for (left) constructive of sequence matrices, with the constraints (s' d ) = -(t' d ) mod d.

We have 8 distinct solutions (s' 15 ) at p = 31 and 16 solutions at sequence p = 61. The opposites are of course also solutions.

The number of solutions depends on the sequence what may not, a priori, be the case when the constraints modulo d are eliminated. The process uses matrices which we call "false modulating" :

-for p k = 3 2 -2 1 Fmod(3) = 2 1 -2 -1 1 1 -for p k = 5 2 -4 1 1 1 2 1 -4 1 1 Fmod(5) = 2 1 1 -4 1 2 1 1 1 -4 -3 1 1 1 1 -for p k 2 -(pk-1) 1 1 … 1 1 2 1 -(pk-1) 1 … 1 1 2 1 1 -(pk-1) … 1 1 Fmod(pk) = … … … … … … … 2 1 1 1 … -(pk-1) 1 2 1 1 1 … 1 -(pk-1) -(pk-2) 1 1 1 … 1 1 P 227/390
Fermat-Catalan abundance factors

Foot-note: The sum of each line of the false modulating matrix is 1. The sum of each column equals 0 except the first column which equals p k .

(b) Implementation of the process

Matrix of dimension d = p i n is deduced from matrix of dimension p i n-1 by using lines 2 to p i n-1 components in lines 2+k.p i n-1 . The p i remaining components are provided by : (Fmod(p i )) n .[s' pi ]

Example 1

Let us start with an example to fix ideas. For p = 7, the successive iteration of the values gives (after a systematic research for d =3) : and for matrix [ns' 27 ] first column (-8, 4, 1, -8, 4, 1, 10, 4, 1, -62, 4, 1, -8, 4, 1, 10, 4, 1, 19, 4, 1, -8, 4, 1, 10, 4, 1) or for matrix [ns' 81 ] first column (127, 4, 1, -8, 4, 1, 10, 4, 1, -62, 4, 1, -8, 4, 1, 10, 4, 1, 19, 4, 1, -8, 4, 1, 10, 4, 1, -116, 4, 1, -8, 4, 1, 10, 4, 1, -62, 4, 1, -8, 4, 1, 10, 4, 1, 19, 4, 1, -8, 4, 1, 10, 4 Hence, for example, for matrix [ns' 25 ] first column (-51, 4, -11, 29, 9, 24, 4, -11, 29, 9, -176, 4, -11, 29, 9, -76, 4, -11, 29, 9, 124, 4, -11, 29, 9) and for matrix [ns' 125 ] first column (-326, 4, -11, 29, 9, 24, 4, -11, 29, 9, -176, 4, -11, 29, 9, -76, 4, -11, 29, 9, 124, 4, -11, 29, 9, 674, 4, -11, 29, 9, 24, 4, -11, 29, 9, -176, 4, -11, 29, 9, -76, 4, -11, 29, 9, 124, 4, -11, 29, 9, 174, 4, -11, 29, 9, 24, 4, -11, 29, 9, -176, 4, -11, 29, 9, -76, 4, -11, 29, 9, 124, 4, -11, 29, 9, -826, 4, -11, 29, 9, 24, 4, -11, 29, 9, -176, 4, -11, 29, 9, -76, 4, -11, 29, 9, 124, 4, -11, 29, 9, 49, 4, -11, 29, 9, 24, 4, -11, 29, 9, -176, 4, -11, 29, 9, -76, 4, -11, 29, 9, 124, 4, -11, 29, 9) We can easily find other distinct solutions by observing that the permutations are equivalent what we did on the first column of the table given higher, then by multiplying the following columns by the method of the modulating matrix : Hence for matrix [ns' 25 ] first column :

(59, -11, 29, 9, 31, -141, -11, 29, 9, 31, -41, -11, 29, 9, 31, 159, -11, 29, 9, 31, -16, -11, 29, 9, 31) P The passage of the matrix of dimension d = 3 n-1 to the matrix of dimension 3 n is done by following procedure : we reuse the components of lines 2 to 3 n-1 in lines 2+k.3 n-1 to 3 n-1 +k.3 n-1 . For the 3 components (bold police) in line 1, 1+3 n-1 and 1+2.3 n- 1 , we adopt the following routine : (ns' 9 ) = (3a+a, b,

(ns' 27 ) = (3(4a)+a, b, c, 3b+a, b, c, 3c+a, b, c, 3(3b+a) b, c, 3b+a, b, c, 3c+a, b, c, 9b+4a, b, c, 3b+a, b, c, 3c+a, b, c, 9c+4a, b, c, 3b+a, b, c, 3c+a, b, c 3(9b+4a)+a, b, c, 3b+a, b, c, 3c+a, b, c, 9b+4a, b, c, 3b+a, b, c, 3c+a, b, c, 9c+4a, b, c, 3b+a, b, c, 3c+a, b, c 3(9c+4a)+a, b, c, 3b+a, b, c, 3c+a, b, c, 9b+4a, b, c, 3b+a, b, c, 3c+a, b, c, 9c+4a, b, c, 3b+a, b, c, 3c+a, b, c) = (40a, b, c, 3b+a, b, c, 3c+a, b, c, 9b+4a, b, c, 3b+a, b, c, 3c+a, b, c, 9c+4a, b, c, 3b+a, b, c, 3c+a, b, c 27b+13a, b, c, 3b+a, b, c, 3c+a, b, c, 9b+4a, b, c, 3b+a, b, c, 3c+a, b, c, 9c+4a, b, c, 3b+a, b, c, 3c+a, b, c 27c+13a, b, c, 3b+a, b, c, 3c+a, b, c, 9b+4a, b, c, 3b+a, b, c, 3c+a, b, c, 9c+4a, b, c, 3b+a, b, c, 3c+a, b, c) ... (ns' 3^n ) = = (3ns' 1 , 3^(n-1) +a, (ns'' 3^(n-1) ), 3ns' 1+3^(n-2) , 3^(n-1) +a, (ns'' 3^(n-1) ), 3ns' 1+3^(n-2) , 3^(n-1) +a, (ns'' 3^(n-1) ))

+a, b, c, 3b+a, b, c, 3c+a, b, c, 3(3c+a)+a, b, c, 3b+a, b, c, 3c+a, b, c) = (13a, b, c, 3b+a, b, c, 3c+a, b, c, 9b+4a, b, c, 3b+a, b, c, 3c+a, b, c, 9c+4a, b, c, 3b+a, b, c, 3c+a, b, c) (ns' 81 ) = = (3.(13a)+a,
Here the three symbols (ns'' 3^(n-1) ) are column vectors (ns' 3^(n-1) ) to which we withdrew the first component which is replaced respectively by 3ns' 1 , 3^(n-1) +a, 3ns' 1+3^(n-2) , 3^(n-1) +a and 3ns' 1+3^(n-2) , 3^(n-1) +a. In this last writing, s' r,k is the r th component of (ns' k ).

Then Then we choose any step relative prime with d = 3.5.7, for example step = 2, and carry out a picking similar to the precedent in a movement modulo 7 in column and modulo 15 in line.

(ns' 105 ) = (696, -16, -13, 29, -16, 52, 29, 96, -13, 29, 64, -13, 29, -16, 78, -116, -16, -13, 29, -16, 52, -174, -16, -13, 29, 64, -13, 29, 96, -13, -116, -16, -13, 29, -16, -312, 29, -16, -13, 29, 64, -13, -174, -16, -13, -116, -16, -13, 29, 96, 52, 29, -16, -13, 29, 64, 78, 29, -16, -13, -116, -16, -13, -174, -16, 52, 29, -16, -13, 29, -384, -13, 29, -16, -13, -116, -16, 78, 29, -16, 52, 29, -16, -13, -174, 64, -13, 29, -16, -13, -116, 96, -13, 29, -16, 52, 29, -16, 78, 29, 64, -13, 29, -16, -13) If, we choose step 11, we fall on the same column vector, but with step 13, we get :

(ns' 105 ) = (696, -13, -16, 29, -13, 64, 29, 78, -16, 29, 52, -16, 29, -13, 96, -116, -13, -16, 29, -13, 64, -174, -13, -16, 29, 52, -16, 29, 78, -16, -116, -13, -16, 29, -13, -384, 29, -13, -16, 29, 52, -16, -174, -13, -16, -116, -13, -16, 29, 78, 64, 29, -13, -16, 29, 52, 96, 29, -13, -16, -116, -13, -16, -174, -13, 64, 29, -13, -16, 29, -312, -16, 29, -13, -16, -116, -13, 96, 29, -13, 64, 29, -13, -16, -174, 52, -16, 29, -13, -16, -116, 78, -16, 29, -13, 64, 29, -13, 96, 29, 52, -16, 29, -13, -16) and The modulating matrix gives us :

(t 105 ) = (48, -1, -1, 2, -1, 4, 2, 6, -1, 2, 4, -1, 2, -1, 6, -8, -1, -1, 2, -1, 4, -12, -1, -1, 2, 4, -1, 2, 6, -1, - 8, -1, -1, 2, -1, -24, 2, -1, -1, 2, 4, -1, -12, -1, -1, -8, -1, -1, 2, 6, 4, 2, -1, -1, 2, 4, 6, 2, -1, -1, -8, -1, -1, -12, -1, 4, 2, -1, -1, 2, -24, -1, 2, -1, -1, -8, -1, 6, 2, -1, 4, 2, -1, -1, -12, 4, -1, 2, -1, -1, -8, 6, -1, 2, -1, 4, 2, -1, 6, 2, 4, -1, 2, -1, -1) 22
43 2 -2 1 37 142 = 2 1 -2 -2 -74 -1 1 1 -35
From that, we draw the column vector :

(ns' 9 ) = (43, -2, -35, 142, -2, -35, -74, -2, -35) For this vector, we have [ns ' 9 ].[ns' 9 ] = 433.9.[t 9 ] where (t' 9 ) = (8, -1, -1, -1, -1, -1, -1, - We choose a step of 5 on the principal diagonal to get :

(ns' 105 ) = (-301, -35, -2, -74, -35, -2, 142, -35, 14, 43, -35, -2, -74, -35, -2, 142, 245, -2, 43, -35, -2, -74, -35, -2, -994, -35, -2, 43, -35, -2, -74, -35, 14, 142, -35, -2, 43, -35, -2, -74, 245, -2, 142, -35, -2, 43, -35, -2, 518, -35, -2, 142, -35, -2, 43, -35, 14, -74, -35, -2, 142, -35, -2 43, 245, -2, -74, -35, -2, 142, -35

(t 105 ) = (56, 1, 1, 1, 1, 1, 1, 1, - 7, -8, 1, 1, 1, 1, 1, 1, -7, 1, -8, 1, 1, 1, 1, 1, -7, 1, 1, -8, 1, 1, 1, 1, -7, 1, 1, 1, -8, 1, 1, 1, -7, 1, 1, 1, 1, -8, 1, 1, -7, 1, 1, 1, 1, 1, -8, 1, -7, 1, 1, 1, 1, 1, 1, -8, -7, 1, 1, 1, 1, 1, 1, 1) Example 2
: d = 3.2 n
For the equality [ns' 3 ].[ns' 3 ] = 3p.[t 3 ], we write first :

a b c 2 2 -1 -1 b c a = 3p. -1 2 -1 c a b -1 -1 2
After development, this equality summarizes, knowing that each line sum of [s' 3 ] is null, with equations : a 2 +a.b+b 2 = 3p c = -(a+b) Supposing this integers system solved, we form the following 2 n columns table :   -

(2 n -1).(ns' 3 ) (ns' 3 ) (ns' 3 ) (ns' 3 ) (ns' 3 ) (ns' 3 ) … (ns' 3 ) -(2 n -1).a a a a a a … a -(2 n -1).b b b b b b … b -(2 n -1).c c c c c c … c
Using a step of 5 on the principal diagonal, we get easily following column vectors : (-7a, c, b, a, c, b, a, c, -7b, a, c,b, a, c, b, a, -7c, b, a, c, b, a, c, b) (ns' 48 ) = (-15a, c, b, a, c, b, a, c, b, a, c, b, a, c, b, a, -15c, b, a, c, b, a, c, b, a, c, b, a, c, b, a, c, -15b, a, c, b, a, c, b, a, c, b, a, c, b, a, c, b) (ns' 96 ) = (-31a, c, b, a, c, b, a, c, b, a, c, b, a, c, b, a, c, b, a, c, b, a, c, b, a, c, b, a, c, b, a, c, -31b, a, c, b, a, c, b, a, c, b, a, c, b, a, c, b, a, c, b, a, c, b, a, c, b, a, c, b, a, c, b, a, -31c, b, a, c, b, a, c, b, a, c, b, a, c, b, a, c, b, a, c, b, a, c, b, a, c, b, a, c, b, a, c, b) The next course can be imagined without pains by taking account of the sequence a, c, b with "carriage return". Before continuing in a new register, we recall however that these examples do not give constructive of sequence matrices.

(ns' 6 ) = (-a, c, -b, a, -c, b) (ns' 12 ) = (-3a, c, b, a, -3c, b, a, c, -3b, a, c, b) (ns' 24 ) =
P 232/390

Fermat-Catalan abundance factors

Test of a prime number

Within the strict framework of our study, the concept of constructive of sequence matrix interests us only when p is a prime number in expression

[s' d ].[s' d ] = p.d.[t d ].
However, we can pose the question of equivalence between the uniqueness of integer solutions of this type of expression, except permutations, and p is a prime number. The construction of the solutions, for a given sequence p, of a construction of sequence matrix of dimension d = ∏p i ni starting from a matrix of dimension p i indicates us immediately that it is necessary to exclude a not prime dimension of such a test. We can give the preceding counterexample p = 61 and d = 15, with at least four distinct solutions : 23242526272829303152 -1 14 -13 -1 -56 -13 -1 14 -13 4 14 -13 -1 14

(s' d=15 ) 1 -
The question of equivalence can then be posed remembering that p = 1 mod d, which requires several choices of d to ensure the cover of prime numbers which is never complete in the absence of d The cover increases only moderately afterwards. This point being cleared up, equivalence remains to be shown what may not be a small matter. We will be satisfied here with some illustrations of the cases which appear during such a test : redundant + non-redundant solutions (14,-7,-7) (13,-11,-2) We see that the solutions types also open interesting fields of study.

Cases Situations (s' 3 ) 1 (s' 3 ) 2 d = 3, p = 7 one solution (5,-4,-1) d

Eigen-elements of the constructive of sequence matrices

Alternative strategy

The eigenvalues can be also used to evaluate elementary constructive of sequence matrices. This can be done only if remarkable identities appear during our investigations. Indeed, on the basis of environment founding matrix eigenvalues and relation In certain cases, the problem shows a "good disposition", elsewhere not. The eigenvalues of the elementary constructive matrices are solutions of an equation with integer coefficients of degree (d i -1)/2. The literal expressions of these coefficients depend on sequences p and are to be determined on a case-by-case basis.

Example d = 12

Eigenvalues matrix of [s 12 ] is :

[σ12](r,s) = ]. [σ' 12 ].[e (-πi/6).(r-1).(s-

]

Example d = 15

We immediately pass to the eigenvalues matrix of [σ' 15 ] : 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 x1 0 0 0 0 0 0 0 0 0 0 0 0 0 y1 0 0 0 0 0 0 0 0 0 0 0 0 0 x2 0 0 0 0 0 0 0 0 0 0 0 0 0 y2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 x3 0 0 0 0 0 0 0 0 0 0 0 0 0 y3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 x4 0 0 0 0 0 0 +i. 0 0 0 0 0 0 0 0 y4 0 0 0 0 0 0 0 0 0 0 0 0 0 x4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 y4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 x3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 y3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 x2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 y2 0 0 0 0 0 0 0 0 0 0 0 0 x1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 y1 0 0 0 0 0 0 0 0 0 0 0 0

The positions of the non-null values correspond to (r-1,15) ≠ 1.

We observe also : p = x i 2 +y i 2 However, none among these relations work with integers, or, a priori, in a simple kind.

Increment of sequence matrix

Let us have respectively the dimension d right [r d ] and left [r' d ] circulant matrices such as : Commutativity between circulant matrices formed thanks to (r d ) and (s d ) is not acquired. Particular vigilance in the use of the relations is essential. Thus :

[t d ].[r' d ] = [r' d ].[t d ] = [s' d ] [r d ].[t' d ] = [s' d ] [r d ].[t d ] = [t d ].[r d ] = [s d ] [r' d ].[t' d ] = [s d ] and [r' d ].[s' d ] = [s' d ].[r' d ] = p.[t d ] [r' d ].[s d ] = [s' d ].[r d ] = p.[t' d ] (76)
[t' d ].[r' d ] ≠ [s d ] [t' d ].[r d ] ≠ [s' d ] and [r d ].[s' d ] = [s d ].[r' d ] ≠ p.[t' d ] [r d ].[s d ] = [s d ].[r d ] ≠ p.[t d ]
The increment of sequence matrix is the right circulant matrix [r d ].

We can choose one of the preceding equalities to build this matrix, for example :

[t' d ].[r d ] = [s' d ] (77)
Foot-note:

The increment matrix for sequence p is obtained by groping as matrices [t' d ] and [s' d ] (like [t d ] and [s d ]) are not invertible (determinant = 0).

Non-uniqueness

For a given constructive of sequence matrix, the increment of sequence matrix is not unique.

Indeed, if [r d ] is solution of [t' d ].[r d ] = [s' d ], then any matrix [r d '] formed from (r d ') = (r d )+(if(k=c mod n,1,0)
) is also solution, c a given integer, n dividing d. This is an immediate consequence of ∑ (s' i,d ) = 0. Thus [r d ] has the property expressed higher concerning the infinity of the solutions of circulant matrices with integer components [r d ] giving Δμ i,d1.d2 = [r d ].μ i,d1 .μ i,d2 /p (with proviso of existence of at least one).

Even powers

The reader will be able to verify the following property which makes it possible to get [r d ] without recourse to [s' d ] :

[t' d ].[r d ] 2n = p 2n .[t' d ] (78)

Property

The term "of increment" is selected because the aforementioned matrix gives access the variations Δμ i,d1.d2 "by integer step". The increment matrix is a change of basis matrix. It makes it possible to get eigenvalues associated with an environment with composite (not only prime) order on the basis of eigenvalues associated environments d i factor of d.

For the implementation of this evaluation, we split d into two relative prime factors repeating the operation as many times as necessary.

Example d = 6

We higher found the forms of the constructive of sequence matrices. Using that, we verify easily what follows (it is a choice among the other possible ones) : family (s 6 ) 1 (s 6 ) 2 (s 6 ) 3 (s 6 ) 4 (s 6 ) 5 (s 6 ) 6 (r 6 ) 1 (r 6 ) 2 (r 6 ) 3 (r 6 ) 4 (r 6 ) 5 (r 6 ) 6

Family (r 6 ) 1 (r 6 ) 2 (r 6 ) 3 (r 6 ) 4 ( r
+2u -m 1 +4v -m 2 -2u -m 3 +2v m 1 -4u m 2 -2v m 3 -2u -m 1 -4v -m 2 +2u -m 3 -2v with Family m 1 m 2 = m 1 +m 3 m 3 1 a+2b 2a+b a-b 2 -(a-b) a+2b 2a+b 3 -(2a+b) -(a-b) a+2b 4 -(a+2b) -(2a+b) -(a-b) 5 a-b -(a+2b) -(2a+b) 6 2a+b a-b - (a+2b)
It is then a question of solving : If we exchange the increment of sequence matrices in the two preceding systems, the result will have nothing to do with the eigenvalues of environment d = 12 cardinal matrix (at sequence 193). This shows well that to find a matrix of increment does not suffice. To a given increment of sequence matrix correspond such order of environments d 1 and d 2 eigenvalues.

The good permutation must be selected to get the eigenvalues of the product environment. This obliges to test the result -1 ] until obtaining a solution with integer components.

[P B ].[DA].[P B
Case p = 1+d+2k.d

Let us verify with a numerical example the existence of an increment of sequence matrix without imaginary component.

We choose d = 12 and p = 109. We start from the eigenvalues in environments d Among other choices, an increment of sequence matrix corresponds to it whose first column is :

p r1 r2 r3 r4 r5 r6 r7 r8 r9 r10 r11 r12 109 0 -2 -3 3 15 -9 -7 14 14 -16 -2 7
Let us use this same matrix for the evaluation of the real part and the imaginary part of the eigenvalues of the matrix of environment at sequence p :

Real part :

-9,26328 0 7 -2 -16 14 14 -7 -9 15 3 -3 -2 This matrix corresponds actually to the matrix of environment 12 when g = 6 and p = 109. The reader can note that it is necessary to carry out an adequate permutation of the eigenvalues so that the assembly works :

μ 1,3 μ 1,3 μ 1,4 μ 3,4 μ 2,3 → μ 3,3 μ 2,4 → μ 2,4 μ 3,3 μ 2,3 μ 3,4 μ 1,4 μ 4,4 μ 4,4
The preceding approach is a pretty alternative for the evaluation of eigenvalues associated an environment of great even value because it is not simple to find by trials and errors the characteristic polynomial of a determinant which does not split down to degree 1.

Environment ∏ p i ri

The general case is solved by iteration of operations on the model of case p 1 i1 .p 2 i2 .

Comments on the environment founding, constructive and increment of sequence matrices

At the stage of our study, the constructive of sequence matrices remain difficult to obtain. By trials and errors, we get easily the increment of sequence matrices. It shows that the method presented here is primarily interesting for solving equations P 239/390

Fermat-Catalan abundance factors

Environment p 1 r and eigenvalues variations

We describe here a method of determination of the eigenvalues by successive trials. Number p 1 being a prime, let us have [A] p1^(r-1), p1^(r-1) the cardinal matrix of x p1^(r-1) in environment p 1 (r-1) . We deduce matrix [A] p1^r, p1^r by splitting of the preceding matrix. Thus, for μ an eigenvalue of [A] p1^(r-1), p1^(r-1) , we seek p 1 eigenvalues in the form μ+Δμ i for [A] p1^r, p1^r . Research thus goes on the literal expression of the successive variations of observed eigenvalues. We thus get a pyramid of variations with preceding results.

Environment

Variations # possible variations

Sum of # possible variations

Eigenvalues # Eigenvalues

1 0 p 1 Δμ (1,i) p 1 p 1 Δμ (1,i) p 1 p 1 2 Δμ (2,j) p 1 2 p 1 3 Δμ (1,i) +Δμ (2,j) p 1 2 p 1 3 Δμ (3,k) p 1 3 p 1 6 Δμ (1,i) +Δμ (2,j) +Δμ (3,k) p 1 3 … … …
As the table indicates it, the choice of variations is surplus compared to the number of eigenvalues. All the combinations are not acceptable and are part of the difficulties rising from the suggested method.

Environment 2 k

It may be that this case has a literal solution for any k (in addition to the general form of page 72). We will give a beginning of development corresponding to this assertion within a conjectural framework. We use preceding paragraph method, as the case shows some simplicity owing to the fact that, to an acceptable variation, an acceptable variation of opposite sign corresponds. We go on here thanks to an attentive observation of these eigenvalues variations (the discovering process for these values having been in reality much more chaotic) that leads first to the table :

α k β k γ k (79) (∏Δμ i )/(a k .p 1/2 ) ((∏Δμ 2i+1 ) 2 -(∏Δμ 2i ) 2 )/(2a k .p 1/2 ) ((∏Δμ 2i+1 ) 2 +(∏Δμ 2i ) 2 )/(2a k .p)
The products bear on indices i, 2i+1, 2i included in the interval [1,d] and a k is an integer common divisor.

We have the remarkable identity which rises from the preceding expressions :

p = (α k 2 +β k 2 )/γ k 2 (80)
Numbers α k , β k and γ k turn out to be integers.

Numerical research

Prime numbers p with (p-1)/ Let us note that these tables were obtained starting from the eigenvalues and not starting from the identity p = (α k 2 +β k 2 )/γ k 2 , which would be the most powerful approach. The values a ki show as powers of 2. The signs of α k depend on the choice of the primitive roots g (in the preceding examples, the choice is that of the smallest one).

The literal expressions of the eigenvalues for cm = 2, cm = 4 and cm = 8 are given then by (± pending on g) :

Case p = 1+r.2.2 k -p 1/2 -±2 1/2 .p 1/2 .(γ 1 -β 1 /p 1/2 ) 1/2 -±4.p 1/4 .(γ 2 +β 2 /p 1/2 ) 1/4 .((γ 1 1/2 +(γ 1 -α 1 /p 1/2 ) 1/2 )/(γ 1 1/2 +(γ 1 +α 1 /p 1/2 ) 1/2 ) 1/2 ±4.p 1/4 .(γ 2 +β 2 /p 1/2 ) 1/4 .((γ 1 1/2 +(γ 1 -α 1 /p 1/2 ) 1/2 )/(γ 1 1/2 +(γ 1 +α 1 /p 1/2 ) 1/2 ) 1/2 ±2 1/2 .p 1/2 .( γ 1 -β 1 /p 1/2 ) 1/2 -±4.p 1/4 .(γ 2 +β 2 /p 1/2 ) 1/4 .((γ 1 1/2 -(γ 1 -α 1 /p 1/2 ) 1/2 )/(γ 1 1/2 -(γ 1 +α 1 /p 1/2 ) 1/2 ) 1/2 ±4.p 1/4 .(γ 2 +β 2 /p 1/2 ) 1/4 .((γ 1 1/2 -(γ 1 -α 1 /p 1/2 ) 1/2 )/(γ 1 1/2 -(γ 1 +α 1 /p 1/2 ) 1/2 ) 1/2 +p 1/2 -±2 1/2 .p 1/2 .(γ 1 +β 1 /p 1/2 ) 1/2 -±4.p 1/4 .(γ 2 -β 2 /p 1/2 ) 1/4 .((γ 1 1/2 +(γ 1 -α 1 /p 1/2 ) 1/2 )/(γ 1 1/2 -(γ 1 +α 1 /p 1/2 ) 1/2 ) 1/2 ±4.p 1/4 .(γ 2 -β 2 /p 1/2 ) 1/4 .((γ 1 1/2 +(γ 1 -α 1 /p 1/2 ) 1/2 )/(γ 1 1/2 -(γ 1 +α 1 /p 1/2 ) 1/2 ) 1/2 ±2 1/2 .p 1/2 .( γ 1 +β 1 /p 1/2 ) 1/2 -±4.p 1/4 .(γ 2 -β 2 /p 1/2 ) 1/4 .((γ 1 1/2 -(γ 1 -α 1 /p 1/2 ) 1/2 )/(γ 1 1/2 +(γ 1 +α 1 /p 1/2 ) 1/2 ) 1/2 ±4.p 1/4 .(γ 2 -β 2 /p 1/2 ) 1/4 .((γ 1 1/2 -(γ 1 -α 1 /p 1/2 ) 1/2 )/(γ 1 1/2 +(γ 1 +α 1 /p 1/2 ) 1/2 ) 1/2
Eigenvalues of environment cm = 2 are obtained on the first column of the table. Eigenvalues of environment cm = 4 are obtained by adding the variations of the second column of the table (to the previous one). Eigenvalues of environment cm = 8 are obtained by adding the variations of the third column of the table.

Case p = 1+2 k +r.2.2 k
The last column of the variations shows pure imaginary numbers in this case (same formula otherwise).

Foot-note 1

Let us recall that the order of the eigenvalues must be adapted to generate actually a cardinal matrix (with integer components). It is necessary to use the method of the principal diagonal to secure control of this point, method which can be literally applied while engaging sometimes big-sized calculations.

Foot-note 2

One may guess some general literal forms of eigenvalues variations for cm = 16, 32, etc. starting from the formulas obtained previously for cm = 2, 4 and 8. Unfortunately, the preceding expressions do not apparently contain all the needed information. For example, in the case cm= 16, we may try the following decompositions :

a k .p 1/8 .(γ k ±β k /p 1/2 ) 1/8 .((γ k-1 tk1 ±(γ k-1 ±α k-1 /(η k-1 ±ζ k-1 .p 1/2 ) 1/4 )/((γ k-1 tk1 ±(γ k-1 ±α k-1 /(η k-1 ±ζ k-1 .p 1/2 ) 1/4 ) … ( )/( )
We could not come up to some satisfactory results with this expression.

Primarily interest of the study

Beyond the problems of the eigenvalues order and the choice of acceptable variations (limited to the choice of certain signs), it is necessary to indicate the main point that is the presence and the form of an equation of decomposition. It is written, α, β and γ being integers :

p = (α 2 +β 2 )/γ 2 (81)
This equation, even with the condition (α, β, γ) =1, has an infinity of solutions and this condition is not necessary. Thus, a fine study of the values α(p), β(p), γ(p) in environment 2 k is not a lesser task and remains a quite open question, more especially as the order of the eigenvalues has an incidence. A first hint would be to find conditions on p that lead to some similitudes on (α(p), β(p), γ(p)) that we highlighted above (red police).

In addition, beyond these equations of decomposition whose ultimate form may be more complex than p = (α 2 +β 2 )/γ 2 , we can suppose the existence of matrices of sequence, may be of more complex form, whose expression remains to be discovered.

Real and imaginary contributions

When p = 1+r.2.2 k , all of eigenvalues are real. When 1+2 k +r.2.2 k , the successive contributions Δμ to the eigenvalues are real except the last one which is imaginary. From this only contribution will arise the whole set of imaginary values in a given environment with multiple factors.

Foot-note: p = 3 is a particular case compared to the other odd prime numbers as the value of the discriminant of x 2 +x.y+y 2 is -3. We have (Δ=-3,p=3) = 3, whereas in the other cases, we always have (Δ,p) = 1.

Quadratic equations 2.1. Study framework and notations

We are interested in the equation u.x 2 +v.x.y+w.y 2 = c and we thus must study u.x 2 +v.x.y+w.y 2 = c mod p k

In the case of Waring sums, we observed that the abundance factors modulo p k are determined from modulo p coefficients with a factor of correction which, in the "good cases", is close to 1. This is not self-evident here since the number of unknown factors (x and y) is equal to the degree of the equation. We suppose however that we are in such a case to start with (and will go further later on) : u.x 2 +v.x.y+w.y 2 = c mod p

In a traditional manner, we call discriminant of u.x 2 +v.x.y+w.y 2 , the expression :

Δ = v 2 -4.u.w

Linear translation of columns of a quadratic table

Any column of a two-dimensional table (x,y) formed by the residues c = u.x 2 +v.x.y+w.y 2 is a constant translation of the preceding column if u ≠ 0 mod p. Indeed, let us consider two points of this table (x,y) and (x+a,y+1) and their right neighbours (x+1,y) and (x+a+1,y+1) and let us seek the evolutions of the values of the residues when passing from (x,y) to (x+1,y) and from (x+a,y+1) to (x+a+1,y+1). We get for first passage E1 Thus, all calculations done, E(c) = E1(c)-E2(c) = 2au+v mod p. The evolution of E(c) is null modulo p when 2a.u = -v mod p. Case p =2 is studied separately. We thus suppose here that p ≠ 2 and we note that 2 has always an inverse mod p. Then :

a.u = -v/2 mod p Then, either parabolic case : (u,v) = (0,0) mod p (5) or case of the square : (a,v) = (0,0) mod p (6) or true quadratic case : a = -v/(2u) mod p (7) since u ≠ 0 always has an inverse mod p. This third case is that of the translation, object of this paragraph (other cases being treated in addition).

Quadratic residues tables

We wish to get the variation vr of the residues between two points (x,y) and (x+a,y+1), that is u. When Δ = 0, we have vr = 0 what means that the table of quadratic residues is from one column to another a simple circulant shift (with a shift of -v/(2u) mod p lines). The abundance factors, obtained by enumeration of the residues, are those of the first column (y=0 and thus c = u.x 2 mod p) multiplied by p.

When Δ ≠ 0, vr varies from one column to another. Let us number 1 to p the columns of the table (corresponding with y = 0 to y = p-1 mod p). The successive variations of vr are vr(1) = -Δ/(4u), vr(2) = -3Δ/(4u), …, vr(p) = -(2p-1)Δ/(4u). The collection {1,3,5,…,2p-1} is in fact only a permutation of {1,2,3,…,p-1} and the whole vr(i) set the multiplication by a non-null constant of this collection, which is another permutation {1,2,3,…,p-1}.

When the p columns of the table are described, vr thus describes the whole set of the p values {0,1,2…, p-1} in an order which depends on Δ and u.

Let us note then that the increase between the first and the i th column ar(i) = ∑ vr(i) is given by ar(i) = -(i-1) 2 .Δ mod p ( 9) 4u with a shift of lines of : P 245/390 Quadratic abundance factors and substitutes a = -(i-1).v/(2u) mod p (10) In expression ar(i), as -Δ/(4u) is a simple multiplicative constant of (i-1) 2 , there exists a primitive root g of p such as, for i > 0, the whole set {ar(i)} is a permutation of the classes {g j .g 2k } mod p, while k varies from 0 to p-2 and either j = 0, or j = 1. According to respective values of (u,v,w), the residues of u.x 2 +v.x.y+w.y 2 = c mod p are thus resulting from one of the following two-dimensional tables : {0, g 2 , g 4 , …, g 2(p-1) }+{0, g 2 , g 4 , …, g 2(p-1) } or {0, g 2 , g 4 , …, g 2(p-1) }+{0, g.g 2 , g.g 4 , …, g.g 2(p-1) } or {0, g.g 2 , g.g 4 , …, g.g 2(p-1) }+{0, g.g 2 , g.g 4 , …, g.g 2(p-1) } First and third expressions are equivalent, in regard of enumeration, since each completed two-dimensional table contains only squares. The first expression (and the third) are that of equation x 2 +y 2 = c mod p of which we know the enumeration characteristics already. The second is an alternative x 2 +g.y 2 = c mod p. It remains us to clarify the conditions on (u,v,w) giving one or the other of these alternatives. This study restores, to some extent, already well-known results.

Enumeration of solutions modulo p

Let us examine initially the two cases c = g 0 .g 2u mod p and c = g 1 .g 2u mod p. Let us choose a couple of numbers x ≠ 0 mod p and y ≠ 0 mod p such as c ≠ 0 mod p. Let us have g a primitive root of p. Then (x.g i , y.g i ) are distinct couples such as u.(xg i ) 2 +v.(x.g i ).(y.g i )+w.(y.g i ) 2 = c.g 2i mod p. Thus, if u.x 2 +v.x.y+w.y 2 = c mod p has t distinct solutions (x,y) with given c, then x 2 +v.x.y+w.y 2 = c.g 2i mod p also has t distinct solutions (x,y). Hence, it results, for the enumeration of the targets c, only three cases with distinct cardinals c = 0 mod p, c = g 0 .g 2u mod p and c = g 1 .g 2u mod p.

In the case of integers variables, the cardinal of the residues c of type g 0 .g 2u mod p is also the cardinal of the residues of type g 1 .g 2u mod p (only c = 0 mod p has a different cardinal). Indeed, if c is of the first type, -c mod p is of the second. Let us have (x,y) solution of u.x 2 +v.x.y+w.y 2 = c mod p and (x+a,y+b) another (distinct) solution. Let us have then (x',y') a solution of u.x' 2 +v.x'.y'+w.y' 2 = -c mod p (subject to existence of the residue -c). Then, the cardinals of the two types are equal if we can find a couple (s',b') ≠ (0,0) such as (x'+a',y'+b') is solution of u.x' 2 +v.x'.y'+w.y' 2 = -c mod p. The development of the resulting equations gives then the necessary and sufficient condition : It is however necessary to prove the condition of existence as mentioned earlier. In the case of the configuration x 2 +y 2 = c mod p that is already proven by our study on Waring sums. In the case x 2 +g.y 2 = c mod p, existence of c implies that of -c by the following arguments. We have -1 = g (p-1)/2 mod p and we consider the cases (p-1)/2 even or odd. If (p-1)/2 = 2k then x 2 +g.y 2 = g 2k .c mod p, equivalent to g 2i +g.g 2j = g 2k .c mod p (keeping in memory the cases x or y = 0), is written g 2(i-k) +g.g 2(j- k) = c mod p. In the same way, if (p-1)/2 = 2k-1 then x 2 +g.y 2 = g 2k-1 .c mod p, equivalent to g 2i +g.g 2j = g 2k-1 .c mod p (still keeping in mind the cases x or y = 0), is written g.g 2(i-k) +g 2(j-k) = c mod p. This suffices for our proof.

Factorisation of the quadratic equation

Let us be in the case u ≠ 0 mod p. Each one of these systems has p distinct solutions (x,y), themselves clearly distinct from one system to another when ∆ ≠ 0 mod p except common solution (0,0). Hence the 2p-1 distinct solutions.

For the others target c (c ≠ 0 mod p), let us consider first the expression X.Y = c mod p. This one has obviously p-1 distinct solutions since any class X (≠ 0) has a reverse mod p, noted X -1 and consequently Y = c.X -1 always exists.

Case Δ non-square

We have two alternatives for non-square Δ : p = 1 mod 4 -Δ non-square p = 3 mod 4 -Δ square Indeed, Δ not being a square, for an unspecified primitive root of p, there exists always i such as Δ = g.g 2i , then -Δ = -g.g 2i = g (p-1)/2 .g.g 2i mod p. If p = 1+4k, then Δ = g.g 2(i+k) , while if p = 3+4k, Δ = g 2(1+i+k) .

Lower case p = 3 mod 4

We write then :

(x+y. (v+i√(-Δ)) ).(x+y. (v-i√(-Δ)) ) = (X+iY).(X-iY) = X 2 +Y 2 = c/u mod p (21) 2u 2u
Equation X 2 +Y 2 = c/u mod p is treated elsewhere in this article. For p = 3 mod 4, X 2 +Y 2 = 0 mod p have only the selfevident solution (X,Y) = (0,0) and #{c=0} = 1.

Lower case p = 1 mod 4

In this case, the factorisation of u.x 2 +v.x.y+w.y 2 mod p is possible neither according to the relation (14), nor according to the relation ( 21), therefore c = 0 mod p is a residue only in the self-evident case (0,0).

Solution of a non-complete modular quadratic equation

Let us have u ≠ 0 mod p. We seek to express a condition equivalent to :

 x integer \ u.x 2 = 1 mod p Here p is a prime. There is always a reverse for u modulo p. Hence, we can write x 2 = u -1 mod p. If a solution exists to this equation, then x -1 mod p is solution of x' 2 = u mod p. Thus u is a square modulo p.

Abundance factors of quadratic equations

The tables below summarize the various situations met for the general quadratic modular equation.

Variables (x,y) of integers p = 2

The case p even is treated by building two-dimensional tables : u.x 2 +v.x.y+w.y 2 x = 0 x = 1 y = 0 0 u y = 1 w u+v+w

Hence the truth table :

(u,v,w) #{0} #{1} (0,0,0) mod 2 4 0 (1,0,0) mod 2 2 2 ≡ (0,0,1) mod 2 2 2 (1,0,1) mod 2 2 2 (0,1,0) mod 2 3 1 (0,1,1) mod 2 3 1 ≡ (1,1,0) mod 2 3 1 (1,1,1) mod 2 1 3 p > 2
The situation is simple here and rises immediately from the proofs implemented in paragraph 2.5. Foot-note: or * is exclusive

Conditions

More on proof

So we indicated above, the proof for variables of integers is implemented in paragraph 2.5.

The proof for prime numbers variables is then immediate using the cardinals of integers' variables. We simply withdraw the elements from two-dimensional tables within the first line (x=0) and first column (y=0) while passing from the first to the second case. Thus, the enumeration relates to the residues {c \ w.y 2 = c mod p, y = 1 to p-1}, {c \ u.

Mixed variables

If x is an integers' variable and y a prime numbers' variable, another table can be drawn up that we leave to the reader's initiative.

If c is an odd prime number, we consider two cases. We check the solutions proposed in the table and we show that there are no others what suffice here. We use for that Legendre notation and the law of quadratic reciprocity where c and p are relative primes : Here, as (c-1)/2 is 0 mod 2, it does not matter if (g 4c 2i -1)/2 is even or odd. In the same way : Indeed, the parity of the product (1+2n).(g 4c i -1)/2 depends only on the parity of (g 4c i -1)/2. The primitive g 4c is 3 mod 4 by definition. Thus, if i is odd, (g 4c i -1)/2 is odd and Legendre symbol (g 4c i /c) equals -1. Otherwise, if i is even, (g 4c i -1)/2 is even, while the Legendre symbol (g 4c Indeed, compared to the expression interesting G(c), the last expression presents a multiplication by (-1) (1+2n).c where c is 3 mod 4, that is by (-1) (1+2n).c = -1. Hence the result.

Having checked the inclusion of the candidate families respectively in G(c) and H (c), it remains us to prove that there are no other solutions. Let us consider the case c = 3 mod 4 (where c is a prime number) and the set {g 4c i } mod 4c. Let us have x = g 4c i mod 4c and y = g 4c j mod 4c. If x = y, then g 4c i = g 4c j mod 4c, then g 4c i-j = 1 mod 4c. There is t an integer such as g 4c = g+t.c. Thus (g+t.c) i-j = g i-j +m.c = 1 mod 4c, m an integer. Hence g i-j = 1 mod c. Hence i = j mod p-1. The set {g 4c i } mod 4c thus has c-1 distinct elements. In the same way {2c+g 4c i } mod 4c has c-1 distinct elements. U{G(c),H(c)} gathering all the classes relative prime with 4c, that is φ(4c) = φ( 4)φ(c) =2(c-1), there are no other solutions. In the case of c = 1 mod 4, we can use the same arguments without new difficulties.

If c contains a square, let us have α.β 2 its decomposition into non-square and square. The equation becomes x 2 = α.β 2 mod p. Existence of a solution with this equation rests self-evidently on existence of a solution, pending on p, for equation (x/β) 2 = α mod p knowing that a reverse always exists mod p (if the number is non-null), the equation transforms into y 2 = α mod p what proves the previous suggested simplification.

If c is positive and is the product of two relative primes factors, c = α.β. Then, we study equation x 2 = α.β mod p seeking set G(α.β) = {p} with residue c. We suppose equation x 2 = α mod p solved, that is set G(α) mod 4α for existence and set H(α) mod 4α for non-existence of a residue. In the same way, for x 2 = β mod p, we have sets G(β) mod 4β and H(β) mod 4β. Let us have p a prime number. If there is x such as x 2 = α mod p, i.e. if α is a square modulo p, and if, for any x, we have x 2 ≠ β mod p, i.e. β is not a square modulo p then clearly αβ is not a square modulo p and, for any x, x 2 ≠ αβ mod p (this can be shown rigorously by using the primitive roots of p). The reasoning is the same one when permuting α and β. In addition, self-evidently, if α and β are squares modulo p, the product αβ is a square modulo p. If α and β are non-squares modulo p, then there is a primitive root g of p and an integer i such as α = g.g 2i mod p and a primitive root g' of p and an integer j such as β = g'.g' 2j mod p. We express then g' according to g by g' = g k mod p. A primitive root is expressed according to another with k relative prime with p-1, therefore k odd. It follows β = g k .g 2j.k = g.g 2m mod p. Thus α.β = g 2(1+i+m) mod p is a square modulo p. This argument is summarized according to the table :

 x \ x 2 = α.β mod p  { x \ x 2 =
α mod p and  x \ x 2 = β mod p} or {x, x 2 ≠ α mod p and x, x 2 ≠ β mod p} x, x 2 ≠ α.β mod p  { x \ x 2 = α mod p and x, x 2 ≠ β mod p} or {x, x 2 ≠ α mod p and  x \ x 2 = β mod p} As  x \ x 2 = α mod p is equivalent to p c G(α) and x, x 2 ≠ α mod p is equivalent to p c H(α), ), it follows immediately with our rules for intersections and unions :

G(α.β) = U(∩(G(α),G(β)),∩(H(α),H(β)) H(α.β) = U(∩(G(α),H(β)),∩(H(α),G(β))
In addition, these rules come from modulo 4α and modulo 4β results for prime α and β numbers, results which are naturally prolonged modulo 4αβ to order αβ.

For any c, we withdraw the set of the square factors and get a decomposition in factors c' = ± ∏ f i . We apply the routines to the various factors of this number. It does not matter, except possibly for the size of calculations, the order and the signs used (provided that the product corresponds well to c).

Let us have x n = c mod p. If n has two factors n 1 and n 2 , self-evidently (x n1 ) .n2 = c mod p and (x n2 ) .n1 = c mod p. Let us pose y 1 = x n1 and y 2 = x n2 . Then y 1 n2 = c mod p and y 2 n1 = c mod p. Considering the events { y 1 \ y 1 n2 = c mod p} and { y 2 \ y 2 n1 = c mod p} independent when n 1 and n 2 are relative primes, the frequency of the joint event is the product of the frequencies of each separate events.

Then, with the property of multiplicative function, we can limit us to the examination of equation modulo p :

x q^t = c mod p (30)

Here q is a prime number and t a positive integer exponent.

Condition of existence of a solution

We ignore the case p = 2 in the next course (without incidence on the frequency).

According to relation (11) of exercise 3, x n = c mod p, c ≠ 0, accepts d solutions if and only if c (p-1)/d = 1 mod p where d = (n, p-1), thus : c (p-1)/(q^t,p-1) = 1 mod p (31) 6.2.4. Type "rho" function

Acyclic and cyclic behaviours

For q fixed, when t increases by increment, the evolution of the variation ∆ t of f(q t ,c) is characterized by a cycle after an initial phase depending on c. The extent of the cycle depends on c but the ratio at the preceding step remains constant. Let us have thus ∆ t the variation of the frequency at step t. We have f(q t ,c) = 1-∆ 1 -∆ 2 -…-∆ t and the two types of evolutions of the variations :

acyclic cyclic -∆ 1 , -∆ 2 , …, -∆ k -∆ k /σ, -∆ k+1 /σ, …, -∆ t-1 /σ
Numerical examples: The frequency is expressed then as follows :

c q σ acyclic cyclic 2 2 2 2 -1/2, -
f(q t ,c) = 1-(∆ 1 +∆ 2 +…+∆ k-1 )-(1+1/σ+1/σ 2 +…+1/σ (t-k) ).∆ k That is also :

f(q t ,c) = 1-(∆ 1 +∆ 2 +…+∆ k-1 )- (1-1/σ (t-k+1) ) .∆ k (1-1/σ)
When t increases the frequency tends towards the limit value :

f(q t→∞ ,c) = 1-(∆ 1 +∆ 2 +…+∆ k-1 )- 1 .∆ k (32) (1-1/σ)

Regular behaviour

We will call regular behaviour the case where the acyclic term is limited to ∆ 1 with σ = q 2 and ∆ 1 = 1/q : c regular  ∆ i+1 = (1/q 2 ).∆ i and ∆ 1 = 1/q Then : f(q t ,c reg ) = 1-(1/q).

(1-1/q 2t ) (33

) (1-1/q 2 )
This frequency is written numerically in the form of a fraction. We have : num r+1 = q 2 .num r -1, num 1 = q-1 dom r+1 = q 2 .dom r , dom 1 = q Incidentally, we have also num r+1 = num r + (q 2 -q-1).dom r . In particular, when q = 2, the numerators are Fibonacci series num r+1 = num r + dom r , num 1 = 1, dom 1 = 2 and the denominators are geometrical series dom r+1 = 4.dom r , dom Let us return to the general case of regular behaviour. c q σ acyclic cyclic 3 = 3 1 5 5 2 -1/5 -1/125, -1/625, … 243 = 3 [START_REF] Dieudonné | Abrégé d'histoire des mathématiques[END_REF] 5 5 2 0, -1/25 -1/625, -1/15625, … 847288609443 = 3 25 5 5 2 0, 0, -1/125 -1/3125, -1/78125, … 3 (5^k).d and ≠ 3 (5^(k+1)).d 5 5 2 {0, …, 0}k terms, -1/5 k+1 -1/5 k+3 , -1/5 k+5 , … Case c = r d , q ≠ 2, ( r = 2 or r ≠ 2) c q σ acyclic cyclic r (q^k).d and ≠ r (q^(k+1)).d q q 2 {0, …, 0}k terms, -1/q k+1 -1/q k+3 , -1/q k+5 , … This last case, which shows a general framework when q ≠ 2, is deduced from the preceding cases. It is summarized in a more easy way than case q = 2. The regular case corresponds to k = 0 (and q ≠ 2).

Residues with multiple factors

For residue c = r i , we observe that the value of r has no incidence. We can extent this observation here. Let us have p 1 ε1 .p 2 ε2 …p j εj the prime factors decomposition of a given residue c : Let us have d e the greatest common divisor of the exponents of c :

d e = (ε 1 , ε 2 , …, ε j )
In order to study the function f(q t ,c = p 1 ε1 .p 2 ε2 …p j εj ), we put d e , by extracting all given q factors, in the following form: d e = d.q k Then (conjecture) : σ acyclic cyclic q 2 {0, …, 0}k terms, -1/q k+1 -1/q k+3 , -1/q k+5 , …

Foot-note

The case of 2 does not appear here anymore. Indeed, when several factors are in presence, the odd factors are prevalent and erase this particular case.

Example : c d e q σ k acyclic cyclic Behaviour equivalence 6 = 3 1 .2 1 (1,1) = 1 2 2 2 0 -1/2 -1/8, -1/32, … ≡ 3 12 = 3 1 .2 2 (1,2) = 1 2 2 2 0 -1/2 -1/8, -1/32, … ≡ 3 36 = 3 2 .2 2 (2,2) = 2 2 2 2 1 0, -1/4 -1/16, -1/64, … ≡ 3 2

Case mod p δ

We examine the cases where δ is greater than 1.

Case c = r d , q = 2 c q σ acyclic cyclic 2 infinite -1 0, 0, … Case c = r d , q ≠ 2 c q σ acyclic cyclic ≠ 2 infinite -1/2 0, 0, …

We do not have any more the many cases we found for δ = 1. In fact the limit value when d → +∞ is reached immediately.

Inclusive and exclusive frequencies

A start of explanation of the preceding results requires the following study.

Set of natural integers

Let us have N* = {1,2,3,…,∞} the set of the positive integers. Let us have m a positive integer whose prime factors decomposition is p 1 ε1 .p 2 ε2 …p j εj . We seek the frequency of m in N*. When we are interested in the "inclusive" frequency of m, that is fi(m), we count all the occurrences of m in N*. When we are interested in the "exclusive" frequency of m, that is fe(m), we count the occurrences rejecting those, while the current number is m.p

(m) = φ(m) (36) m 2
Function φ is Euler totient function. Expressions fe(m) and fi(m) are "limits". They are reached periodically. For the exclusive frequency, this occurs when an element x of N* is equal to k.m with k a positive integer. For the inclusive frequency, this occurs when an element x of N* is equal to k.p 1 .p 2 . … .p i .m with k a positive integer. We can notice that fe(m) and fi(m) are multiplicative functions when the reference set is N*. For the exclusive frequency :

(m 1 ,m 2 ) = 1 => fe(m 1 .m 2 ) = fe(m 1 ).fe(m 2 ) where fe(p ε ) = (p-1)/p ε+1 (37)

For the inclusive frequency :

(m 1 ,m 2 ) = 1 => fi(m 1 .m 2 ) = fi(m 1 ).fi(m 2 ) where fi(p ε ) = 1/p ε (38)

The stake of the concept extends to other sets that N* and is presented for that purpose. (m 1 ,m 2 ) = 1 => fi(m 1 .m 2 ) = fi(m 1 ).fi(m 2 ) where fi(p ε ) = 1/((p-1).p ε-1 ) Let us present then the table concerning the two preceding sets as well as intermediate sets :

Reference set fi(m) fe(m) N* 1/m (1/m).∏ (p i -1)/p i N*-U j {0 mod q j εj }

(1/m).∏ q j εj /(q j εj -1)

(1/m).∏ (p i -1)/p i .∏ q j εj /(q j εj -1) PM1

(1/m).∏ p i /(p i -1) 1/m

In the intermediate set, the union sign relates to j elements. To the N* set, we withdrawn the multiples of q j εj where q j are prime numbers and εj positive integers. It is clear that the limits of the expressions fi(m) and fe(m) for these intermediate sets are reached periodically as in the cases of N* and PMl. The factor q j εj /(q j εj -1) corresponds to a simple increase proportional to the frequencies by "lightening" of the basic set. When j describes all the possible values and when starting ad infinitum the εj approach 1, the expressions (1/m).∏ q j εj /(q j εj -1), respectively (1/m).∏ (p i -1)/p i .∏ q j εj /(q j εj -1), approach more and more (1/m).∏ p i /(p i -1), respectively 1/m. In same time the sets N*-{U j 0 mod q j εj } approach more and more the prime numbers set P. However, the limit case itself cannot be reached actually as it makes disappear altogether the reference set. There is some logic in that as, when we consider this limit set, it is clear that :

Reference set fi(m) fe(m) {P} 0 0

However if now, we pass to a translation of this reference set, the result agrees with a little exception.

Let us pose such a transformation t = ∏ p j εj .

Then

Reference set fi(m) fe(m) {P}-t

(1/m).∏ i≠j p i /(p i -1) (1/m).p i /(p i -1)

In this table, we must of course take account of the factors of t as p i -∏ p j εj , where all the p j are different from p i , cannot be multiple of p i . Hence fi(p i ) = 0 (if p i ≠ 1, what is the case, since 1 is not in the list of the prime numbers). Thus only the sets {P}-1 and {P}+1 do not make any minor exception to the rules of frequencies derived from the N* set. The properties concerning the inclusive and exclusive frequencies are found in other sets that those referred to above. In particular, we can look at sets like : {P(x)}-t, P a polynomial, x = 0, 1, 2… {a x }-t, a being some constant, x = 0, 1, 2… We note the regular return of the inclusive frequencies to the limit values fi(3) = 2/3, fi(7) = 2/7, fi(11) = 2/11 and fi(231) = 8/231.

Other elements of proof

These are only possible tracks to follow.

Condition of non-existence of solutions

Let us consider the equation modulo p, n ≠ 0 mod p-1,

x n = c mod p

Let us have S = {x i } the set of classes x solutions of this equation. If this set is empty or if all of the elements of S are primitive roots of p, then :

x n.m = c mod p, m integer >1, (m,p-1) ≠ 1 has no solution. This interests us for the case x q^t = c mod p

The exponent of x is here q t where q is a prime number. As (q t ,q t+1 ) ≠ 1, if the set of solutions of x q^t = c mod p are primitive roots of p, then for i for i strictly positive integer, x q^(t+i) = c mod p has no solution for target c. The reciprocal assertion is false.

Proof

Let us suppose first that set S is empty. The contra-posed proposition of the hypothesis is {y n.m = c mod p} ≠ {void} => {x n = c mod p} ≠ {void}. Let us suppose also the existence of solutions for y n.m = c mod p without solution for x n = c mod p.

We have y n.m = (y .m ) n = c mod p. Which is contradictory? Let us suppose then set S non empty. By hypothesis, any number x solution of x n = c mod p is some primitive root g of p. If y is solution of y n.m = c mod p, we can write y pending on any primitive root of p, for example g. Thus, there is r an integer such as y = g r mod p. Then g r.

Example

The primitive roots of p = 37 are {2, 5,13,15,17,18,19,20,22,24,32,35}. The solutions of x 3 = 14 mod 37 are {5, 13,19} and are all primitive roots. Thus x 3n = 14 mod 37, n > 1 has no solution. To illustrate that the reciprocal is false, let us consider x 3 = 27 mod 37 whose solutions are {3,4,30}. None of these numbers is a primitive root of 37. However x 9 = 27 mod 37 has no solution.

Regular case

Let us suppose, to simplify (other instances being covered by the regular case), that c is an odd prime number and n = q any prime number. We want to enumerate the occurrences c (p-1)/(q,p-1) = 1 mod p. A priori, the result is :

#{c (p-1)/(q,p-1) = 1 mod p} = = 1-1/q-1/q 3 -1/q 5 -1/q 7 -…. = 1-1/(q-1)+1/q/(q-1)-1/q 2 /(q-1)+1/q 3 /(q-1)-1/q 4 /(q-1)+1/q 5 /(q-1)-1/q 6 /(q-1)+1/q 7 /(q-1)-… = 1+∑(-1) i-1 .fe(q i ,PM1)

We get a sum with alternated signs on the exclusive frequencies exponents of q in the numbers of type p-1. This result is elegant, but we unfortunately did not find satisfactory elements of proof.

Case of the polynomials

Generalities

The case of the monomials having been seen previously, we are interested in "true" polynomials i.e. made up of at least two monomials. Let us go thus to the problem : P(x) = c mod p Polynomial P(X) is without constant term (term deferred in c if necessary).

Regulated behaviour

Let us have n the degree of P(X) and its prime factors decomposition : n = ∏ q i εi Using the two relations ( 29) and ( 33), we say that target c has a regular behaviour (c is regular) for the polynomial P when the frequency of the solutions (x) is equivalent to that of c regular for the dominant monomial of P :

Freq(G) = ∏ (1-(1/q i ). (1-1/q i 2εi ) ) (44) i (1-1/q i 2 )
This behaviour is quite usual.

Quadratic cardinal matrices modulo p

These matrices are necessary to undertake the resolution of more complex problems containing the independent group u.x 2 +v.x.y+w.y 2 . (See definition of an independent group at page 303).

We thus consider the two-dimensional table in a modulo p approach :

card(0) card(1) … card(p-1) 0 1 … p-1 #(0) 0 0 1 … p-1 #(1) 1 1 2 … 0 … … … … … … #(p-1) p-1 p-1 0 … p-2
Here #(i) are the cardinals imposed by the addition of a group u.x 2 +v.x.y+w.y 2 , card(j) are the current cardinals and card'(k) the resulting cardinals. We have then : card'(i) = ∑ #(i-j).card(j) (45) j = 0 to p-1 We studied higher #(i) according to (u,v,w) for variables of integers or prime numbers. We have : #(g 2u+1 ) = #(g 2u'+1 ) = #(g) #(g 2v ) = #(g 2v' ) = #(1) From these two points, it follows : card'( 0) #(0) x 0,1 x 0,2 card(0) card'(g 2u ) = #(1) x 1,1 x 1,2 card(g 2u ) card'(g 2u+1 ) #(g) x 2,1 x 2,2 card(g 2u+1 )

The values of #(0), #(g 2u ) and #(g 2u+1 ) depend on the various cases relating to (u,v,w) which we clarified higher.

The other coefficients of the matrix are determined as follows :

x 0,1 = p-1 ∑ #(-g 2u mod p) u = 0 to p-1 x 0,2 = p-1 ∑ #(-g 2u+1 mod p) u = 0 to p-1 x 1,1 = p-1 ∑ #(1-g 2u mod p) u = 0 to p-1 x 1,2 = p-1 ∑ #(1-g 2u+1 mod p) u = 0 to p-1 x 2,1 = p-1 ∑ #(g-g 2u mod p) u = 0 to p-1 x 2,2 = p-1 ∑ #(g-g 2u+1 mod p) u = 0 to p-1
As -1 = g (p-1)/2 , we have ∑ #(-g 2u mod p) = ∑ #(g 2u+(p-1)/2 mod p). Thus, if p = 1 mod 4 then #(g 2u+(p-1)/2 ) = #(g 2v ) = #(1) otherwise if p = 3 mod 4 then #(g 2u+(p-1)/2 ) = #(g 2v+1 = #(g). It follows :

x 0,1 = ((p-1)/2).(if(p = 1 mod 4, #(1), #(g)))

x 0,2 = ((p-1)/2).(if(p = 1 mod 4, #(g), #( 1)))

P 266/390

Quadratic abundance factors and substitutes

Then we have :

-if p = 1 mod 4 x 1,2 -x 1,1 = #(1)-#(0) x 2,1 -x 2,2 = #(g)-#(0) -if p = 3 mod 4 x 1,2 -x 1,1 = #(g)-#(0) x 2,1 -x 2,2 = #(1)-#(0)
Indeed, let us consider first the difference x 1,2 -x 1,1 = #(1-g 2u+1 mod p)-#(1-g 2u mod p). We form the table :   col 1 col 2 col 3 1 #(g 0 ) #(g 0 -g 1 ) #(g 0 -g 0 ) #(g 0 -g 3 ) #(g 0 -g 2 )

… … #(g 2 -g p-2 ) #(g 2 -g p-3 ) 2 #(g 2 ) #(g 2 -g 1 ) #(g 2 -g 0 ) #(g 2 -g 3 ) #(g 2 -g 2 ) … … #(g 2 -g p-2 ) #(g 2 -g p-3 ) … … … (p-1)/2 #(g p-3 ) #(g p-3 -g 1 ) #(g p-3 -g 0 ) #(g p-3 -g 3 ) #(g p-3 -g 2 ) … … #(g p-3 -g p-2 ) #(g p-3 -g p-3 )
The first column is out of the table. For the remainder, this table contains between brackets the set {0, 1,…, p-1} exactly (p-1)/2 times. As #(1-g 2u+1 ) = #(g 2i -g 2i g 2u+1 ) and #(1-g 2u ) = #(g 2i -g 2i g 2u ) for any i, each block of (p-1)/2 lines is the repetition of the block above it.

In addition, let us consider that we have the equality g i = g 0 -g 2u+1 then g j = g j-i -g 2u+1+j-i . Therefore, if j-i is even, g j can be written in the form g 2v -g 2w+1 . If j-i is odd, we write first g j = g 2u+1+j-i+(p-1)/2 -g j-i+(p-1)/2 and g j will be like g 2v -g 2w+1 provided that (p-1)/2 is either even, or p = 1 mod 4. Thus, when p = 1 mod 4, any number can be written as g 2v -g 2w+1 . By the same arguments, when p = 3 mod 4, any number can be written as g 2v -g 2w . It results immediately that, if p = 1 mod 4, the penultimate column contains between brackets the set {1, 2,…,p-1} exactly (p-3)/4 times and {0} exactly (p-1)/2 times and if p= 3 mod 4, the last column contains between brackets the set {1, 2,…,p-1} exactly (p-3)/4 times and {0} exactly (p-1)/2 times. We proceed then by difference. When p = 1 mod 4, respectively p = 3 mod 4, columns 1 and 3, respectively 1 and 2 contain altogether between brackets (p-1)/2-(p-3)/4 times the elements {1,2,…, p-1}. Then, when p = 1 mod 4, respectively p = 3 mod 4, column 3, respectively 2, contains between brackets (p+1)/4 times elements {g 1 , g 3 …, g p-2 } and (p+1)/4-1 times the elements { g 0 , g 2 …, g p-3 }.

Hence

-if p = 1 mod 4 #(1-g 2u+1 mod p) -#(1-g 2u mod p) = #(1) -#( 0) #(g-g 2u mod p) -#(g-g 2u+1 mod p) = #(g) -#(0)

-if p = 3 mod 4 #(1-g 2u+1 mod p) -#(1-g 2u mod p) = #(g) -#( 0) #(g-g 2u mod p) -#(g-g 2u+1 mod p) = #(1) -#( 0)

With the preceding results and knowing that sum of each line of the matrix is (p-1) 2 , we get the transformation matrix [M quad2 ] : 1) #( 0)+((p+(-1) (p+1)/2 )/4).(#(1)+#(g)) +(if(p = 1 mod 4,-#(1),-#(1)-#(g)))

#(0) ((p-1)/2).(if(p = 1 mod 4, #(1), #(g))) ((p-1)/2).(if(p = 1 mod 4, #(g), #(1))) #(
((p+(-1) (p+1)/2 )/4).(#(1)+#(g)) +(if(p = 1 mod 4, 0, -#(1))) #(g) ((p+(-1) (p+1)/2 )/4).(#(1)+#(g)) +(if(p = 1 mod 4, 0, -#(g)))

#( 0)+((p+(-1) (p+1)/2 )/4).(#(1)+#(g)) +(if(p = 1 mod 4, -#(g), -#(1)-#(g)))

So that also (relation ( 46 If #(1) = #(g) (case for prime numbers variables), this matrix can be contracted by addition of the two last columns (and removal of the last line) into the matrix [M quad1 ] (relation ( 47)) : #(0) (p-1).#(1) #( 1) #( 0)+(p-2).#(1)

Cases of the integers variables

The interesting cases are given by (see page 247) :

#(0) #(1) #(g) Δ ≠ 0 mod p Δ c C(p) 2p-1 p-1 p-1 4 Δ c D(p) 1 p+1 p+1 5
Matrix [M quad1 ] is then when :

p \ Δ c C(p) [M quad1 ] = 2p-1 (p-
P = P -1 = (1/p 1/2 ). 1 p-1 1 -1 We have well : (1/p). 1 p-1 p 2 0 1 p-1 2p-1 (p-1) 2 1 -1 0 p 1 -1 = p-1 p 2 -p+1 and (1/p). 1 p-1 p 2 0 1 p-1 1 (p-1)(p+1) 1 -1 0 -p 1 -1 = p+1 p 2 -p-1
We restore actually the concept of common environment matrix.

Case of the prime numbers variables

We undertake a partial study resulting from the page 248 conditions :

#(0) # (

Approach by the equations of primitive roots

This is an alternative of the preceding approach to get the literal components of the cardinal matrices. We use what we call the equations of primitive roots relative to cardinal matrices [B] of monomial y 2 and in peculiar its restriction [B'] (removing first line and first column (r ≥ 1, r ≥ 1)). The reader will refer to relation 24 of exercise 5 to find : m' r,s = d.{#(u,v) / g r-1 =g u.d +g s-1 .g v.d mod p} where d = 2

We suppose that we know already the first column of the matrix of the studied quadratic group, that is #(0), #(1) and #(g).

We have (r ≥ 1, s ≥ 1) :

x r,s = #(0).{#(v) / g r-1 = 0+g s-1 .g v.d mod p } + #(g 0 ).{#(u,v) / g r-1 = g 0 .g u.d +g s-1 .g v.d mod p} + #(g 1 ).{#(u,v) / g r-1 = g 1 .g u.d +g s-1 .g v.d mod p}

That is also after rearranging the last term of the equality second member (and by replacing d by its value) :

x r,s = #(0).{#(v) / g r-1 = 0+g s-1 .g 2v mod p } + #(g 0 ).{#(u,v) / g r-1 = g 0 .g 2u +g s-1 .g 2v mod p} + #(g 1 ).{#(u,v) / g 1 .g r-1 = g 2u +g 1 .g s-1 .g 2v mod p} Thus :

x r,s = #(0).if(r=s, 1, 0)+ #(g 0 ).m' r,s /2+ #(g 1 ).m' r+1,s+1 /2

Let us choose the case of the matrix of prime numbers variables with p = 1 mod 4 and p \ Δ c C(p) : Thus, with #(0) = 2(p-1), #(1) = p-5, #(g) = p-1 :

x 1,1 = 2(p-1).1+(p-5).m' These evaluations come up with our expectations. Concerning x i,0 , an evaluation can be made on the same principle (equations of primitive roots). This evaluation is carried out further within a more general framework.

Quadratic cardinal matrices modulo p δ

To solve a diophantine equation, where an u.x 2 +v.x.y+w.y 2 group is included, the matrix modulo p δ=1 (at sequence p) is not suitable usually. A higher rank, even infinite, may be necessary. The examination of these conditions is the subject of exercise 11 which follows.

For the moment, let us give the general form of these matrices.

We leave to the reader the care to review on a case-by-case basis the p = 2 sequence which is an exception to what follows.

Integers' variables

In the absence of other diophantine groups, we can reason in the environment d = 1.

Case Δ c C(p) (Δ ≠ 0 mod p) This case is that of the x 1 2 +x 2 2 group which elements we can get thanks to the monomial x 2 modulo p δ study given at page 128 (p = 1 mod case 4). The above eigenvalues are simply the squares of the eigenvalues found on this earlier occasion : This case is again that the x 1 2 +x 2 2 group which we get from x 2 modulo p δ case while p = 3 mod 4 :

δ = 1 2p-1 (p-
μ p δ -p δ-1 √p p δ-1 √p p δ-1 p δ-1 -p δ-2 √p p δ-2 √p … … … μ q = μ 2 p 2δ p 2δ-
μ p δ i.p δ-1 √p -i.p δ-1 √p p δ-1 p δ-1 i.p δ-2 √p -i.p δ-2 √p … … … μ q = μ 2 p 2δ -p 2δ-1 p 2δ-2 -p 2δ-3 … ±p δ
It is easy to anticipate the general matrix form when δ increases from these examples. The reader should note however that it is easier to deduct a new matrix in relying on the second last (mod p δ from modulo mod p δ-2 ) due to some parity breaks.

P 271/390

Quadratic abundance factors and substitutes It should also be noted that if we are working in an environment of higher order, one will perform d-1 duplications of the eigenvalues (excluding the eigenvalue associated with 0 Case Δ = 0 mod p

The case is more complex as it depends on the multiplicity of p in Δ with a rho function type behaviour. Hence, we give first the following example. We observe for δ = 7, p = 3, g = 2 :

k 0 0 0 0 1 1 1 1 2 2 2 2 3 3 i 2i 2i+1 Δ # (
int((δ-1)/2) ≥ i int((δ-1)/2) < i k < i k < i k < i k < i k ≥ i k ≥ i k ≥ i k ≥ i Δ = t.p 2i
tp.p δ p δ+int(δ/2) 2p δ+k 0 0 0 (2k-2i+1).(p-1).p δ+i-1 (2k-2i+1).(p-1).p δ+i-1 2(k-i+1).(p-1).p δ+i-1 2(k-i+1).(p-1).p δ+i-1 Δ = t.p 2i+1 p δ+i p δ+int(δ/2) 2p δ+k 0 0 0 2p δ+i 0 2p δ+i 0

Here the abundance factor of target 0 in the Δ = t.p 2i case is tp.p δ where tp is defined by the recursive equations :

tp(p,1,1) = 1 tp(p,δ,1) = tp(p,δ-1,1)+p-1 tp(p,δ,i) = p.tp(p,δ-2,i-1)

In the p = 3 case, we obtain an array of type :

δ \ i For δ = 7, p = 3, we get again the values : 13 = (13), 27 = 3.( 9), 45 = 3 2 .( 5) with bonus 27 = 3 3 .( 1).

The recursive equations have the solution :

tp(p,δ,i) = p (i-1) .(1+(p-1).(δ-2i+1)) Let us pose : m = or(0,1) (choice of 0 or 1) rm = ((1+(-1) m )/2) rm = or(0,1) depending on the value of m rj = ((1+(-1) j )/2) rj = or(0,1) depending on the value of j

We are only interested by the limit case δ → +∞ which summarizing then in : We can find the value of tp by a simple account of the abundance factor knowing that their sum is equal, for two variables of integers, to p 2δ .

P
From these results, we can then build the cardinal matrices modulo p δ using a similar way as in cases Δ ≠ 0 mod p.

Foot-note : It would be interesting to use the eigenvalues approach to build the cardinal matrices of the Δ = 0 mod p cases.

Unfortunately, if eigenvalues are easy to determine, the transition matrix has more the same form, so largely losing the interest of the construction.

Prime numbers variables

We give the expressions in environment d = 2. This case is that of the y 1 2 +y 2 2 group which elements we find from the study of y 2 modulo p δ with p = 1 mod 4. Modulo p results are given at page 81. The above values are the squares of the eigenvalues found on this occasion :

λ p-1 -1-√p -1+√p λ q = λ 2 (p-1) 2 p+1+2p 1/2 p+1-2p 1/2
It is clear that the eigenvalues starting from λq 3 are all null since prime numbers variables reject all multiples of the p sequence in their representatives. This case is new in the sense that we cannot anticipate it with the y 1 2 +y 2 2 , or even a.y 1 2 +b.y 2 2 groups. Indeed, the various cases match certain δ = v²-4u.w discriminant values. However, we have no solution to equation v²-4u.w = -4a.b when v is odd. (p-1) 2 -p+1-2i.p 1/2 -p+1+2i.p 1/2 Case Δ = 0 mod p Again, the multiplicity of p in Δ infers. With the previous example, we observe (δ = 7, p = 3, g = 2) :

k 0 0 0 0 1 1 1 1 2 2 2 2 3 3 i 2i 2i+1 Δ # (
((δ-1)/2) < i k = 0 0 < k < i k < i k < i k < i k = i k = i k > i k > i k ≥ i k ≥ i Δ = t.p 2i 2(p-1)p δ+i-1 (p-1)p δ- 1+int(δ/2)
2(p-2)p δ-1 2(p-1)p δ-1+k 0 0 0 (p-3).p δ-1+i (p-1).p δ-1+i 2(p-1).p δ-1+i 2(p-1).p δ-1+i 2(p-1).p δ-1+i

(δ-1)/2) ≥ i int((δ-1)/2) < i k = 0 0 < k ≤ i k < i k > i k = i k > i Δ = t.p 2i+1 0 (p-1)p δ-1+int(δ/2) 2(p-2)p δ-1 2(p-1)p δ-1+k 0 0 2p δ+i 0 0
Again, we are interested only in limit case δ → +∞ summarized in : #(0) mod p δ #(p m .g j ) mod p δ #(p 2i+m .g j ) mod p δ i > 0 (k = i) #(p 2k+m .g j ) mod p δ 0 < k < i #(p 2k+m .g j ) mod p δ k > i Δ = t.p 2i 2(p-1).p δ+i-1 2rm.rj.(p-2)p δ-1 ((m+1).p+m-2-(-1) (1-m).j ).p δ-1+i 2rm.rj.(p-1)p δ-1+k 2(p-1).p δ-1+i Δ = t.p 2i+1 0 2.rm.rj(p-2).p δ-1 2rj.(p-1+m).p δ+i-1 2rm.rj.(p-1)p δ-1+k 0

The definitions of m, rm and rj are here unchanged (see case of integers' variables). Then, we can (and we leave that to the reader) build the cardinal matrix relying on the comments made in the case of integers' variables.

Combination of groups of variables

With above results, the reader may get the abundance factors of the following diophantine equation where x represents integers' variables and y prime numbers variables: )+…+(y 1m 2 +y 1m .y 2m +y 2m 2 ) = c mod p δ (48) thanks to the product of matrices (outside of the cases where Δ = 0 mod p, that is here p = 3 and p = 2 sequences) :

(
[P δ ].[μq δ ] k .[P δ -1 ].[P δ ].[λq δ ] m .[P δ -1 ] = [P δ ].[μq δ ] k .[λq δ ] m .[P δ -1 ] (49)
9. Overlapping symmetrical variables

Generalities

We are interested here in the evaluation of the cardinal matrices of expressions like :

z 1 n +z 1 n-1 .z 2 +…+z 1 .z 2 n-1 +z 2 n = if(z 1 ≠ z 2 , (z 1 n+1 -z 2 n+1 )/(z 1 -z 2 ), (n+1).z 1 n ) ( 50
)
The variables z 1 and z 2 are either integers or prime numbers variables.

We recall the Waring study for this type of exercise. Indeed, let us pose d = (p-1,n)

If the couple (z 1 ,z 2 ) is a solution of z =c.g u.d mod p. Thus, the cardinal of c is the same one as that of c.g u.d :

#(c.g u.d ) = #(c)
The dimension of the matrix, d+1, depends as in the case of the variables z n on the sequence p with d = (n,p-1).

It is understood that the matrix of environment [P] is already known (and is that of z d ) and that one of the eigenvalues is (p-1) 2 , p 2 or p(p-1) pending on the types of variables (two of prime numbers, two of integers or one of each type).

Primitive roots equations

As #(c.g u.d ) = #(c), we can build, following the example of exercise 5, a two-dimensional table :

card 0 card 1 card 2 … card d-1 0 g 0 .g u.d g 1 .g u.d gd -1 .g u.d #(0) 0 c #(g 0 ) g 0. g u.d #(g 1 ) g 1 .g u.d … … #(gd -1 )
gd -1 .g u.d We recall that components #(0), #(g 0 ), …, #(gd -1 ) are obtained by noting the number of occurrences modulo p of c = z 1 n +z 1 n-1 .z 2 +…+z 1 .z 2 n-1 +z 2 n for overlapping loops of type « For z i = ini to p-1 » where ini = 0 or 1 pending on the type of variables. The cardinal matrix components satisfy then, while using primitive roots equations (r ≥ 1, s ≥ 1), to the expression : d mz r,s = #(0).{#(v) / g r-1 = 0+g s-1 .g v.d mod p}+ ∑ #(g i ).{#(u,v) / g r-1 = g i .g u.d +g s-1 .g v.d mod p} i = 1

In the second member, the first term equals 1 when r = s and 0 if not. For this sum, with indice i, we divide the expression by g i (what leaves unchanged the equation). It follows then : d mz r,s = #(0).if(r=s, 1, 0)+ ∑ #(g i ).{#(u,v) / g r-i-1 = g u.d +g s-i-1 .g v.d mod p} (52) i = 1 That is also : d mz r,s = #(0).if(r=s, 1, 0) + (1/d). ∑ #(g i ).mb' r-i,s-i (53) i = 1

Here mb' r,s are the components of the reduced (of their first line and first column) matrices [B'] of monomial y d . For the first line of the matrix, we have :

d mz 0,s = #(0).{#(v) / 0 = 0+g s-1 .g v.d mod p}+ ∑ #(g i ).{#(u,v) / 0 = g i .g u.d +g s-1 .g v.d mod p} i = 1
That is also, since the first term of the second member is null : d mz 0,s = ∑ #(g i ).{#(u,v) / g i+(p-1)/2 .g u.d = g s-1 .g v.d mod p} i = 1

In this sum, only the case i+(p-1)/2 = s-1 mod d may have solutions. Thus : mz 0,s = #(g s-1+(p-1)/2 ).{#(u,v) / g u.d = g v.d mod p} So that finally : mz 0,s = ((p-1)/d).#(g s-1+(p-1)/2 ) = ((p-1)/d).mz s-1+(p-1)/2, 0

We encounter the fact that it is necessary to take account of the two cases p = 1 mod 2d and p = 1+d mod 2d.

The primitive roots equations help to deduce the whole set of the cardinal matrix components for z 1 n +z (Δcmz i,0 ) = (cmz i,0 )-mz 0,0 .(1) (cmz 1,0 , cmz 2,0 , cmz 3,0 , …,cmz d,0 ) = (mz 1,0 , mz d,0 , mz d-1,0 , …, mz 2,0 )

The preceding notations must be well understood. The columns and lines of the cardinal matrices are noted [Mz n°line-1 n°columne-1 ]. Expression (σ mz ) is a column vector of dimension d (does not include σ mz 0 ). ( 1) is a column vector of dimension d with components all equal to 1. [Δcmz] is a right circulant matrix of dimension d whose first line is (Δcmz i,0 ), i >0. (λ) is the column vector of the eigenvalues of the cardinal matrix of y d (at sequence p) without λ 0 .

Proof

The principle of the proof is simple.

If the trace matrix [σ mz ] is solution of [MZ] = [P].[σ mz ][Q] with [Q] = [P -1 ]
, then the components of the trace of [σ mz ] are indeed the sought eigenvalues. We thus carry out a simple substitution of the presupposed expressions and we verify that they agree. We recall (see page 72) that the cardinal matrix of y d is :

1 λ 0 */d λ 0 */d … λ 0 */d λ 0 0 0 … 0 1 λ 0 /d λ 0 /d … λ 0 /d 1 λ 1 */d λ 2 */d … λ d */d 0 λ 1 0 … 0 1 λ 1 /d λ 2 /d … λ d /d [B] = [mb i,j ] = (1/p). 1 λ 2 */d λ 3 */d … λ 1 */d 0 0 λ 2 … 0 1 λ 2 /d λ 3 /d … λ 1 /d … … … … … … … … … … … … … … … 1 λ d */d λ 1 */d … λ d-1 */d 0 0 0 … λ d 1 λ d /d λ 1 /d … λ d-1 /d where λ 0 * = λ 0 = p-1
Let us develop [B] and let us look at only the reduced matrix (r ≥ 1, s ≥ 1). We get, k describing 1 to d :

σ mz 1 #(g 0 ) -#(0) #(g 1 ) -#(0) #(g 2 ) -#(0) … #(g d-1 ) -#(0) λ 1 σ mz 2 #(g d-1 ) -#(0) #(g 0 ) -#(0) #(g 1 ) -#(0) … #(g d-2 ) -#(0) λ 2 σ mz 3 = (1/d). #(g d-2 ) -#(0) #(g d-1 ) -#(0) #(g 0 ) -#(0) … #(g d-3 ) -#(0) λ 3 … … … … … … … σ mz d #(g 1 ) -#(0) #(g 2 ) -#(0) #(g 3 ) -#(0) … #(g 0 ) -#(0) λ d So that also : d σ mz i = (1/d). ∑ (#(g k-i )-#(0)).λ k (58) k = 1
that we can also write with ∑ λ k = -d :

d σ mz i = #(0)+(1/d). ∑ #(g k-i ).λ k (59) k = 1 Thus : d d d ∑ σ mz i = d.#(0)+(1/d). ∑ ∑ #(g k-i ).λ k i = 1 i = 1 k = 1 Then, with still ∑ λ k = -d : d d ∑ σ mz i = d.#(0)-∑ #(g i ) i = 1 i = 1 P 278/390
Quadratic abundance factors and substitutes

As, in addition, with ve the number of integers' variables and vp the number of prime numbers variables :

-#(0))+(1/p).(1/d 3 ). ∑ i ∑ k #(g i ).λ r-i+k-1 *.λ k .λ s-i+k-1 Then : d d d mz r,s = #(0).if(r=s, 1, 0)+(1/p).((p-1) 2 /d 2 ). ∑ #(g i )+(1/d 3 ). ∑ #(g i ). ∑ λ r-i+k-1 *.λ k .λ s-i+k-1 i = 1 i = 1 k = 1 So that also : d d mz r,s = #(0).if(r=s, 1, 0)+(1/d). ∑ #(g i ).(1/p).((p-1) 2 /d+(1/d 2 ). ∑ λ r-i+k-1 *.λ k .λ s-i+k-1 ) i = 1 k = 1
and we thus get the equations derived from the primitive roots equations relation 57 : d mz r,s = #(0).if(r=s, 1, 0)+(1/d). ∑ #(g i ).m' r-i,s-i i = 1 What it was necessary to proof. We note that this relation is established whatever the number and the type of variables.

Prime numbers variables

Let us consider the expression y

i,0 = p-1-mz 0,0 /(p-1)+s i-k,d +Δs i-k,d (63)
Here -k corresponds to a shift of the components indices (see example d = 3). However, this is useful only if the corrections Δs i,d (which are not the same from one line to the other) have characteristic (possibly self-evident) literal expressions. For the moment, we solved this problem only for d = 2 (see higher) and d = 3 (see low).

Cardinal of zero

We wrote earlier : #(0) = t.(p-1)

Indeed, (0,0) is a solution a priori, but must be withdraw as out of a prime numbers variable domain of definition. Let us have then (y 1 ,y 2 ) a non-self-evident solution of if(y 1 ≠ y 2 , (y 1 n+1 -y 2 n+1 )/(y 1 -y 2 ), (n+1).y 1 n ) = 0 mod p.

Then, either y 1 n+1 = y 2 n+1 mod p with y 1 ≠ y 2 , or (n+1) = 0 mod p. The last case is solved immediately with n = p-1 and d = (n, p-1), so that d = n = p-1 (64)

and the solutions are the couples (y 1 ,y 1 ), y 1 = 1 to p-1.

In the first case (y 1 /y 2 ) n+1 = 1 mod p. The solutions are y 1 /y 2 = g u.d' mod p, where d' = (n+1,p-1) and u = 0 to (p-1)/d'-1.

Here y 2 may describe 1 to p-1 and we have, since y 1 ≠ y 2 :

t = (p-1,n+1)-1 (65)
The two relations ( 64) and ( 65) lead to the distinction of the sequences modulo n(n+1).

Joining together the two cases, we have :

t = if(p = 1+n, 1, (p-1,n+1)-1) (66)
The illustration is concretised by: (5,11,17,23,41) mod 42 1 mod 42 7 or (7,13,19,25,31,37) (9,17,25,33,41) mod 56 or (5,13,37,45,53) mod 56 7 or (3,11,19,23,27,31,39,47,51,55) (7,31,43,67) mod 72 or (11,23,35,47,59,71) mod 72 37 mod 72 or (13,61) mod 72 or (5,29,53) mod 72 1 mod 72 or(25,49) mod 72 or (17,41,65) (10,40,70) mod 90 or (17,23,29,47,53,59,77,83,89) mod 90 or(31,61) mod 90 or (7,13,43,49,67 ;79) , then the couples (g u.(p-1)/d .y 1 , g u.(p-1)/d .y 2 ) are also solutions, u = 0 to d-1. It would be interesting to find congruence relations as in the case c = 0 mod p to evaluate #(g i-1 )/d (if such relations exist).

Numerical example

Let us have n = 11 and p = 67 (g = 2). We have then p-1 = 66, d = (66,11) = 11, (p-1)/d = and t = (66, 11.12)/ (11, 66, 11.12)-1 = 6.11/11-1 = 5. The cardinal matrix is given thereafter by direct calculation :

Variation with m 0,0 Cardinal matrix [M] of ∑z1 n-i .z2 i , y1 and y2 We verify that t is equal to the announced value and that m 0,i = 6.m i,0 (i>0). Then, we have: Hence :

(Δcm i,0 ) λi (1/11
[σvp] = [P].67 Here the third column can be ignored as it is a simple consequence of the preceding ones and of p = v²+w².

The complete resolution of the case corresponding to the group of variables x We use here the same primitive root, for a given p, as in the case of the variables of integers.

The reader can easily prove the simplicity of the passage from two variables of integers to the two variables of prime numbers by the following adjustments of cardinals :

p Δ(#(0)) Δ(#(g 0 )) Δ(#(g 1 )) Δ(#(g 2 )) Δ(#(g 3 )) 1 mod 4 -1 -8 0 0 0 3 mod 4 -1 -4 0 -4 0
Mix of integers and prime numbers variables : c = x 0) 4(p-1) 0 0 4(p-1) 0 #(g 0 ) p-7+x+if(v = 0 mod 5, 0, 4v.(-1) e1 ) p-3+x+if(v = 0 mod 5, 0, 4v.(-1) e1 ) p-3+x+4.(v.(-1) e1 +w.(-1) e2 ) p-5+x p-1+x #(g 1 ) p-3-x-if(v = 0 mod 5, 0, 4v.(-1) e1 ) p+1-x-if(v = 0 mod 5, 0, 4v.(-1) e1 ) p+1-x-4.(v.(-1) e1 +w.(-1) e2 ) p-3-x p+1-x #(g 2 ) p-3+x-if(v = 0 mod 5, 0, 4v.(-1) e1 ) p+1+x-if(v = 0 mod 5, 0, 4v.(-1) e1 ) p+1+x-4.(v.(-1) e1 +w.(-1) e2 ) #(g 3 ) p-3-x+if(v = 0 mod 5, 0, 4v.(-1) e1 ) p+1-x+if(v = 0 mod 5, 0, 4v.(-1) e1 ) p+1-x+4.(v.(-1) e1 +w.(-1) e2 ) Note : Again, we have to use the same primitive root, for a given p, as in integers' variables case.

Here also, the passage from two integers' variables to the mix of variables is basic :

p Δ(#(0)) Δ(#(g 0 )) Δ(#(g 1 )) Δ(#(g 2 )) Δ(#(g 3 )) 1 mod 4 -1 -4 0 0 0 3 mod 4 -1 -2 0 - 2 0
A complete study makes it necessary to determine the cardinal matrices eigenvalues modulo p, and then modulo p δ when δ tends towards infinite (consisting in simples repetitions of eigenvalues of the target different from 0). The peculiar modulo 2 δ and modulo 3 δ cases are also to be treated.

Symmetrical overlapping variables of degrees 8 and 2 n

Let us have the expression : if(z 1 ≠ z 2 , (z 1 9 -z 2 9).(z 2 -z 1 ), 9z 1 8 )

Our study remains rudimentary due to the lack of relevant observations beyond what follows.

We must consider modulo 8 congruencies (in relation to the z 8 group), but also the modulo 18 congruencies (in connection with (z 1 9 -z 2 9). Modulo 144 cases result which simplify modulo 72. We then have opportunity to rankings in the following way (p = 2 and p = 3 cases are peculiar): p mod 72 #(0) #(g 0 ) à #(g We notice here the peculiar case of 17, particularity that did not appear earlier (as 9 is not a prime number and thus is not a sequence). We notice also that there is no intermediate case for #(0) like 8p-7, 4p-3 or 2p-1 when such case (2p-1) can be sampled previously (therefore, it is difficult to guess the mix of #(g i ) for any degree 2 n ).

Easiness leaves place to difficulty when it comes then to find more detailed rules that would be useful to avoid direct calculation of the cardinals #(g i ). Here again, the passage from two variables of integers to two variables of prime numbers is very simple :

p Δ(#(0)) Δ(#(g 0 )) Δ(#(g 1 )) Δ(#(g 2 )) Δ(#(g 3 )) Δ(#(g 4 )) Δ(#(g 5 )) Δ(#(g 6 )) Δ(#(g 7 )) 1 mod 8 -1 -16 0 0 0 0 0 0 0 5 mod 8 -1 -8 0 0 0 -8 0 0 0 3 mod 4 -1 -4 0 -4 0 -4 0 -4 0
Ditto for the passage of two integers' variables to a mix of variables : p Δ(#(0)) Δ(#(g 0 )) Δ(#(g 1 )) Δ(#(g 2 )) Δ(#(g 3 )) Δ(#(g 4 )) Δ(#(g 5 )) Δ(#(g 6 )) Δ(#(g 7 ))

1 mod 8 -1 -8 0 0 0 0 0 0 0 5 mod 8 -1 -4 0 0 0 -4 0 0 0 3 mod 4 -1 -2 0 -2 0 -2 0 -2 0
Let us notice, to finish with, that the passage from a diophantine equation with unit coefficients to one with some other integer coefficients does not change the rules concerning the cardinal Δ(#(g i )) distinctions modulo 2 n (2 n +1). For example, passing from expression z 0)) Δ(#(g 0 )) Δ(#(g 1 )) Δ(#(g 2 )) Δ(#(g 3 )) Δ(#(g 4 )) Δ(#(g 5 )) Δ(#(g 6 )) Δ(#(g 7 )) Δ(#(g 8 )) Δ(#(g 9 )) Δ(#(g 10 )) Δ(#(g 11 )) Δ(#(g 12 )) Δ(#(g 13 )) Δ(#(g )) Δ(#(g We are interested from now on in the expression with two or more variables :

∑ a i .z 1 (e1) .z 2 (e2) …z k (ek) (71)
We restrict ourselves with the cases (e1)+(e2)+…+(ek) = n constant. (e2) .z 3 (e3) …z k (ek) then g u.(p-1)/d .(z 1 ,z 2 ,…,z k ), where d = (p-1,n), is also solution and we have : #(c.g u.d ) = #(c)

If (z 1 ,z 2 ,…,z k ) is a solution of c = ∑ a i .z 1 n-(e2)-(e3)-…-(ek) .z 2

Primitive roots equations

The construction of two-dimensional tables and of the cardinal matrices thanks to the primitive roots equations remains unchanged by posing d = (n, p-1). It follows thus immediately : ) is an integer. The proof of this point is identical to that carried out higher. It is not necessary to review that. Parameter t varies as a sequence p modulo d.(d+1) function. However, a literal formula of this parameter remains to be established.

-We do not seek any more to highlight the constructive of sequence matrix [s d ] here, since the differences m i,0 -m 0,0 are variable with the choice of the diophantine equation. Hence with (p-1)².p = 70².71 =347900 : [P].71 with same rules of writing as in the preceding paragraph.

Foot-note : As d = p-1, we have p = 1+d mod 2d. We are systematically in the case of eigenvalues and eigenvectors with imaginary components.

Matrix form

As d = p-1, the components of the cardinal matrix are, starting from the second column, simple permutation of the components of the first column. The rule for locating the positions is still to establish.

We also notice a trace with equal components value and some symmetries delimited by the features in dotted lines. [

P real ].13 1/2 = 1 1 1 1 1 1 1 1 1 1 1 1 1 1
0,885 0,568 -0,355 -0,749 0,121 -0,971 0,885 0,568 -0,355 -0,749 0,121 -0,971 1 0,568 -0,355 -0,749 0,121 -0,971 0,885 0,568 -0,355 -0,749 0,121 -0,971 0,885 1 -0,355 -0,749 0,121 -0,971 0,885 0,568 -0,355 -0,749 0,121 -0,971 0,885 0,568 1 -0,749 0,121 -0,971 0,885 0,568 -0,355 -0,749 0,121 -0,971 0,885 0,568 -0,355 1 0,121 -0,971 0,885 0,568 -0,355 -0,749 0,121 -0,971 0,885 0,568 -0,355 -0,749 1 -0,971 0,885 0,568 -0,355 -0,749 0,121 -0,971 0,885 0,568 -0,355 -0,749 0,121 1 0,885 0,568 -0,355 -0,749 0,121 -0,971 0,885 0,568 -0,355 -0,749 0,121 -0,971 1 0,568 -0,355 -0,749 0,121 -0,971 0,885 0,568 -0,355 -0,749 0,121 -0,971 0,885 1 -0,355 -0,749 0,121 -0,971 0,885 0,568 -0,355 -0,749 0,121 -0,971 0,885 0,568 1 -0,749 0,121 -0,971 0,885 0,568 -0,355 -0,749 0,121 -0,971 0,885 0,568 -0,355 1 0,121 -0,971 0,885 0,568 -0,355 -0,749 0,121 -0,971 0,885 0,568 -0,355 -0,749 1 -0,971 0,885 0,568 -0,355 -0,749 0,121 -0,971 0,885 0,568 -0,355 -0,749 0,121 This is however only the matrix modulo p. The selected expression will certainly not have particular regularity modulo p δ and the study of the abundance factors in the limit cases p δ→∞ may be far from self-evident (and it must be carried out for all the sequences).

Example modulo p δ

To better illustrate the previous point concerning the evolution of abundance factors with δ, let us consider the cases : c = x 1 2 +x 1 .x 2 +x 2 3 and c = y 1 2 +y 1 .y 2 +y 2 3

We have the following results by direct research : Sequence p = 5, variables of integers Sequence p = 7, variables of prime numbers Sequence p = 7, variables of integers In the previous tables when two expressions i are corresponding to the same number, then maximum cardinal #(i) prevails (example : 343 corresponds to 0+7 3 k and 0+7 2 k and we do select #(343) = max(#(0+7 3 k), #(0+7 2 k))).

Sequence p = 3, variables of prime numbers δ = 1 δ = 2 δ = 3 δ = 4 δ = 5 δ = 6 δ = 7 i #(i) i #(i) i #(i) i #(i) i #(i) i #(i) i #(i) 1 2 1+3k 2.3 1+3k 2.3 2 1+3k 2.3 3 1+3k 2.3 4 1+3k 2.3 5 1+3k 2.3 2 1 2+3k 1.3 2+3k 1.3 2 2+3k 1.3 3 2+3k 1.3 4 2+3k 1.3 5 2+3k 1.3 0+3k 1.3 0+3k 1.3 2 0+3k 1.3 3 0+3k 1.3 4 0+3k 1.3 5 0+3k 1.3 0+3 2 k 1.3 2 0+3 2 k 1.3 3 0+3 2 k 1.3 4 0+3 2 k 1.3 5 0+3 2 k 1.3 0+3 3 k 1.3 3 0+3 3 k 1.3 4 0+3 3 k 1.3 5 0+3 3 k 1.3 0+3 4 k 1.3 4 0+3 4 k 1.3 5 0+3 4 k 1.3 0+3 5 k 1.3 5 0+3 5 k 1.3 0+3 6 k 1.3 0 1 0 1.3 0 1.3 2 0 1.3 3 0 1.3 4 0 1.3 5 0 1.3 Sequence p = 3, variables of integers δ = 1 δ = 2 δ = 3 δ = 4 δ = 5 δ = 6 δ = 7 i #(i) i #(i) i #(i) i #(i) i #(i) i #(i) i #(i) 1 5 1+3k
δ = 1 δ = 2 δ = 3 δ = 4 δ = 5 δ = 6 i #(i) i #(i) i #(i) i #(i) i #(i) i #(i) 1 9 1+5k
δ = 1 δ = 2 δ = 3 δ = 4 δ = 5 δ = 6 i #(i) i #(i) i #(i) i #(i) i #(i) i #(i) 1 5 1+7k
δ = 1 δ = 2 δ = 3 δ = 4 δ = 5 δ = 6 i #(i) i #(i) i #(i) i #(i) i #(i) i #(i)
We can see some logic in these tables with nevertheless the emergence of "trouble-making" values which prevents predictability of results for any given target.

For the logical part, we observe (i > 0) :

#(0+p i k) = (-1+i.(p-1)).p i-1
Otherwise, outside targets with values inferior to p (obviously not predictable), the For variables of prime numbers, the situation is more complex. The r p numbers are integers whose values depend on p. We have r p ≈ if (p = 1 6, 2, 1 mod) with however a lot of exceptions (for example for sequences p = 31 and p = 43, we have r p = 5). Returning back to the initial tables, we also note the existence of false trouble-making values, that is trouble-making values returning to their natural place when δ increases (example : 158, 1873 and 11477 cases for sequence p = 7).

Because of the numerous comments and exceptions described previously, it seems that the frontal attack to evaluate the cardinal matrices of x +c or p = y 1 2 +y 1 .y 2 +y 2 3 +c where major simplifications prevail thanks to variable p (explained by the ubiquity of 0 in the cardinal matrices eigenvalues).

Conclusion

With the preceding reserves, we can now determine cardinal matrices and eigen-elements of any expression. The matrix method allows then to easily take into account the aggregation of recurring (generalized Waring type) or distinct terms while using simple products of eigenvalues matrices (and matrices of environment and their inverses corresponding to the sequences p).

In the exercise which follows, we are interested in simplification of the general cases thanks to a guide and we reconsider the existence of a common matrix of environment (at given sequence and matrix dimension). Consequently, when a complex problem "turns out well", this results in general from some symmetry (the reader will note that admitted values for δs of the preceding table, values ad infinitum, remain to be demonstrated).

However, when these polynomials are crossed in the same expression, the stability is reached from the start (at sequence 3) because of symmetry of p 2 , whereas it is not the case when we cross two non-symmetrical expressions.

(p 2 )+(q 2 +q) (p 2 +p)+(q 2 +2q) c fae(δ = 1) fae(δ = 2) δs c fae(δ = 1) fae(δ = 2) fae(δ = 3) fae(δ = 4) fae(δ = 5) fae(δ = 6) δs 0 A third assembly in the form of monomial could satisfy our search for simplicity in the last case… 13. The peculiar case of sequence p = 2

Generalities

We discussed at length this sequence at exercise 6 for Waring equations modulo p δ , including when δ → +∞. In Fermat Catalan enumerations exercise essentially turned towards the modulo p study, we suggested for the sequence 2 a direct case-by-case study. Despite the already done work, the reader may feel short for more general equations. As we already wrote, among all sequences, p = 2 is the most determinative with a large margin of error between the modulo p and modulo p δ abundance factors evaluation. It is sometimes easy to forecast a limit value for this or that target by observing the modulo 2 δ evolution (when incrementing δ), but not always and it may be quite the opposite...

Waring sums, odd powers

General view

Here, we offer a flexible use of the cardinal matrices method. Let us consider first a Waring sum with k variables based x n where n is odd. We have seen that there are for a given environment (here n) an environment matrix (which is the change of base P matrix) on one hand and the eigenvalues on the other hand with fixed iterated ratios that we covered at page 167 : the joint contributing weighting due to P and P -1 matrices is a 2 -n ratio and the weighting due to the eigenvalues is 2 k ratio, that is a global iterated ratio of 2 k-n . Let us consider equation : a

1 x 1 n +a 2 x 2 n + … +a k x k n = c mod 2 δ
The affine coefficients have the effect to offset eigenvalues. We seek here the rules of these offsets. All offsets can be built thanks to a "small" number of fundamental bricks. For more clarity let us start with example r.x 13 +s. Several comments are to be made about this table :  The "entry" column is analogous to the behaviour of a rho function. The "behaviour" of this sector is not repeated in the following columns. In fact, we placed target 1 in this category to illustrate the concept, although this target can here as well be transferred to loop 1.

The loops include n columns when we examine an equation in environment n. We collect through a classic calculation (method of the overlapping loops) loops 1 and 2 above data. Then, we can anticipate the values of loops 3, 4, and so on, using the following rule : data fae(2 i ) in column i is deduced from column i-n and i-2n values, using the already mentioned ratio 2 k-n (here 2 k-n = 2), by : fae

(2 i ) = fae(2 i-n )+(fae(2 i-n )-fae(2 i-2n ))/2 k-n
We then construct a This table reads downwards and then upwards as follows :

Starting from (r,s,u,v), we seek in common factor (here 2). The multiplicity of 2 in this common factor (here 1) is the column offset index and the multiplicative index of the underneath read data (here 1 column to the right offset and multiplication by 2 1 ). Having obtained (r',s',u',v') = (r,s,u,v)/gcd(r, s, u, v), we withdraw all non-multiples of 2 factors in this quadruplet.) For example as 192 = 3. 2 8 , we do keep only 2 8 = 64. With this new quadruplet, we shall now proceed with successive divisions (not simultaneous divisions) by 2 n (here 2 3 in presence of monomial x 3 ) of the members to arrive at a cited above library quadruplet. Entry data, loops 1 and 2, and following loops as required, are listed in the last line (here case 3 of above library). The order of (r',s',u',v') of course has no effect. The data under the n+1 columns "direct evaluation" are then calculated by the classical method (of overlapping loops). This calculation is relatively fast when n is moderate (as here 3). The "mode" column is equal to the difference between column 1 and column n+1 under the title "direct evaluation". Finally, we supplement the table by adding the "mode" value in column i under the title "deduction" to the i-n (here i-3) column value (including under the title "direct evaluation") on the next line.

We thus determine the abundance factors of targets containing i factors of 2. For the proposed equation, we see that it is difficult to envisage a direct calculation of abundance factors with large 2 multiplicities. Recursive loop begin here at c = 2 18 , that means for 4 variables a 2 72 step routine, which becomes very difficult to manage.

Let us note that this recursivity is fundamental for target c = 0 abundance factor evaluation (that is the target that is divided an infinity of time by 2) obtained for c = 2 δ→∞ . Here, we immediately find when using the value of the target

… … … #{(2 n a).x n +R(x i n ) = 2 i+n mod 2 δ } … … #{(2 n a).x n +R(x i n ) = 2 i+2n mod 2 δ } Line m+1 #{a.x n +R(x i n ) = 2 i mod 2 δ } … … #{a.x n +R(x i n ) = 2 i+n mod 2 δ } … … …
The contribution of variable x in the enumeration #{(x,x i ) \ (2 n a).x n +R(x i n ) = 2 i+n mod 2 δ } is identical to that of x in #{(x,x i ) \ a.x n +R(x i n ) = 2 i mod 2 δ }. Identically, the contribution of variable x in #{(x,x i ) \ (2 n a).x n +R(x i n ) = 2 i+2n mod 2 δ } is that of x in #{(x,x i ) \ a.x n +R(x i n ) = 2 i+n mod 2 δ }. By difference, we obtain the searched result.

Libraries

When n increases, the interest to hold a library of the entry and loops 1 and 2 values is understandable.

For monomial x 3 , with four variables, we recall base library (replacing target c = 1 in loop 1) : We leave the care to the interested reader to find other libraries following his needs.

Case (r,s,u,v) c = 1 c = 2 c = 4 c = 8 c = 16 c = 32 1 (1,1,1,1) 1

Waring sums, even powers

General view

Let us recall the cardinals table associated to x n that we found at exercise 3 : 1) .(#{1}) 1+2.(#{1}).k #{1} other targets 0 other targets 0

Variable of integers Variable of prime numbers

c #{c} c #{c} 0 or 2 δn.n 2 δ-δn-1 / / 2 r..n (1+2.(#{1}).k) 2 r.(n-
When n is even, the target type 2 r..n do not have a unique cardinal as in the case of odd n (where #(1) = 1), but have interdictions each time as c ≠ 2 r..n (1+2.(#{1}).k). That must be taken into account in the libraries. For the rest, the cardinals' construction principles are the same.

Libraries

In the case of the x 2 monomial, we have #{1} = min(2 2 ,2 δ-1 ), so that #{1} = 4 for the limit case. Targets of type 2 r..n (1+2.(#{1}).k) = 4 r .(1+8k) do interest us in particular. We must distinguish targets 4r.(1 mod 8) and 4r.(or (5,3,7) mod 8).

For three variables, the library is as follows :

Case (u,v,w)

Fermat Catalan sums

For Waring sums, the global iterated ratio in column i+n versus column i was 2 k-n where k is the number of variables and n recurrent monomial, that is when reporting to one column an iterated ratio of 2 k/n-1 , that is either (for one column) 2 -1+∑1/n . In this exponent sum ∑1/n refers to the k variables of the problem. When monomials of different exponents are at stake, is sum on the reverse of the monomials exponents. The iterated (unitary) ratio is thus as follows :

P 309/390

General abundance factors ru = 2 -f ( 9) where f = 1-∑1/n i i For example, for r.x 1 2 +s.x 2 3 +u.x 3 3 +v.x 4

5

, the sum is equal to 1/2+1/3+1/3+1/5 = 41/30, the exponent is 41/30-1 = 11/30 and iterated ratio is 2 11/30 . The loops size (number of columns in a loop) is equal to the least common multiple of the exponents (thus recalling the concept of environment cm of exercise 9) : cm = lcm(n i ) (10) In the example, in column 30 where the pattern repeats the first time, the global iterated report is 2 [START_REF] Nathanson | Additive Number Theory[END_REF] . Returning to the abundance factors of a given target c and its mode v(c), we have then Fan(2 cm.i .c) = Fan(c)+if(i = 0,0,(v(c)+v(c).2 f +v(c).2 2f +…+v(c).2 (i-1).f ), that is also :

Fan(2 cm.i .c) = Fan(c)+(2 i.f -1)/(2 f -1).v(c) (11)
Example : c = x 2 -x 3

We prefer this equation to equation c = x 3 +x 2 that is asymptotic only when c tends to infinity (and so in an unverifiable situation). The environment associated with this equation is lcm (2.3) = 6. For x Because of the small number of variables, standardized abundance factors associated with c = x 2 -x 3 mod p δ diverge with multiplicity of 2 in the target (including p = 2). Any non-null abundance factor target (except c = 0) can be written here as 2 6i .c, where c is a loop 1 target. The mode v(c) depends on the class of c. The (unitary) iterated ratio (that is not workable directly) equals 2 -(1-1/2-1/3) = 2 -1/6 and the global ratio (which do make sense) is equal to 2 -1 . Hence :

Fan(2 cm.i .c) = Fan(c)+(2 i -1).v(c)
In the studied case, target 0 abundance factor diverges.

Libraries

We can create libraries related to Fermat Catalan sums. For c = x 2 -x 3 modulo p δ , we have 11 cases to distinguish to master the subject : (1,1), (1,2), (1,2 2 ), (1,2 3 ), (1,2 4 ), (1,2 5 ), (2,1), (2 2 ,1), (2 3 ,1), (2 4 ,1), (2 5 ,1).

Odd sequences libraries

As well as above, we can deploy libraries with odd sequences. However, these tables are related to the resolution of equations like a i = g i .g j.d modulo p (see exercise 7 page 178) increasing the number of distinct cases making them more difficult to manage. Abundance factors are probably as easy to evaluate directly thanks to eigen-elements previous study.

How does the fitting of numbers operate ad infinitum?

We explain here, in coarse terms, the behaviour of a diophantine equation ad infinitum. Let us settle the framework. We are interested in an equation composed of independent groups of which we observe the divergence behaviour when the variables tend towards +∞. The independent groups which diverge towards +∞ are placed on the left-hand member of equation and those which diverge towards -∞ are placed on the right-hand member with target c in which we incorporate possible constants. The equation of the volumes does not interest us for the moment here, but only the proportions between the numbers of solutions when targets c vary.

The "macroscopic" behaviour of a diophantine equation, is imposed, at a given sequence, by the degree of convergence.

The actual behaviour of the diophantine equation merges with the macroscopic behaviour for any prime numbers variable (because the degree of convergence is, in this cases, the degree of stability). A finite correction (not inevitably convergent for target 0 or some other unique target) is necessary in the presence of integers' variables. The correction for target 0 (or some other unique target) is obtained by passage to the limit (infinite degree of stability for this target). The prime numbers' variables override integers' variables, which means that, when only one of these variables is present in the diophantine equation then the degree of stability evolves to the degree of convergence.

It "suffice" then to calculate the normalized abundance factor of selected targets and to carry out the Euler products. Hence, the process which we propose below.

Strategy of research of the abundance factors

The good practice for abundance factors determination results from preceding rules :

-replace symbolically, in the diophantine equation, integers' variables by prime numbers variables to determine for each sequence the degree of convergence of the diophantine equation : it equals 1 for all sequences except a finite number of them (often including p =2) -identify the classes (modulo p) of sequences associated with such targets (pending on primitive roots by c = g i .g u.d mod p) having similar expressions for abundance factors -return to the initial variables to explicitly determine (if possible) the rules (conjectural or self-evident) controlling the abundance factors modulo p δ of the previously found classes of sequences and identify possible lower-cases generated by the passage to the variables of integers -work out the particular case of target 0 (or some other unique target) -carry out the Euler product for selected target (approximate determination or literal formula pending on peculiar case)

The essential point in this strategy is the initial substitution of integers' variables by prime numbers variables which allows to say "approximately" if a target has a number of solutions worth a.V'(c) or 0(a).V'(c).

Let us signal, even if the goal is not here the resolution of non-asymptotic cases, that to pose a problem like the theorem of Fermat (x n +y n = 0+z n ) is difficult from the start, as target c equals 0, value which implies a particular behaviour and treatment, referring to the study made within exercise 5, treatment which has some response only by building a matrix of infinite dimension.

Matrices of environment 17.1. Generalities

Still concerning good practice, a supplement help for abundance factors evaluation is the recourse to cardinal matrices.

At a given sequence p, to each independent group corresponds a cardinal matrix whose dimension is related to this group degree of stability δs. This matrix is similar to a trace matrix via an intra-unitary matrix [P] called matrix of the environment (p,δs). For the global diophantine expression, while duplicating (or while contracting), as many times as necessary, the spectrum of the eigenvalues corresponding to each independent group, within the framework of a common environment, we form the operator [P].∏[σ].[P -1 ], the products [P -1 ].

[P] being annihilated (producing [I]) between the trace matrices (of eigenvalues). We use σ instead of λ or μ here because an independent group can contain at the same time integers' variables and prime numbers variables.

for example, using the Sylvester matrix and the method of the resultants, we determine first of all its discriminant to the variable x :

Δ 1 = 0 mod p Δ 1 c T(p) Δ 1 ≠ 0 mod p Δ 1 c NT(p) Δ 2 = 0 mod p Δ 2 c T(p) Δ 2 ≠ 0 mod p Δ 2 c NT(p)
The quadratic example shows the role of certain combinations of unions and intersections of these boundaries. The many conditions that we can anticipate from this example demonstrates the difficulty of a comprehensive study to greater than degree 2 polynomials. When the number of overlapping variables turns to n, we are called to assess n partial derivatives (one per variable) and therefore as much boundaries resulting from discriminants Δ i (moreover nothing guaranties that formulas from one case Δ i c T(p) to the other remains identical), to which we add the conditions of nullity of studied parametric diophantine equation coefficients (m additional boundaries for m parametric coefficients) and the conditions of nullity of the variables (n boundaries for n variables).

Let us note that the boundaries are source of ambiguities for the targets. Although not involved in the previous discussion, target 0 may be a boundary. . Foot-note This does not mean that the richness of prime numbers in the preceding cases is significantly larger than for example in polynomial x²+x-12276955783. Indeed, when sequence p increases, its discriminating effect on abundance factor becomes increasingly smaller.

Foot-note 3 710 369 067 405 = 3.5.7.11.13.17.19.23.29.31.37

EXERCICE 14 : POLYNOMIAL MONOTONY

We used rather freely the reciprocal functions in the introduction pages of our article. In this exercise, we clarify the difficulties related to a possible not-monotony of the studied polynomial and the skirting of these difficulties.

Let us consider initially P a monotonous crescent polynomial, and axis (x,y) placed in such a way that P(x) passes by the origin (for example P(x) = a.x n , a > 0 a positive integer and P -1 (x) = (x/a) 1/n ). Graphically, we get P -1 from P by simple permutation of the axis :

y Y = x P -1 (x) P(x) x x = y
Intuitively, the number of integers (or prime numbers) intercepted by P(x) on the y axis rise in reverse ratio with the slope of the curve. The weaker the slope is, larger will be the chances of interception and conversely. However the slope of P -1 (x) is the reverse of the slope of P(x). The reader will refer to exercise 19 for a mathematical highlighting on the question, but for the moment, the argument suffices to our purpose. Hence, the proportion :

#{n (or p) / P(n) = c} ≈ k.P -1 (c)

If P is decreasing, then the situation arises in the form :

y Y = x P -1 (x) Y = x P(x) x x = y x = -y Fig 1 Fig 2 Fig3
When x increases (Fig 1) the number of integers (or prime numbers) intercepted on the y axis increases, and P(x) decreases. The P -1 curve is thus followed according to decreasing x (Fig 2). The increasing (with identical account) can be restored by the use of the mirror image of figure 2, that is figure 3. This method is particularly useful for more broken curves, such as :

y Y = x P -1 (x) Y = x P(x) x x = y x = « y »
The "game" thus consists at the singular points to use mirror curve of P -1 curve obtained by permutation of axis. These singular points are located at P'(x) = 0 and P''(x) < 0. The initial form is kept (no mirror effect) at P'(x) = 0 and P''(x) > 0.

Convergent or divergent iterative function

Let us consider expression (a > 0) q i+1 = q i +a.ln(q i ), q 1 = 2 (5)

Ad infinitum, the resulting density for q i is :

1/(a.ln(q i ))

As 1/ln(p i ) is the density of prime numbers ad infinitum, if a > 1, on average the representative curve of expression ( 5) is above the representative curve of the set of prime numbers and the difference of ordinate will diverge ad infinitum with a positive variation. Conversely if a < 1, on average the representative curve of the expression is below the representative curve of prime numbers set and the difference of ordinate will diverge ad infinitum with a negative variation.

The curves below give an example (set of p i and iterations q i+1 = q i +a.ln(q i ) with a=0.9, a=1, a=1.1) :

If we suppose a > 1, then there exists a number σ 1 such as  i > σ, q i > p i .

If we suppose a < 1, then there exists a number σ 2 such as  i > σ, q i < p i . We deduce : If a = 1 and q i+1 = q i +a.ln(q i ) where q 1 = 2, then a number σ such as ( i > σ, q i > p i ) or ( i > σ, q i < p i ) does not exist.

LOGARITHMIC ITERATION VERIFICATION OF VARIATION ITERATIVE FORMULA P(p i+1 ) = P(p i )+P'(p i ).Ln(p i )

Thus, when i →∞, p i oscillates around the co-ordinates defined by q i+1 = q i +ln(q i ), q 1 = 2.

The asymptotic concept of oscillations can be considered only with an infinity of times.

Lemma

The prime numbers set oscillates around the coordinates defined by the iterative expression : q i+1 = q i +Ln(q i ), q 1 = 2

We could not insist enough here on the originality of the approach. The concept is not a local approach of the notion of difference between prime numbers but a global approach with an absolute reference. Then, it is interesting to get the difference between the i th prime number and the i th number resulting from the iteration.

Mesh error

Self-evidently, integers position is given by the iterative formula n i+1 = n i +1. In this case, the mesh error is null. It is thus a concept which does not intervene. We locate the position of prime numbers by the iterative formula q i+1 = q i + Ln(q i ), q 1 = 2 and a difference margin to this theoretical mesh. Let us have thus e i = p i -q i and p i = q i + e i . Then we have :

Proposition

There exists a finite number s such as the mesh error to order i is lower than the logarithm of i th prime number power this very number :

 s > 0 /  i, | e i | < Ln s (p i ) (6)
Indeed, according to the preceding lemma, because of asymptotic oscillation, a family of differences {e i } such as e i is positive for any consecutive i is necessarily of finite cardinal. In the same way for negative variations {e i }. Thus the proposition.

The difficult point to reach the proof is here that the number of families is not finite.

The proposition does not affirm anything on the value s which can, a priori, be of considerable size (quasi infinite, but not infinite). The graph below gives for s an order of magnitude of 4 to 5, but nothing guarantees it (increase in curve). However, that s be about 10 10 would not change anything to what follows. -20 000 000 40 000 000 60 000 000 80 000 000 100 000 000 Pi (pi-qi)/(ln(qi))^2 (pi-qi)/(ln(qi))^3 (pi-qi)/(ln(qi))^4 (pi-qi)/(ln(qi))^5

EVOLUTION DE L'ECART RELATIF DE MAILLE

Stochastic distribution of prime numbers

The reader will refer to [14] for more details on this paragraph subject. He will find in particular Cramer's conjecture lim sup (p i+1 -p i )/ Ln(p i )) 2 = 1 i → +∞ and In this approach, it is interesting to note that implicitly i is deduced from p i with a margin of error, whereas in our approach, we refer to an absolute (not relative) framework but in which it is not possible, a priori, to evaluate its value because of its iterative construction. Prime numbers have art to maintain their secrecy.

Maximum differences between expressions of prime numbers

Mesh error defines a variation of position of a prime number compared to an absolute position. We immediately deduce from it the relative maximum difference between two prime numbers. With order i, it equals | e i | +| e i+1 |, which is also Ln s (p i ) + Ln s (p i+1 ). This value is equivalent to 2Ln s (p i ). Moreover, if we reason on a group of consecutive prime numbers all attached to an absolute position keeping in mind the mesh error, we can use the average error :

| e i | < Ln s (p i ) ( 7)

Cardinals of prime numbers in an interval

Let us have F(x) a monotonous function derivable in an interval I of length 1 (n not necessarily integer) and F'(x) its derived function. Let us have then I = [n,n+1].

We seek the prime numbers cardinal ranging between F(n) and F(n+1).

Average cardinal on I

Far from the origin, the average cardinal is equal to the increase of function F by the average difference between prime numbers, that is roughly (F(n+1)-F(n))/Ln(n+1/2) or also : The volume on the surface (between c-1/2 and c+1/2) which interests us is given by the partial derivative with respect to variable c, the product of the integrals remaining unchanged (because a constant) : k V'(c) = (k/n).c k/n-1 . П I(n,i) (7) i = 1

The condition of divergence of this term is : k > n (8)

Concerning the asymptotic enumeration of the solutions of ( 1), it is then necessary to adjust the result coming from ( 5) with method given in exercise 5 : k lim #{(x 1 , x 2 , …, x k ) / x 1 n + x 2 n +…+ x k n = c} = Fan(c).(П I(n,i)).(k/n).c (k/n-1) . ( 9)

c →∞ i = 1
where Fan(c) is target c abundance factor expressed at exercises 9 and 10.

The reader can examine in [11] p121 Hardy and Littlewood circle method which gives an asymptotic expression of diophantine equation x 1 n + x 2 n +…+ x k n = c enumeration as : lim #{(x 1 , x 2 , …, x k ) / x 1 n + x 2 n +…+ x k n = c} = S(c).(Г(1+1/n)) k /Г(k/n).c (k/n-1) + O(c k/n-1-δ ) (10) c →∞ S(c) is called singular series in [11], Г is the gamma function. To establish the correlation between expressions ( 9) and ( 10), we use the remarkable identity which we find in [11] p 134 : p h → +∞ and by : M (n,q) = #{x 1 n + x 2 n +…+ x k n = c (mod q)} whose evaluation is the object of exercise 6.

We thus have indeed the identity between the abundance factor and fudge factor.

General Waring case

Let us have a i > 0, i ranging 1 to k. The general case is : a 1 .x 1 n + a 2 .x 2 n +…+ a k .x k n < c (12) a 1 .x 1 n + a 2 .x 2 n +…+ a k .x k n = c (13) We need to evaluate the following integral :

x 1 = (c/a 1 ) (1/n) x 2 = ((c-a 1 .x 1 n )/a 2 ) (1/n) x 3 = ((c-a 1 . The condition of divergence is identical to the preceding : k > n ( 16)

Then, we get ad infinitum enumeration of solutions after proper weighting :

P 332/390
Integer enumeration in a hypervolume and on its surface ).(k/n).c (k/n-1) (17) c →∞ i = 1

Hyperplane case

We use the preceding result, with n = 1, for the diophantine inequality and equality (x i positive integers ) :

x 1 + x 2 +…+ x k < c (18) x 1 + x 2 +…+ x k = c (19) . This leads us, as I(1,i) = 1/i, to V(c) = c k /k! (20)

V'(c) = c (k-1) /(k-1)! ( 21)

The condition of divergence (when c tends towards infinite) of the last term is k > 1and the sought cardinal is (as Fan(c) = 1 for any c) : #{x 1 + x 2 +…+ x k = c} ≡ c (k-1) /(k-1)! ( 22)

Hypersphere case

We propose here a polar alternative for the evaluation of the integrals to establish the cardinals of (x i positive integers) :

x 1 ² + x 2 ² +…+ x n ² < c (23) x 1 ² + x 2 ² +…+ x n ² = c (24)

Let us start by drawing the following picture : Here the account of all solutions (at c = 70 for example), roughly amounts to calculate relative surface occupied by bold police characters compared to the square defined by the value of c and to multiply by (c 1/2 ) 2 . This surface is that of a circle quadrant. The ratio equals π/4 and the sought result is (π/4).c.

To pass from the 2-dimension of this table to a 3-dimension table, the reader will refer to [13] (p 967) for the arguments concerning curvilinear frame constitution and in particular spherical co-ordinates. Thus, these co-ordinates and Cartesian co-ordinates are related by equations :

x = ρcos(θ)sin(φ), y = ρsin(θ)sin(φ), z = ρcos(φ) The metric (ds)² = (dx)²+(dy)²+(dz)² = (dρ)²+ ρ²sin²(φ)(dθ)²+ ρ²(dφ)² leads to the scale factors : h ρ ² = 1, h θ ² = ρ²sin²(φ), h φ ² = ρ² and to the differential element of volume : dV = ds ρ ds θ ds φ = h ρ dρh θ dθh φ dφ = ρ²sin(φ)dρdθdφ

The attentive reader will be able to generalize this type of frame references to the case of higher dimension spaces. Thus, let us pose for example for a 5-dimension space :

x 1 = ρsin(θ)sin(φ) sin(φ') sin(φ'') x 2 = ρcos(θ)sin(φ) sin(φ') sin(φ'') x 3 = ρcos(φ) sin(φ') sin(φ'') x 4 = ρcos(φ')sin(φ''), x 5 = ρcos(φ'')

This system of co-ordinates provides relation x 1 ² + x 2 ² +x 3 ² + x 4 ² + x 5 ² = ρ² and the scale factors : h ρ = 1, h θ = ρsin(φ)sin(φ')sin(φ''), h φ = ρsin(φ')sin(φ''), h φ' = ρsin(φ''), h φ'' = ρ and then the differential element of volume : dV = ds ρ ds θ ds φ ds φ' ds φ'' = ρ 4 sin(φ)sin²(φ')sin 3 (φ'')dρdθdφdφ'dφ'' More generally the element of volume in a n-dimension space is : dV = ρ n-1 sin(φ 1 )sin²(φ 2 )…sin n-2 (φ n-2 )dρdθdφ 1 dφ 2 …dφ n-2

With that tool, we seek the ratio of the hypersphere volume to the circumscribe hypercube in a n-dimension space. The hypercube of radius 2R has a volume (2R) n . The volume of hypersphere is obtained by integration thanks to the element of volume in hyperspheric co-ordinates of which all the variables are separable. The integration boundaries are : 0 ≤ ρ≤R, 0≤θ≤2π, 0≤ φ 1 ≤π, 0≤ φ 2 ≤π, …, 0≤ φ n-2 ≤π and volume is equal to : V = ∫ ρ n-1 dρ ∫dθ ∫sin(φ 1 )dφ 1 ∫sin²(φ 2 )dφ 2 …∫sin n-2 (φ n-2 )dφ n-2

On interval [0, π], it is elementary to show by integration by parts that π π ∫ sin n (t)dt = (n-1)/n ∫ sin n-2 (t)dt 0 0

By recurrence, we get the two following cases : π π ∫ sin 2n (t)dt = (2n )(2n-3)...3 (2n) (2n-2)...4

∫ sin 2 (t)dt = (2n-1)(2n-3).. The ratio of the hypersphere to the hypercube is thus :

-in dimension 2n : π n (25) Here, we may be astonished by the rapid evolution of the ratio towards limit zero. But the representative picture of the phenomenon with n dimensions is that of a sea urchin with 2 n prickles (built with the planes between axis and the planes at the end of axis units vectors) and a central body (the inscribed hypersphere) which approaches in proportion more and more a pinhead.

In fact, not only the ratio r(t) tends towards 0, but r(t)/r(t-1) also. Indeed, using ( 25) or ( 26), we verify that the ratio r(t)/r(t-2) equals π/4n. Then, by Stirling's formula [2] n ! → (n/e) n .(2π n) 1/2 when n → +∞ and by the fact that (1+i) 1/i → e when i → 0 (according to [13]), we get the limit : r(t) → √π .

r(t-1) 2 √n

This is true as well for even t as for odd t and is can be also written : r(t) → √π . We can see in this table that values u(t) close to ratio r(t)/r(t-1) limit values (according to 27) are obtained rather quickly. In addition, when t → +∞, we have r(t)/r(t-1) → 0. This indicates in high dimension t a drastic reduction of our sea urchin pinhead to each incrementing t → t+1.

Let us note that ( 25) and ( 26) can be written in a single way in dimension n : π (n-i)/2 ((n-i)/2)! (28) (n).(n-i).(n-2).(n-2-i).(n-4).(n-4-i)... (4). (4-i). (2). (2-i) where i = 0 if n is even and i = 1 if n is odd, That is also i = (1+(-1) n+1 )/2, and leads for (28) to : That we considered the ratio of volumes of hypersphere to circumscribed hypercube or simply the ratio of the one of the quadrants is the same. Quadrant length equals c ½ , thus we have volume c n/2 in dimension n. In such a case, the abundance factors values collapse when a sequence p = 3 mod 4 is reached for a given divisor of c (for example the abundance factor of target 10008 collapse for sequence p = 463. Without going in details, we observe at first sight that a great number of targets have in theory normalized abundance factors much smaller than 1. This potential deficit of solutions must be made up by some others target for an average value of normalized abundance factors about unit. On this extract, this compensation is symbolized by target c = 100010 whose abundance factor is superior to 10 (being thus equivalent to 10 targets).

For k > 2, the order of magnitude of normalized abundance factors is the unit (but with current decade variations). We observe easily the convergence of the normalized factors towards a limit (whereas the number of variable increased only one unit compared to the preceding case). For as much, we must remember here that the evaluations count for asymptotic cases i.e. when c →∞, which is not the case here. Even if all of normalized abundance factors converge apparently quickly, we remain here far from the account in the proportions of the actual chosen targets solutions that we present below (mod p). The improvement appears only by considering several targets in arithmetic progression. We distinguish here the power of the last variable (m instead of n).

Case of a boundless volume

As posed, the studied (asymptotic) case allows a fine evaluation. We deduce from previously studied Waring general case :

x k+1 k V(c) ≡ ∫ (c+a k+1 .x k+1 m ) k/n .dx k+1 П I(n,i)/a i 1/n (35) i = 1 i = 1

We get the derivative of this expression by deriving the expression inside the integral : < q+c} = a.(π/8).q 2 /Ln(q) p < q lim #{x 1 2 + x 2 2 = q+c} = a.(π/4).Fan(c).q/Ln(q) p < q

The numerical application leads for this last equation to : The adjustment of coefficient "a" was made to get equal for preceding targets, on the average, the sum of actual solutions cardinals and the sum by literal calculations. When this is carried out, we realize that the difference between the number of the solutions of a given target and its computed value improves in general at each decade (with episodic exceptions). In this table, we verify values of the coefficient "a" close to 1. After an increase between the case p < 100 and the case p < 1000, the value of a converges slowly towards 1 (p < 10000).

We also note that if the differences between the actual number of solutions and the value obtained by literal evaluations are relatively ill-matched in the case p < 100, the asymptotic behaviour is getting close to what we expect when p increases with a standard deviation of the variations of about 3% for the 40 targets we present here.

Case of the powers of degree n, prime numbers variables

We examine the enumeration of the diophantine solutions of following inequality and equality :

Numerical application 1 lim #{y 1 2 < p+c} = (4/3).p 3/2 /Ln 2 (p) p →∞ lim #{y 1 2 = p+c} = 2.Fan(c).p 1/2 /Ln 2 (p) p →∞ For p finite, we introduce the factors of correction a : lim #{y 1 2 < q+c} = (4/3).a 2 .q 3/2 /Ln 2 (q) p < q lim #{y 1 2 = q+c} = 2a 2 .Fan(c).q 1/2 /Ln 2 (q) p < q Hence, for the second equation : We observe a good correspondence between the exact number of the solutions and the calculated cardinals for the majority of the targets. Interdictions are respected also (a single solution for one square target does not put into question an abundance factor equal to zero). We note for the square targets that the abundance factor when it is not null from the start by the rules of interdiction modulo (1 mod 2 and 2 mod 3) tends slowly towards zero (case c = 36). The case c = 0 is a remarkable exception. Indeed, the abundance factor of this target tends towards infinite whereas the equation p² = q, p and q prime numbers is obviously without solutions. This result is comprehensible for p²+q = 0 as the abundance factor although the same one is multiplied by a volume which is null. It is not the case here where volume is q 1/2 /ln 2 (q) and also tends towards infinite. The target c = 0 is of "rich target" type. For p finite, we introduce a factor of correction a.

lim #{y 1 2 + y 2 2 < q+c} = a 3 .(π/2).q 2 /Ln 3 (q) p < q lim #{y 1 2 + y 2 2 = q+c} = a 3 .π.Fan(c).q/Ln 3 (q) p < q

The numerical application gives for the second equation : We note a monotonous progression towards the asymptotic behaviour, with an initial awaited increase of factor a, then a slow fall of this factor (which is observed here only starting at p < 1000000) towards 1.

By drawing a table of modulo 2 solutions, the reader will easily verify that cases c = 0 mod 2 are impossible except y 2 = 0 mod 2, that is y 2 = or y 2 = 2 (as y 2 represents prime numbers). In the same way, cases c = 2 mod 3 are impossible except y 2 = 0 mod 3, that is y 2 = 0 or y 2 = 3. This leads to the null abundance factor of targets 0 mod 2 or 5 mod 6 in the preceding table. In these cases, the enumerations of asymptotic branches, are not like p/Ln 3 (p) but like p The observation is as awaited. We initially see an increase of the adjustment factor a, which then slowly tends towards its asymptotic value 1. In the same way, the standard deviations are tightened when p increases (with oscillations).

Case of a boundless hypervolume, generalized Waring, variables mix

We are interested, to finish the exercise, with relatively general diophantine equations solutions enumeration, namely : a x1 .x 1 (x1) + a x2 .x 2 (x2) +…+ a xk .x k (xk) + a y1 .y 1 (y1) + a y2 .y 2 (y2) +…+ a ym .y m (ym) = c + a zr .z r (zr)

(28)

We go on simply using previous arguments :

-We divide by a zr the two members of the equation.

-We make the changes of variables X k (xk) = (a xk /a zr ).x k (xk) , so that x k = (a xk /a zr ) (1/(xk) .X k for each variable except z r . -We pose : a k+m = Π ( a zr ) 1/(xi) . Π ( a zr )

(31) i

In addition, for prime numbers variables, we must also treat integrals like : Y i = (c/a zr +z r (zr) -r(X,Y)) (1/(yi)) 0,99247 0,99462 1,00323 1,01398 0,99140 1,00323 0,99462 0,99462 0,99892 1,00645 1,00645 0,99462 1,00968 0,98710 0,99785 1,00323 0,99174 0,98348 1,00300 0,99024 0,99399 1,01126 0,83333 0,88889 1,33333 0,83333 0,72222 1,33333 0,83333 0,72222 1,33333 0,83333 0,72222 1,33333 0,83333 0,88889 1,33333 0,83333 5 0,83333 0,88889 1,33333 0,83333 0,68611 1,33333 0,83333 0,72222 1,33333 0,79167 0,72222 1,33333 0,83333 0,88889 1,26667 0,83333 7 0,73413 0,93122 1,36508 0,77381 0,73512 1,49207 0,77381 0,63624 1,39683 0,81052 0,67063 1,42857 0,93254 0,82540 1,11587 0,87302 0,73413 0,93122 1,36508 0,77381 0,73512 1,49207 0,77381 0,63624 1,39683 0,81052 0,67063 1,42857 0,93254 0,82540 1,11587 0,87302 0,66824 0,85959 1,32133 0,82837 0,75397 1,39643 0,78373 0,68110 1,49532 0,81052 0,64484 1,52930 0,91461 0,75132 1,03004 0,84504 0,66824 0,85959 1,32133 0,82837 0,75397 1,39643 0,78373 0,68110 1,49532 0,81052 0,64484 1,52930 0,91461 0,75132 1,03004 0,84504 0,65261 0,87969 1,34451 0,84290 0,75838 1,38826 0,78373 0,66716 1,46471 0,79630 0,66181 1,50248 0,91461 0,75572 1,00293 0,86480 0,65261 0,87969 1,34451 0,84290 0,75838 1,38826 0,78373 0,66716 1,46471 0,79630 0,66181 1,50248 0,91461 0,75572 1,00293 0,86480 0,65261 0,87969 1,34451 0,84290 0,75838 1,38826 0,78373 0,66716 1,46471 0,79630 0,66181 1,50248 0,91461 0,75572 1,00293 0,86480 0,64770 0,87496 1,34885 0,85468 0,75186 1,39274 0,77951 0,66357 1,46313 0,80144 0,66608 1,49439 0,92346 0,74597 1,00077 0,86760 0,64235 0,86051 1,35289 0,84634 0,74734 1,40842 0,77951 0,65710 1,45874 0,79361 0,66358 1,49664 0,93317 0,74541 1,00979 0,86043

The products of the normalized factors being relatively stabilized at sequence p = 37, we can use these values for the numerical application. We get the following The difference in this equation with the preceding one is small, but the behaviour is rather different. Indeed, we observe that a correction is necessary as early as p = 2 (δ = 3 instead of δ = 1).

The volume integral of is the same one as previously. Thus : lim #{2x 1 At sequence 2, it is necessary to make evaluations mod 2 3 , because the proportions between the abundance factors for δ = and δ = 2 are not stabilized (equality of all the factors) as we known from exercise 11. 0,5000 1,5000 0,5000 0,5000 1,5000 0,5000 1,5000 1,5000 0,5000 1,5000 0,5000 0,5000 1,5000 0,5000 1,5000 1,5000 3 0,6667 2,0000 0,1667 0,6667 2,0000 0,1667 2,0000 2,0000 0,1667 2,0000 0,6667 0,1667 2,0000 0,6667 0,5000 2,0000 5 0,6667 2,0000 0,1667 0,6667 2,0000 0,1667 2,0000 2,0000 0,1667 2,0000 0,6667 0,1667 2,0000 0,6667 0,5000 2,0000 7 0,6667 2,0000 0,1905 0,6984 1,7143 0,1508 2,0952 2,0000 0,1667 2,2857 0,6984 0,1429 1,8095 0,6984 0,5000 2,0000 11 0,6667 2,0000 0,1905 0,6984 1,7143 0,1508 2,0952 2,0000 0,1667 2,2857 0,6984 0,1429 1,8095 0,6984 0,5000 2,0000 13 0,6667 2,1538 0,1807 0,6447 Certain variations still remain beyond 10% (c = 2, c = 3, c = 8). This is certainly related, not to an ill-matched abundance factor evaluation, but rather to the number of actual solutions, which close to the origin do not reflect yet the distribution ad infinitum. It would be useful to supplement calculations at p < 10 5 , p < 10 6 , p < 10 7 (what would require more than several weeks of calculations with the tools at our disposal).

For the second equation (surface equation) : We get well what we may expect : reduction of the standard deviations when q increases and adjustment coefficients a tending towards 1.

P 374/390

Generation of prime numbers by quadratic sums ans analogues

The numerical application leads for prime numbers variables to two other tables :

For the first equation q (p < q) 100 500 The value of a moves away initially from the asymptotic value 1, then converges, when q increases (while possibly oscillating).

For the second : We get the anticipated result again : reduction of the standard deviations when q increases and adjustment coefficients a tending slowly towards 1 (because of the logarithm in denominator with an exponent 3).

Generation of squares

Translated squares generation by a quadratic expression will give us the opportunity to use the tools that we made up in this article. This kind of diophantine equations is thus interesting as it allows also coming back to the concept of "rich targets".

Squares generation general case would be here equation u.z 1 2 +v.z 1 .z 2 +w.z 2 2 = a.z 3 2 +c, with u, v, w, a constants and z 1 , z 2 and z 3 integers or prime numbers variables. A comprehensive study would result in many cases, which can all be treated the same way as the particular case we propose to study here using items developed in exercises 6 and 10. Let us have thus the integer variables diophantine equation : We rewrite this equation x 1 2 +x 1 .x 2 +x 2 2 -x 3 2 = c. For the factor abundance evaluation, we are interested in the two independent groups x In a more general way, he have at order δ (when i > 1) : #(0) δ p 2δ +(p-1).(p 2δ-2 +p 2δ-3 +…+ p 2δ-1-int(δ/2) ) #(p i .g 2u ) δ p 2δ +(p-1).(p 2δ-2 +p 2δ-3 +…+p 2δ-1-int(i/2) )+(-1) i .p 2δ-1-int((i+1)/2) #(p i .g.g 2u ) p 2δ +(p-1).(p 2δ-2 +p 2δ-3 +…+p 2δ-1-int(i/2) )-p 2δ-1-int((i+1)/2)

To normalize three variables of integers needs to multiply by p δ /(p δ ) 3 . Then at order δ, after summing factors of (p-1) (when i > 1), we get :

The limit case δ → +∞ is then (while conducting the same work as we did for p = 1 mod 12 for the other cases) :

Case p = 1 mod 12 and 11 mod 12 Case p = 5 mod 12 and 7 mod 12 fan(0) 1+1/p fan(g 2u ) 1-1/p fan(g.g 2u ) 1+1/p fan(p.g 2u ) = 1-1/p 2 fan(p.g.g 2u ) 1-1/p 2 fan(p i>1 .g 2u ) 1+1/p-1/p (int(i/2)+1) -1/p (1+int((i+1)/2)) fan(p i>1 .g.g 2u ) 1+1/p-1/p (int(i/2)+1) +(-1) i /p (1+int((i+1)/2))

For the diophantine equation x 1 2 +x 1 .x 2 +x 2 2 = x 3 2 +c volume law evaluation, we have to use the following integral : V'(c) = f.x 3 We draw from this : lim #{x 1 2 +x 1 .x 2 +x 2 2 < x 3 2 +c} = (π/(9√3)).x 3 3

x 3 →∞ and fan(0) δ 1+p -1 -p -(int(δ/2)+1) fan(p i .g 2u ) δ = 1+p -1 -p -(int(i/2)+1) +(-1) i .p -(1+int((i+1)/2)) fan(p i .g.g 2u ) δ

1+p -1 -p -(int(i/2)+1) -p -(1+int((i+1)/2)) fan(0) 1+1/p fan(g 2u ) 1+1/p fan(g.g 2u ) 1-1/p fan(p.g 2u ) = 1-1/p 2 fan(p.g.g 2u ) 1-1/p 2 fan(p i>1 .g 2u ) 1+1/p-1/p (int(i/2)+1) +(-1) i /p (1+int((i+1)/2)) fan(p i>1 .g.g 2u ) 1+1/p-1/p (int(i/2)+1) -1/p (1+int((i+1)/2)) lim #{x 1 2 +x 1 .x 2 +x 2 2 = x 3 2 +c} = (π/(3√3)).Fan(c).x 3 x 3 →∞

Numerical application

Abundance factors of sequences p = 2 and p = 3

To begin with, we evaluate by direct calculation (method of the overlapping loops), the abundance factors for sequences p = 2 and p = 3. The limit value for target 0 is got making the following observation : This time, the limit value for target 0 is got with recursive relations : δ fae(δ) 1 15 0 mod 2 9.fae(δ-1) 1 mod 2 9.fae(δ-1)+2.3 (3δ-1)/2

Rich targets

The diophantine equation x 1 2 +x 1 .x 2 +x 2 2 -x 3 2 = c admits for c = 3r 2 (r a constant) solutions like (x 1 = 2r, x 3 -x 2 = r) and (x 2 = 2r, x 3 -x 1 = r) contributing immediately to 2x 3 solutions. They are not the only ones. The existence of these degenerate forms which add to none degenerate kinds will express asymptotically in divergent abundance factor ∏ (1+1/p).

Table of results

Number of solutions

Numerical evaluation Variation c Fan(c) x 3 <100 x 3 <300 x 3 <1000 x 3 <3000 x 3 <100 a=1,035

x 3 <300 a=1,021

x 3 <1000 a=1,000

x 3 <3000 a=1,002

x 3 <100 x 3 <300 x 3 <1000 x 3 <3000 We observe that the adjust coefficient a is close to 1 even near the origin.

For the targets c = 3r 2 , we have given artificially, to the abundance factor, a value va :

x 3 < 100 x 3 < 300 x 3 < 1000 x 3 < 3000 va 4, 6 5,6 6,6 7,6 Of course, there is no good reason for a uniform value va between all these c = 3r 2 kind targets. Especially, for c = 0, definitely peculiar, we observe variations with the chosen hypothesis. 3, 30, 31, 100, 119, 184, 203, 208, 315, 338, 360, 381, 387 Fermat 1, 188, 207, 308

They are quoted

  [a,+∞[ or ]-∞, b] Let us consider the example 2x 2 -3z = c. This leads to evaluate 2x 2 -3z < c. Let us consider the situation for target c = 1000 :

  k So that at the limit (by abusive writing) :∞ #{(x, y, …) / R(x, y, …) = c modulo 2 ∞ .3 ∞ … p ∞ ∞ } = ∏ #{(x, y, …) / R(x, y, …) = c modulo p m ∞ } (2) m

  p-1) 1/(p-1) 1/(p-1) 1/(p-1) 1/(p-1) (i.e. ∑ = 1)

  *g-p * Int(gi *g/ p)

  For x = 0 To cm -1 gx =g* gx -p * Int(g * gx / p) gy = gpmun For y = 0 To cm -1 gy =g* gy -p * Int(g * gy / p) gud = 1 For u = 1 To (p -1) / d gud = gud * gd -p * Int(gd * gud / p) gvcm = 1 For v = 1 To (p -1) / cm gvcm = gvcm * gcm -p * Int(gcm * gvcm / p) If gx = gud + gy * gvcm -p * Int((gud + gy * gvcm) / p) Then Range("A1").Offset(x + 2, y + 2) = Range("A1").Offset(x + 2").Offset(1, i * d + 2 + ((p -1) / 2) -d * Int((p -1) / 2 / d)) = (p -1) / cm * d Next i End Sub

  by analogy with the usual case d=2)  c is a non-n-residue mod p p

  us see this on two examples (n = 2, n = 3) obtained by direct resolution with δ = 16 :

Fan

  y+1+(p-1)/2 mod d y'-1 = x-y+(p-1)/2 mod d x'-1 = y-1+(p-1)/2 mod d y'-1 = x-1+(p-1)/2 mod d x'-1 = -x+y mod d y'-1 = -x+1-(p-1)/2 mod d

  choose two among these transformations, namely T2 and T4, to find the whole set of identities between matrix B

  is equivalent to : (x'+0) = -(x+0)+2 mod d (y'+a) = -(x+0)+(y+a)+1+(p-1)/2 mod d Thus if : T2 T2 (x,y) →(x',y') then (x+0,y+a) →(x'+0,y'+a) This means the redundancy of the variables indices from lines x to x' such as x+x' = 2 mod d, and thus of the corresponding equations. Let us observe that in the case x = x' = 1, it concerns the same line. P 68/390 Modulo pi abundance factors for Waring sums

  Having established cardinal matrix A, the evaluation of A k , k unspecified, imposes the research of the eigenvalues and eigenvectors of A. Let us have P A the matrix of eigenvectors of A and D A the eigenvalues matrix. Then D A =

  study made in exercise 4, the eigenvalues of such a matrix [CI d (c 0 ,c 1 , …, c p-1

  /d λ 3 */d … λ 1 */d Let us pose [P'] the matrix obtained by the withdrawal of the first line and first column of [P]. Then :

2 )

 2 p+1)/3) 2 +(x 1 -(p+1)/3).(x 3 -(p-2)/3)+(x 3 -(p-2)/3) 2 -p/3 = 0 (65) (x 2 -(p-2)/3) 2 +(x 2 -(p-2)/3).(x 3 -(p-2)/3)+(x 3 -(p-2)/3) 2 -p/3 = 0

  2 .(s(A).m max +or(0,1)) 2 +3)/4 (97)Numerical evaluationLet us have p m the prime number such that m = (Δx 1 -Δx 1 mod9)/9 of a given value is reached for the first time. We give underneath on the left side the numerical values for range -16 ≤ m ≤ 15 and on the right the graph for a wider range -67 ≤ m ≤ 66.

  are disjointed in pairs according to the horizontal boundaries 6/π and π/3. Moreover, let us take into account the value of m. As we have seen above, the sign of m separates families 3, 3', 4, 4', 6 and 6'from families 1, 1', 2, 2', 5 and 5'. Hence the cloud of points, obtained for values of p ≤ 821551, below :On the other hand, families are not totally disjointed according to the vertical boundary m = 0. Indeed, for m = 0, we have the following cases (blending between families 5, 5' and 6

  * = λ 0 , λ 3 = λ 1 *, λ 4 = λ 2 *, λ 3 * = λ 1 and λ 4 * = λ 2 .

5 P

 5

2 )

 2 2d (matrices of type 1), then the eigenvalues of matrices A and B is all real and Σ Re(μ i = p.d 2 +p.(p-d) pour la matrix A (μ 0 included) and Σ Re(λ i 2 ) = p.d 2 +(p-1).(p-1-d) pour la matrix B (λ 0 included).

  = P A .[μ].P A -1 ). [μ'] : dimension d eigenvalues square matrix, obtained by removing first line and first column (i.e. by removing μ 0 = p) [λ] : dimension d+1eigenvalues square matrix of [B] = [A]-[I], trace type matrix. [λ'] : dimension d square matrix, resulting of eigenvalues matrix [λ] by removing first line and first column (i.e. by removing λ 0 = p-1) such as defined on page 74 [λ'*] : dimension d square matrix, conjugate of [λ']. [I] = I (d,d) : dimension d identity square matrix. [J] = [0…0,1,0…0] (d,d) : dimension d square matrix of which all components are null except one column of components equal to 1, this column being at position 1+(p-1)/2 mod d. [J'] = [0…0,1,0…0] (1,d) : line matrix of dimension (1,d) of which all components are null except one to 1, the position being given by above rule. [K'] = [K'] (d,1) : column matrix of dimension (d,1) of which all components are null except first line whose component is equal to 1. [U] =[1] (d,d) : dimension d square matrix, all components being 1. [U'] = [1…1] (1,d) : line matrix of dimension (1,d) of which all components are equal to 1. [U"] = [1…1] (d,1) : dimension d column matrix of which all components are equal to 1.

  n and, without need of more development, we can write : addition, all so simply : [A] = [P B ].[μ].[P B -1 ] and [A'] = [P B ].[μ'].[P B -1 ] then [A].[A'] = [P B ].[μ].[μ'].[P B -1 ] = [P B ].[μ'].[μ].[P B -1 ] = [A'].[A] (163) 31.5. Example : Case n = 2

  p).(p k+j +(1/2).(i) k+j .p (k+j)/2 .(-(-1) k -(-1) j +i.p 1/2 .(-(-1) k +(-1) j )))

Case 3

 3 Let us suppose : c = p x1 .g x2 and x1 = y1 = k.n+r, 0≤r<n and P 123/390 Modulo p Δ abundance factors correction x2 = y2 Then a x1,x2,y1,y2 = ∑ #(c-x) = ∑ #(p x1 .g x2 -p y1 .g y2 .(g 0 +v(y1).

  [A'] = p-d and card line i≠1 [A'] = p and on other hand to the literal expressions of further matrices : Line c = 0 p δ-δn-1 +Ф(δ-δn)+Ф(δ-δn+1)+…+Ф(δ-1)+Ф(δ) = p δ-δn-1 .(p-1)+ p δ-δn .(p-1)+…+ p δ-1 .(p-1) = p δ Line c = p x1.n .g 0 d.p x1.(n-1) +d.Ф(x1.(n-1)+1)+d.Ф(x1.(n-1)+2)+…+d.Ф(δ-x1.n-1) +p δ-x1.n-1 .card line1 [A']+ Ф(δ-x1.n+1)+…+Ф(δ) = d.p x1.(n-1) +d.p x1.(n-1) .(p-1)+d.p x1.(n-1)+1 .(p-1)+…+d.p δ-x1.n-2 .(p-1) +p δ-x1.n-1 .(p-d)+p δ-x1.n .(p-1)+…+p δ-1 .(p-1) = p δ Line c = p x1.n .p x2 g 0 0+p δ-x1.n-1 .card linei≠1 [A']+Ф(δ-x1.n+1)+…+Ф(δ) = p δ-x1.n-1 .(p)+ p δ-x1.n .(p-1)+…+p δ-1 .(p-1) = p δ Then, we can completely describe case p∤n with following definitions : [A] = A 1 : square matrix of dimension d+1, solution of case δ = 1. [A'] = A' 1 : square matrix of dimension d, resulting from [A] by removing first line and first column. [λ] : [B} = [A]-[I] eigenvalues square trace matrix of dimension d+1 (with adequate eigenvalues order, namely such as [B] k = P A .[λ] k .P A -1 ). [λ'] : square matrix of dimension d resulting of [λ] by removing first line and first column (i.e. by removing λ 0 = p-1) [λ'*] : square matrix of dimension d conjugated of [λ']. [I] = I (d,d) : identity square matrix of dimension d. [J] = [0…0,1,0…0] (d,d) : square matrix of dimension d of which all components are null except one column with components 1, this column being in position 1+(p-1)/2 mod d. [J'] = [0…0,1,0…0] (1,d) : line matrix of dimension (1,d) of which all components are null except one at 1, the position being given by above rule. [K] = [K] (d,d) : square matrix of dimension d of which all components are null except first line whose components are equal to 1. [K'] = [K'] (d,1) : column matrix of dimension d of which all components are null except first line whose component is equal to 1. [U] = [1] (d,d) : square matrix of dimension d, all components being 1. [U'] = [1…1] (1,d) : line matrix of dimension (1,d) of which all components are equal to 1. [U''] = [1…1] (d,1) : column matrix of dimension d of which all components are equal to 1. [0] = [0] (d,d) : square matrix of dimension d, all components being 0. [μ δ ] = [M] (d.δ,d..δ) : trace type square matrix of dimension d.δ+1 made up with [A δ ] eigenvalues {μ δ } clarified below. [P δ ] = [P δ ] (d.δ,d..δ) : square matrix of dimension d.δ+1 made up with [A δ ] eigenvectors clarified below. [P δ -1 ] = [P δ -1 ] (d.δ,d..δ) : square of a nature d.δ+1, reverse matrix of [P δ ], clarified below.

  (p-1) p t+1 (p-1) p t+2 (p-1) p t+2 (p-1) p t+3 (p-3)/2 p t+3 (p-1)

0 1

 0 /p 2t+1 (p-1)/2p 2t+1 (p-1)/2p 2t+1 (p-1)/2p 2t (p-1)/2p 2t (p-1)/2p 2t-1 (p-1)/2p 2t-1 … (p-1)/2p 4 (p-1)/2p 4 (p-1)/2p 3 (p-1)/2p 3 (p-1)/2p 2 (p-1)/2p 2 (p-1)/2p (p-1)/2p 1/2p 2t+1 (p-1)/4p 2t+1 (p-1)/4p 2t+1 (p-1)/4p 2t (p-1)/4p 2t (p-1)/4p 2t-1 (p-1)/4p 2t-1 … (p-1)/4p 4 (p-1)/4p 4 (p-1)/4p 3 (p-1)/4p 3 (p-1)/4p 2 (p-1)/4p 2 (-1-√p)/4p (-1+√p)/4p 1/2p 2t+1 (p-1)/4p 2t+1 (p-1)/4p 2t+1 (p-1)/4p 2t (p-1)/4p 2t (p-1)/4p 2t-1 (p-1)/4p 2t-1 … (p-1)/4p 4 (p-1)/4p 4 (p-1)/4p 3 (p-1)/4p 3 (p-1)/4p 2 (p-1)/4p 2 (-1+√p)/4p (-1-√p)/4p 1/2p 2t (p-1)/4p 2t (p-1)/4p 2t (p-1)/4p 2t-1 (p-1)/4p 2t-1 (p-1)/4p 2t-2 (p-1)/4p 2t-2 … (p-1)/4p 3 (p-1)/4p 3 (p-1)/4p 2 (p-1)/4p 2 -(p+1)/4p (p-1)/4p 0 1/2p 2t (p-1)/4p 2t (p-1)/4p 2t (p-1)/4p 2t-1 (p-1)/4p 2t-1 (p-1)/4p 2t-2 (p-1)/4p 2t-2 … (p-1)/4p 3 (p-1)/4p 3 (p-1)/4p 2 (p-1)/4p 2 (p-1)/4p -(p+1

  (p-1) p t+1 (p-1) p t+2 (p-1) p t+2 (p-1) p t+3 (p-1)/2 p t+3 (p-3)

0 1

 0 /p 2t+1 (p-1)/2p 2t+1 (p-1)/2p 2t+1 (p-1)/2p 2t (p-1)/2p 2t (p-1)/2p 2t-1 (p-1)/2p 2t-1 … (p-1)/2p 4 (p-1)/2p 4 (p-1)/2p 3 (p-1)/2p 3 (p-1)/2p 2 (p-1)/2p 2 (p-1)/2p (p-1)/2p 1/2p 2t+1 (p-1)/4p 2t+1 (p-1)/4p 2t+1 (p-1)/4p 2t (p-1)/4p 2t (p-1)/4p 2t-1 (p-1)/4p 2t-1 … (p-1)/4p 4 (p-1)/4p 4 (p-1)/4p 3 (p-1)/4p 3 (p-1)/4p 2 (p-1)/4p 2 (-1+i√p)/4p (-1-i√p)/4p 1/2p 2t+1 (p-1)/4p 2t+1 (p-1)/4p 2t+1 (p-1)/4p 2t (p-1)/4p 2t (p-1)/4p 2t-1 (p-1)/4p 2t-1 … (p-1)/4p 4 (p-1)/4p 4 (p-1)/4p 3 (p-1)/4p 3 (p-1)/4p 2 (p-1)/4p 2 (-1-i√p)/4p (-1+i√p)/4p 1/2p 2t (p-1)/4p 2t (p-1)/4p 2t (p-1)/4p 2t-1 (p-1)/4p 2t-1 (p-1)/4p 2t-2 (p-1)/4p 2t-2 … (p-1)/4p 3 (p-1)/4p 3 (p-1)/4p 2 (p-1)/4p 2 -(p+1)/4p (p-1)/4p 0 1/2p 2t (p-1)/4p 2t (p-1)/4p 2t (p-1)/4p 2t-1 (p-1)/4p 2t-1 (p-1)/4p 2t-2 (p-1)/4p 2t-2 … (p-1)/4p 3 (p-1)/4p 3 (p-1)/4p 2 (p-1)/4p 2 (p-1)/4p -(p+1

  [I] and [U] : [λ'].[λ'*] = (p.d-d+1).d.[I] -(p-1).d.([U]-[I]) (14) Then trivially : [λ'].r.[I].[λ'*] = r.((p.d-d+1).d.[I] -(p-1).d.([U]-[I])) Hence, with blocks (d,d) corresponding to eigenvalues of type r.[I], we may not use [λ'] and [λ'*] in [P] and [P -1 ], but for example -d.[I] and (p.d-d+1).[I]-(p-1).([U]-[I]). This alternative was suggested in some examples d = 2 before.

Example 4 :

 4 p = 5, n = 5, δ = 3 :g= 2, p δ = 125, g[START_REF] Dieudonné | Abrégé d'histoire des mathématiques[END_REF]

  p = 5, n = 25, δ = 3 :g= 2, p δ = 125, g 5 = 32, g 25 = : When "generating" column is not on head of blocks (above trace blocks) as in previous examples, we get the following example (matrix of the type 2 where d = (n, p-1) even and p = 1+d mod 2d) : Example 6 : p = 3, n = 6, δ = 3 (d = 2, p = 3 mod 4) :g= 2, p δ = 27, g 6 = 8, g 25 = 57

  x 7 then :(λ-8)/8 = α = x 4 /x 0 = -x 0 /x 4 = x 5 /x 1 = -x 1 /x 5 = x 6 /x 2 = -x 2 /x 6 = x 7 /x 3 = -x 3 /x

  x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 x0 -x0 -x0 -x0 -x0 -x0 -x0 -x0 -x0 -x0 -x0 -x0 -x0 -x0 -x0 -x0 -x0 x0 x0 x0 x0 x0 x0 x0 x0 -x0 -x0 -x0 -x0 -x0 -x0 -x0 -x0 x8 x8 x8 x8 x8 x8 x8 x8 x8 x8 x8 x8 x8 x8 x8 x8 x0 x0 x0 x0 x0 x0 x0 x0

  x0 -x0 -x0 x2 x2 x2 x2 -x2 -x2 -x2 -x2 -x2 -x2 -x2 -x2 x10 x10 x10 x10 x10 x10 x10 x10 x10 x10 x10 x10 x10 x10 x10 x10 x0 x0 -x0 -x0 x2 x2 x2 x2 x2 x2 x2 x2 x2 x2 x2 x2 x2 x2 x2 x2 x2 x2 x2 x2 x2 x2 x2 x2 x2 x2 x2 x2 x0 x0 -x0 -x0 -x2 -x2 -x2 -x2 x6 x6 x6 x6 x6 x6 x6 x6 x6 x6 x6 x6 x6 x6 x6 x6 x6 x6 x6 x6 x6 x6 x6 x6 x0 x0 -x0 -x0 -x2 -x2 -x2 -x2 -x6 -x6 -x6 -x6 -x6 -x6 -x6 -x6 x14 x14 x14 x14 x14 x14 x14 x14 x14 x14 x14 x14 x14 x14 x14 x14 x0x0 -x0 -x0 -x2 -x2 -x2 -x2 x6 x6 x6 x6 x6 x6 x6 x6 -x6 -x6 -x6 -x6 -x6 -x6 -x6 -x6 -x6 -x6 -x6 -x6 -x6 -x6 -x6 -x6 x0 x0 -x0 -x0 -x2 -x2 -x2 -x2 -x6 -x6 -x6 -x6 -x6 -x6 -x6 -x6 -x14 -x14 -x14 -x14 -x14 -x14 -x14 -x14 -x14 -x14 -x14 -x14 -x14 -x14 -x14 -x14 x0 -x0 x1 x1 -x1 -x1 -x1 -x1 -x5 -x5 -x5 -x5 -x5 -x5 -x5 -x5x13 x13 x13 x13 x13 x13 x13 x13 x13 x13 x13 x13 x13 x13 x13 x13 x0 -x0 x1 x1 x1 x1 x1 x1 -x1 -x1 -x1 -x1 -x1 -x1 -x1 -x1 x9 x9 x9 x9 x9 x9 x9 x9 x9 x9 x9 x9 x9 x9 x9 x9 x0 -x0 x1 x1 -x1 -x1 -x1 -x1 x5 x5 x5 x5 x5 x5 x5 x5 -x5 -x5 -x5 -x5 -x5 -x5 -x5 -x5 -x5 -x5 -x5 -x5 -x5 -x5 -x5 -x5 x0 -x0 x1 x1 x1 x1 x1 x1 x1 x1 x1 x1 x1 x1 x1 x1 -x1 -x1 -x1 -x1 -x1 -x1 -x1 -x1 -x1 -x1 -x1 -x1 -x1 -x1 -x1 -x1 x0 -x0 x1 x1 -x1 -x1 -x1 -x1 -x5 -x5 -x5 -x5 -x5 -x5 -x5 -x5 -x13 -x13 -x13 -x13 -x13 -x13 -x13 -x13 -x13 -x13 -x13 -x13 -x13 -x13 -x13 -x13 x0 -x0 x1 x1 x1 x1 x1 x1 -x1 -x1 -x1 -x1 -x1 -x1 -x1 -x1 -x9 -x9 -x9 -x9 -x9 -x9 -x9 -x9 -x9 -x9 -x9 -x9 -x9 -x9 -x9 -x9 x0 -x0 x1 x1 -x1 -x1 -x1 -x1 x5 x5 x5 x5 x5 x5 x5 x5 x5 x5 x5 x5 x5 x5 x5 x5 x5 x5 x5 x5 x5 x5 x5 x5 x0 -x0 x1 x1 x1 x1 x1 x1 x1 x1 x1 x1 x1 x1 x1 x1 x1 x1 x1 x1 x1 x1 x1 x1 x1 x1 x1 x1 x1 x1 x1 x1 x0 -x0 -x1 -x1 x3 x3 x3 x3 x3 x3 x3 x3 x3 x3 x3 x3

  x11 x11 x11 x11 x11 x11 x11 x11 x11 x11 x11 x11 x11 x11 x11 x11 x0 -x0 -x1 -x1 -x3 -x3 -x3 -x3 -x7 -x7 -x7 -x7 -x7 -x7 -x7 -x7 x15 x15 x15 x15 x15 x15 x15 x15 x15 x15 x15 x15 x15 x15 x15 x15 x0 -x0 -x1 -x1 x3 x3 x3 x3 x3 x3 x3 x3 x3 x3 x3 x3 x3 x3 x3 x3 x3 x3 x3 x3 x3 x3 x3 x3 x3 x3 x3 x3 x0 -x0 -x1 -x1 -x3 -x3 -x3 -x3 x7 x7 x7 x7 x7 x7 x7 x7 x7 x7 x7 x7 x7 x7 x7 x7 x7 x7 x7 x7 x7 x7 x7 x7 x0

  Let us have two couples of real numbers (a,b) and (c,d) such as (a+b.i) 2 = c+d.i (i²=-1). Then a 2 -b 2 +2a.b.i = c+d.i, that is c = a 2 -b 2 and d 2 = 4a 2 .b 2 , and then 4a 4 -4c.a 2 -d 2 = 0 having a couple of solutions a = (2.(c 2 +d 2 ) 1/2 +2c) 1/2 /2 and b = (2.(c 2 +d 2 ) 1/2 -2c) 1/2 /2. As c+d.i is a root of unit (1), we have c 2 +d 2 = 1, then : a = (2+2c) 1/2 /2 and b =(2-2c)

For

4 .

 4 Cardinal matrices for Waring integers' variables enumeration, boundless volumes, unit mesh 4.1. Case p odd (p >2) and p∤ n x 1 n + x 2 n + … + x k n = c + x k+1 n + x k+2 n + … + x k+i n modulo p δ (25) We examined this case for δ = 1 at the end of exercise 5. The argument does not change here : P 162/390 Modulo p Δ abundance factors correction

  t[A'] : square matrix of dimension d, transposed of matrix[A'] resulting from[A] by removing first line and first column. [λ' d/2 ] : square matrix of dimension d resulting from [λ'] while shifting eigenvalues by d/2 (mod d) positions. [λ'* d/2 ] : square matrix of dimension d conjugate of [λ' d/2 ]. [I] = I (d,d) : square matrix identity of dimension d. [J 0 ] = [1,0…0] (d,d) : square matrix of dimension d of which all components are 0 except first column with components equal to 1. [J' 0 ] = [1,0…0] (1,d) : line matrix of dimension d of which all components are null except first component equal to 1. [K' d/2 ] : column matrix of dimension d of which all components are null except line at position 1+d/2 whose component is equal to 1. [R] = [R] (d,d) : square matrix of dimension d of which all components are equal to p-1 except trace of which all components equal p.(1-d)-1. [μ δd/2 ] : trace square matrix of dimension d.δ+1 obtained with eigenvalues {μ δ } and shifting of d/2 (mod d) positions.

  y 2 ± … ± y m = c mod p δ is self-evident starting from y 1 + (±y 2 ) + … + (±y m ) = c mod p δ

32) 10 . 1 .

 101 Evaluation of Waring abundance factors in integers' variables enumeration p = 2, 2∤ n

  or empty Multiplication of line vector [P δ ].[λ δ /(p δ )] k by column vector ['/p] k +(p 1 +…+p n-1

  d).((p i.(n-k) -1)/(p n-k -1)).((p-1).∑(μ'/p) k .[I]+(p n -p).[I/p] k )).[1] +(1/d).p i.(n-k) .[λ'*].[μ'/p] k .[1] 4 i ≠ 0, j > 0, n ≠ k (1+(1/d).if(n = k, i, (p i.(n-k) -1)/(p n-k -1)).if(i=0, [0], (p-1).∑(μ'/p) k .[I]+(p n -p).[I/p] k ) +(1/d).p i.(n-k) . if(j=0, [λ'*].[μ'/p] k , (p-1).∑(μ'/p) k .[I] +p.((1-d).p j-1 -1).[I/p] k )).[1] 5 i ≠ 0, j = 0, n = k (1+(1/d).(i.((p-1).∑(μ'/p)

  ']+(1/d).[ λ '*].[λ'/(p-1)] m .[U"]

  '].((p-1)/d).(p 0.n .[λ'/p] m +p 1.n .[λ'/p 2 ] m +…+p i.n .[λ'/p i+1 ] m +...).[U''] Thus : fan δ=∞ (0) = 1+((p-1)/d).[U'].[λ'/p] m .[U''].(p 0.n +p 1.(n-m) +…+p i.(n-m) +...) For c = 0, i is rejected ad infinitum, so that : fan δ=∞ (0) = 1+((p-1)/d).[U'].[λ'/p] m .[U''].

  Let us have R (…) = c mod p an equation for which we already obtained a cardinal matrix [M]. When we pass to equation R (…)+t = c mod p where t is a constant, occurrences enumeration of each element of {0, 1,…, p-1} undergoes a elementary permutation. The new matrix is thus [M].[MP] where [MP] is an elementary permutation of identity [I]. (only one element 1 per line and column and all the other elements with 0

For

  gcd (Δ,p i ) = 1 #{(x) / ax²+bx = c mod p i , x = 0 to p i -1} = 1

  [A] d,cm =[B] d,cm +[I] (8) Proof First line of matrix By division, we get for the characteristic equation of the first line : c(1,y) = #{(u,v) \ -1 = g (p-1)/2 = g y-2 .g (v.cm/d-u).d mod p} It follows y-2+(v.cm/d-u).d = (p-1)/2 mod p-1, then y = 2+(p-1)/2 + (v.cm/d-u).d mod p-1. As d divides p-1, we deduce immediately y = 2+(p-1)/2 mod d and (v.cm/d-u) = 0 mod (p-1)/d. This last equation has an equal number of solutions whatever y (because independent of y) and thus (p-1)/d solutions (u,v), hence : c(1,2+(p-1)/2 mod d) = (p-1)/d c(1,y≠2+(p-1)/2 mod d) = 0 First column of matrix From c(x,1) = #{(u,v) / g x-2 = g u.d mod p}, it follows x = 2+u.d mod p-1. The equation has only one solution at x = 2 +k.d with d choice for k (k = 0 mod (p-1)/d). So that : c(x,1) = if(x=2 mod d, d, 0) Matrix blocks

  j,cm = [P B ] j,cm .[DB] j,cm .[P B -1

  cardinal matrices to Fermat-Catalan cardinal matrices is done by elementary Cramer operations. Eigenvectors matrix [P] will thus be unchanged.

  } = p i+j .(p-1) k+m -[M1] i .[M2] j .([M1]-[I]) k .([M2]-[I]) m . 0 #{g.g 2u } p i+j .(p-1) k+m 0 Here [M1] and [M2] are respectively the above matrices applying to monomials x 4 and x 2 when cm = 4 taking account of p modulo 8 congruence. [I] is the identity matrix of dimension cm+1. The use of the eigenvalues and eigenvectors of [M1], [M2], [M1]-[I] and [M2]-[I] matrices simplifies, of course, literal calculation of this case which can be carried out with exercise 5 data and left to the interested reader.

  3) -z 4 = (p-1)/2-x 4 -x 8 (p-5)/2-z 1 -z 4 = (x 2 +x 5 )/2+1 (p-1)/2-x 2(3) +z 2 -z 3 = (p-1)/4-(x 3 -x 4 +x 5 )/2 -(p-7)/2+x 2(3) +z 1 +z 4 = x 8 So that : P 203/390 Fermat-Catalan abundance factors

  b and c are given by the study of the case d = 3, done at exercise 5

  (a-b) 2 +(b-c) 2 +(c-a) 2 = 18 and (a-b).c+(b-c).a+(c-a).b = 0 and the intermediate expressions :

  a k .λ j = (a i .a k .b j +(-1) j .a i .a k +(-1) i .a j .a k +(-1) i+j .a k ).p+((-1) i+j .a k .b i .b j -(-1) i .a k .b i -(-1) j .a k .b j +a k ) +p 1/2 .((-1) j .a i .a k .b j +(-1) i .a j .a k .b i +(-1) i+j .a k .(b i +b j )-a i .a k -a j .a k -(-1) i .a k -(-1) j .a k .

  these data can be used, to find the expressions of [M1] and [M2] using the property of the averages.

  1 mod 12 what is awaited. It remains to discover the values of integers x1, x2, x3, x4, x5 and x8 as functions of p using eigenvalues. Again, we can propose a conjectural expression of eigenvalues on the basis of decomposition p = 3y 2 +z 2

  2 and 16.3 given at exercise 5 : 4d). ∑ r k ∑ (-1) m .a j .a j+k+m-1 +a j+k .a j+m-

  r 1 -∑ r k s 2j = (d/2).(r 2j -r d+2-2j ) s 2j+1 = (d/2).(r 2j+1 +r d+1-2j )-∑ r k Hence: r d+2-2j = r 2j -(2/d).s 2j or still r 2j+1 = r d+1-2j -(2/d).s d+1-2j Then : s 2j+1 = (d/2).(r d+1-2j +(2/d).s d+1-2j -r d+1-2j )-∑ r k = d.r d+1-2j -s d+1-2j -∑ r k = d.(r d+1-2j -r 1 )-s d+1-2j -∑ r k +d.r 1 and also : r d+1-2j -r 1 = (1/d).(-s 1 +s 2j-1 +s d+1-2j ) That is : r 2j -r 1 = (1/d).(-s 1 +s d+2-2j +s 2j ) Then, using r 2j+1 = r d+1-2j -(2/d).s d+1-2j , we get : r 2j+1 -r 1 = (1/d).(-s 1 +s 2j+1 +s d+1-2j )

  ) )+∑ [r k ].(μ j mod d,(d) .μ j mod 2

i2

  Contrary to the preceding case, environment 2 r .d does not have particular simplicity. We thus are interested in the case of two prime factors d 1 (d1.d2) = μ m mod d1,(d1) +μ m mod d2,(d2) +∑ r k .μ m+k-1 mod d1,(d1) .μ m+k-1 mod d2,(d2)

i = 0 .

 0 If matrix [r d ] exists, there is then an infinity of such matrix with integer components. Any variation [r d '] = [if(k = a mod t,b,0)]+[r d ] is also solution, a and b being given integers. Here t is either d 1

  We know how to evaluate eigenvalues of circulant matrices of standard cardinal matrices and then of Fermat Catalan matrices. This does not prevent from being interested in what follows. We seek a general method to get matrix [r d ] relevant to the components r k in expression k = d 1 .d 2 μ m,(d1.d2) = μ m mod d1,(d1) +μ m mod d2,(d2) +∑ r k .μ m+k-1 mod d1,(d1) .μ m+k-1 mod d2,(d2) than two relative prime factors are obtained by simple duplication of the preceding elementary operation.NotationsWe designate by (x d ) the first column of a dimension d right circulant matrix [x d ] or of a dimension d left circulant matrix [x' d ] (the accent ' is allotted to left matrices) and by (x i,d ) the components of that first column, i = 1 to d. This definition of (x d ) must be kept in memory because confusion with the first line would make inoperative the next course of this text.

  similitudes [x d ].[x d ] = d.[x d ] with integer components exist, similitudes whose possible interest was not explored. 21.6. Non-entire ring Equation [x d ].[y d ] = [x d ].([x d ]-d.[I d ]) = 0 admit self-evident solutions [x d ] = [0] and [x d ] = d.[I d ] which do not interest us. The ring of the matrices of dimension d is not an entire ring (integral domain) and other solutions exist such as environment founding matrices. It is easy to find examples of non-integrity of square matrices ring thanks to environment founding matrices. Let us have [t d1 ] and [t d2 ] two of them such as (d 1 ,d 2 ) = 1. Let us have d = d 1 .d 2 . We build the square matrices [t d1,d ] and [t d2,d ] of dimension d by assembly of d/d 1 by d/d 1 blocks of the first matrix and d/d 2 by d/d 2 blocks of the second matrix. Then : [t d1,d ].[t d2,d ] = [0] (52)

  [me(r,s)] = [e (2πi/d).(r-1).(s-1) ] is a solution for eigenvectors matrix (r = 1 to d, s = 1 to d). The corresponding eigenvalues are [σ r,s ] = [if(r≠s, 0, if((r-1,d) = 1, 0, d))]. Consequently, we have d-(d 1 -1).(d 2 -1)…(d k -1) eigenvalues 0 and (d 1 -1).(d 2 -1) )…(d k -1) eigenvalues d, d i being divisors of dbetween founding matrices and cardinal matrices eigenvalues of disjoined environments Let us have two disjoined environments d 1 and d 2 and product d = d 1 .d 2 . Let us have μ d = (μ 1,d , μ 2,

  (s' d ) = (t' d ) mod d (67) Let us note that there are matrices solutions of [s' d ].[s' d ] = p.d.[t d ] which do not respect relation (s' d ) = -(t' d ) mod d (or (s' d ) = (t' d ) mod d) or/and constraint p = 1 mod d.

9 .

 9 Conservation of components averages Let us have an environment d not prime. Let us have d 1 a divisor of d and such as (d/d 1 ,d 1 ) ≠ 1. The average of the sums modulo d 1 of environment d matrix components gives environment d/d 1 matrix. For example, let us consider the environment d = 12 and d 1 = 6.

  t

  [s' d ].[s' d ] = p.d.[t d ], we can say that [s' d ] has d-(d 1 -1).(d 2 -1) eigenvalues 0 and (d 1 -1).(d 2 -1) eigenvalues whose products two by two are d 2 p (d i being divisors of d). The difficulty is the determination of these eigenvalues independently of the systematic use of [s' d ].[s' d ] = p.d.[t d ].

  The reader will easily get the matrices [r] for families 2 to 6.It follows the numerical examples :

  [t'12].(ri,d) = ∑ (s d,i )2 = 13896 = 12.3.2.193 = d.(d 1 -1).(d 2 -1).p and we meet again a calculation step containing a decomposition equation with adapted constraints.

  [s' d ] 2 = p.d.[t d ] for case 2d with d even.

  (c) = (u.(x+1) 2 +v.(x+1).y +w.y 2 )-(u.x 2 +v.x.y+w.y 2 ) and for the second E2(c) = (u.(x+a+1) 2 +v.(x+a+1).(y+1)+w.(y+1) 2 )-(u.(x+a) 2 +v.(x+a).(y+1) +w.(y+1) 2 ).

  (x+a) 2 +v.(x+a).(y+1) +w.(y+1) 2 -(u.x 2 +v.x.y+w.y 2 ) = u.a 2 +2a.u.x+v.a+v.x+v.a.y+w.(2y+1). Using 2au+v = 0 mod p and Δ = v 2 -4.u.w, it follows :

  u.a 2 +w.b 2 +(2.u.a+v.b).x+(2.w.b+v.a).y+v.a.b = u.a' 2 +w.b' 2 +(2.u.a'+v.b').x+(2.w.b'+v.a').y+v.a'.b' mod p This amounts to ask whether a quadratic equation of unknowns (a,b) defined by t = u.a 2 +w.b 2 +(2.u.a+v.b).x+(2.w.b+v.a).y +v.a.b mod p which has one solution has a second one, which is true since of degree 2.

  meaning only if √Δ exists (sign √ is used for its practical side, √Δ being an integer modulo p), i.e. if Δ is a square modulo p or a quadratic residue of x 2 modulo p. In this case : u.(x+y. (v+√Δ) ).(x+y. (v-√Δ) ) = 0 mod p (12) 2u 2u admits two systems of solutions : x+y.(v+√Δ/(2u)) = 0 mod p (13

  which the non-null determinant -(√Δ)/u ensures one and only one solution to each system. These solutions are distinct (self-evident ). Hence p-1 solutions in (x,y) Case Δ = 0 The factorisation leads to (x+y.v/2u) 2 = c/u mod p (18) which has solution only if c/u is a square, giving then x+y.v/2u = (c/u) 1/2 mod p (19) or x+y.v/2u = -(c/u) 1/2 mod p (20) each one of these equations having p distinct solutions, themselves distinct from one to the other of equations (#c = 2p) except if c = 0 mod p (redundant equation) (#c = p).

6 .

 6 Classes of squares and non-squaresWe will indicate by C(p) ≡ {c} the set of the non-null classes which are squares modulo p, i.e. the set of the numbers c such as  x \ c = x 2 mod p. We indicate by D(p) ≡ {c'} the set of the classes which are non-squares modulo p, i.e. the set of the numbers such as  x, c' ≠ x 2 mod p. We write this as : c c C(p) and c' c D(p)

  x 2 = c mod p, x = 1 to p-l} and {c =0} unless line or/and column presents zeros right away (case u or/and w = 0 mod p). The cardinals are thus induced by the below table which gives the residues enumerations of the first line and column :

  in addition to preceding remarks, we use the primitive roots fundamental property :( g ) = -1 mod p p As g 4c =g+t.c, t an integer, it follows :( g 4c +t.c ) = ( g 4c ) = -

  factors decomposition of a positive integer m. The exclusive frequency fe(m), respectively inclusive frequency fi(m), of m in the set PM1 = {1, 2, 4, 6, 10, …, p i -1, …, +∞}, where p i describes all the prime numbers, is given by : that fe(m) and fi(m) are multiplicative functions in the reference set PM1. For the exclusive frequency :(m 1 ,m 2 ) = 1 => fe(m 1 .m 2 ) = fe(m 1 ).fe(m to find divisor 231 in the list {p i -1}, i = 1 to +∞ (p i = 2, p 2 = 3…). As 231 = 3.7.11, we have fe(231) = 1/231 and fi(231) = 1/((3-1).(7-1).(11-1)) = 1/120.

  n.m = c = g n mod p. It follows (r.m-1).n = 0 mod p-1, then r.m = 1 mod p-1. If (m,p-1) = d, then r.m = d mod p-1 according to Bachet/Bézout theorem, therefore r.m ≠ 1 mod p-1 what is contradictory.

  -1)/2).(if(p = 1 mod 4, #(1), #(g))) ((p-1)/2).(if(p = 1 mod 4, #(g), #(1))) #(1) #(0)+(p/4).(#(1)+#(g)) +(1/4).if(p = 1 mod 4, -(5.#(1)+#(g)), -3.(#(1)+#(g))) (p/4).(#(1)+#(g))+(-1) (p+1)/2 )/4).(#(1)+#(g)) +(1/4).if(p = 1 mod 4, -#(1)-#(g), -3.#(1)+#(g)) #(g) (p/4).(#(1)+#(g)) +(1/4).if(p = 1 mod 4, -#(1)-#(g), #(1)-3.#(g)) #(0)+(p/4).(#(1)+#(g)) +(1/4).if(p = 1 mod 4, -(#(1)+5.#(g)), -3.(#(1)+#(g)))

  p+1) (p-1).p 3 .(p+1) (p-1).p.p 3 .(p+1) (p-1

  1).p 3 .(p-1) p 4 .((p-1)/2).(p-5) p 4 .((p-1)/2).(p-1) 2p 2 .(p-1) (p-1).p 2 .(p-1) (p-1).p 2 .(p-1) (p-1).p 3 .(p-1) (p-1).p 3 .(p-1) p 4 .((p-1)/2).(p-5) p 4 .((p-1)/2).(p-1) 2p 2 .(p-1) (p-1).p 2 .(p-1) (p-1).p 2 .(p-1) (p-1).p 3 .(p-1) (p-1).p 3 .(p-1) p 4 .((p-1)/2).(p-5) p 4 .((p-1)/2).(pp-5) ((p-1)/2).p 2 .(p-5) ((p-1)/2).p 2 .(p-5) ((p-1)/2).p 3 .(p-5) ((p-1)/2).p 3 .(p-5) p 4 .(p 2 -2p+9)/2 p 4 .(p-1).(p-3)/2 p 2 .(p-1) ((p-1)/2).p 2 .(p-1) ((p-1)/2).p 2 .(p-1) ((p-1)/2).p 3 .(p-1) ((p-1)/2).p 3 .(p-1) p 4 .(p-1).(p-3

  p-1) ((p-1)/2).2p.(p-1) ((p-1)/2).2p.(p-1) p 4 .((p-1)/2).(p-1) p 4 .((p-1)/2).(p-5) 2p.(p-1) ((p-1)/2).2p.(p-1) ((p-1)/2).2p.(p-1) p 4 .((p-1)/2).(p-1) p 4 .((p-1)/2).(p-5) 2p.(p-1) ((p-1)/2).2p.(p-1) ((p-1)/2).2p.(p-1) p 4 .((p-1)/2).(p-1) p 4 .((p-1)/2).(p-5) p.(p-5) ((p-1)/2

  p-5) ((p-1)/2).p 2 .(p-5) ((p-1)/2).p 2 .(p-5) ((p-1)/2).p 3 .(p-5) ((p-1)/2).p 3 .(p-5) p 4 .(p 2 -2p+5)/2 p 4 . (p 2 -4p+7)/2 p 2 .(p-1) ((p-1)/2).p 2 .(p-1) ((p-1)/2).p 2 .(p-1) ((p-1)/2).p 3 .(p-1) ((p-1)/2).p 3 .(p-1) p 4 .(p 2 -4p-1p+1) ((p-1)/2).p.(p+1) ((p-1)/2).p.(p+1) p 2 .(p+1).(p-3we find the elements from the study of y 2 modulo p δ with p = 3 mod 4. Modulo p results are given at page81. The above values are the squares of the values found on this earlier occasion :

  mb r,s = (1/p).(((p-1)/d).λ 0 +(1/d 2 ).∑ k λ r+k-1 *.λ k .λ s+k-1 ) and by change of indices : mb r-i,s-i = (1/p).((p-1) 2 /d+(1/d 2 ).∑ k λ r-i+k-1 *.λ k .λ s-i+k-1 ) (57) From (σ mz ) = [Δcmz].(λ)/d, we deduce by developing the definition of [Δcmz] :

  d 1 λ d /d λ 1 /d … λ d-1 /d It follows : σ mz 0 σ mz 1 .(p-1)/d σ mz 2 .(p-1)/d … σ mz d .(p-1)/d 1 (p-1)/d (p-1)/d … (p-1)/d σ mz 0 σ mz 1 .λ 1 */d σ mz 2 .λ 2 */d … σ mz d .λ d */d 1 λ 1 /d λ 2 /d … λ d /d [MZ] = (1/p). σ mz 0 σ mz 1 .λ 2 */d σ mz 2 .λ 3 */d … σ mz d .λ 1 */d σ mz 1 .λ d */d σ mz 2 .λ 1 */d … σ mz d .λ d-1 */d 1 λ d /d λ 1 /d … λ d-1 /d Then the sum being on k = 1 to d :σ mz 0 +((p-1)/d).∑ k σ mz k ((p-1)/d).(σ mz 0 +(1/d).∑ k σ mz k .λ k ) … ((p-1)/d).(σ mz 0 +(1/d).∑ k σ mz k .λ k+d-1 ) σ mz 0 +(1/d).∑ k λ k *.σ mz k ((p-1)/d).σ mz 0 +(1/d 2 ).∑ k σ mz k .λ k *.λ k … ((p-1)/d).σ mz 0 +(1/d 2 ).∑ k σ mz k .λ k *.λ k+d-1 [MZ] = (1/p). σ mz 0 +(1/d).∑ k λ k+1 *.σ mz k ((p-1)/d).σ mz 0 +(1/d 2 ).∑ k σ mz k .λ k+1 *.λ k … ((p-1)/d).σ mz 0 +(1/d 2 ).∑ k σ mz k .λ k+1 *.λ k+d-d).∑ k λ k+d-1 *.σ mz k ((p-1)/d).σ mz 0 +(1/d 2 ).∑ k σ mz k .λ k+d-1 *.λ k … ((p-1)/d).σ mz 0 +(1/d 2 ).∑ k σ mz k .λ k+d-1 *.λ k+d-1We examine the four usual cases : Component column 1 and line 1 mz 0,0 = (1/p).(σ mz 0 +((p-1)/d).∑ k σ mz k ) = (1/p).(σ mz 0 +p.#(0)-σ mz 0 ) (while using relation 61) = #(0) This is a self-evident identity. Components line 1 (except column 1) mz r,0 = (1/p).((p-1)/d).(σ mz 0 +(1/d).∑ k σ mz k .λ k+r-1 ) = ((p-1)/d).(1/p).(σ mz 0 +(1/d).∑ k σ mz k .λ* k+r-1+(p-1)/2 ) (while using λ* k = λ* k+(p-1)/2 ) = ((p-1)/d).m mz 0,r+(p-1)/2 We find the usual relation between the components of the first line and those of the first column. Components column 1 (except line 1) mz 0,s = (1/p).(σ mz 0 +(1/d).∑ k λ k+s-1 *.σ mz k ) = (1/p).(σ mz 0 +(1/d).∑ k λ k+s-1 *.(#(0)+(1/d).∑ m #(g m-k ).λ m )) We use then the properties (see page 57) of the eigenvalues of the matrix of y d : Σ λ i *.λ i = d.(p.d-p+1) = p.d 2 -d(p-1) Σ λ k+1-i *.λ k+1-j = -d.(p-1)With the change of indices m-k = i, it follows :mz 0,s = (1/p).(σ mz 0 -#(0)+(1/d 2 ).∑ k ∑ m #(g m-k ).λ k+s-1 *.λ m ) = (1/p).(σ mz 0 -#(0)+(1/d 2 ).∑ k ∑ i #(g i ).λ k+s-1 *.λ k+i ) = (1/p).(σ mz 0 -#(0)+(1/d 2 ).∑ i #(g i ).∑ k λ k+s-1 *.λ k+i ) P 279/390Quadratic abundance factors and substitutes= (1/p).(σ mz 0 -#(0)+(1/d 2 ).∑ i #(g i ).(if(i =s-1,1,0)p.d 2 -d.(p-1)) = (1/p).(σ mz 0 -#(0)-(p-1/d).∑ i #(g i )+p.∑ i #(g i ).if(i =s-1,1,0) = (1/p).(σ mz 0 -#(0)-(p-1/d).∑ i #(g i )+p.#(g s-1 ) = #(g s-1 )while using relation 60 This is actually the definition of mz 0,s . Components except line 1 and column 1 mz r,s = (1/p).(((p-1)/d).σ mz 0 +(1/d 2 ).∑ k σ mz k .λ k+r-1 *.λ k+s-1 ) = (1/p).(((p-1)/d).σ mz 0 +(1/d 2 ).∑ k (#(0)+(1/d).∑ m #(g m-k ).λ m ).λ k+r-1 *.λ k+s-1 ) = (1/p).(((p-1)/d).σ mz 0 +(1/d 2 ).#(0).∑ k λ k+r-1 *.λ k+s-1 +(1/d 3 ).∑ k ∑ m #(g m-k ).λ m .λ k+r-1 *.λ k+s-1 ) We still use the property : Σ λ k+1-i *.λ k+1-j = -d.(p-1)+if(i=j,1,0).p.d² Then mz r,s = (1/p).(((p-1)/d).σ mz 0 +if(r=s, 1, 0).#(0).p-((p-1)/d).#(0)+(1/d 3 ).∑ k ∑ m #(g m-k ).λ m .λ k+r-1 *.λ k+s-1 ) = (1/p).(((p-1)/d).(σ mz 0 -#(0))+p.if(r=s, 1, 0).#(0)+(1/d 3 ).∑ k ∑ m #(g m-k ).λ m .λ k+r-1 *.λ k+s-1 ) By change of indices m-k → i, then while shifting by -i, it follows : mz r,s = (1/p).(((p-1)/d).(σ mz 0 -#(0))+p.if(r=s, 1, 0).#(0)+(1/d 3

  verify, with the matrix of environment below, that [M] = [P].[σ vp ][P -1 ] (here [P] = [P -1 ]).

  p-11+x+if(v = 0 mod 5, 0, 4v.(-1) e1 ) p-7+x+if(v = 0 mod 5, 0, 4v.(-1) e1 ) p-7+x+4.(v.(-1) e1 +w.(-1) e2 ) p-7+x p-3+x #(g 1 ) p-3-x-if(v = 0 mod 5, 0, 4v.(-1) e1 ) p+1-x-if(v = 0 mod 5, 0, 4v.(-1) e1 ) p+1-x-4.(v.(-1) e1 +w.(-1) e2 ) p-3-x p+1-x #(g 2 ) p-3+x-if(v = 0 mod 5, 0, 4v.(-1) e1 ) p+1+x-if(v = 0 mod 5, 0, 4v.(-1) e1 ) p+1+x-4.(v.(-1) e1 +w.(-1) e2 ) #(g 3 ) p-3-x+if(v = 0 mod 5, 0, 4v.(-1) e1 ) p+1-x+if(v = 0 mod 5, 0, 4v.(-1) e1 ) p+1-x+4.(v.(-1) e1 +w.(-1) e2 ) Note :

  s = #(0).if(r=s, 1, 0)+ (1/d). ∑ #(g i ).m' r-i,sFrom primitive roots equations, it results for [M] the cardinal matrix with same rules on notations :[M] = [m i,j ] = [P].[σ gr ][P -1 case, we always observe (what can be proved without particular difficulty) t = #(0)/(p-1

  verify, with underneath environment matrix, that [M] = [P].[σ gr ][P -1 ] (here [P] = [P -1 ]).

[

  ) = a.y6 .(9a.d(3a.d-2b.c)+b 2 .(4b.d-c 2 )+4a.c 3 )We deduce immediately discriminant to the variable y: Disc (y) = a.x6 .(9a.d(3a.d-2b.c)+c 2 .(4a.c-b 2 )+4b 3.d) Let us have Δ 1 = 9a.d(3a.d-2b.c)+b 2 .(4b.d-c 2 )+4a.c 3 and Δ 2 = 9a.d(3a.d-2b.c)+c 2 .(4a.c-b 2 )+4b 3 .d and also the definitions of two sets T(p) and NT(p) c c T(p)   x integer \ c = x 3 mod p c c NT(p)   x integer \ c ≠ x 3 mod pThe domains to take in account, modulo p, will then, at least, depend on the following boundaries :

  p i+1 -p i << (Ln(p i ))2 and a correction of the model by Maier's theorem hypothesis has an elementary form : ε > 0, π(x) = li(x) + O(x 1/2+ε ) so that also  ε > 0, i = li(p i ) + O(p i 1/2+ε )

  right near the origin, since we choose (within the framework of a conjecture however) an s such as relation (7) is true everywhere, we have :# min {(p) on I} >| F(n+1)-F(nus to write, in particular : n 0 \  n > n 0 ,  p a prime number \ n 2 < p < (n+1) 2 and n 0 \ e > 1,  n > n 0 ,  p a prime number \ n e < p < (n+1

1 ∫ 0 1 B 0

 1010 (z) b-1 .(1-z) a-1 dz = B(a,b) = Γ(a).Γ(b)/Γ(a+b)Let us carry out the change of variables z = t n , so that dz = nt n-1 dt, then :1 (a,b) = ∫ n. t n-1 (t n ) b-1 .(1-t n ) a-1 dt = n ∫ t nb-1 .(1-t n ) a-1 dt 0 Then, we choose the couple (a,b) in such manner to identify B(a,b) with I(n,i), that is : i) = (1/n).B(1+(i-1)/n,1/n) = (1/n).Γ(1+(i-1)/n).Γ(1/n)/Γ(1+i/n) = ∫ (1-t n ) (i-1)/n .dt 0We also have an equivalent expression, which can be useful later on :1 ∫ (1-t n ) s .dt = (1/n).Γ(1+s).Γ(1/n)/Γ(1+s+1/n) = I(n,s.n+1) 0Let us return to our purpose :k k k П I(n,i) = П (1/n).B(1+(i-1)/n,1/n) = (1/n) k П Γ(1+(i-1)/n).Γ(1/n)/Γ(1+i/n) =(1/n) k .Γ(1).(Γ(1/n)) k /Γ(1+k/n) ahypervolume and on its surface The last equality is obtained by two by two elimination in numerator and denominator of Γ expressions in preceding product. Using the remarkable factorisation property of the gamma function Г(x+1) = x.Г(x) and Γ(1) = 1, we get : k П I(n,i) = (Γ(1+1/n)) k /Γ(1+k/n) = (n/k).(Γ(1+1/n)) k /Γ(k

  with separable variables like previously. To get the expression of this integral, we study first the integral :y = ((c-r(x))/a) (1/n) I = ∫ (c-r(x)-a.y n ) (k/n) dy 0We use the change of variable y → Y where Y n = a.y n . Thus dy = (y/Y).dY = (1/a)(1/n) .dY and :Y = (c-r(x)) (1/n) I = (1/a) (1/n) ∫ (c-r(x)-Y n ) (k/n) dY 0This integral identical to previously studied integral (4) but multiplicative factor (1/a)(1

  2n (t)dt ... ∫ sin 2 (t)dt ∫ sin(t)dt = π n

  -(1+(-1) n+1 )/2)/2)! (29) n П (t -(1+(-1) n+1 )/2) t = 2

Facteur

Facteur

0 lim

 0 c) ≡ ∫ ((c+a k+1 .x k+1 m ) k/n )'dx k+1 П I(n,i)/a i 1c) ≡ (k/n). П I(n,i)/a i 1/n ∫ (c+a k+1 .x k+1 m ) k/n-1 .dx k+1 (36) enumeration ad infinitum of the solutions after weighting :k lim #{(x 1 ,…,x k+1 )/a 1 x 1 n +…+a k x k n = c+a k+1 x k+1 m }=Fan(c). k ( П ( a k+1 )1/n .I(n,i)) . x k+1 m(k/n-1)+1 (38) a k+1 .(m.k-m.n+n) a i x k+1 →∞, given c i = 1 If m(k/n-1)+1 =

  1,3333 0,3333 1,3333 1,3333 0,3333 1,3333 1,3333 0,3333 1,3333 1,3333 0,3333 1,3333 1,3333 0,3333 1

  deriving first x 3 2 +c inside the integral with respect to c) :

  Abundance factors (fae(δ)) Normalized factors (fan(δ) = fae(δ)/3 2δ ) c

  } = (π/(3√3)).Fan(3r 2 ).x 3 x 3 →∞ the value of Fan(c) will increase with x 3 with rapid evolution near the origin and lower pace further. P 380/390 Generation of prime numbers by quadratic sums ans analogues

Table 1

 1

	Truth table 1

Table 2

 2

	Truth table 2

table :

 :

	sequence \ y j n	0	1	2 n	3 n	4 n	5 n	6 n	7 n	8 n	9 n	10 n	11 n …	∞
	2	ε 0

table (

 (

	modulo 23) :
	i	g 0	i	g 1	i	g 2	i	g 3	i	g 4	i	g 5	i	g 6	i	g 7	i	g 8	i	g 9	i
	1	5		7		10	11	14	15	17	19	20	21
	3	10	21	11	20	7		17	14	5		19	15
	5	20	17	19	5		15	7		21	11	10	14
	7	17	5		14	7		19	11	20	15	21	10
	9	11	15	20	19	21	14	7		10	5		17
	13	21	20	15	17	11	5		10	7		14	19
	15	19	14	5		10	17	21	15	20	11	7
	17	15	19	17	14	20	10	11	21	7		5
	19	7		11	21	15	10	19	5		14	17	20
	21	14	10	7		21	5		20	19	17	15	11

  [START_REF] Tenenbaum | Que sais-je ? n° 571 Les nombres premiers[END_REF]

  table 1 appearing in a square block (under card r+1 ), that is to say the size of this square multiplied by the uniform multiplicity d of each target and divided by redundancy (p i -1)/d i of the targets (identical enumeration of 5, 12, 8 and 1 in given example, of which only 5 for example is evaluated in the matrix). Thus the aforementioned column sum equals ((p i -1)/d i ).((p i -1)/d i ).d i /((p i -1)/d i ) = p i -1 for the matrices type B. For A, we add 1 of course. Obviously, this logic does not function for target 0 because neither its multiplicity (1 instead of d), nor its redundancy (1 instead of (p i -1)/d i ) do correspond. However, we know that except first column of table 1, only one square of the table contains this element. We reason on these two corresponding columns of matrix B. The sum of the whole set of matrix components is (p i -1)². By difference to the other columns which have all sum (p

i -1), the two columns which interest us have sum 2(p i -1). As the sum of first column is d i , that of other column is 2.(p i -1)d i . In case of matrix A, the sum of the first column is d i +1, that of the other column is 2p i -1-d i .

15.1.1. General framework For

  sequence p equal to 2, calculation is carried out directly. Let us consider odd sequences. We then find an general expression in the form of a sum with increasing number of terms when p increases. Because of the importance of what follows and for more clarity, we make simultaneous evaluation of literal expressions and of the numerical example of page 54.

	Cardinal matrix [C] associated to operator x n at sequence p is build starting with a circulant matrix [CI] :
	P 69/390	Modulo pi abundance factors for Waring sums

15.1.2. Case odd d

				λ 3 * … λ 1 *
	Thus :
	[P'] = (1/(d.p 1/2 )).[λ'*]	and	[P' -1 ] = (1/(d.p 1/2 ))[λ']
	σ k = if(ve=1,1,0)+d.( ∑ cos((2π/p).g	k-1+u.d )-i.sin((2π/p).g	k

18. Case of the hyperplanes, first quadrant, unit mesh and logarithmic mesh

				).(p-2)	(p-1).(p 2 -3p+3)	(p-1).( p 3 -4p 2 +6p-4)
	#{g u }	1	p-2	p 2 -3p+3	p 3 -4p 2 +6p-4	p 4 -5p 3 +10p 2 -10p-5
	It acts of equations x 1 + x 2 + … + x k = c and y 1 + y 2 + … + y m = c which we already solved. Here n = 1 thus d = 1. The
	matrices A and B are with two dimensions and thus identical to that obtained in the preceding paragraph. The reader will be
	able to compare the enumeration y 1 + y 2 + … + y m = c by relation (45) and by relation (5) of this exercise to note their perfect identity after normalization by multiplication by p/(p-1) k of the terms of the relation (45).

19. Case of hyperspheres, unit mesh

  The analysis is carried out modulo p i . p i describes the set of prime numbers and each variable is replaced by its representative [0 2 , 1 2 , 2 2 , …, (p i -1) 2 ] to build a table with k dimensions. Elements of the table are obtained by modulo p i sums.

	x 1 2 + x 2 2 + … + x k 2 = c	(46)
	Here x i are positive integers.
	(application to case n =2)

  1 2 +t 1 t 2 +t 2 2 = (1+α+α 2 ).t 1 2 , it follows sin 2 (θ) = (3/4).t 2 /p = (3/4).(t/t 1 ) 2 /(1+α+α 2 ). Hence the following cases :

	Families	5, 5', 6 and 6'	1, 1', 3 and 3'	2, 2', 4 and 4'
	t	t 2	t 1	t 1 +t 2
	t	α.t 1	t 1	(1+ α).t 1

  In this table, target p x1 .g x2 has, at preceding stage (whose results are posted on the first two lines), cardinal card x1,x2 and p x1 .g x2 .g x3.d has equal cardinal card x1,x2

).d } are u(y1) numbers with same cardinal card y1,y2 at stage k. We reason by recurrence to check this hypothesis.

  Matrix A δ (matrix is built easily while starting on right and bottom) (first column of this table does not belong to matrix)

However, if we adopt such a form, matrices [P δ ] which we are brought to handle regularly, would be of less practical use. P 125/390 Modulo p Δ abundance factors correction Case p ∤ n

. Recapitulative table, function of n, of values of ε

  1/2 /2By recurrence, we get the following table, where figure one of the "generative" roots of unit (the term "generative" is placed between brackets for whole set is generated only with additional use of the sign -) :

	Matrix	t "generative" root
	size
	1	1
	2	0 -1
	4	1 i
	8	2 √2/2(1+i)
	16	3 (√(2+√2))/2+i.(√(2-√2))/2
	32	4 (√(2+√(2+√2)))/2+i.(√(2-√(2+√2)))/2
	64	5 (√(2+√(2+√(2+√2))))/2+i.(√(2-√(2+√(2+√2))))/2
	128	6 (√(2+√(2+√(2+√(2+√2)))))/2+i.(√(2-√(2+√(2+√(2+√2)))))/2
	256	7 (√(2+√(2+√(2+√(2+√(2+√2))))))/2+i.(√(2-√(2+√(2+√(2+√(2+√2))))))/2
	…
	2 t+1	t α t = (√(2+√(2+√(2+…√2))…)/2+i.(√(2-√(2+√(2+…√2))…)/2
	Whole set of roots is then :
		{α t , -α t , α t 3 , -α t	3 , α t 5 , -α t 5 , …,α t	2^t-1 , -α t 2^t-1 }
	By direct elementary calculation, we verify :
			α t .α t * = 1
	so that :
		(α t ) i . (α t *) i = 1
	1.2.8.5
	P 153/390			Modulo p Δ abundance factors correction

  ).

	Recurrent elements ε	Eigenvalues λ # Eigenvectors	Eigenvectors
	(except two first)		(any x i whose indices
			are not indicated
			are equal to 0)
	P 156/390		Modulo p Δ abundance factors correction

  1 (p-1) at sequence p, Stage 2 -restore sum to p δ by multiplication of factors with p δ .

			k	m
	Fan{p δ ,p,δ}				p δ
	Fan{p δ-1 .g 0 ,p,δ}				0
	Fan{p δ-1 .g 1 ,p,δ}				0
	…				...
	Fan{p δ-1 .gd -1 ,p,δ}				0
	...				...
	Fan{p 1 .g 0 ,p,δ} Fan{p 1 .g 1 ,p,δ}	=	AN δ	BN δ	0 0
	…				...
	Fan{p 1 .gd -1 ,p,δ}				0
	Fan{p 0 .g 0 ,p,δ}				0
	Fan{p 0 .g 1 ,p,δ}				0
	…				...
	Fan{p 0 .gd -1 ,p,δ}				0
	Thus :
			[AN δ ] = 1/p δ .[A δ ]
	and for k variable of integers :
				k
	Fan{c=p δ ,p,δ}			p δ
	Fan{p δ-1 .g 0 ,p,δ}			0
	Fan{p δ-1 .g 1 ,p,δ}			0
	…			...
	Fan{p δ-1 .gd -1 ,p,δ}		0
	...			...
	Fan{p 1 .g 0 ,p,δ} Fan{p 1 .g 1 ,p,δ}	=	AN δ	0 0
	…			...
	Fan{p 1 .gd -1 ,p,δ}			0
	Fan{p 0 .g 0 ,p,δ}			0
	Fan{p 0 .g 1 ,p,δ}			0
	…			...
	Fan{p 0 .gd -1 ,p,δ}			0
	In the same way :
			[BN δ ] = (1/(p δ-1 .(p-1))).[B δ ]
	and for m variable of prime numbers :
				m
	Fan{c=p δ ,p,δ}			p δ
	Fan{p δ-1 .g 0 ,p,δ}			0
	Fan{p δ-1 .g 1 ,p,δ}			0
	…			...
	Fan{p δ-1 .gd -1 ,p,δ}		0
	...			...
	Fan{p 1 .g 0 ,p,δ} Fan{p 1 .g 1 ,p,δ}	=	BN δ	0 0
	…			...
	Fan{p 1 .gd -1 ,p,δ}			0
	Fan{p 0 .g 0 ,p,δ}			0
	Fan{p 0 .g 1 ,p,δ}			0
	…			...
	Fan{p 0 .gd -1 ,p,δ}			0
	P 161/390		Modulo p Δ abundance factors correction

In the case of a mix of variables of integers and prime numbers, we carry out following matrices product :

  Δμ i,d ] 2 = p.d.[t d ] by building Δμ i and the products Δμ i .Δμ i+m .

					p.d.[t d ] = [	μ i,d1 .μ i,d2 ] 2 = [Δμ i,d ] 2 √p	(63)
	Example d = 12 = 4.3 (at sequence p = 97).
	We verify [Δμi	Δμi+1	Δμi+2	Δμi+3	Δμi+4	Δμi+5	Δμi+6	Δμi+7	Δμi+8	Δμi+9	Δμi+10	Δμi+11
	20,859	-2,7717	10,1169	-51,098	-24,089	-22,9	4,56743	41,5448	3,22987	25,6715	-14,684	9,55313
	Δμi.Δμi	Δμi.Δμi+1	Δμi.Δμi+2	Δμi.Δμi+3	Δμi.Δμi+4	Δμi.Δμi+5	Δμi.Δμi+6	Δμi.Δμi+7	Δμi.Δμi+8	Δμi.Δμi+9	Δμi .Δμi+10	Δμi .Δμi+11
	435,099	-57,814	211,028	-1065,9	-502,47	-477,67	95,2721	866,585	67,3719	535,483
				) :
	P 219/390							Fermat-Catalan abundance factors

  integer components of matrix [s d ] are evaluated by trials and errors :

		p	s 1	s 2	s 3	s 4		s 5		s 6	s 7	s 8		s 9	s 10	s 11	s 12
		337	-30	23	-61	2		27		47	98	-13		3	-70	-37	11
		409	42	-1	71	50	-33		-25	-118	-13		-9	26	47	-37
		433	66	59	71	50	-69		-13	-70	11		3	-46	-1	-61
		457	-6	83	23	74	-33		11	26	-1		39	-94	-49	-73
		577	42	-37	-37	-142		-9		-13	50	59		-33	50	-13	83
		601	-6	71	59	122	-33		-1	-46	-25		39	-70	-13	-97
		1129	42	83	107	146	-69		-13	-118	-25		27	-70	11	-121
		1153	-30	-121	23	-70		39		-73	-94	11		-9	194	71	59
		s mod 12	6	11	11	2		3		11	2	11		3	2	11	11
		t mod 12	6	1	1	10		9		1	10	1		9	10	1	1
		s+t mod	0	0	0	0		0		0	0	0		0	0	0	0
		12
	Let us apply modulo 6 spaced averages:
		p	(s1+s7)/2 (s2+s8)/2 (s3+s9)/2 (s4+s10)/2 (s5+s11)/2 (s6+s12)/2 (s7+s1)/2 (s8+s2)/2 (s9+s3)/2 (s10+s4)/2 (s11+s5)/2 (s12+s6)/2
		73	10	17	7	-10	-17		-7	10	17		7	-10	-17	-7
		97	-14	-19	-5	14		19		5	-14	-19		-5	14	19	5
		193	-2	23	25	2		-23		-25	-2	23		25	2	-23	-25
		241	-14	17	31	14		-17		-31	-14	17		31	14	-17	-31
		313	22	-13	-35	-22	13		35	22	-13		-35	-22	13	35
		337	34	29	-5	-34	-29		5	34	29		-5	-34	-29	5
		409	-38	-31	7	38		31		-7	-38	-31		7	38	31	-7
		433	-2	-37	-35	2		37		35	-2	-37		-35	2	37	35
		457	10	-31	-41	-10	31		41	10	-31		-41	-10	31	41
		577	46	35	-11	-46	-35		11	46	35		-11	-46	-35	11
		601	-26	-49	-23	26		49		23	-26	-49		-23	26	49	23
		1129	-38	-67	-29	38		67		29	-38	-67		-29	38	67	29
		1153	-62	-7	55	62		7		-55	-62	-7		55	62	7	-55
	We find actually the components of case d = 6. Let us pass to the variations (Δsi = if (si+si+6)/2):
			p	Δs1	Δs2	Δs3	Δs4		Δs5	Δs6	Δs7	Δs8	Δs9	Δs10	Δs11	Δs12
			73	32	6	16	12		-16	6	-32	-6	-16	-12	16	-6
			97	-16	18	-8	36		8	18	16	-18	8	-36	-8	-18
			193	-28	24	-14	48		14	24	28	-24	14	-48	-14	-24
			241	-16	-30	-8	-60		8	-30	16	30	8	60	-8	30
			313	-52	-24	-26	-48		26	-24	52	24	26	48	-26	24
			337	-64	18	-32	36		32	18	64	-18	32	-36	-32	-18
			409	80	6	40	12		-40	6	-80	-6	-40	-12	40	-6
			433	68	24	34	48		-34	24	-68	-24	-34	-48	34	-24
			457	-16	42	-8	84		8	42	16	-42	8	-84	-8	-42
			577	-4	-48	-2	-96		2	-48	4	48	2	96	-2	48
			601	20	48	10	96		-10	48	-20	-48	-10	-96	10	-48
			1129	80	54	40	108		-40	54	-80	-54	-40	-108	40	-54
			1153	32	-66	16	-132		-16	-66	-32	66	-16	132	16	66
	Obviously (within a conjectural framework), the variations arise on these sequences in a very simple way :
				Δs 1	Δs 2	Δs 3	Δs 4		Δs 5	Δs 6	Δs 7	Δs 8	Δs 9	Δs 10 Δs 11 Δs 12
			p	4u	2v	2u	4v		-2u	2v	-4u	-2v	-2u	-4v	2u	-2v
	We apply then p.d.(d 1 -1).(d 1 -1) = ∑ (s d ,i ) 2 to the case d = 12 = 3.4 and d = 6 = 3.2. It follows 72p = ∑ (s 12 ,i ) 2 and 24p = ∑ (s 6,i ) 2 and table :
	Fam.	(s 12 ) 1 = (s 6 ) 1 +Δs 1	(s 12 ) 2 = (s 6 ) 2 +Δs 2	(s 12 ) 3 = (s 6 ) 3 +Δs 3	(s 12 ) 4 = (s 6 ) 4 +Δs 4	(s 12 ) 5 = (s 6 ) 5 +Δs 5	(s 12 ) 6 = (s 6 ) 6 +Δs 6	(s 12 ) 7 = (s 6 ) 1 +Δs 7	(s 12 ) 8 = (s 6 ) 2 +Δs 8	(s 12 ) 9 = (s 6 ) 3 +Δs 9	(s 12 ) 10 = (s 6 ) 4 +Δs 10	(s 12 ) 11 = (s 6 ) 5 +Δs 11	(s 12 ) 12 = (s 6 ) 6 +Δs 12
		a+2b+4u 2a+b+2v	a
		p	s 1	s 2	s 3	s 4		s 5		s 6	s 7	s 8		s 9	s 10	s 11	s 12
		73	42	-1	-1	2		-9		23	-22	-13		-33	-22	23	11
		97	-30	23	11	50		3		-1	2	-13		27	-22	-13	-37
		193	-30	-1	-37	50		39		47	26	-49		-9	-46	11	-1
		241	-30	-61	-25	-46		39		-13	2	-1		-9	74	23	47
		313	-30	11	-13	-70		-9		-37	74	59		39	26	-61	11
	P 223/390										Fermat-Catalan abundance factors

  relative prime with d = 15. One of the solutions (s' 15 ) at sequence p = 61 is the first column of the table which follows. The other solutions result with the steps indicated.

	Steps	1	2	4	7	8	11	13	14
		-23	-23	-23	-23	-23	-23	-23	-23
		-31	-16	-1	29	-16	44	-16	-16
		-16	-1	-16	-16	-31	29	44	-16
		32	-13	-13	-13	17	32	17	-13
		-1	-16	-31	-16	-16	-16	29	44
		4	19	4	4	19	19	4	19
	(s'15)	-13 29	-13 -16	17 -16	-13 -1	32 44	-13 -16	32 -31	17 -16
		-16	-31	-16	44	-1	-16	-16	29
		17	32	-13	32	-13	17	-13	-13
		19	4	19	19	4	4	19	4
		44	29	-16	-16	-16	-31	-16	-1
		-13	17	32	17	-13	-13	-13	32
		-16	44	29	-31	-16	-16	-1	-16
		-16	-16	44	-16	29	-1	-16	-31

22.12.1. Case d = p i ni , p i > 2 22.12.1.1. Alternative 1 : False modulating matrices (a) Presentation

	p = 31							p = 61
	7	7	7	7	7	7	7	7	-23 -23 -23 -23 -23 -23 -23 -23 -8	-8	22 22 22 22 52 52
	-31 -16 -16 -1	14 14 14 29	-31 -16 -16 29 -16 -16 -1	44 -31 29 -16 -1	-1	29	-1	14
	-16 -1	29 14 -31 -16 14 14	-16 -16 -31 -16 -1	44 -16 29 -31 29 29 -16 -1	-1	14	-1
	-13	2	17 -13 -13 17 -13	2	32 -13 17 -13 -13 17 -13 32	2	2	2	2	-58 32 -13 -13
	29 14 14 14 -16 -1 -16 -31	-1	44 -16 -16 -16 29 -31 -16 -31 29	-1	29 -16 -1	-1	14
	4	-11 -11	4	-11	4	-11	4	4	19 19	4	19	4	4	19	4	4	-11 -11 -11 -11 -56	4
	17 -13	2	-13 -13	2	17 -13	-13 17 32 -13 -13 32 17 -13	2	2	32	2	2	-58 -13 -13
	14 14 -16 -31 14 29 -16 -1	29 -16 44	-1 -16 -31 -16 -16 29 -31 29 -16 -1	-1	-1	14
	14 14 -31 -16 29 14	-1 -16	-16 29	-1	44 -31 -16 -16 -16 -31 29	-1	-1	29 -16 14	-1
	-13 17 -13	2	2	-13 -13 17	17 -13 -13 32 32 -13 -13 17	2	2	2	-58 32	2	-13 -13
	-11	4	4	-11	4	-11	4	-11	19	4	4	19	4	19 19	4	4	4	-11 -11 -11 -11	4	-56
	14 29 14 14	-1 -16 -31 -16	44	-1 -16 -16 29 -16 -16 -31 29 -31 -16 -1	-1	29 14	-1
	2	-13 -13 17 17 -13	2	-13	-13 32 -13 17 17 -13 32 -13	2	2	-58 32	2	2	-13 -13
	-1 -16 14 29 -16 -31 14 14	-16 -16 -16 -31 44	-1	29 -16 29 -31 -1	-1	29 -16 -1	14
	-16 -31 -1 -16 14 14 29 14	-16 -31 29 -16 -16 -16 44	-1	29 -31 -1	29 -16 -1	14	-1
	22.12. Non-coherent solutions
	We highlight matrices solutions of [ns'

d ].[ns' d ] = p.d.[t d ] which do not respect relation (ns' d ) = (t' d ) mod d (or (ns' d ) = -(t' d ) mod d). The reader will note that matrices of this type cannot be used to get matrices like [r d ]. What follows is given thus on a purely "free of charge" basis. It is relatively easy to work out evidence that [ns' d ].[ns' d ] = d.p.[t d ] according to the diagrams suggested below. However, we will not extend on these proofs since [ns' d ] like matrices are not appropriate to the required purpose.

  Let us start from [ns' d ].[ns' d ] = p.d.[t d ] for d = 3. One solution is matrix [ns' 3 ] = [s' 3 ] :

	(b) Case d = p k n , p k > 3	-174 78	29 -13	29 -13	29 -13	29 -13		29 -13	29 -13
		96	-16	-16	-16	-16		-16	-16
	The illustration spreads while using :
		ns' 1 , pi^n		ns' 1 , pi^(n-1)		1		ns' 1 , pi	a1
	ns' 1+pi^(n-1), pi^n …	= pi.	ns' 1+pi^(n-2) , pi^(n-1) …	+a1.	1 …	with	ns' 2 , pi	=	a2 …
	ns' 1+2.pi^(n-1), pi^n		ns' 1+2.pi^(n-2) , pi^(n-1)		1		ns' pi , pi	a pi
	22.12.2.2. Case d = ∏p i	ni p i > 2
	We find false constructive of sequence matrices of dimension d starting from the matrices of dimension d i = p i ni factors of
	d.
	Let us proceed initially by giving examples
	Example 1 : d = 15, p = 211
		29	-16	-13	2	2		-1	-1
		-16	-13	29	= 3.211.	-1		2	-1
		-13	29	-16		-1		-1	2

: [ns' 3^n ].[ns' 3^n ] = 3 n .p.[t 3^n ] P 229/390 Fermat-Catalan abundance factors

  we verify that actually [ns' 105 ].[ns' 105 ] = 105.211.[t 105 ] where first column of [t 105 ] is :

.12.2.3. Case d = 2 n .∏p i ni

	There are, a priori, no false (or true) constructive of sequence matrices in case d = 2 n . This does not mean that it is so for d
	= 2 n .∏p i ni , p i > 2, quite to the contrary. Let us give some examples.
	Example 1 : d = 72, p = 433
	We have d = 3 2 .2 3 . We start with [ns' 3 ].[ns' 3 ] = 433.3.[t 3 ] for which we find by direct research
	37	-2	-35	2	2	-1	-1
	-2	-35	37	= 3.433.	-1	2	-1
	-35	37	-2		-1	-1	2

  16 being an odd integer

		2897	181	1		44			31	1		1		3971		-2272	85
		3089	193	1		8			55	1		2		196		-197	5
		3121	195	1		-40			39	1		2		-1620		19	29
		3217	201	1		56			-9	1		2		2268		-1973	53
		3313	207	1		8		-57	1		2		36		11339	197
		3697	231	1		36		-49	1		81		49		36	1
		3761	235	1		-56		-25	1		2		-448		-3961	65
		3793	237	1		52		-33	1		1		117		-6712	109
		3889	243	1		60		-17	1		9		735		764	17
		4049	253	1		32			55	1		8		100		-3371	53
	Prime numbers p with (p-1)/32 being an odd integer
				cm = 4						cm = 8					cm = 16
	p	(p-1)/32	ak1/2 1	α1	β1	γ1		ak2/2 8	α2	β2	γ2	ak2/2 23	α23	β23	γ23
	97	3	1	4	-9	1		1	9	4	1	2	113	2292	233
	353	11	1	8	-17	1		2	16	319	17	2	-7349	1276	397
	673	21	1	-12	23	1		1	-243	-712	29	2	23	12	1
	929	29	1	-20	23	1		1	-125	376	13	2	70403	451400	14989
	1249	39	1	32	15	1		8	36	173	5	1	335648	-2577975	73561
	1697	53	1	-4	-41	1		1	-121	5972	145	2	2911249	-11230916	281641
	1889	59	1	-40	-17	1		2	-500	-1157	29	1	10047328 175431271 4042985
	2017	63	1	-44	-9	1		1	-891	-3712	85	2	2349457	-4761108	118217
	2081	65	1	20	-41	1		1	845	7112	157	2	7340899	5151320	196589
	2273	71	1	-8	47	1		2	-256	-769	17	2	158357	277468	6701
	2593	81	1	48	-17	1		16	243	76	5	1	13340416	25529529	565673
	2657	83	1	16	-49	1		16	49	16	1	1	11634752	-609511	226025
	3041	95	1	-4	55	1		1	-289	-1348	25	2	-209437	-35800	3853
	3169	99	1	-12	55	1		1	-675	-15128	269	2	-23236447 -18795804	530905
	3361	105	1	56	15	1		2	1008	3631	65	2	-40764627 -20274380	785317
	3617	113	1	44	-41	1		1	3971	11672	205	2	3470863	-1638212	63817
	4001	125	1	40	-49	1		2	20	13979	221	1	242788864 -209468777 5069465
	4129	129	1	60	23	1		1	1215	9756	153	2	3285249 -147314668 2293145
	4513	141	1	48	47	1		16	27	-1948	29	1	248396544 287453529 5655177
	5153	161	1	68	23	1		1	833	-29348	409	2	51219997 115299688 1757549
	5281	165	1	60	-41	1		25	135	1228	17	2	-156123305 -17685492 2162113
	5857	183	1	76	-9	1		1	13851	-176	181	2	63949257	22913132	887617
	6113	191	1	28	-73	1		1	1183	-1556	25	2	-770258893 2760698152 36658109
	6689	209	1	80	-17	1		8	40	-3353	41	4096	-395032	299999	6065
	7393	231	1	72	47	1		2	1296	8239	97	2	-371964437 -5970640884 69574765
	7457	233	1	-76	-41	1		1	-15979	23072	325	2	-702832913 -69095348 8178217
	7649	239	1	68	55	1		1	6137	-54316	625	2	-48336875 1268367928 14513029
	7841	245	1	-40	79	1		2	-2420	859	29	1	13616480 125999111 1431209
	8161	255	1	40	-81	1		2	4500	19451	221	1	32888992 1565402295 17332057
	Other examples
							193	449		577			257
					p			=			=		=			=
							1+3.2 6	1+7.2 6		1+9.2 6			1+2 8
					ak			2 1			2 1		2 1			2 1
		p	(p-1)/16 cm = 4	ak1/2 1	α1 β1	α1	cm = 4	12 7	β1	γ1	20 7	ak2/2 8	24 -1	α2	cm = 8	β2 16 -1	γ2
		17 113	1 7	1 1	γ1 ak	4 8		1 2 8	-1 7	1 1	1 2 8	1 1	1 2 9	1 8		4 106 1 2 12	1 10
		241 337 401 433 593 881 977 1009 1201 1297 1361 1489 1553 1777 1873 2129 2161 2417 2609	15 21 cm = 8 25 27 37 55 cm = 16 61 63 75 81 cm = 32 85 93 97 111 cm = 64 117 133 135 151 cm = 128 163	1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1	α2 β2 γ2 ak α3 β3 γ3 ak α4 β4 γ4 ak α5 β5 γ5 ak α6 β6 γ6	4 16 -20 -12 -8 16 -4 -28 24 36 20 20 -32 -16 28 40 -44 4 -20	15 -9 -1 -17 27 64 5 2 24 23 -25 -7 12 31 1 15 2 64 -25 -1 -2011 13704 31 997 -33 / 23 39 / / -33 / 23 / 15 -49 / / 47 /	1 1 -784 5 1 37 1 2 24 1 1 -275 16 1 13 1 2 64 1 1 -323275 -952664 1 47477 1 / 1 1 / / 1 / 1 / 1 1 / / 1 /	1 16 1 1 2 16 1 1 1 81 1 1 8 8 1 2 1 1 1	9 9 -245 -27 -64 25 -121 -567 432 1 845 405 -4 -72 567 1280 -2475 -137113 108 -293 13 2 23 -63136 -24873 2825 2 64 -22584 5785 / / / / / 49 / / -5 /	-388 16 1 16 88 1 936 2 23 -409 16 -1088 -2759 -772 185 556 2 71 751 36 -128 1087 -1072 113 -296 2 168 197 -713 1289 105411184 7292 6575369 -1393 2 384 1384 4 1 -16432 664 1025	25 1 13 45 17 1 25 25 25 1 37 13 5 17 169 41 61 1 13
		2801	175	1		-20		-49	1		25		-5		688	13
		2833	177	1		48			23	1		8		216		-1313	25
	P 240/390 P 241/390										Fermat-Catalan abundance factors Fermat-Catalan abundance factors

  since it is equivalent to the existence of arithmetic series in the prime numbers set which enumeration answers exactly to relation (41).Let us see however how we can arrive to this result as a limit case. We will initially deduce inclusive enumeration from exclusive enumeration. Let us have m as product p ε , where p is a prime number. Then fe(p ε ) = 1/p ε , fe(p ε+1 ) = 1/p ε+1 ,… fe(p ε+i ) = 1/p ε+i . Thus fi(p ε ) = ∑ 1/p ε+i , with sum indices varying from 0 to +∞.

							#		#				#	#
				pi		i	exclusive	i/231	inclusive	i/120	pi	i	exclusive	i/231	inclusive	i/120
							division		division				division	division
		10627	1296	9	5,61	12	10,80	53593	5460	25	23,64	42	45,50
		11551	1392	10	6,03	13	11,60	54517	5547	26	24,01	43	46,23
		13399	1589	11	6,88	14	13,24	54979	5589	44	46,58
		14323	1680	12	7,27	15	14,00	55441	5629	45	46,91
		16633	1924			16	16,03	55903	5675	46	47,29
		18481	2117	13	9,16	17	17,64	56827	5763	47	48,03
		19867	2248	14	9,73	18	18,73	57751	5851	27	25,33	48	48,76
		22639	2529			19	21,08	61909	6223	28	26,94	49	51,86
		23563	2621			20	21,84	66067	6596	50	54,97
		25411	2802			21	23,35	66529	6633	51	55,28
		25873	2847			22	23,73	67453	6720	29	29,09	52	56,00
		27259	2984	15	12,92	23	24,87	69763	6916	30	29,94	53	57,63
		28183	3073	16	13,30	24	25,61	70687	7003	54	58,36
		29569	3209	17	13,89	25	26,74	72073	7134	55	59,45
		30493	3292			26	27,43	72997	7218	31	31,25	56	60,15
		32341	3469			27	28,91	73459	7254	57	60,45
		32803	3519	18	15,23	28	29,33	74383	7339	58	61,16
		34651	3701			29	30,84	75307	7420	32	32,12	59	61,83
		36037	3829			30	31,91
	Proof
	It is acquired Hence fi(p ε ) = (1/p ε )(1/(1-1/p)) = 1/((p-1).p ε-1 ).
	Let us have then m as product p 1 summon all of the terms of fi(p 1 ε1 .p 2 ε1 .p 2 ε2 , where p 1 and p 2 are prime numbers. We then build the following table allowing to ε2 ) :
	fi(p 1	ε1 .p 2 ε2 )		p 1 ε1		p 1 ε1+1	…		p 1 ε1+i	Totals i →∞, j→∞
	p 2 p 2 ε2+1 ε2	1/(p 1 ε1 .p 2 ε2 ) 1/(p 1 ε1 .p 2 ε2+1 ) 1/(p 1 1/(p 1 ε1+1 .p 2 ε2 ) ε1+1 .p 2 ε2+1 )	… …	1/(p 1 ε1+i .p 2 ε2 ) 1/(p 1 ε1+i .p 2 ε2+1 )	1/((p 1 -1).p 1 ε1-1 .p 2 ε2 ) 1/((p 1 -1).p 1 ε1-1 .p 2 ε2+1 )
		…
	p 2	ε2+j	1/(p 1 ε1 .p 2 ε2+j ) 1/(p 1 ε1+1 .p 2 ε2+j )	…	1/(p 1 ε1+i .p 2 ε2+j )	1/((p 1 -1).p 1 ε1-1 .p 2 ε2+j )
													1/((p 1 -1).p 1 ε1-1 .(p 2 -1).p 2 ε2+j )
	Hence fi(p 1	ε1 .p 2	ε2 ) = fi(p 1 ε1 ).fi(p 2 ε2 ). By successive iterations, this result spreads in the form (with of course  k ≠ m, (p k ,p m )
	= 1) :
								fi(p 1 ε1 .p 2 ε2 …p i εi ) = fi(p 1 ε1 ).fi(p 2 ε2 )…f(p i εi )
	Thus function fi is a multiplicative function (provided function fe is one) :
							#		#				#	#
				pi		i	exclusive	i/231	inclusive	i/120	pi	i	exclusive	i/231	inclusive	i/120
							division		division				division	division
			463	90	1	0,39	1	0,75	37423	3962	33,02
		2311	344	2	1,49	2	2,87	39733	4177	19	18,08	34,81
		3697	516	3	2,23	3	4,30	42043	4397	36,64
		4159	573			4	4,78	42967	4492	37,43
		4621	624	4	2,70	5	5,20	43891	4568	20	19,77	38,07
		6007	784	5	3,39	6	6,53	46663	4822	21	20,87	40,18
		6469	839			7	6,99	48049	4950	22	21,43	41,25
		7393	939	6	4,06	8	7,83	48973	5033	23	21,79	41,94
		8317	1044			9	8,70	50359	5168	24	22,37	43,07
		8779	1094	7	4,74	10	9,12	50821	5205	43,38
		9241	1146	8	4,96	11	9,55	51283	5247	43,73
	P 262/390										Quadratic abundance factors and substitutes

  It is not the place here to open such a study. We give only one illustration of this point.

	x²-1	# Div 3	# Div 7	# Div 11	# Div 231	3.fi(3)	7.fi(7)	11.fi(11)	231.fi(231)
	215295	310				2,00431	2,00647	2,01509	8,46336
	…	…	…	…	…	…	…	…	…
	477480	461				2,00145	1,99566	1,98987	7,68886
	478863	462	198	126	24	2,00289	2,00289	2,00289	8,01156
	480248					2,00000	2,00000	2,00000	8,00000
	Numerical example: x ² -1
	x²-1	# Div 3	# Div 7	# Div 11	# Div 231	3.fi(3)	7.fi(7)	11.fi(11)	231.fi(231)
	0	1	1	1	1	3	7	11	231
	3	2				3,00000	3,50000	5,50000	115,50000
	8					2,00000	2,33333	3,66667	77,00000
	15	3				2,25000	1,75000	2,75000	57,75000
	24	4				2,40000	1,40000	2,20000	46,20000
	35		2			2,00000	2,33333	1,83333	38,50000
	48	5				2,14286	2,00000	1,57143	33,00000
	63	6	3			2,25000	2,62500	1,37500	28,87500
	80					2,00000	2,33333	1,22222	25,66667
	99	7		2		2,10000	2,10000	2,20000	23,10000
	120	8				2,18182	1,90909	2,00000	21,00000
	143			3		2,00000	1,75000	2,75000	19,25000
	168	9	4			2,07692	2,15385	2,53846	17,76923
	195					2,14286	2,00000	2,35714	16,50000
	224		5			2,00000	2,33333	2,20000	15,40000
	255					2,06250	2,18750	2,06250	14,43750
	288					2,11765	2,05882	1,94118	13,58824
	323					2,00000	1,94444	1,83333	12,83333
	360					2,05263	1,84211	1,73684	12,15789
	399		6			2,10000	2,10000	1,65000	11,55000
	440			4		2,00000	2,00000	2,09524	11,00000
	483		7			2,04545	2,22727	2,00000	10,50000
	528			5		2,08696	2,13043	2,39130	10,04348
	575					2,00000	2,04167	2,29167	9,62500
	624					2,04000	1,96000	2,20000	9,24000
	675					2,07692	1,88462	2,11538	8,88462
	728		8			2,00000	2,07407	2,03704	8,55556
	783					2,03571	2,00000	1,96429	8,25000
	840		9			2,06897	2,17241	1,89655	7,96552
	899					2,00000	2,10000	1,83333	7,70000
	960					2,03226	2,03226	1,77419	7,45161
	1023			6		2,06250	1,96875	2,06250	7,21875
	1088					2,00000	1,90909	2,00000	7,00000
	1155		10	7	2	2,02941	2,05882	2,26471	13,58824
	…	…	…	…	…	…	…	…	…
	52899	154	66	42	8	2,00870	2,00870	2,00870	8,03478
	53360		66	42		2,00000	2,00000	2,00000	8,00000
	53823	155	67	43	9	2,00431	2,02155	2,03879	8,96121
	…	…	…	…	…	…	…	…	…
	211599	307				2,00217	1,99348	1,98478	7,53261
	212520	308	132	84	16	2,00434	2,00434	2,00434	8,01735
	213443					2,00000	2,00000	2,00000	8,00000
	214368	309	133	85	17	2,00216	2,01080	2,01944	8,48164

  ). For example, in the environment d = 2, for δ = 3 (and Δ c D(p)) :

	μq 0		p 6	changes in	μq 0	p 6
	μq 1 μq 2	=	-p 5 p 4		μq 1,1 μq 21,2	-p 5 -p 5
	μq 3		-p 3		μq 2,1	=	p 4
					μq 2,2	p 4
					μq 3,1	-p 3
					μq 3,2	-p 3

  ).∑ k ∑ m #(g i ).λ k+i .λ k+r-1 *.λ k+s-1 )

= #

(0)

.if(r=s, 1, 0)+(1/p).(p-1)/d).(σ mz 0

  ).

	-176	-264	-286	-286	-286	-286	-308	-286	-253	-242	-286	-18,992		79,217
	-286	-176	-264	-286	-286	-286	-286	-308	-286	-253	-242	53,660		700,216
	-242	-286	-176	-264	-286	-286	-286	-286	-308	-286	-253	23,508		323,892
	-253	-242	-286	-176	-264	-286	-286	-286	-286	-308	-286	-24,442		326,429
	-286	-253	-242	-286	-176	-264	-286	-286	-286	-286	-308	32,468	=	873,337
	-308	-286	-253	-242	-286	-176	-264	-286	-286	-286	-286	-19,167		109,795
	-286	-308	-286	-253	-242	-286	-176	-264	-286	-286	-286	2,364		242,933
	-286	-286	-308	-286	-253	-242	-286	-176	-264	-286	-286	-7,967		183,413
	-286	-286	-286	-308	-286	-253	-242	-286	-176	-264	-286	1,684		241,431
	-286	-286	-286	-286	-308	-286	-253	-242	-286	-176	-264	-31,126		-160,956
	-264	-286	-286	-286	-286	-308	-286	-253	-242	-286	-176	-22,989		39,293

  The reader may try with the few numerical examples that follow concerning degree 8.

	p p p	p mod 72 p mod 72 p mod 72	g g g	v v v	w w w	#(0) #(0) #(0)	#(g 0 ) #(g 0 ) #(g 0 )	#(g) #(g) #(g)	#(g 2 ) #(g 2 ) #(g 2 )	#(g 3 ) #(g 3 ) #(g 3 )	#(g 4 ) #(g 4 ) #(g 4 )	#(g 5 ) #(g 5 ) #(g 5 )	#(g 6 ) #(g 6 ) #(g 6 )	#(g 7 ) #(g 7 ) #(g 7 )
	37 991 1259	37 55 35		1	6	289 7921 1	32 984 1240	24 984 1280	24	40
	109 1063 47	37 55 47		3	10	865 8497 1	96 1096 40	104 1016 56	120	88
	181 1279 191	37 55 47	19	9	10	1441 10225 1	160 1208 200	168 1336 184	152	216
	397 1423 263	37 55 47		19	6	3169 11377 1	384 1368 264	392 1464 264	408	376
	541 7 479	37 7 47	13	21	10	4321 13 1	648 8 536	544 4 424	496	448
	613 79 839	37 7 47	11	17	18	4897 157 1	672 64 824	584 92 856	600	568
	757 151 911	37 7 47	17	9	26	6049 301 1	728 144 824	640 156 1000	768	864
	829 223 983	37 7 47		27	10	6625 445 1	800 232 1032	888 212 936	696	904
	1117 367 59	37 7 59		21	26	8929 733 1	920 368 48	1184 364 72	1056	1280
	13 439 131	13 7 59	15	3	2	25 877 1	24 400 160	8 476 104	8	8
	157 727 347	13 7 59		11	6	313 1453 1	208 768 352	144 684 344	128	144
	229 1087 419	13 7 59		15	2	457 2173 1	216 1064 400	224 1108 440	248	224
	373 1231 491	13 7 59		7	18	745 2461 1	384 1240 464	376 1220 520	416	312
	661 1303 563	13 7 59		25	6	1321 2605 1	592 1304 560	736 1300 568	672	640
	733 1447 1283	13 7 59		27	2	1465 2893 1	728 1456 1280	760 1436 1288	680	760
	877 31 1427	13 31 59		29	6	1753 61 1	864 28 1408	920 32 1448	800	920
	1021 103 71 Examples (for two integers' variables) 13 10 11 31 71 1093 13 33 463 31 359 71	30 2	2041 205 1 2185 925 1	1024 100 56 1144 492 384	1032 104 88 1168 432 336	960 1048	1064 1008
	p 1237 607 431 1381 751 503	13 31 71 p mod 13 31 71	g	9 v 15	34 w 34	2473 1213 1 #(0) 2761 1501 1	1336 612 496 #(g 0 ) 1320 788 488	1248 600 368 #(g) 1432 712 520	1272 #(g 2 ) 1304	1088 #(g 3 ) 1464	#(g 4 )	#(g 5 )	#(g 6 )	#(g 7 )
	2 1453 823 647 3 61 967 719 73 277 1039 863 433 349 1327 1151 577 421 1399 1223 937 709 1471 1367 1009 1153 1297 97 241 313 457 601 673 1033 1249 1321 193 337 409 769 1129 1201 17 89 233 449 521 593 809 881 953 1097 41 113 257 401 617 761 977 1049 1193 1409 1481 137 281 353 569 641 857 929 1217 1289 1361 1433 919 1187 631 971 487 827 271 683 199 467 127 251 1459 179 1171 107 883 1319 811 1103 739 1031 523 887 379 743 307 599 163 383 19 311 1277 239 1061 167 773 23 701 1451 557 1307 269 1163 197 1091 53 1019 1181 947 1109 659 821 587 677 443 461 227 389 83 317 11 173 1291 101 859 29 787 1373 643 1301 571 1229 499 1013 283 941 211 797 139 653 67 509 1483 293 1123 149 1051 5 907 1429 691 1213 619 1069 547 997 331 853 43 1439	2 13 31 71 3 61 31 71 1 61 31 71 1 61 31 71 1 61 31 71 1 61 31 71 1 1 1 55 35 55 35 55 35 55 35 55 35 55 35 19 35 19 35 19 23 19 23 19 23 19 23 19 23 19 23 19 23 19 23 53 23 53 23 53 23 53 11 53 11 53 11 53 11 53 11 29 11 29 11 29 11 29 11 29 11 29 11 29 11 29 67 29 67 29 67 5 67 5 67 5 67 5 67 5 67 5 67 5 67 5 43 5 43 5 43 5 43 61 43 61 43 61 43 61 43 61 43 71	2 11 5 5 17 5 13 5 11 5 10 5 7 10 13 7 5 5 7 13 5 10 21 11 11 11 3 3 3 3 3 3 3 3 3 3 6 3 3 3 3 6 3 3 3 3 3 3 3 3 3 3 3 3 3 6 3 3 13 14 17 11	3 5 3 9 17 5 1 15 19 15 15 33 1 9 15 13 21 5 23 3 15 5 7 9 3 25 27 25 1 5 13 7 11 23 5 25 13 29 5 7 1 1 19 19 31 5 13 25 35 11 5 17 13 25 29 23 31 35 31 37 11 31 17 5 19 13 1 7 5 25 25 1 19 17 11 13 1 5 37 25 35 23 29 11 13 5 17 7 1 23 27 13 31 23	38 6 8 14 12 18 24 14 24 22 28 8 36 4 4 12 4 24 12 32 32 36 12 16 20 12 20 24 4 8 8 20 20 8 28 16 28 16 4 8 16 20 16 20 4 32 32 28 16 4 16 8 20 4 4 20 16 8 20 8 34 10 22 26 14 10 14 2 34 22 14 26 10 10 14 2 10 2 2 26 2 22 10 26 22 22 2 10 2 30 22 30 6 18	1 2905 1645 1 3 121 1933 1 577 553 2077 1 3457 697 2653 1 4609 841 2797 1 7489 1417 2941 1 8065 9217 10369 193 481 625 913 1201 1345 2065 2497 2641 385 673 817 1537 2257 2401 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 7345 1 5041 1 3889 1 2161 1 1585 1 1009 1 11665 1 9361 1 7057 1 6481 1 5905 1 4177 1 3025 1 2449 1 1297 1 145 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2581 1 1717 1 1573 1 1285 1 1141 1 997 1 565 1 421 1 277 1 133 1 2965 1 2245 1 2101 1 1813 2857 1381 2425 1237 2137 1093 1993 661 1705 85 1	1448 812 672 3 6 88 1052 736 104 288 1124 912 480 344 1340 1120 496 448 1428 1200 800 688 1436 1440 1032 1144 1184 216 392 320 792 592 816 1144 1192 1232 272 328 360 928 1384 1296 72 120 232 536 552 520 776 904 968 1112 104 136 280 568 616 680 840 1128 1176 1224 1752 184 488 328 568 568 808 952 1144 1320 1368 1400 968 1224 648 928 456 792 248 648 184 456 120 264 1456 200 1136 104 880 1352 832 1144 768 1080 560 888 400 776 272 552 144 344 16 312 1340 232 988 168 716 24 732 1456 588 1296 284 1104 300 1104 60 1024 1252 960 1044 608 964 608 612 416 436 288 404 80 276 16 148 1276 84 772 36 828 1428 612 1188 596 1188 540 1140 284 980 180 772 124 628 68 532 1472 260 1080 164 1048 20 856 1560 664 1088 608 1144 544 968 304 808 56 1392	1344 832 624 32 880 704 0 48 312 952 816 352 296 1312 1184 560 464 1368 1248 784 664 1504 1296 848 1328 1120 48 192 464 272 576 784 1024 1296 1200 144 448 512 816 1008 1168 0 112 224 384 368 624 640 912 976 1168 16 64 176 432 608 544 1072 1152 1296 1456 1248 192 208 240 496 688 944 784 1216 1168 1120 1328 856 1152 600 1016 504 864 280 720 200 480 120 240 1448 160 1192 112 872 1288 776 1064 696 984 472 888 344 712 328 648 168 424 8 312 1184 248 1104 168 832 24 608 1448 592 1320 288 1224 136 1080 80 1016 1216 936 1232 712 624 568 720 472 416 168 384 88 384 8 144 1304 96 944 16 744 1408 672 1392 544 1344 456 1008 280 896 240 752 152 704 64 480 1492 352 1164 128 1052 0 956 1320 716 1144 628 1080 548 1032 356 896 28 1488	1448 56 32 256 496 328 480 336 976 784 880 944 1360 96 208 336 368 640 624 1104 1216 1200 224 384 432 576 1136 1296 24 88 264 424 776 584 664 952 1176 1208 16 88 216 360 568 832 1000 1128 1048 1560 1512 216 280 392 696 648 968 888 1208 1448 1544 1528 1260 1100 764 780 444 268 140 44 1044 964 916 692 516 388 292 164 132 20 1380 1396 1172 996 804 884 644 548 276 180 4 1288 1248 1000 984 872	1568 64 64 248 352 424 576 432 960 696 944 1328 1344 96 224 272 480 544 704 912 1168 1360 160 320 416 704 1104 1024 16 64 272 416 464 544 1040 832 960 1072 48 144 240 368 640 960 1072 1040 1136 1360 1536 96 240 448 528 640 944 944 1328 1328 1408 1392 1328 1056 784 688 608 240 216 32 1216 1200 784 688 480 384 320 240 96 48 1280 1232 1216 912 1088 784 640 480 288 128 0 1544 1368 1048 1000 832	88 352 720 928 1208 1096 1312 56 200 224 472 464 624 1016 1128 1520 176 296 424 704 1096 1200 0 80 256 400 448 576 960 800 896 976 64 96 304 400 736 736 928 1152 1232 1408 1328 80 224 320 592 656 928 1136 1104 1216 1264 1520	32 464 608 896 1152 1024 1296 128 224 400 480 752 576 1168 1616 1296 256 416 384 912 1200 1248 16 96 160 624 432 640 816 1024 864 1184 48 160 272 368 576 800 1008 1024 1232 1392 1536 128 272 368 592 656 656 1104 1184 1040 1664 1520	112 480 624 1088 1120 1184 1280 48 224 272 448 576 656 992 1136 1360 128 208 384 736 992 1104 16 80 224 352 608 736 784 848 1040 976 24 80 240 352 720 776 832 896 1360 1456 1376 112 240 384 656 576 832 752 1296 1344 1344 1520	48 432 496 1008 1264 1184 1456 1264 880 784 704 432 352 304 96 1568 1424 1072 880 1072 768 480 368 336 144 16 1088 752 784 800 528 528 464 240 80 0 1264 1104 768 352 288 176 1392 1232 896 592 656 336 208 256 80 1424 1120 832

  According to our usual rules, variable x 1 is of the integers type (ve = 1), while variables y 1 and y 2 are of the prime numbers type (vp = 2). By direct evaluation, the cardinal matrix of this group, at the sequence p = 71, presents as follows :

	10.4. Example
	Let us have equation :
				c = x 1 .(4.x 1 6 +7.y 1 6 -6x 1 .y 1	5 + x 1 .y 1 .y 2 3 .(2x 1 -y 1 ))
	Difference with m 0,0	Cardinal matrix [M]
	(0)		9730	47390	47390	52710	47670	47670		47600	47740
	-4991		4739	53634	49175	48538	47593	48055		48125	48041
	-4991		4739	49175	52605	48076	49098	48055		48615	47537
	-4459		5271	48538	48076	52549	47593	48629		48594	48650
	-4963		4767	47593	49098	47593	53025	49112		48104	48608
	-4963		4767	48055	48055	48629	49112	53543		47600	48139
	-4970		4760	48125	48615	48594	48104	47600		53529	48573
	-4956		4774	48041	47537	48650	48608	48139		48573	53578
	Then :
		(Δcm i,0 )								λi		σgr i
		-4991	-4991	-4459	-4963	-4963		-4970	-4956	23,708		2984,030
		-4956	-4991	-4991	-4459	-4963		-4963	-4970	-32,742		6480,714
		-4970	-4956	-4991	-4991	-4459		-4963	-4963	-27,686		5174,330
	(1/7).	-4963 -4963	-4970 -4963	-4956 -4970	-4991 -4956	-4991 -4991		-4459 -4991	-4963 -4459	17,306 3,143	=	6002,175 4484,997
		-4459	-4963	-4963	-4970	-4956		-4991	-4991	15,499		6618,733
		-4991	-4459	-4963	-4963	-4970		-4956	-4991	-6,228		2548,021

11. Asymmetrical overlapping variables of any degrees 11.1. Generalities

		1	10	10	10	10	10	10	10
		1	3,387	-4,680	-3,960	2,472	0,449	2,214	-0,890
		1	-4,680	-3,960	2,472	0,449	2,214	-0,890	3,387
	1/2 =	1 1	-3,960 2,472	2,472 0,449	0,449 2,214	2,214 -0,890	-0,890 3,387	3,387 -4,680	-4,680 -3,960
		1	0,449	2,214	-0,890	3,387	-4,680	-3,96	2,472
		1	2,214	-0,890	3,387	-4,680	-3,96	2,472	0,449
		1	-0,890	3,387	-4,68	-3,96	2,472	0,449	2,214
	In the most general cases, d is equal to p-1 and cardinal matrix dimension will be p (in a study modulo p).
	The primitive roots equations remain of topicality. The matrix of environment [P] answers the traditional situation (with d
	= p-1) and cardinal matrix eigenvalues research is now standard. Hence :
				[M] = [m i,j ] = [P].[σ gr ][P -1 ]	(77)
	where
				(σ gr ) = [Δcm i,0 ].(λ)/d	(78)
	and
					d = p-1

  The interest of this second representation of the cardinal matrix is in what follows. Indeed, when we ar interested in the eigenvalues, we build :

	The reader can verify, with the matrix of environment underneath, that [M] = [P].[σ gr ][P -1 ] (here [P] = [P real ]+i.[P imag ]).
		(Δcmi,0)	λréel		λimag		σgr réel		σgr imag
		8 14 18 -2 16 -16 20 10 36 0 38 -12 10,625		-5,577		54,470		40,586
		-12 8 14 18 -2 16 -16 20 10 36 0 38	6,817		-9,876		-110,489		-0,409
		38 -12 8 14 18 -2 16 -16 20 10 36 0	-4,255		-11,220		12,466		18,704
		0 38 -12 8 14 18 -2 16 -16 20 10 36 -8,982		7,957		-31,663		-30,396
		36 0 38 -12 8 14 18 -2 16 -16 20 10	1,446		-11,913		28,561		-16,974
	(1/12).	10 36 0 38 -12 8 14 18 -2 16 -16 20 -11,651 20 10 36 0 38 -12 8 14 18 -2 16 -16 10,625	+i.	-2,872 5,577	=	-18,345 54,470	+i.	-1,335 -40,586
		-16 20 10 36 0 38 -12 8 14 18 -2 16	6,817		9,876		-110,489		0,409
		16 -16 20 10 36 0 38 -12 8 14 18 -2 -4,255		11,220		12,466		-18,704
		-2 16 -16 20 10 36 0 38 -12 8 14 18 -8,982		-7,957		-31,663		30,396
		18 -2 16 -16 20 10 36 0 38 -12 8 14	1,446		11,913		28,561		16,974
		14 18 -2 16 -16 20 10 36 0 38 -12 8 -11,651		2,872		-18,345		1,335

  table below gives a sample of troublemaking values t that arises modulo p 2 . Their values are apparently random or even anachronistic (case p = 11 and t = 7) :Let us however supplement this piece of study. If we follow the research to modulo p 3 trouble-making values, we observe a correlation with the previous rank of values :The correlation between (t(mod p 3 )-t(mod p 2 ))/p 2 values and sequence p remain however obscure. Trouble-making values modulo p 2 cardinals #(t) are, a priori, easy to speculate for variables of integers :

	p	5	7	11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83
	t(mod p 2 ) 7	11	7 160 97 61 60 329 109 1144 751 107 496 1411 975 2696 4250 1692 3565 5562 2695
	This forbids at this point the exact evaluation of cardinal matrices of x 1 2 +x 1 .x 2 +x 2 3 and y 1 2 +y 1 .y 2 +y 2	3 (which size is infinite)
	and corresponding eigenvalues.
	p			5	7	11	13	17	19	23	29	31
				57	158	1217	2019	1831	2588	10111	7057	8758
	t(mod p 3 )	=	=	=	=	=	=	=	=	=
				7+2.5 2	11+3.7 2	7+10.11 2	160+11.13 2	97+6.17 2	61+7.19 2	60+19.23 2	329+8.29 2	109+9.31 2
					p	#(t) prime numbers var., mod p 2	#(t) integers var., mod p 2
				2 or 3		/			/
				1 mod 4		p.(2p-2-rp)			2p.(p-1)
				3 mod 4		p.(2p+1-rp)			2p 2

  These values do not interest us because not stabilized (as we can observe it on initial tables). Stabilized values are, a priori, obtained modulo p 3 (for modulo p 2 trouble-making values t) in accordance with the following conjecture that is very close the previous one :

	p	5	7	11	13	17	19	23	29	31
	t(mod p 3 ) 57	158	1217	2019	1831	2588 10111 7057	8758
	#(t)/p 2	12	8	12	36	48	20	24	84	32
	That is (for not stabilized first instances, evaluation of stabilized cardinals is left to the motivated reader) :
	p		#(t) prime numbers var., mod p 3	#(t) integers var., mod p 3
	2 or 3			/				/
	1 mod 4		p 2 .(3p-3-rp)			3p 2 .(p-1)
	3 mod 4		p 2 .(p+1-rp)			p 2 .(p+1)
	p		#(t) prime numbers var., mod p 2	#(t) integers var., mod p 2
	2 or 3			/				/
	1 mod 4		p.(2p-3-rp)			p.(2p-3)
	3 mod 4		p.(2p+1-rp)			p.(2p+1)
	P 302/390						Quadratic abundance factors and substitutes

Trouble-making values modulo p 3 cardinals are again, a priori, easy to speculate for integers' variables with :

  table of the following type (for (r,s,u,v) = (384,160,416,2)) :

					Mode c	Direct evaluation	Deduction
						2 0	2 1	2 2	2 3	2 4	2 5	2 6	2 7	2 8	2 9 2 10 2 11 2 12 2 13 2 14 2 15 2 16 2 17 2 18 2 19 2 20
	r	s	u	v		16. fae(c)
	384 160 416 2		0	32	0	0 128 96 96 224 208 208 272 272 256 288 288 280 296 296 292 300 300
	r'	s'	u'	v'
	192 80 208 1
	64 16 16	1	48	16	0	0	64 48 48 112 104 104 136 136 128 144 144 140 148 148 146 150 150
		16 16	1	48	16	0	0	64 56 56 88 88 80 96 96 92 100 100 98 102 102
		16 16	1	24	16	8	8	40 40 32 48 48 44 52 52 50 54 54
		2	16	1	8	16	16	8	24 24 20 28 28 26 30 30
		2	2	1		16	16 12 20 20 18 22 22

  3 , not null abundance factors targets are 2 3r like. For x 2 , not null abundance factors targets are 2 2r .(1 mod 8) and 2 2r .(≠1 mod 8) like. When both variables are put together, nothing imposes a homogeneous behaviour within the class (≠ 1 mod 8). We must distinguish classes 5 mod 8, 3 mod 8 and 7 mod 8. Equal to 6 environment and 4 classes of targets cause loops including 24 items. Loops 1 and 2 are represented below :

	c	fae(c)	fan(c)	c	fae(c)	fan(c)	Mode loop2 / loop1	Global iterated ratio
	0	122880
	1 mod 8	40960	2,5	2 6 .(1 mod 8)	90112	5,5	3	2 -1
	5 mod 8	8192	0,5	2 6 .(5 mod 8)	24576	1,5	1	2 -1
	3 mod 8	8192	0,5	2 6 .(3 mod 8)	24576	1,5	1	2 -1
	7 mod 8	8192	0,5	2 6 .(7 mod 8)	24576	1,5	1	2 -1
	2.(1 mod 8)	8192	0,5	2 6 .2.(1 mod 8)	24576	1,5	1	2 -1
	2.(5 mod 8)	8192	0,5	2 6 .2.(5 mod 8)	24576	1,5	1	2 -1
	2.(3 mod 8)	8192	0,5	2 6 .2.(3 mod 8)	24576	1,5	1	2 -1
	2.(7 mod 8)	8192	0,5	2 6 .2.(7 mod 8)	24576	1,5	1	2 -1
	2 2 .(1 mod 8)	40960	2,5	2 6 .2 2 .(1 mod 8)	90112	5,5	3	2 -1
	2 2 .(5 mod 8)	8192	0,5	2 6 .2 2 .(5 mod 8)	24576	1,5	1	2 -1
	2 2 .(3 mod 8)	24576	1,5	2 6 .2 2 .(3 mod 8)	57344	3,5	2	2 -1
	2 2 .(7 mod 8)	24576	1,5	2 6 .2 2 .(7 mod 8)	57344	3,5	2	2 -1
	2 3 .(1 mod 8)	24576	1,5	2 6 .2 3 .(1 mod 8)	57344	3,5	2	2 -1
	2 3 .(5 mod 8)	24576	1,5	2 6 .2 3 .(5 mod 8)	57344	3,5	2	2 -1
	2 3 .(3 mod 8)	24576	1,5	2 6 .2 3 .(3 mod 8)	57344	3,5	2	2 -1
	2 3 .(7 mod 8)	24576	1,5	2 6 .2 3 .(7 mod 8)	57344	3,5	2	2 -1
	2 4 .(1 mod 8)	40960	2,5	2 6 .2 4 .(1 mod 8)	90112	5,5	3	2 -1
	2 4 .(5 mod 8)	40960	2,5	2 6 .2 4 .(5 mod 8)	90112	5,5	3	2 -1
	2 4 .(3 mod 8)	8192	0,5	2 6 .2 4 .(3 mod 8)	24576	1,5	1	2 -1
	2 4 .(7 mod 8)	8192	0,5	2 6 .2 4 .(7 mod 8)	24576	1,5	1	2 -1
	2 5 .(1 mod 8)	8192	0,5	2 6 .2 5 .(1 mod 8)	24576	1,5	1	2 -1
	2 5 .(5 mod 8)	8192	0,5	2 6 .2 5 .(5 mod 8)	24576	1,5	1	2 -1
	2 5 .(3 mod 8)	8192	0,5	2 6 .2 5 .(3 mod 8)	24576	1,5	1	2 -1
	2 5 .(7 mod 8)	8192	0,5	2 6 .2 5 .(7 mod 8)	24576	1,5	1	2 -1

3 710 369 067 405(x d -x d/2 )+c

	c	p i	d
	-12 250 397	271	8
	-12 250 397	71	4
	-12 250 397	71	2
	-11 755 519	281	8
	-11 755 519	97	4
	-11 755 519	41	2
	1 922 153	263	16
	1 922 153	229	8
	1 922 153	151	4
	1 922 153	59	2
	1 998 181	283	32
	1 998 181	239	16
	1 998 181	37	8
	1 998 181	37	2
	3 303 809	281	16
	3 303 809	71	4
	3 303 809	67	2
	The uncut sequences of 3 710 369 067 405(x d -x d/2 )+c are unchanged for exponents d higher than those of the table (we
	checked until d = 16384).

  This means very different limits according to factors dividing c. Let us illustrate the example with some targets and corresponding normalized factors until sequence 1500.

	Targets	Divisors 2	Divisors	Divisors
			1 mod 4	3 mod 4
	100001			11, 9091
	100002	2	2381	3, 7
	100003			100003
	100004	2 2		23, 1087
	100005		5, 113	3, 59
	100006	2	1613	31
	100007		97	1031
	100008	2 3		3 3 , 463
	100009		13, 157	7 2
	100010	2	5, 73, 137
	100011		17, 37, 53	3
	100012	2 2	2273	11
	.

  1/2 /Ln 2 (p), what we do not develop more here. lim #{x 1 2 +x 2 4 < p+c} = (1/2).(Γ(1/2).Γ(5/4)/Γ(7/4)).(4/7).(p+c) 7/4 /Ln(p+c) (26) mod p, we get by direct research the following table, which allows the normalized abundance factors evaluation (we can also use results of exercise 9) : In addition, we can form the following table :

			p →∞
	and
			lim #{x 1 2 +x 2	4 = p+c} = (1/2).(Γ(1/2).Γ(5/4)/Γ(7/4)).(Fan(c).p 3/4 /Ln(p)	(27)
			p →∞
	As Γ(1/2) = π 1/2 = 1.77245385091, Γ(5/4) ≈ 0.906402477055, Γ(7/4) ≈ 0.919062526849 :
							lim #{x 1 2 +x 2 4 < p+c} ≈ 0,5.p 7/4 /Ln(p)
							p →∞
	and
						lim #{x 1 2 +x 2 4 = p+c} ≈ 0,874.Fan(c).p 3/4 /Ln(p)
						p →∞
	For x 1	2 +x 2 4 = p+c Abundance factors at sequence p		Product of normalized abundance factors at sequence p
		p	c = 0		c = 1	c = 2	c = 3	c = 4	c = 0	c = 1	c = 2	c = 3	c = 4
		2	2		2		2		2	2	1,00000	1,00000	1,00000	1,00000	1,00000
		3	8		5		5		8	5	1,33333	0,83333	0,83333	1,33333	0,83333
		5	16		19		17		25	23	1,06667	0,79167	0,70833	1,66667	0,95833
		7	48		41		41		41	41	1,21905	0,77282	0,69147	1,62698	0,93552
		11	120		109	109	109	109	1,32987	0,76579	0,68518	1,61219	0,92701
		13	144		163	153	163	151	1,22757	0,80015	0,67201	1,68454	0,89730
		17	256		275	271	265	275	1,15536	0,80898	0,66953	1,64118	0,90720
		19	360		341	341	341	341	1,21617	0,80661	0,66758	1,63638	0,90454
		23	528		505	505	505	505	1,26905	0,80502	0,66626	1,63315	0,90276
		29	784		803	809	809	823	1,22529	0,79610	0,66380	1,62712	0,91499
		31	960		929	929	929	929	1,26481	0,79524	0,66308	1,62537	0,91400
		…	…		…		…		…	…	…	…	…	…	…
		157	24336	24515	24505	24471	24471	1,25318	0,80235	0,66257	1,62865	0,91722
			Number of solutions			Numerical evaluation			Variation
		c	p<1000	p<10000 p<100000 p<800000	p<1000 a=1,0750	p<10000 a=1,0974	p<100000 a=1,1212	p<800000 a=1,1055	p<1000 p<10000 p<100000 p<800000
		0	33	133		614	2406		30,0	130,2	599,5	2382,8	9,1%	2,1%	2,4%	1,0%
		1	13	78		368	1468		19,2	83,4	383,8	1525,6 -47,7% -6,9%	-4,3%	-3,9%
		2	15	64		301	1190		15,9	68,9	316,9	1259,8 -5,7% -7,6%	-5,3%	-5,9%
		3	41	168		791	3165		39,0	169,2	779,1	3096,7	4,9%	-0,7%	1,5%	2,2%
		4	24	104		444	1780		22,0	95,3	438,7	1744,0	8,5%	8,4%	1,2%	2,0%
	Sums all targets	547	2518	10009							0%	0%	0%	0%
	Max											9,10% 8,35%	2,37%	2,16%
	Min										-47,73% -7,58% -5,29% -5,87%
	Standard deviation											23,98% 6,62%	3,59%	3,71%

  table :The results are overall in conformity with awaited values even if local variations are noted. Coefficient a is close to the asymptotic value 1. The standard deviations decrease regularly with the increase of p max.

				Number of solutions	Numerical evaluation		Variation
	c	fan	p<100	p<1000	p<10000	p<100 a = 1,08224	p<1000 a = 1,05524	p<10000 a = 1,03104	p<100	p<1000	p<10000
	9	0,79361	11	80	1006	10,7		94,8	950,9	-2,58%	18,48%	-5,47%
	10	0,66358	9	71	772	9,0		79,3	795,1	-0,44%	11,62%	3,00%
	11	1,49664	16	170	1771	20,2		178,7	1793,3	26,31%	5,14%	1,26%
	12	0,93317	17	141	1182	12,6		111,4	1118,2	-25,88%	-20,96%	-5,40%
	13	0,74541	8	78	824	10,1		89,0	893,2	25,82%	14,13%	8,40%
	14	1,00979	14	105	1200	13,6		120,6	1210,0	-2,61%	14,86%	0,83%
	15	0,86043	12	124	1060	11,6		102,8	1031,0	-3,18%	-17,13%	-2,74%
	Sums of all targets		206	1822	18280	206		1822	18280
	Max									-25,88%	-20,96%	-7,04%
	Min									90,47%	18,48%	8,40%
	Standard
	deviation									30,43%	11,28%	4,19%
	Numerical application 2
	Let us have to solve :
					2x 1 2 +3x 2 3 +5x 3	4 = p 2 +c
				Number of solutions	Numerical evaluation		Variation
	c	fan	p<100	p<1000	p<10000	p<100 a = 1,08224	p<1000 a = 1,05524	p<10000 a = 1,03104	p<100	p<1000	p<10000
	0	0,64235	8	75	828	8,7		76,7	769,7	8,42%	2,29%	-7,04%
	1	0,86051	15	110	1042	11,6		102,8	1031,1	-22,54%	-6,57%	-1,05%
	2	1,35289	22	159	1576	18,3		161,6	1621,1	-16,96%	1,62%	2,86%
	3	0,84634	6	98	1015	11,4		101,1	1014,1	90,47%	3,14%	-0,09%
	4	0,74734	12	94	893	10,1		89,3	895,5	-15,91%	-5,05%	0,28%
	5	1,40842	19	171	1735	19,0		168,2	1687,6	0,09%	-1,63%	-2,73%
	6	0,77951	14	95	913	10,5		93,1	934,0	-24,82%	-2,00%	2,30%
	7	0,65710	6	69	740	8,9		78,5	787,4	47,88%	13,73%	6,40%
	8	1,45874	17	182	1723	19,7		174,2	1747,9	15,87%	-4,28%	1,45%

P 367/390 Generation of prime numbers by Fermat-Catalan sums and analogues

  ,1638 0,1753 0,6194 1,5997 0,1487 2,1891 2,1077 0,1549 2,2107 0,7190 0,1518 1,8548 0,7196 0,5309 1,9263The products of the normalized factors being relatively stabilized at the sequence p = 73, we can use these values for the numerical application. We have then the following table :

						1,6264 0,1489 2,2295 2,0256 0,1560 2,2857 0,7163 0,1502 1,8559 0,6984 0,5385 1,8974
	17	0,6667 2,1538 0,1807 0,6447 1,6264 0,1489 2,2295 2,0256 0,1560 2,2857 0,7163 0,1502 1,8559 0,6984 0,5385 1,8974
	19	0,6667 2,1538 0,1754 0,6258 1,5978 0,1515 2,1904 2,0256 0,1569 2,2456 0,7331 0,1502 1,8668 0,7147 0,5227 1,9418
	23	0,6667 2,1538 0,1754 0,6258 1,5978 0,1515 2,1904 2,0256 0,1569 2,2456 0,7331 0,1502 1,8668 0,7147 0,5227 1,9418
	29	0,6667 2,1538 0,1754 0,6258 1,5978 0,1515 2,1904 2,0256 0,1569 2,2456 0,7331 0,1502 1,8668 0,7147 0,5227 1,9418
	…	…	…	…	…	…	…	…	…	…	… …	…	…	…	…	…
	73	0,6667 2Number of solutions		Numerical evaluation			Variation
	c		Fan(c)	p<100		p<1000	p<8000		p<100 a = 1,0908	p<1000 a = 1,0436	p<8000 a = 1,0278		p<100	p<1000		p<8000
	0		0,667			61	620		9,3	76,2		623,1		-28,53%	24,84%		0,50%
	1		2,164			253	2121		30,2	247,2	2022,4		-16,23%	-2,31%		-4,65%
	2		0,175	4		21	183		2,4	20,0		163,8		-38,92%	-4,65%		-10,47%
	3		0,619	5		57	520		8,6	70,7		578,9		72,64%	24,12%		11,33%
	4		1,600			192	1511		22,3	182,7	1495,2		11,47%	-4,83%		-1,05%
	5		0,149	2		12	132		2,1	17,0		139,0		3,62%	41,54%		5,29%
	6		2,189			252	2013		30,5	250,0	2046,0		12,99%	-0,78%		1,64%
	7		2,108			270	2005		29,4	240,7	1969,9		-18,41%	-10,83%		-1,75%
	8		0,155	2		19	161		2,2	17,7		144,7		7,92%	-6,89%		-10,09%
	9		2,211			259	2060		30,8	252,5	2066,3		46,72%	-2,50%		0,30%
	10		0,719			90	660		10,0	82,1		672,0		-37,38%	-8,75%		1,81%
	11		0,152	2		17	132		2,1	17,3		141,9		5,79%	2,01%		7,50%
	12		1,855			206	1721		25,9	211,9	1733,6		-7,68%	2,85%		0,73%
	13		0,720	3		67	647		10,0	82,2		672,5		234,28%	22,67%		3,95%
	14		0,531	8		76	538		7,4	60,6		496,2		-7,51%	-20,21%		-7,77%
	15		1,926			197	1742		26,8	220,0	1800,4		-0,57%	11,69%		3,35%
	Sums of all targets		250		2049	10346		250	2049	10346
	Max													234,28%	41,54%		11,33%
	Min													-38,92%	-20,21%		-10,47%
	Standard deviation												65,11%	16,34%		6,02%

1 1 -2 -3 1 -2 1 -3 -2 1 1

+u.x 3 3 +v.x

 3 = c mod 2 δ . As usually, the abundance factors depend on the multiplicity of the sequence (here 2) in target c. The stabilized values of the factors are got for δ greater than some value. The following table represents the abundance factors (multiplied by 16) for the proposed equation for a few specific cases :

1 1 0,5 0,33333 0,25 0,2 0,16667 0,14286 0,125 0,11111 2 1 0,78540 0,66667 0,58905 0,53333 0,49087 0,45714 0,42951 0,40635 3 1 0,88332 0,80613 0,75 0,70666 0,67178 0,64286 0,61832 0,59714 4 1 0,92704 0,87402 0,83304 0,8 0,77253 0,74916 0,72891 0,71111

4,878 2,137 -2,222 2,952 -1,742 0,215 -0,724 0,153 -2,830 -2,090 1 4,878 2,137 -2,222 2,952 -1,742 0,215 -0,724 0,153 -2,830 -2,090 -1,727 1 2,137 -2,222 2,952 -1,742 0,215 -0,724 0,153 -2,830 -2,090 -1,727 4,878 1 -2,222 2,952 -1,742 0,215 -0,724 0,153 -2,830 -2,090 -1,727 4,878 2,137 1 2,952 -1,742 0,215 -0,724 0,153 -2,830 -2,090 -1,727 4,878 2,137 -2,222 1 -1,742 0,215 -0,724 0,153 -2,830 -2,090 -1,727 4,878 2,137 -2,222 2,952 1 0,215 -0,724 0,153 -2,830 -2,090 -1,727 4,878 2,137 -2,222 2,952 -1,742 1 -0,724 0,153 -2,830 -2,090 -1,727 4,878 2,137 -2,222 2,952 -1,742 0,215 1 0,153 -2,830 -2,090 -1,727 4,878 2,137 -2,222 2,952 -1,742 0,215 -0,724 1 -2,830 -2,090 -1,727 4,878 2,137 -2,222 2,952 -1,742 0,215 -0,724 0,153 1 -2,090 -1,727 4,878 2,137 -2,222 2,952 -1,742 0,215 -0,724 0,153 -2,830 Thus with (p-1)².p = 12.12.13 = 1872 : P 299/390 Quadratic abundance factors and substitutes (σgr) = 1872 54,470+i.40,586 -110,489-i.0,409 12,466+i.18,704 -31,663-i.30,396 28,561-i.16,974 -18,345-i.1,335 54,470-i.40,586 -110,489+i.0,409 12,466-i.18,704 -31,663+i.30,396 28,561+i.16,974 -18,345+i.1,335

EXERCISE 3 : PRIMITIVE ROOTS AND RESIDUES

The asymptotic enumeration of Waring sums cannot be done without studying first the simplest equation :

x n = c mod p δ With this intention, an examination of primitive roots is essential. The reader familiar of these objects will be able of course to fly over the following paragraphs. For our part, having found few elements on the subject in the mathematical literature, we built the exercise primarily by our own means. We entitle "lemmas" results which we wanted to highlight and who are possibly new. These lemmas are essential to the study of the cases with multiple variables x 1 -1 Let us pose it = int(i/n), m = i mod n. So, we get for line c\2 i and c∤2 i+1 : # c = 2 k.δ +(2 1 +2 2 +…+2 n-1 ).2 k.(δ-1) +(2 1 +2 2 +…+2 n-1 ).2 n .2 k.(δ-2) +…+(2 1 +2 2 +…+2 n-1 ).2 (it-1).n .2 k. (δ-it) +r Here the expression of r depends on its congruence relative to n : if i = 0 mod n r = 0 if i ≠ 0 mod n r = -2 (i+1-i mod n) .2 it.(n-k) .2 k. (δ-1) Details with r imports us however little, since we seek results for δ → +∞. Then : # c = 2 k.δ .(1+ (2 n -2)/2 n .(2 n-k +2 2(n-k) +…+2 it.(n-k) )+(2 (i mod n)+1 -2).2 it.(n-k) /2 k )

So that for normalized factors :

P 168/390

Modulo p Δ abundance factors correction Fan(c,p,δ→∞) = 1+(1-1/2 n-1 ).(2 n-k +2 2.(n-k) +…+2 it.(n-k) )+(2 (i mod n)+1 -2).2 it.(n-k) /2 k Thus 2 i \c and 2 i+1 ∤c  Fan(c,2,δ→∞)) = 1+if(n=k,it.(1-1/2 n-1 ),((2 n-1 -1)/2 k-1 ). (2 (n-k).it -1)/(2 (n-k) -1))+(2 (i mod n)+1 -2).2 it.(n-k) /2 k When n = 1, we get self-evidently without condition on c : Fan(c,p,δ→∞) = 1 = Fan(c,p,δ=1) If c is not multiple of 2 n , we get it = int(i/n) = 0 and : 2 n ∤c  Fan(c,2,δ→∞) = 1 = Fan (c,2,δ=1) This last result is one that we seek to establish in priority since it shows that "often" correction mod p δ is not even to be considered.

Case n = 2

This cases makes exception to other even n cases. However, to reduce the text, we will suppose that the result is in conformity with general result of n odd cases.

Case n even

All the elements were reviewed higher to get abundance factors in this case. For as much, abundance factors general evaluation remains for us a difficult task. We will admit here, when taking everything into account, ultimate expression is reduced to the same expression as in odd n cases.

So that : 2 i \c and 2 i+1 ∤c  Fan(c,2,δ→∞) = 1+if(n=k,it.(1-1/2 n-1 ),((2 n-1 -1)/2 k-1 ). (2 (n-k).it -1)/(2 (n-k) -1))-2 i+1-i mod n .2 it.(n-k) /2 k For k < n, a correction is essential. For k = n, it is necessary. For k > n, we are in a correction which becomes negligible with increase of k.

Evaluation of abundance factors in Waring integers' variables enumeration p>2, p∤ n 10.2.1. Generalities

We are interested in cases where δ tends towards infinite. After the study of various cardinal matrices cases and ways of calculating abundance factors, in particular the necessity for recourse to normalization, we now examine concrete evaluation of these coefficients. We carry out this calculation per blocks as we did in case δ = 1 at exercise 5). We have got eigenvalues and eigenvectors expressions when p > 2. We use [A δ ] k = [P A(δ) ].[μ δ ] k .[P -1 A(δ) ] [AN δ ] = (1/p δ ). [A δ ] to write normalized abundance factors (here [p δ ] is column vector (p δ , 0, 0, …, 0)) :

Fan([c]=[p i.n+j .g r .g u.d ],p,δ) = [AN δ ] k [p δ ] = 1/p k.δ .[P A(δ) ].[μ' δ ] k .[P -1 A(δ) ].[p δ ] = [P A(δ) ].[μ' δ /p δ ] k .[P - 1 A(δ) ].[p δ ]

We are interested here in a set of targets whose generic form is p i.n+i .g r .g u.d , i and j being obtained by Euclidean division by n (exponent of x k ). Here [p δ ] is column vector [p δ , 0, 0,…, 0]. We deal here for a given target ([c]) with the product of a line vector resulting from [P A(δ) ], of a trace vector resulting from[μ' δ /p δ ] with exponent k and of a column vector resulting from [P - 1 A(δ) ].[p δ ].

Case c ≠ 0

We give successively these three expressions which multiply "column with column" while starting with trace vector. We found previously eigenvalues [μ' δ ] of matrix A :

Eigenvalues sum

The sum of environment matrix eigenvalues is null (relation 16.1 of exercise 5). It follows immediately by the average property that the sum of Fermat-Catalan eigenvalues is null.

∑ λ i,d,cm = 0 (26) i = 0 to d-1

11. Method of principal diagonal.

The eigenvalues are supposed to be known here.

Cardinal matrices evaluation starting from the eigenvalues requires a well given order. For an environment cm, there are cm! combinations of these eigenvalues where only cm.φ(d) are adapted. For a compound environment cm = ∏ p i k , these combinations are thus very rare. However, a good arrangement among others can be found with method without examining all the possible combinations. The principle is successively to create two-dimensional tables. The eigenvalues of the environment matrix p i 2 are classified thanks to those of the environment p i j and p i , j varying from 1 to k-1 giving the eigenvalues in a correct order within the matrix of environment p i k . When all the eigenvalues associated with environment p i k are thus classified in a good order, we continue the process with p i1 k1 and p i2 k2 (p i1 ≠ p i2 ), then p i1 k1 .p i2 k2 and p i3 k3 (p i3 ≠ p i1 , p i3 ≠ p i2 ), … until obtaining a good order of eigenvalues associated with environment ∏ p i k .

The method of the principal diagonal is as follows. For two environments d 1 and d 2 , whose eigenvalues are classified in a good order (thus in such way that [A] ] d1 and that [A] d2 = [P B ] d2 . [DA] d2 .[P B -1 ] d2 ), we form a twodimensional table with, on the first axis, the eigenvalues in the order of environment d 1 matrix, on the second axis, the eigenvalues in a good order of environment d 2 matrix. Table itself is supplemented with environment d 1 .d 2 matrix eigenvalues, the averages according to lines and columns having respectively to give the eigenvalues placed on the axis. The placement of the eigenvalues of environment d 1 .d 2 matrix on the two-dimensional table is carried out by initially seeking a good combination along the first axis, then along the second axis. The eigenvalues associated with product d 1 .d 2 environment are obtained while following the principal diagonal of the table. When we leave the table at position n, we start again at position n+1 at parallel diagonal with principal diagonal on top of table. When (d 1 ,d 2 ) = 1, after d 1 .d 2 steps, we describe the whole table only once and arise on principal diagonal. We then have the sought environment matrix eigenvalues in a good order. When (d 1 ,d 2 ) ≠ 1, when we arise on principal diagonal, after d 1 .d 2 /(d 1 ,d 2 ) steps, on bottom of table and we take again the process with a shift of one column. After (d 1 ,d 2 ) shifts, we will have a priori eigenvalues in a good order.

Example : d = 12

An example is treated for sequence p = 193. The standard cardinal matrix is : Eigenvalues of the matrix are roughly (without taking a priori their order in account) : 193, 58.723704, 54.48156, -53.31949, 17.898724, -50.92077, 57.09326, -5.542397, -88.18578, -7.964582, 2.3828543, -24.33113, 39.684041 The standard cardinal matrix for d 1 = 4 is : Eigenvalues of the matrix, in an adapted order, i.e. such as [A] 193, -0.053882, 37.985892, -27.73101, -10.201004 The standard cardinal matrix for d 2 = 3 is:

Fermat-Catalan abundance factors 16. Environment cm = 6.

Problem exposition

Let us have to study for example : x 1 2 +x 2 3 +x 3 6 = c The environment of this case is cm = 6. The matrices to be studied relate to the divisors of 6, i.e. d = 1, d = 2, d = 3 and d = 6. At sequence p, we have cm = lcm ((p-1,6) ,2) ,3),6). Thus, we have three principal cases and two lower cases as we will see further : The cardinal matrix immediately shows the equality between abundance factors :

For p = 5 mod 12

We recover the cardinal matrix corresponding to d = 2 and p = 1 mod 4 for monomials x 2 ([M1]) and x 6 ([M3]) and d = 1 for monomial x 3 ([M2]) which we only need to multiply by (1,0,0) to get the abundance factors :

For p = 11 mod 12

The principle is identical. We recover the cardinal matrix corresponding to d = 2 and p = 3 mod 4 for the monomials x 2 ([M1]) and x 6 ([M3]) and d = 1 for the monomial x 3 ([M2]) which we only need to multiply by (1,0,0) :

1 p-1 p-1 0 = p 2 #{g.g 2u } 0 (p+1)/2 (p-1)/2 1 p-1 p-1 0 p 2

For p = 1 mod 12

We must consider respectively three matrices for x 2 [M1], x 3 [M2] and x 6 [M3].

We use all of the symmetry properties resulting from the relation g x-2 = g u.d + g y-2 .g v.cm mod p while starting with [M3] which is the "standard" matrix of environment 6. We have for [M3] :

We find all calculations made : y = 0 mod 6 y = 2 mod 6 y = 4 mod 6 Δx 1 (3) (-2z+2)/3 (-3y+z+2)/3 (3y+z+2)/3 Δx 2 (3) (3y+z-1)/3 -(2z+1)/3 (-3y+z-1)/3 Δx 3 (3) (-3y+z-1)/3 (3y+z-1)/3 -(2z+1)/3 and : y = 0 mod 6 y = 2 mod 6 y = 4 mod 6 Δz 1 -(4z+5)/3 -(z+5)/3 -(z+5)/3 Δz 2 (9y+z-1)/6 (3y+z-1)/6 (-3y+z-1)/6 Δz 3 (3y+z-1)/6 (3y+z-1)/6 (3y-5z-1)/6 Δz 4

(-3y+2z-2)/3 (-3y+2z-2)/3 (-3y-z-2)/3

These elements make it possible to completely characterize matrix [M2] components as functions of y and z (with a choice of g such as z 2 -z 6 (equivalent to z 2 -z 3 ) is sign of y) :

Components \ Conditions y = 0 mod 6 y = 2 mod 6 y = 4 mod 6 z 1 +1

(p-1)/6-2(2z+1)/3 (p-1)/6-(z+2)/3 (p-1)/6-(z+2)/3 z 2 (p-1)/6+3y/2+(z-1)/6 (p-1)/6+y/2+(z-1)/6 (p-1)/6-y/2+(z-1)/6 z 3

(p-1)/6+y/2+(z-1)/6 (p-1)/6+y/2+(z-1)/6 (p-1)/6+y/2-(5z+1)/6 x 1(3) -3-z 1 (p-1)/6+2(z-1)/3 (p-1)/6-y+2(z-1)/3 (p-1)/6+y+2(z-1)/3 x 2(3) -z 2 (p-1)/6-y/2+(z-1)/6 (p-1)/6-y/2-(5z+1)/6 (p-1)/6-3y+(z-1)/6 z 6 = x 3(3) -z 3 (p-1)/6-3y/2+(z-1)/6 (p-1)/6+y/2+(z-1)/6 (p-1)/6-y/2+(z-1)/6 z 4 +1

(p-1)/6-y/2+(2z+1)/3 (p-1)/6-y+(2z+1)/3 (p-1)/6-y-(z-1)/3 (p-1)/2-x 3( 3 We see that the expressions of [M2] components are linear in y and z, expressions varying according to congruence modulo 3 of y/2. It is the same for matrix [M3] since product [P B ]. [D [M3] ].[P B -1 ] (=[M3)] calls to the same expressions, in particular ∑ a i 3 , ∑ a i 2 .a i+1 and ∑ a i

Systematic research of constructive of sequence matrices by increment

An incremental systematic research by tests and errors of the constructive of sequence matrices is simple by the principle but difficult in the application because the number of matrices to be tested increases exponentially with d. When d is odd, it is more practical to use the eigenvalues (see preceding paragraph). When d is even, other alternatives are to be sought.

Property of cancellation

Per heritage of the cancellation property (relation 51 of this exercise) of environment founding matrix, the sum of n positions spaced components of a constructive of sequence matrix is null when n is a divisor of environment d. Thus (proof still to be established) : ∑ (s' i,d ) = 0 (75) i = c mod n n \ d

Case d = 6

It is about the simplest case to be considered and, indeed, it is so. The elements of the first column of [t 6 ], environment founding matrix, are :

We have d = 3.5 avec (3,5) = 1. The multiplicative factor to pass from d = 3 to d = 15 is f = 5. We use column vector (ns' 3 ) = (29, -16, -13) in the form :

-(f-1). (ns '5) (ns'5) The sum of each line is null. We then choose any step relative prime with d = 3.5, for example step = 2, and we pick one after the other the met numbers while following the principal diagonal starting with the first element of preceding matrix (in fact any diagonal, right or left, and any starting point can be selected) in a modulo 5 in column and modulo 3 on line movement. We thus get the vector column (ns' 15 ) = (-116, -13, -16, 29, -13, 64, 29, -13, -16, 29, 52, -16, 29, -13, -16) and verify that :

-116 -13 -16 29 -13 64 29 -13 -16 29 52 -16 29 -13 -16 2 8 1 1 -2 1 -4 -2 1 1 -2 -4 1 -2 1 1 - 13 -16 29 -13 64 29 -13 -16 29 52 -16 29 -13 -16 -116 1 8 1 1 -2 1 -4 -2 1 1 -2 -4 1 -2 1 -16 29 -13 64 29 -13 -16 29 52 -16 29 -13 -16 -116 -13 1 1 8 1 1 -2 1 -4 -2 1 1 -2 -4 1 -2 29 -13 64 29 - 13 -16 29 52 -16 29 -13 -16 -116 -13 -16 -2 1 1 8 1 1 -2 1 -4 -2 1 1 -2 -4 1 -13 64 29 - 13 -16 29 52 -16 29 -13 -16 -116 -13 -16 29 1 -2 1 1 8 1 1 -2 1 -4 -2 1 1 -2 -4 64 29 -13 -16 29 52 -16 29 -13 -16 -116 -13 -16 29 -13 -4 1 -2 1 1 8 1 1 -2 1 -4 -2 1 1 -2 29 - 13 -16 29 52 -16 29 -13 -16 -116 -13 -16 29 -13 64 -2 -4 1 -2 1 1 8 1 1 -2 1 -4 -2 1 1 -13 -16 29 52 -16 29 -13 -16 -116 -13 -16 29 -13 64 29 =15.211. 1 -2 -4 1 -2 1 1 8 1 1 -2 1 -4 -2 1 -16 29 52 -16 29 -13 -16 -116 -13 -16 29 -13 64 29 -13 1 1 -2 -4 1 -2 1 1 8 1 1 -2 1 -4 -2 29 52 -16 29 -13 -16 -116 -13 -16 29 -13 64 29 -13 -16 -2 1 1 -2 -4 1 -2 1 1 8 1 1 -2 1 -4 52 -16 29 - 13 -16 -116 -13 -16 29 -13 64 29 -13 -16 29 -4 -2 1 1 -2 -4 1 -2 1 1 8 1 1 -2 1 -16 29 - 13 -16 -116 -13 -16 29 -13 64 29 -13 -16 29 13 -16 -116 -13 -16 29 -13 64 29 -13 -16 29 -13 -16 -116 -13 -16 29 -13 64 29 -13 -16 29 -16 -116 -13 -16 29 -13 64 29 -13 -16 29 52 -16

Example 2 : d = 105, p = 211

We have d = 3.5.7 = 15.7 with (15,7) = 1. We can use previously found matrix [ns' 15 ] and we have for the multiplicative factor -(f-1) = 6. We use column vector (ns' 15 ) = (-116, -13, -16, 29, -13, 64, 29, -13, -16, 29, 52, -16, 29, -13, -16) to form the table :   -

We seek a solution among others for this system of equations, for example :

The reader will easily verify that these expressions are actually integers with the conditions of definition of a and b for the six preceding families.

Detail of example d = 12, p = 193

For [s' d ] :

-30 -1 -37 50 39 47 26 -49 -9 -46 11 -1 -30 -1 -37 50 39 47 26 -49 -9 -46 11 -1 6 1 1 -2 -3 1 -2 1 -3 -2 1 1 -1 -37 50 39 47 26 -49 -9 -46 11 -1 -30 -1 -37 50 39 47 26 -49 -9 -46 11 -1 -30 -37 50 39 47 26 -49 -9 -46 11 -1 -30 -1 -37 50 39 47 26 -49 -9 -46 11 -1 -30 50 39 47 26 -49 -9 -46 11 -1 -30 -1 -37 50 39 47 26 -49 -9 -46 11 -1 -30 -1 -37 -2 1 1 6 1 1 -2 -3 1 -2 1 -3 39 47 26 -49 -9 -46 11 -1 -30 -1 -37 50 39 47 26 -49 -9 -46 11 -1 -30 -1 -37 47 26 -49 -9 -46 11 -1 -30 -1 -37 50 39 47 26 -49 -9 -46 11 -1 -30 -1 -37 50 39 1 -3 -2 1 1 6 1 1 -2 -3 1 -2 26 -49 -9 -46 11 -1 -30 -1 -37 50 39 47 . 26 -49 -9 -46 11 -1 -30 -1 -37 50 39 47 =12.193 .

- 23 -49 -9 -46 11 -1 -30 -1 -37 50 39 47 26 -49 -9 -46 11 -1 -30 -1 -37 50 39 47 26 1 -2 1 -3 -2 1 1 6 1 1 -2 -3 -9 -46 11 -1 -30 -1 -37 50 39 47 26 -49 -9 -46 11 -1 -30 -1 -37 50 39 47 26 -49 -3 1 -2 1 -3 -2 1 1 6 1 1 -2 -46 11 -1 -30 -1 -37 50 39 47 26 -49 -9 -46 11 -1 -30 -1 -37 50 39 47 26 -49 -9 -2 -3 1 -2 1 -3 -2 1 1 6 1 1 11 -1 -30 -1 -37 50 39 47 26 -49 -9 -46 11 -1 -30 -1 -37 50 39 47 26 -49 -9 -46 1 -2 -3 1 -2 1 -3 -2 1 1 6 1 -1 -30 -1 -37 50 39 47 26 -49 -9 -46 11 -1 -30 -1 -37 50 39 47 26 -49 -9 -46 11

For [r d ] (second matrix) : 3 1 -2 1 -3 -2 1 1 6 1 1 -2 -8 0 0 0 -7 5 -8 7 0 5 0 0 39 50 -37 -1 -30 - 1 11 -46 -9 -49 26 47 1 -2 1 -3 -2 1 1 6 1 1 -2 -3 0 -8 0 0 0 -7 5 -8 7 0 5 0 47 39 50 -37 -1 -30 - 1 11 -46 -9 -49 26 -2 1 -3 -2 1 1 6 1 1 -2 -3 1 . 0 0 -8 0 0 0 -7 5 -8 7 0 5 = 26 47 39 50 -37 -1 -30 -1 11 -46 -9 -49 1 -3 -2 1 1 6 1 1 -2 -3 1 -2 5 0 0 -8 0 0 0 -7 5 -8 7 0 -49 26 47 39 50 -37 -1 -30 -1 11 -46 -9 -3 - 2 1 1 6 1 1 -2 -3 1 -2 1 0 5 0 0 -8 0 0 0 -7 5 -8 7 -9 -49 26 47 39 50 -37 -1 -30 -1 11 -46 - 2 1 1 6 1 1 -2 -3 1 -2 1 -3 7 0 5 0 0 -8 0 0 0 -7 5 -8 -46 -9 -49 26 47 39 50 -37 -1 -30 -1 11 1211 -46 -9 -49 26 47 39 50 -37 -1 -30 -1 125 -8 7 0 5 0 0 -8 0 0 0 -7 -1 11 -46 -9 -49 26 47 39 50 -37 -1 -30

The reader will be able to check that with these values -1 ] equals (with an excellent precision when 14 decimals after the comma are used) :

EXERCISE 10 : QUADRATIC ABUNDANCE FACTORS AND SUBSTITUTES

Quadratic equations are here of the type c = u.x 2 +v.x.y+w.y 2 . These cases are important because they have "overlapping variables". They open the field on more general cases which we will consider too. The abundance factors of this type of equations are obtained by two-dimensional tables method, but the two variables x and y must be worked out simultaneously.

Abundance factors modulo p of a quadratic equation with unit coefficients

Let us have p a prime number. Let us pose : p = x 2 +x.y+y 2 (1)

A two-dimensional table of x and y modulo 6 values shows that p can only be 0, 1, 2, 3 or 4 mod 6. The number p being a prime number, it remains p = 2, p = 3 or p = 1 mod 6 (with uniqueness of the couple (x,y) conjecture for given p except self-evident permutation of x and y). This distinction between modulo classes is met also during the evaluation of abundance factors. Foot-note : p = 2 is a peculiar case as there is no primitive root and the instances for c summarizes to c = 0 mod 2 and c ≠ 0 mod 2 (c = 1 mod 2).

We use the method employed at the beginning of exercise 7 to count asymptotically (q being a prime numbers variable)

The corresponding equations are of the type :

x 2 +x.y+y 2 = q+c mod p (4)

In the case of integers' variables, it suffices to withdraw to p 2 the cardinals to the initial table. In the case of prime numbers variables, we withdraw to (p-1) 2 the cardinals to the initial table. Thus :

Prime numbers variables (x,y) #(c = 0 mod p) #(c = g 0 .g 2u mod p) #(c = g Here q is a prime numbers' variable and x and y are variables either of integers, or of prime numbers.

The asymptotic enumeration of prime numbers solutions by an expression u.x 2 +v.x.y+w.y 2 requires the study of : u.x 2 +v.x.y+w.y 2 = q+c mod p (23)

As we proved at the beginning of exercise 7, in the case of integers' variables, it suffices to withdraw to p 2 the cardinals of the initial table. In the case of prime numbers' variables, it suffices to withdraw to ((p-1) 2 the cardinals to the initial table. Hence :

Variables (x,y) of integers

Inventory of the classes

We observe the following cases for equation c = x 2 mod p :

When the target is formed of more than two relative primes' factors, the last relation is put in the form :

This writing means that G, respectively H, are the union of the set of the t combinations of classes at the same time of type G i and H j provided that the number of selected H j families is even, respectively odd.

The cases deduced from the decomposition c = α.β with gcd(α,β) = 1 (using in particular β = -1) are :

The first table calls for several observations concerning the use of the operators of union (U), of intersection (∩) and the term with indice g 4c .

The intersection of two families of numbers G(α) modulo q and G(β) modulo r is obtained as follows. Let us have t the greatest common multiple of q and r. We then express G(α) and G(β) in an equivalent way modulo t. The intersection is obtained by the choice of the elements common to G(α) modulo t and G(β) modulo t. The union is carried out on the same model except that all the elements modulo t are taken into account.

The table below illustrates the point.

P 251/390

Quadratic abundance factors and substitutes 3 11,13,23,25,35,37,47,49,59} mod 60 H(3) = {5,7} mod 12 = {5,7,17,19,29,31,41,43,53,55} mod 60 5 G(5) = {1,9} mod 10 = {1, 9,11,19,21,29,31,39,41,49,51,59} mod 60 H(5) = {3,7} mod 10 = {3, 7,13,17,23,27,33,37,43,47,53,57} mod 7,11,17,43,49,53,59} mod 60 Let us explain now the meaning of g 4c . We used until now notation g for a prime number p primitive root. The property of primitive roots is to generate by exponentiation the whole set of non-null classes modulo p (1 to p-1). In the case of g 4c , where c is a prime number, we are interested by a behaviour modulo 4c, namely (c prime number replaced by letter p) the values of g 4p i mod 4p, i = 1 to 4p-1. The primitive roots modulo 4p are those which generate p-1 distinct classes including 1 and 4p-1.

Primitive roots modulo 4p

Definition

The set of the primitive roots modulo 4p is the set of the primitive roots modulo p carried to 3 modulo 4 by successive additions of p (mod 4p).

Properties

The set {g 4p i } mod 4p is a cyclic group of order p-1. We have g 4p p-1 = 1mod 4p. If p = 1 mod 4 then g 4p (p-1)/2 = -1+2p mod 4p. If p = 3 mod 4 then g 4p (p-1)/2 = -1 mod 4p. Only primitive roots modulo 4p have all of these properties. For each preceding numbers {3,11,43,75,79,83,115,55}, we have well g 4p (p-1)/2 = 4p-1 (but not for example for p = 7) and g 4p p-1 = 1 when p = 31. The even numbers of this table cannot, self-evidently, generate 1 mod 4p. The numbers n = {65,73,105,13,17,21,53,117} are such as n p-1 = 1, but n (p-1)/2 = 2p-1 (for p = 31). In addition among {1,3,5,…,123}, we also find the family n = {7,19,51,59,71,103,107,111} such as n p-1 = 1, but n (p-1)/2 = 2p+1. After ascending the numbers, only the family {3,11,43,55,75,79,83,115} is correct. By exponentiation, all the generated classes g 4p i , i = 1 to p-1 is identical to the preceding one whatever the chosen generator in the family of the primitive roots modulo 4p. Here with g 4p = 3, we get : {3,9,27,81,119,109,79,113,91,25,75,101,55,41,123,121,115,97,43,5,15,45,11,33,99,49,23,69,83,1} Let us have in ascending order : G(31) = {1, 3,5,9,11,15,23,25,27,33,41,43,45,49,55,69,75,79,81,83,91,97,99,101,109,113,115,119,121,123} The other classes modulo 4p relative prime (which are always odd) to 4p are : 13,17,19,21,29,35,37,39,47,51,53,57,59,61,63,65,67,71,73,77,85,87,89,95,103,105,107,111,117} Foot-note: Concepts of primitive roots are possible modulo 2 n p. For lack of a particular utility here, we do not develop this point here. However, it is useful to say, that for current interest, we must work modulo 4p and not modulo 2p. Let us have to prove if p = 1 mod 4 then g 4p (p-1)/2 = -1+2p mod 4p, otherwise if p = 3 mod 4 then g 4p (p-1)/2 = -1 mod 4p. We have g 4p p-1 = 1 mod 4p as well if g 4p = 1 mod 4 or if g 4p = 3 mod 4. Indeed, g 4p = g+t.p where t is an integer between 0 and 3. Thus g 4p =g mod p, hence g 4p p-1 = g p-1 = 1 mod p what involves self-evidently g 4p p-1 = 1 mod 4p. In addition g (p-1)/2 = -1 mod p, hence g (p-1)/2 = -1+m.p mod p, m an integer ranging between 0 and 3. Let us calculate g 4p (p-1)/2 = (3+4u) (p-1)/2 = (-1) (p-1)/2 +4n = (g+t.p) (p-1)/2 = g (p-1)/2 +k.p = -1+(k+m).p mod 4p Then if (p-1)/2 is odd, (k+m).p = -4n+r.4p = 4q, then k+m is multiple of 4, henceg 4p (p-1)/2 = -1 mod 4p. If (p-1)/2 is even, 1+4n = -1+(k+m).p+r.4p, hence 2 = (k+m).p+4q. Only k+m = 2 mod 4 is a solution of this equation. Henceg 4p (p-1)/2 = -1+2p mod 4p. Then, let us show that {g 4p i } mod 4p is cyclic of order p-1 exactly. We have g 4p i = (3+4u) i = (-1) i +4n = (g+t.p) i = g i +k.p mod 4p and g i ≠ 1+m.p mod 4p if 0 < i < p-1 for any m. Hence g 4p i ≠ 1+(m+k).p mod 4p if 0 < i < p-1. If at this time g 4p i = 1 mod 4p, that involves 0 ≠ (m+k).p mod 4p, so that m+k ≠ 0 mod 4 for any m what is false. Hence g 4p i ≠ 1 mod 4p for any i, 0 < i <p-1. As g 4p p-1 = 1 mod 4p, {g 4p i } mod 4p is cyclic of order p-1 exactly.

Proofs

Proof 2

We like to answer the question of existence (or not) of a solution to equation x 2 = c mod p pending on the value of c. The case c = 1 is self-evident since x = 1 is always solution, therefore any number p is appropriate for given c. The case c = -1 is written x 2 = -1 = g (p-1)/2 mod p which has a solution if (p-1)/2 is square, hence p = 1 mod 4. The case c = 2 results from Legendre relation (2/p) = (-1) (p^2-1)/8 mod p. If p = 1 mod 8 or p = 7 mod 8 then (2/p) = 1 mod p (existence of a residue), if not if p = 3 mod 8 or p = 5 mod 8 then (2/p) = -1 mod p (non-existence of a residue).

Let us approach the passage of c with -c when c is a prime number. We use G(-c) = U(∩(G(c),G(-1)),∩(H(c),H(-1)) mod 4c and H(-c) = U(∩(G(c),H(-1)), ∩(H(c),G(-1)) mod 4c. The table for these cases is easily written recalling that g 4c = 3 mod 4 :

What shows our matter completely.

Examples of evaluation

We illustrate our study for the following examples. For c = 3, prime number we have only one primitive root modulo 3 which is g= 2. We reason then modulo 12 (4c = 12). Then g 4c = 2+3p = 11 equals 3 mod 4, then G(c) = {11 0 ,11

For H(-3) We can form in the same way G(5) = {1,9,11,19} mod 20 = {1,9} mod 10 and G(-5) = {3,7,13,17} mod 20 = {3,7} mod 10. We can then evaluate G (15) and H (15), either by considering that 15 = 3.5 or that 15 = (-3). (-5).

Let us adopt this last way.

For G (15) -3 G(-3) = {1,7} mod 12 = {1, 7,13,19,25,31,37,43,49,55} mod 60 H(-3) = {5,11} mod 12 = {5,11,17,23,29,35,41,47,53,59} mod 60 -5 G(-5) = {1,3,7,9} mod 20 = {1, 3,7,9,21,23,27,29,41,43,47,49} mod 60 H(-5) = {11,13,17,19} mod 20 = {11,13,17,19,31,33,37,39,51,53,57,59} mod 7,11,17,43,49,53,59} mod 60 For H( 15) 7,13,19,25,31,37,43,49,55} mod 60 H(-3) = {5,11,17,23,29,35,41,47,53,59} mod 60 -5 H(-5) = {11,13,17,19,31,33,37,39,51,53,57,59} mod 60 G(-5) = {1,3,7,9,21,23,27,29,41,43,47,49} mod ,19,23,29,31,37,41,47} mod 60 The set G(15) U H(15 contains well the set of relative prime numbers to 15. The reader will be able to verify that we get the same families by using such other decompositions of c = 15 : 15 = 3.5 = -1.3.-5…

If we continue then with the case c = 105 = 3.5.7, we proceed at the stage c = (3.5).7 knowing that 3.5 and 7 are relative primes, that is using the general formula of intersections and unions : G(105) = U(G(3)∩G( 5)∩G (7),G(3)∩H( 5)∩H (7),H(3)∩G( 5)∩H (7),H(3)∩H( 5)∩G( 7)) mod 420 H(105) = U(G(3)∩G( 5)∩H (7),G(3)∩H( 5)∩G (7),H(3)∩G( 5)∩G (7),H(3)∩H( 5)∩H( 7)) mod 420 P 255/390

Quadratic abundance factors and substitutes

Generalization of the quadratic reciprocal classes concept

We make this study although non-essential to the initial objective.

Let us have the expression a.x 2 +b.x+c and its discriminant Δ = Δ = b 2 -4a.c. We ask the question  ? x integer \ a.x 2 +b.x+c = 0 mod p (24)

and we seek a satisfactory condition on p for a given triplet (a,b,c). We note that when x 2 = c mod p, x 2 -c = 0 mod p and thus Δ = 4c. Hence, by analogy, the condition :

Let us note first that Δ being form b 2 -4ac, Δ = 0 mod 4 and Δ = 1 mod 4 are the only cases to be considered. That results from an elementary observation on the parity of b which implies Δ ≠ 2 mod 4 and Δ ≠ 3 mod 4. In particular, there is no case Δ = -1. If Δ is positive, then either Δ = 4k, or Δ = 1+4k, k > 0. If Δ is negative, either Δ = -4k, or Δ = -3-4k, k > 0. If Δ is a prime number of positive or negative sign, it can be only 1 mod 4.

The alternatives generated by Δ are then the following ones:

Let us have P(x) = a.x 2 +b.x+c = 0 mod p and Δ = b 2 -4a.c, a ≠ 0.

We rewrite P(X) as :

This means that Δ is a square mod p.

Let us have G the prime numbers p such as the equation a.

Illustration of this law

We give an illustration of this law for Δ ranging between -10 and 10 , 3, 9, 19, 25, 27 mod 28 5, 11, 13, 15, 17, 23 mod 28 -8 1, 3, 9, 11, 17, 19, 25, 27 mod 32 5, 7, 13, 15, 21, 23, 29, 31 mod 32 8 1, 7, 9, 15, 17, 23, 25, 31 mod 32 3, 5, 11, 13, 19, 21, 27, 29 mod 32 -9

1 mod 4 3 mod 4 91 mod 1 void - 10 1, 7, 9, 11, 13, 19, 23, 37 mod 40 3, 17, 21, 27, 29, 31, 33, 39 mod 40 10 1, 3, 9,13, 27, 31, 37, 39 mod 40 7, 11, 17, 19, 21, 23, 29, 33 mod 40 Let us consider, for example, equation 23x 2 +10x+1 = 0 mod p. We have Δ = 8. Thus, when p = 1 mod 4, the equation has a solution and when p = 3 mod 4, it does not have any. It will be the same with any equation of degree 2 with discriminant 8.

Reciprocity : Reciprocal classes for polynomials

Generalities

Overflowing of the quadratic framework, we put here the question of the generalization of the research of all classes p having a solution(s) x (or not) for a given polynomial :

We will call "reciprocity" the research of prime numbers sequences {p} as : p c G(P) or p c H(P)

Here G and H are to be determined like functions of the coefficients of polynomial P, the set G(P) being that with solutions x and the set H (P) that without solutions x. We fly over this vast problem which has certainly interesting ramifications in other fields that ours. Some proofs will not be led to end.

Case of the monomials

Let us start by examining the case x n = c mod p (27)

We seek the prime numbers p for which the residue c, fixed by advance, exists and rather, failing this ambition, the frequency of such incidences.

Limit frequency

Let us have f(n,c) the frequency of existence of a given residue c when p describes the set of prime numbers {p} for the suggested equation : f(n,c) = lim i→∞ #{i \  x and x n = c mod p i }/i (28)

Let us pose one moment the question of the existence of a limit. This point is not acquired, in particular as the numerical applications show an undulation around the thereafter proposed values, undulations which take force randomly (as in the majority of the problems relating to the prime numbers). The existence of a limit is thus admitted.

Multiplicative frequency

Conjecture

The frequency is a multiplicative function :

The limit value of the function is : f(q t→∞ ,c reg ) = 1-q (q 2 -1) Hence, the numerical applications : q 1-q/(q-1)/(q+1) 2 1/3 Let us note that when the prime number q is large, the frequency of the incidences is close to 1, i.e. a solution to x n = q t mod p exists almost all the time.

Incidence of the exponent divisibility

If we admit that frequency function f is actually multiplicative, with regular behaviour targets c, we deduce that there is no continuity of the frequency when exponent n (in x n = c mod p) increases, quite to the contrary. When n increases, between n which might be a prime number and n+1 its neighbour which might be a number with many divisors, frequency f(n,c) tends towards 1 in the first case and towards 0 in the second case :

The situation is quite the same when c is not regular.

Classification of the residues

Residues with one prime factor

Function f(q t ,c) is not multiplicative towards c. We can however determine, within a conjectural framework, the behaviour of the residues. Let us start with the residues having one prime factor. Quadratic abundance factors and substitutes c q σ acyclic cyclic 8 = 2 3 3 3 2 0, -1/9 -1/81, -1/729, … 64 = 2 6 3 3 2 0, -1/9 -1/81, -1/729, … 4096 = 2 [START_REF] Peter | Advanced Engineering Mathematics[END_REF] 3 3 2 0, -1/9 -1/81, -1/729, … 32768 = 2 [START_REF] Waldschmidt | Valeurs zêta multiples[END_REF] 3 3 2 0, -1/9 -1/81, -1/729, … 2 3d and ≠ 2 9d 3 3 2 0, -1/9 -1/81, -1/729, … c q σ acyclic cyclic 512 = 2 [START_REF] Koninck | Introduction à la théorie des nombres[END_REF] 3 3 2 0, 0, -1/27 -1/243, -1/2187, … 262144 = 2 18 3 3 2 0, 0, -1/27 -1/243, -1/2187, … 2 9d and ≠ 2 27d 3 3 2 0, 0, -1/27 -1/243, -1/2187, … c q σ acyclic cyclic 8 = 2 3 3 3 2 0, -1/9 -1/81, -1/729, … 512 = 2 [START_REF] Koninck | Introduction à la théorie des nombres[END_REF] 3 3 2 0, 0, -1/27 -1/243, -1/2187, … 134217728 = 2 27 3 3 2 0, 0, 0, -1/81 -1/729, -1/6561, … c q σ acyclic cyclic 2 (3^k).d and ≠ 2 (3^(k+1)).d 3 3 2 {0, …, 0}k terms, -1/3 k+1 -1/3 k+3 , -1/3 k+5 , … 3 3 3 2 0, -1/9 -1/81, -1/729, … 729 = 3 6 3 3 2 0, -1/9 -1/81, -1/729, … 531441 = 3 [START_REF] Peter | Advanced Engineering Mathematics[END_REF] 3 3 2 0, -1/9 -1/81, -1/729, … 3 3d and ≠ 3 3^d 3 3 2 0, -1/9 -1/81, -1/729, … c q σ acyclic cyclic 3 = 3 1 3 3 2 -1/3 -1/27, -1/243, … 27 = 3 3 3 3 2 0, -1/9 -1/81, -1/729, … 19683 = 3 [START_REF] Koninck | Introduction à la théorie des nombres[END_REF] 3 3 2 0, 0, -1/27 -1/243, -1/2187, … 7625597484987 = 3 27 3 3 2 0, 0, 0, -1/81 -1/729, -1/6561, … c q σ acyclic cyclic 3 (3^k).d and ≠ 3 (3^(k+1)).

. Cardinal of zero

We have now adding solution (0,0), only new solution, for #(0) :

The factor t is identical to the prime numbers variables preceding t factor.

Cardinal of #(g i-1 )

We still have mz i,0 = #(g i-1 ) = 0 mod d

Number of classes modulo p

We found, in the generalities and in the study of the cardinal of 0, the importance of both (n, p-1) and (n+1, p-1). This comes from the exponents we find in :

The cardinal matrices (and the eigenvalues) of this group are obtained using those of the z n group as we proved earlier. The modulo p classes, which differ, are thus numbering (n.(n+1), p-1) (67)

However, the evaluation of the first column of the cardinal matrix does only require the study of half of these classes :

(n.(n+1), p-1) / 2 (68)

We will get back the total number of classes when we will perform the multiplication with the eigenvalues(σ mz ) = [Δcmz].(λ)/d, distinctions introduced by z n (see examples degree 4 and 8 below).

Symmetrical overlapping variables of degree 2

Let us have :

Let us apply relation 59. It follows for the prime numbers variables (mzb i,0 = #(g i-1 )) :

and for variables of integers (mza i,0 = #(g i- 1)) :

P 283/390

Quadratic abundance factors and substitutes σ ve 1 = (mza 1,0 .λ 1 +mza 2,0 .λ 2 )/2+mza i0,0 σ ve 2 = (mza 2,0 .λ 1 +mza 1,0 .λ 2 )/2+mza i0,0 recalling also that σ ve 0 = p ve and σ vp 0 = (p-1) vp .

Variables of integers

The two cases p \ Δ c C(p) on one hand and the two cases p \ Δ c D(p) on the other end give the same eigenvalues and we get again the results of paragraph 8.1

Variables of prime numbers p = 1 mod 4 and p \ Δ c C(p)

We get of course the results of paragraph 8.2.

Symmetrical overlapping variables of degree 3

Let us have the expression : z 1 3 +z 1

2

.z 2 +z 1 .z 2 2 +z 2 3

Cardinal matrices

The cardinal matrices, deduced from primitive roots equations in case d = 3 (except for p = 2 and p = 3), are described below in regard of sequences.

P 284/390

Quadratic abundance factors and substitutes

From (σ gr ) = [Δcm i,0 ].(λ)/d, in case d = 3, we get using relation 59 and posing mzb i,0 = #(g i-1 ) for prime numbers variables matrices : σ vp 1 = (mzb 1,0 .λ 1 +mzb 2,0 .λ 2 + mzb 3,0 .λ 3 )/3+mzb i0,0 σ vp 2 = (mzb 3,0 .λ 1 +mzb 1,0 .λ 2 + mzb 2,0 .λ 3 )/3+mzb i0,0 σ vp 3 = (mzb 2,0 .λ 1 +mzb 3,0 .λ 2 + mzb 1,0 .λ 3 )/3+mzb i0,0 and posing mza i,0 = #(g i ) for the integers' variables matrices :

σ ve 1 = (mza 1,0 .λ 1 +mza 2,0 .λ 2 + mza 3,0 .λ 3 )/3+mza i0,0 σ ve 2 = (mza 3,0 .λ 1 +mza 1,0 .λ 2 + mza 2,0 .λ 3 )/3+mza i0,0 σ ve 3 = (mza 2,0 .λ 1 +mza 3,0 .λ 2 + mza 1,0 .λ 3 )/3+mza i0,0

Below, the eigenvalues will appear on the right of each matrix.

We use the components s 1 , s 2 and s 3 of constructive of sequence matrix of dimension 3 resulting from relations 71 of exercise 5. We also consider the following decomposition of p :

Let us note that we already used this decomposition at the time of Fermat-Catalan matrices study in environment d = 6 (relation 33 of exercise 9). We pose then :

Number i depends on the choice of g and p following the conjecture :

Important foot-note Expression p-1 (here (p-1)/2) distinguishes itself particularly here. Indeed, it allows differentiating two classes of primitive roots. For other environments, this kind of distinction could possibly spread (constituting an interesting study case).

We illustrate underneath the value of i according to g (and p).

We have then (the eigenvalues are on the right of the matrices) :

Quadratic abundance factors and substitutes

Sequence p = 7 mod 12

Sequence p = 11 mod 12 p p 2 -p p 2 p p 2 -p 0

Variables of prime numbers : c = y 1 3 +y 1 2 .y 2 +y 1 .y 2 2 +y 2 3 .

Sequence p = 2

Sequence p = 1 mod 12

Sequence p = 5 mod 12

with the two tables :

We use here the decomposition p = t 1 2 +t 1 t 2 +t 2 2 (t 1 > t 2 ) In the cases of sequences 1 mod 12 and 7 mod 12, the correction, with the parameters u 1 , u 2 and u 3 , remains identical.

Corrections of indices (conjecture)

The illustration of i indices is presented as follows. Only cases p = 1 mod 6 are relevant. 6, 24, 96, 57, 10, 40, 51, 95, 53, 103, 85, 13, 52, 99, 69, 58, 14, 56 0 37, 39, 47, 79, 98, 65, 42, 59, 18, 72, 70, 62, 30, 11, 44, 67, 50, 91 0 127 3 10 3, 92, 12, 112, 114, 48, 67, 65, 14, 46, 6, 56, 57, 97, 101, 96, 7, 23 0 116, 109, 86, 83, 55, 78, 93, 106, 58, 118, 43, 91, 45, 39, 110, 53, 29, 85 0 157 6 7 5, 96, 77, 34, 119, 24, 84, 137, 87, 69, 6, 21, 152, 61, 80, 123, 38, 133, 73, 20, 70, 88, 151, 136 0 142, 26, 91, 83, 55, 114, 85, 62, 60, 53, 139, 94, 15, 131, 66, 74, 102, 43, 72, 95, 97, 104, 18 , 61, 18, 13, 28, 50, 51, 48, 57, 44 1 32, 20, 7, 46, 63, 12, 31, 41, 11, 34 2 73 4 5 59, 31, 20, 60, 34, 29, 14, 42, 53, 13, 39, 44 1 5, 15, 45, 62, 40, 47, 68, 58, 28, 11, 33, 26 2 79 5 2 3, 54, 37, 34, 59, 35, 77, 43, 63, 28, 30, 66 1 6, 29, 48, 74, 68, 39, 70, 75, 7, 47, 60, 53 2 97 4 7 5, 40, 29, 38, 13, 7, 56, 60, 92, 57, 68, 59, 84, 90, 41, 37 1 21, 71, 83, 82, 74, 10, 80, 58, 76, 26, 14, 15, 23, 87, 17, 39 2 103 1 10 35, 48, 87, 84, 74, 75, 44, 78, 54, 77, 85, 43, 6, 20, 101, 62 1 5, 51, 67, 86, 12, 40, 99, 21, 70, 96, 11, 71, 65, 45, 88 ,53 2 139 5 8 2 , 128, 130, 119, 110, 90, 61, 12, 73, 85, 19, 104, 123, 88, 72, 21, 93, 114, 68, 111, 15, 126 1 32, 102, 134, 92, 50, 3, 53, 56, 109, 26, 135, 22, 18, 40, 58, 98, 17, 115, 132, 108, 101, 70 6, 133, 54, 140, 52, 146, 15, 106, 48, 7, 130, 63, 114, 111, 120, 93, 82, 56, 134, 51 1 77, 115, 109, 102, 119, 13, 141, 89, 129, 12, 14, 117, 61, 104, 71, 108, 126, 96, 112, 30, 35 2, 128, 42, 80, 67, 50, 103, 72, 44, 45, 109, 130, 7, 122, 147, 117, 153, 12, 116, 89, 154, 76, 137, 129, 106, 101, 107 1 32, 92, 20, 139, 94, 148, 18, 11, 52, 68, 114, 124, 112, 159, 70, 79, 3, 29, 63, 120, 19, 75, 73, 108, 66, 149, 82 2 181 2 13 2, 128, 47, 112, 98, 118, 131, 58, 96, 171, 84, 127, 179, 53, 134, 69, 83, 63, 50, 123, 85, 10, 97, 54 1 57, 28, 163, 115, 78, 105, 23, 24, 21, 77, 41, 90, 124, 153, 18, 66, 103, 76, 158, 157, 160, 104, 140, 91 5, 153, 127, 142, 22, 17, 57, 123, 174, 152, 135, 78, 148, 167, 15, 73, 188, 40, 66, 51, 171, 176, 136, 70, 19, 41, 58, 115, 45, 26, 178, 120 1 37, 90, 52, 163, 47, 10, 113, 61, 91, 44, 34, 114, 53, 155, 111, 77, 156, 103, 141, 30, 146, 183, 80, 132, 102, 149, 159, 79, 140, 38, 82, 116 2 199 1 14 3, 197, 134, 176, 148, 34, 110, 192, 71, 120, 119, 186, 75, 149, 166, 22, 118, 54, 163, 183, 77, 15, 189, 73, 84, 143, 170, 152, 164, 95 1 44, 108, 127, 48, 167, 154, 30, 179, 146, 168, 87, 105, 129, 113, 190, 6, 195, 69, 153, 97, 68, 185, 142, 38, 41, 39, 173, 150, 99, 133 2 In the preceding table, the primitive roots g, respectively g', are of the type g 1 1+6u , respectively g' 1 1+6v

, where g 1 , respectively g 1 ', is one of the primitive roots and we have g' 1 = g 1 -1+6w , for some integer w (u and v are integers also). When i = 0, there is no ambiguity related to the choice of g on the values of the abundance factors. When i is different from 0, we reversed the positions of g, when necessary, to get a homogeneous table for the indices (indice 1 first, 2 afterwards).

In the following table, we include the preceding primitive roots while indicating, in bold police, those roots corresponding to a class of the type (or ((p-1)/2,-(p-1)/2))) (1/(2^n) 6, 18, 54, 59, 55, 43, 7 35, 44, 10, 30, 26, 17, 51, 31 , 61, 18, 13, 28, 50, 51, 48, 57, 44 ((p-1)/2) 1/2 32, 20, 7, 46, 63, 12, 31, 41, 11, We added in the preceding table some numbers between brackets {}. Indeed, these values do not correspond to primitive roots (g), but are of the type g 1+6u (see line p = 151, 75 = (151-1)/2, 23 = (-(151-1)/2) 1/2 ). These additions are necessary to get always solutions with the suggested conjecture. We can notice that research g 1+6u = (or((p-1)/2,-(p-1)/2))) (1/(2^n)) is often successful starting at n = 0 or n = 1, but comprises exceptions (here for p = 73). The reader will notice also the change of sign in front of (p-1)/2 when the solutions are on both sides of the table and the balance of the number of solutions classes.

Variation with the constructive components

Variables of integers.

If we reconsider the expression formulated higher for integers variables

the values of Δs i-k,d in the case d = 3 are :

Variables of prime numbers.

If we reconsider the expression formulated higher for prime numbers variables In the self-evident cases, the values 0 under Δs i,d are in fact those of (s i,d =0)+Δs i,d .

Prime numbers generation abundance factors

We examine first the diophantine equation :

The abundance factors mod p for this problem are obtained by posing the products of matrices

and recording the first columns values.

The matrices commutate and the simplest approach is to choose the last indicated expression.

We have then:

The normalized abundance factors are obtained by multiplication by p/(p².(p-1)) = 1/(p.(p-1)).

For target zero :

Variables of prime numbers

The normalized abundance factors are obtained by multiplication by p/(p-1) 3 . Hence :

Thus there exists very few prime numbers of type z

Abundance factors for differences between pure cubes and overlapping cubes

We are interested this time in the diophantine equation :

The abundance factors of c mod p for this problem are obtained by posing the products of matrices

and again recording the values of the first columns.

The matrices commutate and the simplest thing is to choose the indicated order. We have then :

p 2 +(e 1-i -4).p+4-s 1-i -2s 3 p 2 +(e 1-i +2).p+4+s 1-i -2s 3

The normalized abundance factors are obtained by multiplication by p/(p-1) 3 .

We have :

The divergence of the expression remains conjectural since in any rigour z 2 = 0 is not a prime number.

Let us note that the previous abundance factors evaluations were done modulo p and not modulo p δ→∞ . Stronger conclusions relative to their convergences or divergences would be worth a more thorough study that we leave to the reader initiative.

Symmetrical overlapping variables of degree 4

Let us consider :

Cardinal matrices

The sequence decomposition which plays a crucial role in the evaluation of the first column of the cardinal matrices is here (with v odd and w even) :

We have to consider modulo 8 congruencies (due to z 4 ), but also modulo 10 congruencies (due to (z 1 5 -z 2 5)/(z 1 -z 2 ) when z 1 ≠ z 2 ). Modulo 40 cases result for there. In fact a modulo 20 simplification appears that we exploit.

We will limit us next to the evaluation of the components #(g i ) of the cardinal matrices first column :

Variables of integers : c = x

In the cases of p = 3 mod 20, p = 7 mod 20, p = 11 mod 20 and p = 19 mod 20, there is no decomposition in two squares sums (as p = 3 mod 4). For p = 13 mod 20 and p = 17 mod 20, we have always v ≠ 0 mod 5 (which is not the case when p = 1 mod 20 or p = 9 mod 20).

Note 2 :

According to the primitive g choice, the values #(g 1 ) and #(g 3 ) may have to be permuted.

The exponents e1 and e2 are given by the following (conjectural) rules : For e1 : By direct evaluation, the cardinal matrix of the preceding group, at the sequence p = 13 (g = 2), presents as follows, with the usual order of targets (0, g 0 , g 1 , etc.) : We are interested here in the practice of abundance factors evaluation in the most general cases. It is necessary for that to give a description of the equations to be solved.

Types of variables

Integers' variables are noted x i and prime numbers variables are noted y j . If one or the other type is concerned, we will write it z k .

Degree of the equation

All the equations are posed in their developed form and constants are included within the target c. Thus (y+2).(x+3) = c is rewritten xy+2x+3y = c'. The diophantine expression is then an addition of monomials of one or more variables.

Independent groups

When several variables are overlapping in a diophantine expression, the method of multi-dimensional tables must be applied per block (and not variable by variable and monomial by monomial). We will call independent group of monomials a block of expressions which does not have common variables with an another block of expressions. We will call composing group each monomial. For example, in x 1 2 y 1 3 y 2 +x 1 3 +y 2 +x 2 2 -y 3 5 = c, the independent groups are x 1 2 y 1 3 y 2 +x 1 3 +y 2 , x 2 2 and -y 3 [START_REF] Dieudonné | Abrégé d'histoire des mathématiques[END_REF] and the composing groups are x 1 2 y 1 3 y 2 , x 1 3 , y 2 , x 2

2

. We note respectively these groups Gi k and Gc k . The diophantine equation is written either ∑ Gi k = c, or ∑ Gc k = c.

Economy of means

To solve the problem R

which is to find relative weights of c for each element of

when each variable is replaced by its representative [0,1,2,…,p δ -1] or [g 0 ,g 1 ,g 2 ,…,g φ(p^δ)-1 ] pending the peculiar case.

In theory, the solution is obtained for the limit δ → +∞. However, if the weighting of elements modulo p δ+1 is the same one as the weighting of the elements modulo p δ starting from some exponent δ, it is clear that we save ourselves an infinity of calculations.

It is this strategy which we adopt here.

Base of convergence

We will call base of convergence of n at sequence p, noted δb, the smallest positive integer δ for which the greatest common divisor of n and p δ-1 (p-1) remains unchanged.

Degree of convergence of a prime numbers monomial

The degree of convergence δc of a monomial is the minimal value δ, such as the normalized representative of the monomial modulo p δ does not evolve, when the exponent δ increases. It is obviously related to the base of convergence.

Let us consider the independent group y n . Let us have sequence p and d δ = (n,Φ(δ)) = (n,p δ-1 .(p-1))

We have then:

For more clarity, we give the example of the monomial of prime numbers variable y 6 .

P 304/390

General abundance factors [1,3,5,7,9,11,13,15] 1,2,4,5,7,8] mod 9 ≡ [1,2] mod 3 y 6 = [1,1,1,1,1,1] mod 9 ≡ [1] mod 9 Evolution1 y = [1,2,4,5,7,8,10,11,13,14,16,17,19,20,22,23,25,26] mod 27 ≡ [1,2] mod 3 y 6 = [1,10,19,19,10,1,1,10,19,19,10,1,1,10,19,19,10,1,1,10,19,19,10,1] mod 27 ≡ [1,10,19] [1,2,3,4,6,7,8,9,11,12,13,14,16,17,18,19,21,22,23,24] mod 25 ≡ [1,2,3,4] mod 5 y 6 = [1,14,4,21,6,24,19,16,11,9,9,11,16,19,24,6,21,4,14,1] mod 25 ≡ [1,4,6,9,11,14,16,19,21,24] mod 25 ≡ [1,4]

The sign of equivalence is use here each time the proportion of the occurrences of each element remains the same one. This example illustrates further considerations.

Proof

According to the argument developed within exercise 3, the representative of a variable y n , at sequence p ≠ 2, g being a primitive root of p and with d = (n,p δ-1 .(p-1)), is {g 0.d , g 1.d , g 2.d , …, g (Φ(δ)/d-1).d } mod p δ . For p = 2, the representative is {5 0.d , 5 1.d

« Rho » function

The degree of convergence separates, relatively to independent groups modulo p δ behaviour, two zones : a zone δ < δc "non-forecastable" to be inspected on a case-by-case basis and a zone δ ≥ δc having the characteristics observed for δ = δc.

Definition of the degree of convergence of a monomial (of prime numbers or not)

By definition, the degree of convergence of the integers' monomial x n at sequence p coincides with the degree of convergence of the prime numbers monomial y n at sequence p. By extension (and definition), the degree of convergence of a diophantine expression, at sequence p, is the degree of convergence of this expression after substitution of integers' variables by prime numbers variables.

Beyond the degree of convergence, the abundance factors of integers' variables do not undergo new interdictions. For a given target, we can estimate convergence or divergence of the number of solutions pending on a non-null constant. Proof of this point remains open.

Degree of stability

We will call degree of stability of an expression at sequence p, the minimal value of δ, noted δs, for which the normalized abundance factor remains unchanged modulo p δ .

Rules of stability

The degree of stability of prime numbers variables is finite and equals the degree of convergence. The degree of stability of integers' variables is infinite. The degree of stability depends on the independent groups and the studied sequence : δs = δs(p, Gi 1 , …, Gi k )

Composite monomials

It acts of monomials including several variables. For the degree of convergence, we have :

This results self-evidently from the two-dimensional tables (or multi-dimensional tables) method, since when the entries are stabilized, the contents of the table are stabilized. More interesting is the following result : δc(p, y 1 n1 y 2 n2 …y k nk ) ≤ δc(p, y gcd(d1,d2,…,dk) ) where d i = (n i , Φ(δ)) (5) Indeed, the structure of a representative of a monomial at sequence p is {g 0.d1 , g 1.d1 , g 2.d1 , …, g (Φ(δ)/d1-1).d1 }.. Considering z 1 n1 and z 2 n2 two variables with representatives {g 0.d1 , g 1.d1 , g 2.d1 , …, g (Φ(δ)/d1-1).d1 } mod p δ and {g 0.d2 , g 1.d2 , g 2.d2 , …, g (Φ(δ)/d2-1).d2 } mod p δ , where d 1 and d 2 correspond to convergent cases. We form the two-dimensional table of the product z 1 n1 z 2 n2 . The elements of this table are products like g i.d1 g j.d2 , that is also g i.d1+j.d2 . According to Bézout theorem, there is a couple (i,j) such as i.d 1 +jd 2 = gcd(d 1 ,d ). Thus, when couples (i,j) cover their field of definition, then numbers g i.d1+j.d2 cover g k.gcd(n1,n2) . By iteration on all of the variables, it follows relation (5).

Arbitrary compositions

By the same argument, we get immediately :

The attentive reader will see that all is said with relation (6).

The more variables are overlapping, paradoxically (but not as much as that), the more abundance factors research is likely to be simple. In particular, when a prime numbers variable (type y) figures in a diophantine equation by itself and with exponent 1, abundance factors evaluation can be led modulo p (the case of integers variable degree 1 is self-evident giving a normalized factor identically equal to 1 for all targets).

Other outstanding examples are : δs(p, z 1 .z 2 ) = δs(p, z 1 ) (7) δs(p, z 1 2 +z 1 .z 2 +z 2 2 ) = δs(if(p ≠ 3, z 1 , z 1 2)) (8)

Case of the polynomials and influence of the target

In polynomials, we cannot guess the factor of convergence (and of stability of a given target) like the following table illustrates it, table where we emphasize the simplicity of case p 2 and the complexity of case p 2 +p (prime numbers variable) at sequence 3. The matrices of environment [P] rise from the study made in exercise ( 5) p 72 and concretizes with exercise ( 6) p 125 in a particular case (p does not divide n). Only are then to find the eigenvalues. The eigenvalues obey a diagram requiring only the search for cm values on a case by case basis :

The configuration of an independent group matrix is related with the configuration of monomial x cm matrix, we got at exercise (6). The matrices are built starting from right and bottom. Trace blocks n° 1 mod n (starting from bottom) result from the initial trace block with the same position. Trace blocks n° ≠ 1 mod n are equal on n-1 blocks with components to determine. The blocks above the trace blocks are worked out vertically and are equal. The blocks of n° 1 mod n are based on the values from the matrix δ = 1 in identical position with adapted factor p k . Trace blocks ≠ 1 mod n are to be determined. Blocks below trace blocks are worked out horizontally. Blocks of n° 1 mod n are based on the values from the matrix δ = 1 in identical position with adapted factor ((p-1)/n).p k . Trace blocks ≠ 1 mod n are to be determined but seem to be related to the block in left top of δ = 1 matrix. It is good practise to verify at the end of construction that the sum of each line is p δ for an integers' variables matrix, (p-1) δ for a prime numbers variables matrix or a combination of both in the event of variables of two kinds.

Characterization of environment matrices

To show the existence of a common matrix of environment, the simplest way is to completely characterize it. Indeed, let us return to relation (29) of exercise 5 related method where the reader will refer for more details. In general, the cardinal matrix [C] of an independent group, with k integers' variables and m prime numbers variables, satisfies the equation :

Here #(i-j) are the components at position (i,j) of the cardinal matrix. When several independent groups are present, the reasoning is extended to the lowest common multiple of n i , that is n = lcm(n i ). In this "not contracted" form, the cardinal matrix is a right circulant matrix :

and of eigenvectors matrix, the matrix of the special complex unitary group of dimension n (SU n ()) : 1) This or some equivalent matrix can be used as change of base matrix. Gathering targets with equal cardinals, sorting #(p k .g i .g u.cm ) according to increasing i and decreasing k (or some other choice) rearranges this matrix and its inverse exactly in the same way (for a given choice) for a given environment cm. Hence the result.

Singularities and resultants

The study of the quadratic equation u.x 2 +v.x.y+w.y 2 shows the existence of several conditions relative to variables coefficients (u,v,w) and discriminant (Δ) giving domains where abundance factors have same literal formulas (same literal functions of parameter "sequence"). When the equation becomes more complex (3 and higher degrees), we can expect to have even more different situations where coefficients and discriminants play a role. For equation a.x 3 +b.x 2 .y+c.x.y 2 +d.y 3 ,

EXERCISE 12 : EQUITABLE DIAGRAM AND INTERDICTION DIAGRAM

We seek the criteria reinforcing (or not) the richness of prime numbers production for a given diophantine expression and a selected target. We limit ourselves to an approach modulo p of abundance factors.

Equitable diagram

An expression R(…) = c of one (or several) variables of integers (or prime numbers) will be called equitable if the abundance factors of R are equal for any target c. This means that fan(c) = 1 for any c.

According to exercise 3, it is easy to deduce that a monomial x n will be equitable if it gives a complete system of residues to all sequences p i . It is the case if d = (n,p i -1) = 1 for any p i , and thus if and only if n = 1. Thus P(x) = x is the only equitable monomial (ax, a an integer > 1, is not). For a polynomial, some good sense leaves to admit their absence.

Thus, strict equitable behaviour does not have much interest. We will rather speak of equitable diagram to evoke a tendency to the production of systematically close to 1 abundance factors.

Equitable sequence

The sequences p i evolving from 2 to +∞, we call equitable sequence the prime number p i until which a polynomial is equitable : fae(c,p j ) = fae(c-1,p j ) = 1, p j = 2 to p i , c = 1 to p j Any polynomial whose sum of coefficients is odd induces an equitable diagram at sequence p i = 2.

Polynomials with equitable diagram

Let us have P(x) = x² = c mod p i . Quadratic residues theory tells us, that except for the sequence p i = 2 for which only one non-null residue is to be considered (namely 1), we cannot get complete system of residues. Thus polynomial n² is equitable at sequence 2 and only for this sequence. This result is the same one for n = 2 i , i an integer.

According to exercise 3, we know that if (n,p i -1) = 1 then x n admits a complete system of residues. Let us carry out the prime factors decomposition of n and let us have m the smallest factor of n (n = Пm i j , m = min(m i )). Then, the equitable sequence of x n is carried at least to the sequence p i = m. Thus, for example, the monomials P(x) = x n are increasingly equitable for increasing prime numbers n. We will see this point through the histograms presented further.

Diagram of interdiction

A diophantine expression induces an interdiction diagram for target c, if for a given sequence p i , part of the targets have a null abundance factor : fae(c,p i ) = 0 Then fan(c) = 0.

Sequence of interdiction

We can define this sequence as the prime number p i for which there is a target c (which will be lower than p i ) such as fae(c,p i ) = 0

Interest of the notion

The absence of production of some target modulo p i increases the abundance of the others target modulo p i . The game is to seek to prohibit the greatest possible number of targets. The concept of sequence is important as more prohibition intervenes early, larger will be the proportion of prohibitions and thus the effectiveness of prohibition. Let us note that any polynomial, whose sum of the coefficients is even, armature a diagram of prohibition at sequence p i = 2.

Polynomials with diagram of interdiction

We are interested here in prime numbers generation by a polynomial. Thus, we write P(n) +c = p, n an integers variable and p a prime numbers variable. We seek an expression of P(n) inducing some interdiction. Let us choose P(n) in the form P(n) = 2.3… p i .Q(n), Q(n) any unitary polynomial. Then P(n) = 0 mod p j for p j = 2 to p i . The two-dimensional table built with P(n) (then equal to 0) and -p giving the targets to sequence p j includes then targets 1 to p j appearing 1 time exactly at sequence p j , which means the interdiction of target 0 from sequence 2 to sequence p i , and by consequence, of all targets 0 mod p j from sequences p j = 2 to p i . Alternatives can be proposed. The laws of congruence intrinsic to the method involve a P 314/390 Equitable diagram and interdiction diagram prohibition of a target c multiple of p k starting from sequence p k for any monomial P(n) = p 1 .p 2 …p k …p z .Q(n), where product p 1 .p 2 …p k …p z may contain p k several times. Thus polynomial P(n) = 3 2 .11 4 .n 3 "prohibits" targets 0 (mod 3) and 0 (mod 11) starting from sequences 3 and 11 respectively.

Another alternative is a research on the coefficients of a unit polynomial (coefficient dominating equal to 1). A system of equations modulo 2 to modulo p i is then to be solved. There is no limit with this process of enrichment except the number of polynomial coefficients necessary to satisfy all posed equations. This has an immediate effect on the degree of the polynomial which implies, even if the richness relating to other, of same degree, randomly chosen polynomials may be large, a low number of actual solutions due to the degree reached by the polynomial we actually built (the order of magnitude of the enumeration is n (1/d) /ln(n), d dominating degree of the polynomial). As example, for a polynomial of the form P(n) = n 3 +bn²+cn, we get interdictions 0 mod 2 and 0 mod 3 for b+c = 1 mod 2, b = 0 mod 3 and c = -1 mod 3. Prohibition 0 mod 5, simultaneously with preceding prohibitions, cannot be obtained. For a polynomial of the form P(n) = n 4 +bn 3 +cn²+dn, we get the prohibitions 0 mod 2 and 0 mod 3 for b+c+d = 1 mod 2, b+d = 0 mod 3 and c = -1 mod 3. Interdiction 0 mod 5, simultaneously with the preceding ones, is also impossible. With degree 5, it is possible to reconcile the simultaneous prohibitions of 0 mod 2, 0 mod 3 and 0 mod 5, but not 0 mod 7.

Histograms of density of distribution

The histogram of normalized abundance factors clearly reveals the nature of a given polynomial relative to prime numbers distribution : polynomial with marked diagram of interdiction or equitable diagram. At sequence p i , we can get exactly this distribution by carrying out calculations until this sequence with 2.3… p i consecutive targets. The number of targets to be taken into account quickly reaches millions (510510 for p i =17 and 9699690 for p i =19). Thus, it is out of question to do an exhaustive study. So, we suppose randomness of the behaviour. We lean on the theory of probability. It suffices, in general, to use a small sample to represent a population correctly. On modulo basis, we can deduce an infinity of others target c of uncut sequence p i (at least), namely c (mod 2.3…p i ).

Prime numbers prolific polynomials

With preceding method, we recover R.Ruby and G.Fung polynomials, which are prolific under other reports (see following paragraph). These polynomials offer large uncut sequences. However, from considerations relative to the diagrams of interdiction, we can exhibit polynomials with much higher uncut sequences. In fact, the capacity to deal with large integer of our data-processing tool is the only limit to get higher terms (see below).

In addition, we draw attention to François Dress and Michel Olivier article [4]. It refers to Schinzel conjecture, (whose simplest case is prime numbers twins conjecture), which can be stated as follows : if there is no prime number which, for any m, divides at least one of the integers P(m), P(m+1),…, P(m+n-1) then there is an infinity of m such as P(m), P(m+1),…, P(m+n-1) are all prime numbers. In this article, the so called periodic prime divisor (ppd) is any prime number p which, for any m, divides at least one of the integers P(m), P(m+1),…, P(m+p-1). We observe that the uncut sequence of P (of value p i ) coincides with the value p = p i+1 (the uncut sequence precedes the ppd).

Polynomials generating successive prime numbers

These polynomials are intuitively related to prolific polynomials. The account of the records is done on the basis of prime number of different absolute values. Thus, polynomial n²-79n+1601 with prime numbers results for n = 0 to 79 has same "statute" that Euler polynomial n²-n+41 with prime values for n = 0 to 40. Transformation n into -n+40 allows to pass from one to the other : (-n+40)²-(-n+40)+41 = n²-79n+1601. We get also easily intermediate polynomials.

HISTOGRAMM OF MULTIPLICATIVE FACTORS PRODUCTS

Equitable monomials (prime degrees) (p i = 167)

Records

It can be "playful" to establish records of uncut sequences. We find the followings among which a case with p i = 283 (in [4] where the matter is a little different from ours, the greatest uncut sequence corresponds to p i = 139 for polynomial x²+x-12276955783) :

EXERCISE 13 : LOGARITHMIC MONOTONY

Our preceding exercises enabled us to largely study the arithmetic law giving the asymptotic proportions of various targets.

We are interested from now on in the geometrical law of evaluation of the volume.

Integration

We like to show a property of the logarithmic function under an integral. This property, remarkable by itself, is essential for asymptotic evaluations. We initially carry out a development by integral parts (u = t, v = Ln n (t), u' = 1, v' = n.Ln n-1 (t)/t) :

∫ Ln n (t)dt = [t. Ln n (t)] -n.∫ Ln (n-1) (t)dt

Then, we can write the succession of equalities :

By successive eliminations between the last term of these expressions and the first term of the following expression multiplied by the adequate factor, it follows :

If n is null, the expression does not include logarithms and does not interest us here. If n is a positive integer, the development admits a finite number of terms, if not the succession of terms is infinite. In these two cases however : The second integral proves that when c tends towards infinite, the surface under the curve of function Ln n (t) is quasiidentical from origin (c 0 > 2) to c/2 and from c/2 to c. The only monotonous continuous functions having same integral from 0 to c/2 and from c/2 to c are the constants. The logarithm function being monotonous (whatever the sign of n), that means that this function has, under an integral, the behaviour of a constant. However a constant can be extracted from the integral. The value of this constant is given by ( 1) after division over the length of integration (that is c). This constant is thus :

It is there a remarkable result which involves exceptional facilities of calculations. Its use is justified with logarithmic function under an integral as well in numerator and in denominator. The form in denominator is that which interests us in general. If several logarithms of independent variables arise in a multiple integral, they will be treated one after the other taking care of the upper boundary (or more exactly the boundary than we make tend towards infinite which in certain exercises may be the lower boundary).

To support our statement, we carry out the development of a more general expression of the problem.

P 320/390

Logarithmic monotony

We carry out the development by integral parts (u = F(t) where F is the primitive of the function f, an integrable function (in general a polynomial), v = Ln n (t), u' = f(t), v' = n.Ln n-1 (t)/t ) :

∫ f(t).Ln n (t)dt = [F(t).Ln n (t)] -n.∫ f(t)Ln (n-1) (t)dt Then, we can write the succession of equalities :

By successive eliminations between the last term of these expressions and the first term of the following expression multiplied by the adequate factor, it follows :

∫ f(t

We get while tending c towards infinite : This expression shows again that logarithms can be extracted from integrals taking value of the divergent upper boundary of this integral.

For the practical use of the previous result, there should be no singular points neither for f(t), nor for ln(t) in the interval of integration. In particular, the replacement of the indefinite boundary by a definite boundary c 0 should not involve a divergence (hence c 0 = 2 in general).

Restrictive note :

The passage of relation ( 4) to relation (5) supposes that the limit of a product is the product of the limits what is not proofed, (we think), in the case general ! However, the numerical examples confirm it systematically.

Derivation

We can carry the same reasoning while derivating.

We have :

If f(t) is a polynomial expression, when t tends towards infinite, f(t)/t/f'(t) tends towards a constant and the members of this equation tend towards 1. Thus, the logarithm can be extracted from the derivative as in the preceding case.

EXERCISE 15 : GAPS BETWEEN NUMBERS

Thermodynamics and integers

We offer this paragraph as recreation for the reader. Let us consider all collection of prime numbers and a law of composition (.) called multiplication. Let us call S k (p) the entropy of p and pose : S k (p) = k.Ln(p) Then for n = ∏ p i , we get : S k (n) = k.∑iLn(p) = ∑ i.S k (p) and so S k (n) -∑ i.S k (p) = 0

Integers are generated by prime numbers thanks to the law of composition (.) with a null-difference of entropy, i.e. without increase in disorder.

We can express that as (multiple factors are broken up into as many factors who contribute in an equivalent way) :

Proposition : A number is equivalent to the sum of the disorders of his factors.

There is obviously no increase in entropy by multiplication with neutral element :

Prime numbers are the most primitive forms of numbers. If we consider the process of "creation" of prime numbers, an approach, at the same time simple and satisfactory, is the following (the reader will refer for this relation later in the text) :

Boltzmann constant equals 1 here, which is a priori an ideal value.

A prime number is created by adding the precedent one and its disorder. Thus, as p i  ∑ln(p j ), we can express the point by : Proposition : The value of a prime number is "equivalent" to the disorder of his prime predecessors.

According to [6] (p 98 and pages following), the concept of entropy can be also transcribed in terms of lack of information. The lack of information mi is equivalent to the logarithm of a probability :

This value is based thus naturally and without discrimination on the lacks of information which it receives from the collection of prime numbers already created. The quantum of information k equals 1, reassuring result in binary theory which is the base of information theory (0 = no signal, 1 = reception of an information).

Proposition : The value of a prime number is "equivalent" to the lack of information of his prime predecessors. (this is amazing knowing that each one of these numbers is perfectly defined and knowing the regularity of the construction of prime numbers by Eratosthenes sieve)

Another concept being attached to the entropy is that of energy. We can say that :

Proposition : Integers multiplication takes place with a minimal energy need.

By extension of the principle :

Proposition : Numbers multiplication takes place with a minimal energy need.

It is natural that a collection of elements obeying a physics with minimal energy borrows from Boltzmann statistics. What about Fermi statistics ?

A little bit of philosophy

Preceding paragraph dealt with a physical approach of prime numbers. But, which is really the first one : the physical world or the world of numbers ? If it is the second, then the reasoning is to be made reverse. The physical phenomena result from the laws of numbers and entropy phenomena is just another expression of the nature of numbers.

Differences between expressions of integers

Let us return to more serious purpose.

Let us have F a function. Let us write the Taylor development to order 1 of F : F(n i+1 ) ≡ F(n i ) + (n i+1 -n i ).F'(n i ) Thus we have : n i+1 = n i + 1 ( 1)

Average differences between expressions of prime numbers

Jacques Hadamard and Charles-Jean de la Vallee Poussin, thanks to Riemann's precursory work [3], proved in 1896 the theorem of rarefaction of prime numbers and gave the asymptotic expression for the i th prime number π(p i ) ≡ p i /Ln(p i ).

From #{(p) < p i } → p i / Ln(p i ), when p i tends towards infinite, we draw for "large" p i , #{p < p i } ≡ p i /Ln(p i ) and #{p < p i+1 } ≡ p i+1 /Ln(p i+1 ), so that :

This expression induces a density of average probability 1/Ln(x) at x or a distance between a prime number p i and its successor p i+1 of Ln(p i ). Thus :

Let us note here that the selected origin value (here p 1 =2) is not essential because the logarithm has an effect of very fast "absorption" which brings back the initial terms Ln(p i ) to a negligible value compared to the whole set (which is in fact infinite).

Let This means that the density of numbers F(x), x a prime number, is 1/(F'(x).Ln(x)) at position x. We observe the "footbridge" between integers' variables (relations ( 1) and ( 2)) and prime numbers variables (relations ( 3) and ( 4)) by the simple addition of a logarithm in the equations.

We could not limit us to the developments of Taylor of the first degree :

However this correction has, a priori, only little or no incidence on the asymptotic cases and we will not return to it.

We find on the graph below, limiting us to first degree Taylor development, a good agreement between iterative values and actual values. The graphs in correspondence, indicated by sharp colours on the one hand and pastels close tones on the other hand, slightly shifted close to the origin but meet when x-coordinate increases.

EXERCISE 16 : INTEGER ENUMERATION IN A HYPERVOLUME AND ON ITS SURFACE

Waring case

We tackle here the problem of enumeration of diophantine equation solutions (x i positive integers ) :

Let us recall, in preamble, that the enumerations are asymptotic only when target c tends towards infinite. Nothing however prohibits making calculations out of this framework, using results later on in an adapted framework. To answer the present problem, we write first :

The reader will find theoretical arguments and evaluation of volumes for double and triple integrals in [13]. The reasoning is prolonged in dimension n cases satisfying the following integral :

This apparently complex integral is with separable variables and is resolved in a product of simple integrals.

To get the expression of this integral, we study first integral :

The variable in this expression is y and x = r(x 1 , x 2 , …) can be regarded temporarily as a constant. Adopting the change of variable z = y/(c-r(x)) (1/n) , we get dz = dy / (c-r(x)) (1/n) , so that dy = (y/z)dz.

It is clear that the successive use of this relation in the multiple integral leads gradually to a product of simple integrals, with a coefficient c to the power k/n.

The I(n,i) integrals do not have, a priori, any literal formulation out of the cases n = 1 and n = 2 (see further the literal formulas for these). However, they are independent of c and can be calculated with an excellent precision for any given n without difficulty. The following tables give some numerical illustrations of them. For I(n,i)

The volume is then :

-in dimension 2n :

The derivative with respect to c of the last expression is :

Finally for the enumeration, we use the results of exercise 5 relation ( 52) : Thus, and after normalization (by multiplication by p

(k+1)(pi-1)/4 .(1+(-1) k+1 )) and T(c,(u (n,i) , v (n,i) , w (n,i) )) = {sorting function of (u (n,i) , v (n,i) , w (n,i) )) according to increasing g i r , r describing integers of [0,p i -1]}

Examples

The table below gives abundance factors for a number of variables k ranging between 1 and 5 :

For k = 1, there exists always a sequence p for which c = k. g 2u mod p. The factors product is then 0. Thus the case triviality.

For k = 2, « c = k. g.g 2u mod p » and « c = k. g 2u mod p » gives the same result. Hence:

Fan(c) = Π (1-1/p) . Π (1+1/p) . Π (2-1/p) . Π (1/p) p = 1 mod 4 p = 3 mod 4 p\c p = 1 mod 4 p\c p = 3 mod 4

The product Π p=1mod4 (1-1/p). Π p=3mod4 (1+1/p) equals roughly 1,2732 and the normalized abundance factor follows :

The condition of divergence for V'(c) is then :

When m(k/n-1)+1 = 0, we also have a divergence of the number of solutions, but this divergence is logarithmic, therefore very slow. Let us note that the divergence of V'(c) is not exactly the same as the divergence of target c solutions. It is also necessary to take account of the behaviour of Fan(c) (who can be null).

Let us give some examples : +(-x 3 ) 3 = c}, the abundance factor is given by the same expression as in the case of equation {x 13 +x 2 3 +x 3 3 = c}. Thus we have :

Here the term f is that we found at exercise 5 for degree 3 Waring equations. Evaluation of Fan (0) shows that this term tends towards zero when sequences increase. For c a cube, the Fan(c) term increases regularly (tending towards infinite).

We recall that x 1 3 + x 2 3 = x 3 3 is a Fermat theorem example shown without solutions by Ernst Kummer. In addition, for c a cube, x 1 3 +x 2 3 = c+x 3 3 solutions increase linearly with x 3 and Fan(c) cannot thus be other than an infinite term. Otherwise, the number of solutions increases as a logarithmic with x 3 . followed by Roger Frye with the smallest solution 95800 4 +217519 4 +414560 4 = 422481 4 . This is however not contradictory with Fan(c) who tends towards zero when sequence p increases.

Example 3 : Euler conjecture

The Swiss mathematician conjectured that there is no solution whenever n = m, k = n-1 and c = 0. This is invalidated by Noam Elkies example. Our study shows that for c = 0, the number of solutions is o(Ln(x)) thus rare but not necessarily finite. In addition, for c ≠ 0, the number of solutions tends towards infinite in logarithmic ways -if fan(c) converges towards a non-null limit what is the case in general but we do not demonstrate this here -.

EXERCISE 17 : PRIME ENUMERATION IN A HYPERVOLUME AND ON ITS SURFACE

Case of the hypervolume

We tackle here the problem of the solutions enumeration of diophantine equality (y i positive prime) y 1 n + y 2 n +…+ y m n = c (1) using as usually y 1 n + y 2 n +…+ y m n < c ( 2)

As for the previous exercise, the enumeration corresponding to the equality will be rigorous only when c tends towards infinite. We write the following integral (with the usual reserves on the boundaries (c to be replaced by c-2))

(dy m /Ln(y m )).(dy m-1 /Ln(y m-1 ))…(dy 1 /Ln(y 1 )) ( 3)

To get this integral expression, we study first the following one :

The variable of this expression is y and x can be regarded temporarily as a constant.

Recalling the property of the logarithm under an integral, we write as c tends towards infinite : [START_REF] Dieudonné | Abrégé d'histoire des mathématiques[END_REF] 2

Here, we consider the case of multiple integrals (we know that the property is true in the case of the simple integral). Then the upper boundary (c-r(x)) (1/n) will take values such as on the average :

1/Ln (c-r(x)) (1/n) = 1/Ln(e i .(c) (1/n) ) =1/(Ln(e i )+Ln((c) (1/n) )) = <a i >/Ln(c 1/n ) ( 6)

where e i is ranging in interval ]0,1[. Then, adopting the change of variable z = y / (c-r(x)) (1/n) , we get dz = dy / (c-r(x)) (1/n) so that also dy = (y/z)dz.

ε tends towards 0 when c tends towards infinite. By calculating V(c) gradually, this expression with m variables will be written :

where a m = <a 1 >.<a 2 >…<a m >.

Let us pose according to the notations of the preceding exercise :

Prime enumeration in a hypervolume and on its surface When c tends towards infinite, in expression 1/(Ln(e i )+Ln((c) (1/n) )) = <a i >/Ln(c 1/n ), e i being finite, Ln(e i ) becomes negligible compared to Ln(c (1/n) ), thus each <a i > will tend towards 1 and it will be the same for a.

Of course, we have to do here with logarithms. This means an extremely slow convergence towards the aforementioned limit as the graph shows it below. In this graph, we represented the values taken by coefficient a defined as the ratio of the real solutions to asymptotic theoretical solutions, that is :

Schematically, we observe, when n increases (with m constant), a "translation with elongation" towards the right of the graphics of the representative curves of a(n,m) and when m increases (with n constant), a "translation with contraction" towards the top of the graph. For a(1,1), we deal with the problem, actually solved, of the rarefaction of the prime numbers. We know, according to Hadamard and de la Vallee Poussin, that in this case a tends towards 1. We observe that, close to the origin, a takes initially values lower than 1, increasing later on to join a zone of oscillations (a not being represented here for all unit increments of c, the whole set of oscillations do not appear) before dropping gradually. This phenomenon of oscillations close to the origin are found in the other examined cases, but "close to the origin" can now mean targets c of already big sizes. Thus for a (3,3), the zone of downward trend is hardly started on the graph and for a (4,4) it is not reached yet whereas c equals 10 10 (that is c 1/4 ≈ 316 what brings back well in the zone of oscillation of a (1,1)). The order of magnitude c > 500 n is more or less, the x-coordinate origin for the zone of a downward trend of coefficient a(n,m).

Let us return now to the derivative of expression V(c) :

This expression takes the limit value :

The condition of divergence of this term is : m > n (11) Concerning the enumeration ad infinitum of the solutions of ( 1), it is necessary to weight the result obtained at (8) according to the method given in exercise 5 :

Evolution of coefficient a

Prime enumeration in a hypervolume and on its surface

General hypervolume case

Let us have : a 1 .y 1 n + a 2 .y 2 n +…+ a m .y m n < c (13) a 1 .y 1 n + a 2 .y 2 n +…+ a m .y m n = c (14) We studied in the preceding paragraph the effect of the change of variables (y i ) → (Y i ) where Y i n = a i .y i n on the integral of enumeration of the function volume of the diophantine inequality and equality with integer variables. This change of variables operates exactly in the same way here but we must hold account in addition of the behaviour of logarithmic expressions. We have

According to the development leading to relation ( 8), we deduce that c must be replaced by c/a i in our new expression. Thus :

When c tends towards infinite Ln((c/a i ) 1/n ) is equivalent to Ln(c 1/n ). However, for any numerical checking, it is convenient to preserve the initial expression.

Using then a m → 1, we get the limiting value of V' (c) :

The divergence condition remains : m > n (17) As for the enumeration ad infinitum of the solutions of ( 9), it is then necessary to weight the result obtained at (16) according to the method given in exercise 2 :

Case of the hyperplane

We use the preceding result, with n = 1, as well for the diophantine inequality and the equality (x i positive integers) :

The divergence condition of the last term is m > 1 and the sought cardinal, using the results of exercise 5, is :

#{y 1 + y 2 +…+ y m = c} ≡ Fan(c).c (m-1) /((m-1)!Ln m (c)) ( 23) with :

Prime enumeration in a hypervolume and on its surface

We find this expression asymptotic in [2] p 66 as :

#{y 1 + y 2 +…+ y m = c} = Sm(c).c (m-1) /((m-1)!Ln m (c)).(1+O(ln ln(c)/ln(c)) ( 25) where

and we showed at exercise 5 relation ( 9) the equality of Fan(c) and S m (c) which follows out here from ( 23) and ( 25).

General case of the hyperplane

The general case is solved without difficulty using afore studied elements. We can write directly : m #{(y 1 , y 2 , …, y m ) / a 1 y 1 + a 2 y 2 +…+ a m y m < c} ≡ 1/m!. П (c/a i )/Ln((c/a i ))

or(and(p i ∤c,p i \a i ),and(p\c,p∤a i )) p i ∤c, p i ∤a i p i \c, p i \a i

We give here a numerical example for the diophantine equation :

The use of relations ( 27) and ( 28) lead, with the proviso of using the factor of correction a 3 = (1.23) Here p, q and r describe the prime numbers including p = 2, which explains the occurrences for the even values of c. Of course, these occurrences become negligible in front of those obtained for the odd values of c, when c tends towards infinite. The volume function of cardinals expression being common to the even and odd values of c, fan(c) = 0 is coherent with c an even target (it would be also the case while withdrawing p = 2 of the list of prime numbers).

Let us note also that Fan(c) is obtained at sequence p i = 1613 with an excellent precision (the 8 th and more decimals after the comma vary imperceptibly).

We present another example below for lesser quantity of variables (m = 2 instead of m = 3). The dispersion of the enumerations is larger than in the preceding case.

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Prime enumeration in a hypervolume and on its surface

Goldbach problem

In the previous case, for k = 2, we had (when c tends towards infinite)

#{y 1 + y 2 = c} ≡ Fan(c).c/Ln 2 (c) (29) with :

Problem of Vinogradov

In the previous case, for k = 3, we had #{y 1 + y 2 + y 3 = c} ≡ Fan(c).c 2 /(2.Ln 3 (c)) (31) with :

We can get in [11] the asymptotic enumeration of Vinogradov (proven result) in the form of :

where the fudge factor S is written :

The reader will verify without difficulty, thanks to (32) and ( 34), the equality :

Case of the hypersphere

We are interested by (y i positive prime numbers) : Prime enumeration in a hypervolume and on its surface

Finally for the enumeration of (36), we use the results of exercise 5 at (59), ( 60) and ( 62): Thus, and after normalization by multiplication by p i /(p i -1) n lim #{y 1 ² + y 2 ² +…+ y n ² = c}= ( П T(c,(u (n,i)

with, for p i = 2

with, for p i = 1 mod 4, if n = 0 mod 2 : 2) .sh((n+1).Ln((p i 2) .sh((n-1).Ln((p i 1/2 +1)/(p i -1) 1/2 )))

with, for p i = 1 mod 4, if n = 1 mod 2 :

with, for p i = 3 mod 4 : u (n,i)

(1/p i ).((p i -1) n +(p i -1).(p i +1) n/2 .cos(n.arcos (-1/(p i +1) 1/2 )))

v (n,i) = (1/p i ).((p i -1) n +(p i +1) (n+1)/2 .cos((n-1).arcos (-1/(p i +1) 1/2 ))) w (n,i)

(1/p i ).((p i -1) n +(p i +1) (n+1)/2 .cos((n+1).arcos (-1/(p i +1) 1/2 )))

T (c, (u (n,i) , v (n,i) , w (n,i) )) = {sorting function of (u (n,i) , v (n,i) , w (n,i) )) with increasing g i r , r describing integers range [0,p i -1]} w (n,i) is used only with p i ≥ 3

Case n > 2

Here, we could use again the earlier studied cases n = 3, 4, 6 or 12 (explicit formula), what we do not make to reduce the already cumbersome text. We distinguish here the power of the last variable (r instead of n).

Case of a boundless volume

We got above for a 1 .y 1 n + a 2 .y 2 n +…+ a m .y m n < c :

So that :

Then, replacing V m (c), it follows :

Prime enumeration in a hypervolume and on its surface m y m+1 m V(c) ≡ a m . П I(n,i) ∫ 1/Ln(y m+1 ). П ((c+a m+1 .y m+1 r )/a i ) 1/n /Ln(((c+a m+1 .y m+1 r )/a i ) 1/n ).dy m+1 (41) i = 1 2 i = 1

Asymptotically c is negligible, so that : m y m+1 m V(c) ≡ a m . П I(n,i) ∫ 1/Ln(y m+1 ). П ((a m+1 .y m+1 r )/a i ) 1/n /Ln(((a m+1 .y m+1 r )/a i ) 1/n ).dy m+1 i = 1 2 i = 1

Then, as Ln(((a m+1 .y m+1 r )/a i ) 1/n ) = (r.Ln(y m+1 )+Ln(a m+1 /a i ))/n ≈ (r/n).Ln(y m+1 ) m y m+1 m V(c) ≡ a m .(a m+1 /a i ) m/n .(n/r) m . П I(n,i). ∫ 1/Ln(y m+1 ). П y m+1 r/n /Ln(y m+1 ).dy m+1 i = 1 2 i = 1

Still asymptotically, we can extract the logarithms from the integral (the lower limit 2 for y m+1 can also be replaced by 0). A new coefficient of correction a', which tends towards 1, is then leaned against a m , which is equivalent to rewrite the product as a m+1 where a tends towards 1 on the asymptotic branch : m y m+1

V(c) ≡ a m+1 .(a m+1 /a i ) m/n .(n/r) m /Ln m+1 (y m+1 ). П I(n,i). ∫ y m+1 r.m/n .dy m+1 i = 1 0

Thus finally (r.m/n is positive and thus different from -1) : m V(c) ≡ 1/(r.m/n+1).a m+1 .(a m+1 /a i ) m/n .(n/r) m .( П I(n,i). ).(y m+1 r.m/n+1 /Ln m+1 (y m+1 )) (

For the evaluation of V'(c), it is necessary to return to the relation (41). By derivation with respect to variable c, we get : m y m+1 m V'(c) ≡ a m+1 . П I(n,i).( ∫ 1/Ln(y m+1 ). П ((c+a m+1 .y m+1 r )/a i ) 1/n /Ln(((c+a m+1 .y m+1 r )/a i ) 1/n ).dy m+1 )'

The derivation can be carried inside the integral : m y m+1 m V'(c) ≡ a m+1 . П I(n,i). ∫ (1/Ln(y m+1 )).( П ((c+a m+1 .y m+1 r )/a i ) 1/n /Ln(((c+a m+1 .y m+1 r )/a i ) 1/n ))'.dy m+1 i = 1 2 i = 1

We can extract the constant multiplicative factor due to the coefficients a i out of the integral and asymptotically neglect these same coefficients in the logarithm. Also let us note, that asymptotically, the logarithm can be extracted from a derivative, as we do for integrals. Hence :

So that : m y m+1

V'(c) ≡ a m+1 . П I(n,i)/a i 1/n . ∫ (1/Ln(y m+1 )).((c+a m+1 .y m+1 r ) m/n /Ln m ((c+a m+1 .y m+1 r ) 1/n ))'.dy m+1 i = 1 2

As (u/v)' = u'/v-u.v'/v², we have, for the variable c, ((P(c)) m /Ln m (P(c)))' = m.(P(c))'.(P(c)) m-1 /Ln m (P(c)).(1-1/Ln(P(c))).

Thus with P(c) = (c+a m+1 .y m+1 r ) 1/n and P'(c) = (1/n).(c+a m+1 .y m+1 r ) 1/n-1 , it follows :

. П I(n,i) ∫1/Ln(y m+1 ).m/n.(c+a m+1 .y m+1 r ) 1/n-1+(m-1)/n /Ln m ((c+a m+1 .y m+1 r ) 1/n ).(1-Ln -1 (c+a m+1 .y m+1 r ) 1/n ).dy m+1 i = 1 2

That is :

Prime enumeration in a hypervolume and on its surface m y m+1

V'(c) ≡ a m+1 .m.n m-1 .a i -m/n

. П I(n,i). ∫(1/Ln(y m+1 )).(c+a m+1 .y m+1 r ) m/n-1 /Ln m (c+a m+1 .y m+1 r ).(1-(Ln -1 (c+a m+1 .y m+1 r ))/n).dy m+1 i = 1 2

Asymptotically 1/n/Ln(c+a m+1 .y m+1 r ) is negligible in front of 1, just as c in front of a m+1 .y m+1 r . We have also Ln(a m+1 .y m+1 r ) ≈ Ln(y m+1 r ) = r.Ln(y m+1 ). It follows : m y m+1

V'(c) ≡ a m+1 .m. n m-1 .(1/a i ) m/n . П I(n,i). ∫ (a m+1 .y m+1 r ) m/n-1 /Ln m (y m+1 r )/Ln(y m+1 ).dy m+1 i = 1 2 Hence m y m+1

V'(c) ≡ a m+1 .a m+1 -1

.(m/n).(n/r) m .(a m+1 /a i ) m/n . П I(n,i). ∫ y m+1 r(m/n-1) /Ln m+1 (y m+1 ).dy m+1 i = 1 2

Extracting the logarithm, we get :

.(m/n).(n/r) m .(a m+1 /a i ) m/n /Ln m+1 (y m+1 ). П I(n,i). ∫ y m+1 r(m/n-1) .dy m+1 i = 1 2

Two cases are to be examined as in the exercise concerning prime numbers variables.

If r(m/n-1)+1 ≠ 0 m V'(c) ≡ a m+1 .a m+1 -1

.(m/(r.m-r.n+n)).(n/r) m .(a m+1 /a i ) m/n .( П I(n,i)). y m+1 r(m/n-1)+1 /Ln m+1 (y m+1 )

The divergence condition of V'(c) is thus here the strict inequality : r(m/n-1)+1 > 0

In peculiar, when r = n, this condition is written : m > n-1

EXERCISE 18 : GENERATION OF PRIME NUMBERS BY WARING SUMS

Case of degree n powers, variables of integers

We are interested here by the enumeration of the solutions of the diophantine equation (x i positive integers) :

To answer this problem, we write initially :

The usual reasoning leads us to write the following integral :

This integral with separable variables is solved as a product of simple integrals. By the previously used method, we can give his expression directly :

We can neglect c in this expression.

Hence:

Concerning the enumeration ad infinitum of the solutions of (1), it is necessary to weight the result we obtained in (3) with the normalized abundance factors we get thanks to relation (21) of exercise 8 and multiplication by p i /((p i -1).p i k :

p i (A) is the cardinal matrix studied in exercise 5 and the notations of this exercise are that adopted afore.

Case of degree 2 sums

p →∞ i = 1

From exercise 8 relations ( 22) and ( 24), we get after normalization (multiplication by p i /(p i k .(p i -1))) :

P 348/390

Generation of prime numbers by Waring sums

For p i = 1 mod 4 :

For p i = 3 mod 4 :

Thus per examples :

In addition for p i = 2 (the reader will easily verify this case), after normalization the abundance factor is 1.

We can rewrite the sorting function T(c, (u ki1 , v ki1 , w ki1 )) and T(c, (u ki3 , v ki3 , w ki3 )) defined previously. Indeed, we know already that u ki1 and u ki3 correspond to the case p i \c. For v ki1 and v ki3 , they correspond to the cases where c = g i x.di

.g i 0 mod p i , x an integer, g i a primitive root of p i and d i = (n,p i -1) = (2,p i -1) = 2. For w ki1 and w ki3 , they correspond to the cases where c = g i x.di .g i 1 mod p i , x an integer, g i a primitive root of p i and

d .g i 1 mod p i then c (pi-1)/2 = g i (pi-1)/2 = -1 mod p i per definition of the primitive. Then we use the condition x is an integer to get the abundance factor : In addition (see exercise 16 relation (29

We can use again all the preceding results by replacing matrix A by matrix B (while introducing for numerical verifications the coefficient a of exercise 17 which we know to tend very slowly towards 1). Then, according to relation (10) +∞

p i

Here (B) = (A)-(I) is the cardinal matrix of prime numbers variables cases.

Case of degree 1 sums, prime numbers variables

We directly give the result deduced from former work :

we deduce of relation ( 5) of exercise 5, by incrementing m (as we know that p and -p have same representative, thus y 1 + y 2 +…+ y m = p + c becomes y 1 + y 2 +…+ y m -p = c and then y 1 + y 2 +…+ y m + y m+1 = c).

Case of twin prime numbers

We begin again in detail.

We write for the equality and the inequality :

The enumeration of ( 16) answers the following integral :

where

The integration of (18) gives, after logarithm extraction while neglecting the lower boundary, V 1 (c) = y 1 /ln(y 1 ) avec y 1 = c+y 2 then, y 2 V(c) = ∫ (c+y 2 ).dy 2 /(ln(c+y 2 ).ln(y 2 )) V(c) = 1/(ln(c+y 2 ).ln(y 2 )) ∫ (c+y 2 ).dy 2 2 So that :

V(c) = (y 2 2 /2+cy 2 )/(ln(c+y 2 ).ln(y 2 )) ( 19)

Asymptotically, the value of c becomes negligible and the enumeration of ( 16) is :

For the enumeration of ( 15), we must return to (19) and derive with respect to c while keeping variable y 2 constant. Thus :

V'(c) = y 2 /(ln(c+y 2 ).ln(y 2 )) + (y 2 2 /2+cy 2 ).((ln(c+y 2 ))/y 2 +(ln(y 2 ))/(c+y 2 ))/((ln(c+y 2 ).ln(y 2 )) 2 )

Here, asymptotically, c is again negligible, so that :

As the second term is negligible at the limit, finally :

Let us note however that the convergence of ( 21) towards ( 22) is very slow (because logarithmic). The asymptotic enumeration of ( 15) becomes thus, as in mathematical literature : lim #{y 1 -y 2 = c} ≡ Fan(c).y 2 /Ln 2 (y 2 ) (23) y 2 →∞ with :

(p i -1) 2 (p i -1)

Case of degree 2 powers, prime numbers variables

p →∞ i = 1

From exercise 8 relations ( 22) and ( 24), we get after normalization (multiplication by p i /((p i -1) m .(p i -1))) :

For p i = 1 mod 4 :

For p i = 3 mod 4 :

We give, as examples, the following For

Let us note that :

-for p = 2, Fan({1}) = 0, that is any target 1 mod 2 has a null abundance factor, -for p = 3, Fan({2}) = 0, that is any target 2 mod 3 has a null abundance factor.

Then :

We still have : m In 2004, Terence Tao and Ben Green presented a prepublication showing that there exist arbitrarily long prime numbers arithmetic series. Recently, this work was extended to the polynomials. The object of this exercise is to supplement the proof of existence of solutions by their actual enumeration.

Generation of integers by a polynomial

We examine a simple problem bringing an interesting lighting on what will follow. Let us consider a polynomial with integer coefficients increasing on whole studied interval [x 0 ,x[, that is P(x). Variable x takes integer values and we want to determine the number of integer solutions observed in the interval [y 0 , y[ = [P(x 0 ), P(x)[ by P(x). Self-evidently, there are exactly x -x 0 , that is also P -1 (P(x))-P -1 (P(x 0 )), or P -1 (y)-P -1 (y 0 ). Thus (replacing y by x) :

#{n c [x 0 , x[ / P(n) = c, c an integer, n an integer} = P -1 (x)-P -1 (x 0 )

For x 0 = 0 and P(0

The highlighting is thus given on the importance and the way in which the reciprocal function of P appears.

Generation of prime numbers by a polynomial on a boundless volume

Let us write the diophantine equation :

where n is a variable of integers and p a variable of prime numbers.

With the usual strategy, we write initially the integral representative of volume P(n) < p+c :

Here p is not limited by the value c. We can then write, as p is independent of n :

Derivation with respect to c gives than p V'(c) ≈ (1/Ln(p)). P -1 (c+p) -(1/ p.Ln 2 (p)). ∫ P -1 (c+p)dp 2

The asymptotic expression of P(n) and P -1 (n) is obtained thanks to the dominant monomial. Thus P(n)=a.n k and P -1 (n)=(n/a) (1/k) .

We have also asymptotically P -1 (c+p) ≈ P -1 (p). The derivative becomes then V'(c) ≈ P -1 (c+p)/Ln(p). (1-k/((k+1).Ln(p)) which reduces asymptotically to lim V'(c) = lim P -1 (c+p)/Ln(p) = lim P -1 (p)/Ln(p)

(3) p → +∞ p → +∞ p → +∞ Then, using preceding results (exercise 8) :

lim #{(n,p) / P(n) -p = c} = lim Fan(c).P -1 (p)/Ln(p) (4) p → +∞ p → +∞ with a normalized abundance factor identical to the case of finite volume (the passage of p to -p does not have an effect on these factors) :

Generation of prime numbers by polynomials Let us have k and a respectively the degree and the dominant coefficient of P. Then asymptotically P(x) → a.x k and P -1 (x) → (x/a) (1/k) . Thus : lim #{(n,p) / P(n) -p = c} = lim Fan(c). (p/a) (1/k) /Ln(p) (5) p → +∞ p → +∞

In general, close to the origin, a corrective factor (common to all targets) is necessary to match the theoretical and the empirical curves. However, we can trace asymptotic evolution curves, while freeing us possibly of a correction factor, (preserving c in the expression if c is "large"), by using, either for monomials (Δp i = p i -p i-1 ) :

∑ Δp i . Fan(c).(1/k).(1/a) (1/k) .(p i +c) (1/k-1) /Ln(p i ) (6) i → +∞ or for polynomials, thanks to graphic approximation of P -1 (inversion of axis) : ∑ Δp i . Fan(c).(P -1 (p i +c))'/Ln(p i ) (7) i → +∞ The two examples, chosen for the fact that they give a relatively high number of solutions, show the good adequacy between the number of solutions obtained by direct evaluation and literal evaluation.

3. Case of a degree 2 polynomial. Hardy-Littlewood conjecture.

The case is solved immediately using above results and results of exercise 8.

Then for c = -1 :

Replacing Legendre symbol by its literal expression :

We recover here the usual expression of Hardy-Littlewood conjecture.

The convergence towards this limit value is slow. Thus, for n²+1 lower than 10 8 , a correction of about 10% is necessary, for HARDY LITTLEWOOD CONJECTURE (cardinal of prime numbers of type n 3 +n 2 -1544n+c) P 358/390 Generation of prime numbers by polynomials example, to match theoretical and empirical curves (see graphic illustrations). However, the expression below, pending to expression (6), does not require such an important factor of correction close to the origin (Δp i = p i -p i-1 ).

∑ Δp i . Fan(c).(1/2).(1/a) (1/2) .(p i +c) (-1/2) /Ln(p i ) (9) i → +∞

Generation of prime numbers by identical polynomials sum

Let us consider the diophantine equation :

(P(x 1 )) n + (P(x 2 )) n +…+ (P(x k )) n = p + c (10) where P(x) = a d .x d + a d-1 .x d-1 +…+ a 0 Asymptotically P(x) ≡ a d .x d and the diophantine equation admits then for V(c) and V'(c), the equations resulting from the study of the Waring sums case. However, to get the abundance factors, it is necessary to return to the initial equations. We will undertake literal evaluations within a broader framework thereafter.

EXERCISE 20 : GENERATION OF PRIME NUMBERS BY FERMAT-CATALAN SUMS AND ANALOGUES

Case of a finite hypervolume, unit mesh

To begin with, we tackle the problem of diophantine equation solutions enumeration in the case of positive integers' variables and a given target c :

Here the exponent of the variable x i is not equal to i but a positive integer noted (i).

To solve this problem, we write according to the usual way

whose approximate enumeration is given by :

x 1 = c (1/(1)) x 2 = (c-x 1 (1) ) (1/(2)) x 3 = (c-x 1 (1) -x 2 (2) ) (1/(3)) x k = (c-x 1

This integral is with separable variables and is solved in a product of simple integrals.

To get the expression of this integral, we study first the integral :

The variables in this expression are x i and r(x) does not depend of these variables (all x i are independent variables) and can be regarded temporarily as a constant. Adopting the change of variable z = x i /(c-r(x)-x i-1 (i-1) ) (1/(i)) , we get dz = dx i /(c-r(x)-x i-1 (i- 1) ) (1/(i)) , so that also dx i = (x i /z)dz. Moreover (c-r(x)-x i-1

It is clear that the successive use of this relation in the multiple integral leads gradually to a product of simple integrals as follows :

Step

Thus :

The derivative results easily :

The divergence condition for V'(c) is :

1/(k)+1/(k-1)+…+1/(2)+1/(1) >1 ( 6)

Expressions ( 3) and ( 4) can be also written as Г functions. For that, let us proceed initially to the change of variable z = t (i) , then dz = (i).t (i)-1 dt (recalling again that (i) is an indice representing a positive integer), so that :

Generation of prime numbers by Fermat-Catalan sums and analogues 1 1

∫ (1-t (i) ) (1/(k)+1/(k-1)+…+1/(i+1)) dt = 1/(i) ∫ z (1/(i)-1) .(1-z) (1/(k)+1/(k-1)+…+1/(i+1)) dz 0 0

We recall the identities given at exercise 16 for Beta and Gamma functions : As with exercise 16, combining here ( 6) and ( 7), we get mutual elimination of Γ expressions in numerator and denominator of the whole product. It results from it :

Thus using gamma function factorisation property Г(x+1) = x.Г(x), we get (here (i) describes all the values (1) to (k)) :

V

V'(c) = ∑(

).

(i

These two expressions generalize Waring case (indices (i) of same value) given at exercise 16.

We use thereafter, in a preoccupation with writings simplification, the two expressions :

Euler and Weierstrass formulas Γ(s) = ∑ n→∞ n!.n s /(s.(1+s)…(n+s)) and Γ(s) = (e -γ.s /s).∏ n→∞ (1+s/n) -1 .e s/n are essential tools for numerical applications. Euler-Mascheroni constant γ is roughly 0,57721566490153286). Incidentally, we may also use, the Euler reflexion formula (called also formula of the complements) Γ(s).Γ(1-s) = π/sin(π.s) and the duplication formula Γ(s).Γ(s+1/2) = 2 1-2s .π 1/2 .Γ(2s), s complex number, 0 < Re(s) <1.

The application to two variables x and y with (1) = p, (2) = q gives: 1 V'(c) = (1/p+1/q)c 1/p+1/q-1 ∫(1-t p ) (1/q) dt ( 14) 0

Case of a boundless hypervolume, unit mesh

We tackle now the problem of diophantine equation solutions enumeration (x i positive integers) in the case : (1) + x 2 (2) +…+ x k (k) < c + x k+1 (k+1) + x k+2 (k+2) +…+ x k+i (k+i) (16) Let us pose : r = x k+1 (k+1) + x k+2 (k+2) +…+ x k+i (k+i)

Enumeration of ( 16) answers to the integral :

where x 1 = (c+r) (1/(1)) x 2 = (c+r-x 1 (1) ) (1/(2)) x 3 = (c+r-x 1 (1) -x 2 (2) ) (1/(3)) x k = (c+r-x 1

This last expression is managed like previously until the step k. Thus :

To slightly simplify, we place ourselves in the case : r = x k+1 (k+1)

Then :

x k+1

V(c) = cti. ∫ (c+x k+1 (k+1) ) sti .dx k+1 0 and while deriving inside the integral with respect to c :

x k+1

V'(c) = cti.sti. ∫ (c+x k+1 (k+1) ) sti-1 .dx k+1 0 Asymptotically, c are negligible, therefore :

x k+1

V(c) = cti. ∫ x k+1 (k+1).sti .dx k+1 0 and :

x k+1

V'(c) = cti.sti. ∫ x k+1 (k+1). (sti-1) .dx k+1 0 Then :

V(c) = (cti/((k+1).sti+1)).x k+1 (k+1).sti+1 and (k+1).(sti-1) ≠ -1 V'(c) = (cti.sti/((k+1).(sti-1)+1)).x k+1

The condition of divergence of V'(c) is thus : (k+1).(sti-1) ≥ -1 (18)

Fermat-Catalan conjecture

We want to verify that if 1/p + 1/q + 1/r < 1 then x p + y q = z r has a finite number of solutions. Let us pose :

x p + y q < c + z r (19)

We thus have according to the preceding notations (1) = p, (2) = q, (k+1) = r, sti = 1/p+1/q, cti = Γ(1+1/p).Γ(1+1/q)/ Γ(1+1/p+1/q). The divergence condition of the general case gives r.(1/p+1/q-1) ≥ -1, so that also 1/p+1/q+1/r ≥ 1. Conversely when 1/p+1/q+1/r < 1, the formulas of the general case show that the number of solutions does not diverge. This is in accordance with Fermat-Catalan conjecture.

More precisely, we get :

Generation of prime numbers by Fermat-Catalan sums and analogues

If 1/p+1/q+1/r ≠ 1 lim #{x p + y q = z r } = Fan(c).

(1/r)(1/p+1/q) . Γ(1+1/p).Γ(1+1/q) . z r.(1/p+1/q+1/r-1) (20) z → +∞

(1/p+1/q+1/r-1) Γ(1+1/p+1/q) If 1/p+1/q+1/r = 1 lim #{x p + y q = z r } = V'(c) = Fan(c).(1/p+1/q) Γ(1+1/p).Γ(1+1/q) . ln(z) (21) z → +∞ Γ(1+1/p+1/q) Foot-note:

We have to verify however, to avoid an undecided case, that the normalized abundance factor does not diverge.

Pillai conjecture

According to Catalan Mihailescu theorem, the only solution, except trivial solutions (1,0) and sometimes (-1,0), of equation We proceed by simple application of the preceding results. We have k = 1, (k+1) = b and sti = 1/a and the divergence condition (18) becomes b.(1/a-1) ≥ -1 so that 1/a+1/b ≥ 1, which is impossible when a and b are integers strictly greater than 1 except for a = b =2. In this last case, we have (x 1 +x 2 ).(x 1 -x 2 ) = c which resolves in a finite number of solutions for a given c value (pending on the number of odd or even factors of c) and resolves in (x 1 ,x 2 ) = (±1,0) when c = 1.

Linear increase

Let us consider the case 1/(1)+1/(2)+…+1/(k) = 1 (23)

for the diophantine equation x 1 (1)

Then :

Thus the dependence to variable x k+1 is linear whatever the exponent of this variable in the studied diophantine equation.

In particular for x 2 + y 2 = c + z r , we have an expression independent of r :

V'(c) = (π/4).z

Taking account of relative abundances, we get : #(c) = Fan(c).(π/4).z (25)

Case of a finite hypervolume, logarithmic mesh

The study of the diophantine equation in the case of prime numbers variables is led in a same way as in exercise 16 while using exercise 13. As usually, to get asymptotic enumeration, we extract logarithms from integrals. We can then use the evaluations made in the case of the unit mesh. We do not give immediately the results as we will give some more general further.

Case of a boundless hypervolume, logarithmic mesh

The preceding remarks still apply.

Friedlander and Iwaniec enumeration

Friedlander and Iwaniec proved in 1996 the infinity of the prime numbers of type x 1 2 +x 2 4 . Using previously quoted results, we have the following asymptotic enumerations : I = ∫ ((c/a zr +z r (zr) -r(X,Y)-Y i (yi) ) (m) )/Ln((c/a zr +z r (zr) -r(X,Y)) (1/(yi)) ).dY i (32) 2The logarithm extraction, when Y i tends towards infinite, will yield : Y i = (c+a zr .z r (zr) -r(X,Y)) (1/(yi)) I ≡ (1/Ln(e.(c/a zr +z r (zr) ) (1/(yi)) ) ∫ (c/a zr +z r (zr) -r(X,Y)-Y i (yi) ) (m) dY i (33) 2 with

(1/Ln(e.(c/a zr +z r (zr) ) (1/(yi)) ) = <a>/Ln((c/a zr +z r (zr) ) (1/(yi)) ) ≈ <a>/Ln((z r (zr) ) (1/(yi)) ) = <a>/(((zr)/(yi)).Ln(z r ))

where, by the same arguments that those developed at (9) in exercise 17, e is comprised in the interval ]0,1[, a tends towards 1 when z r tends towards infinite (c finite and (zr) et (yi) are the exponents of the variables z r and y i ).

We pose thus : The divergence condition is here (zr).(stxy-1) > -1 (42)

Case of a boundless hypervolume, sum of separate polynomials

We finish this exercise by a remark. When we have a diophantine expression composed of polynomials with separate variables (of integers or prime numbers), we can for the evaluation of the hypervolume return to the preceding problem. It suffice to retain only the higher degrees monomials (of each variable). However, abundance factors evaluation is done while returning to the general method (exercise 2).

Numerical application 1

Let us have to solve :

Here, we suppose that x 1 , x 2 , x 3 are integers' variables and p a prime numbers variable.

For the evaluation of V(c) and V'(c), we can neglect p in p 2 +p, thus 2x

Elementary quadratic forms

We are interested here by the diophantine equation :

The transformation of the quadratic form ∑ i ∑ j α ij .x i .x j to λ 1 .y 1 2 +λ 2 .y 2 2 +…+λ n .y n 2 is explained with some numerical examples in [12] (p 747). There, it is mentioned the theorem of the principal axis. We seize this elegant point of view to give the literal approach for u.x 1 2 +v.x 1 .x 2 +w.x 2 2 .

Let us pose

[X] =

x 1 x 2 Let us write in matrix form the quadratic expression :

When [U] is symmetrical, the theorem of the principal axis applies in the terms which follow. Let us have λ 1 and λ 2 the eigenvalues of [U]. The orthogonal matrix [Q] which diagonalizes [U] allows the change of co-ordinates :

x 2 Thus we have :

-sin(θ) cos(θ) It follows : λ 1 = cos 2 (θ).u-sin(θ).cos(θ).v+sin 2 (θ).w λ 2 = sin 2 (θ).u+sin(θ).cos(θ).v+cos 2 (θ).w 0 = sin(θ).cos(θ).(u-w)+(cos 2 (θ)-sin 2 (θ)).v/2 From last equation, we get : tan

Using some elementary trigonometric identities and by choosing k= 0, it follows :

These last equations give the studied conics principal axis. As

Equation λ 1 .y 1 2 +λ 2 .y 2 2 = c, c a constant, is of elliptic type when λ 1 and λ 2 are of same sign (Δ < 0) and of hyperbolic type when λ 1 and λ 2 are opposite signs (Δ > 0). Thus, Δ = 0 is the borderline between these two types of curves. The conditions v = 0 and u.w = 0 form other borderlines involving on principal axis orientation. The principal cases (self-evident linear case is not given here) pending on Δ, u.w and v signs are represented below : P 370/390 Generation of prime numbers by quadratic sums ans analogues Case (Δ < 0, u.w < 0) is not represented as it is impossible.

To limit the further integral boundaries situations, we are interested thereafter only in cases 1 and 3 where u.w > 0 and v > 0 and still more simply with u > 0,v > 0 and w > 0, however admitting borderline cases which do not appear here.

Thus we suppose Δ ≠ 0, u ≥ 0, v ≥ 0 and w ≥ 0.

We begin with the evaluation of "volume" V(c) delimited by the target c when :

u.x 1 2 +v.x 1 .x 2 +w.x 2 2 < c (8)

Here x 1 and x 2 are variables, u, v and w parameters.

The field of definition of the variables is the first quadrant x 1 ≥ 0 and x 2 ≥ 0.

The integration of this expression requires distinguishing positive and negative cases for Δ.

Case Δ < 0

The change of variable x 1 = (4w.c/(4u.w-v 2 )) 1/2 .sin(θ) gives after some elementary handlings : P 371/390

Generation of prime numbers by quadratic sums ans analogues V(c) = arcsine((-Δ/(4u.w)) 1/2 ) .c (10) (-Δ) 1/2 Case Δ > 0

In the same way, the change of variables x 1 = (4w.c/(4u.w-v 2 )) 1/2 .sh(θ) leads to : V(c) = arcsh((Δ/(4u.w)) 1/2 ) .c (11) (Δ) 1/2 Let us pose (with Δ ≠ 0) : f = if(Δ < 0, arcsine((-Δ/(4u.w)) 1/2 ) , arcsh((Δ/(4u.w)) 1/2 ) ) ( 12) (-Δ) 1/2 (Δ) 1/2 In the case u = 1, v = 0, w = 1, (v 2 -4u.w < 0), we get V(c) = (Arcsine(1)/2).c = (π/4).c, case we studied at exercise 18.

Generation of prime numbers

A quadratic equation of the preceding type (positive u, v and w) has only a finite number of solutions. This type of problem does not interest us. However, with V(c) expression, we can settle an equation of the following type where x and y are integers' variables and p a prime number variable :

u.x 1 2 +v.x.x 2 +w.x 2 2 < p+c (13) Then, boundless volume is given thanks to integral :

p = 2 x 1 = 0 x 2 = 0 Thus : p V(c) = f. ∫ ((p+c)/ln(p)).dp [START_REF] Waldschmidt | Valeurs zêta multiples[END_REF] p = 2 At this step, c being negligible in front of p, asymptotic evaluation gives :

and (by deriving first p+c inside integral ( 15)) :

V'(c) = f.p/ln(p) We draw from this : lim #{u.x 1 2 +v.x 1 .x 2 +w.x 2 2 < p+c} = (f/2).p 2 /Ln(p) (16) p →∞ and lim #{u.x 1 2 +v.x 1 .x 2 +w.x 2 2 = p+c} = f.Fan(c).p/Ln(p) (17) p →∞

For finite p, we introduce a correction factor a (which is not equal in the two following equations). lim #{u.x 1 2 +v.x 1 .x 2 +w.x 2 2 < p+c} = (a.f/2).q 2 /Ln(q) (18) p < q and lim #{u.x 1 2 +v.x 1 .x 2 +w.x 2 2 = p+c} = a.f.Fan(c).q/Ln(q) (19) p < q Alternatively, we can proceed as at exercise 17, to get an enumeration with y 1 and y 2 prime numbers variables :

u.y 1 2 +v.y 1 .y 2 +w.y 2 2 < p+c (20)

Then, with the same conditions on Δ, u, v and w that previously, the volume is given by the double integral : p y 1 = ((c+p)/u) +v.y 1 .y 2 +w.y 2 2 < p+c} = 2.a 3 .f.q 2 /Ln 3 (q) p < q and lim #{u.y 1 2 +v.y 1 .y 2 +w.y 2 2 = p+c} = 4.a 3 .f.Fan(c).q/Ln 3 (q) p < q

Numerical application

For u = 1, v = 1, w = 1, we get f = π/(3√3). Thus (the values of a are distinct in each equation, but all tend asymptotically towards 1) : lim #{x 1 2 +x 1 .x 2 +x 2 2 < p+c} = (a.π/(6√3)).q 2 /Ln(q) p < q and lim #{x 1 2 +x 1 .x 2 +x 2 2 = p+c} = (a.π/(3√3)).Fan(c).q/Ln(q) p < q and lim #{y 1 2 +y 1 .y 2 +y 2 2 < p+c} = (2a 3 .π/(3√3)).q 2 /Ln 3 (q) p < q and lim #{y 1 2 +y 1 .y 2 +y 2 2 = p+c} = (4a 3 .π/(3√3)).Fan(c).q/Ln 3 (q) p < q

We use the abundance The value of a converges gradually towards 1 when q increases. using the common environment matrix to x 1 2 +x 1 .x 2 +x 2 2 and -x 3 2 (that of x 2 ) and using the eigenvalues associated to these groups.

Eigenvalues relative to -x 3 2

Eigenvalues relative to -x 3 2 are deduced from eigenvalues relative to x 3 2 distinguishing two cases (see page 116 ). If p = 1 mod 2d (that is here p = 1 mod 4 in environment d = 2), the values are unchanged. If p = 1+d mod 2d (that is p = 3 mod 4 here), the eigenvalues with strictly greater than zero indices are shifted by d/2 (that is here simply a permutation of μ 1 and μ 2 ). Case p = 2 is treated separately. Hence,

It should be noted however that the choice of g is arbitrary. When we are moving from the literal calculation to the actual calculation of abundance factors, it will be necessary to make good matches.

Eigenvalues relative to x 1 2 +x 1 .x 2 +x 2 2

The discriminant of x 1 For p = 1 mod 3 : (p δ-1 .p, -p δ-1 .p 1/2 , p δ-1 .p 1/2 , p δ-1 .1, p δ-1 .1, -p δ-2 .p 1/2 , p δ-2 .p 1/2 , p δ-2 .1, p δ-2 .1…) (p 2δ , p 2δ-1 , p 2δ-1 , p 2δ-2 , p 2δ-2 , p 2δ-3 , p 2δ-3 , p 2δ-4 , p 2δ-4 …) p 3δ , -p 3δ-3/2 , p 3δ-3/2 , p 3δ-3 , p 3δ-3 , -p 3δ-9/2 , p 3δ-9/2 , p 3δ- 6 , p 3δ-6 … 5 mod 12 (p, -p 1/2 , p 1/2 ) (p 2 , -p, -p) (p δ , -p δ-1/2 , p δ-1/2 , p δ-1 , p δ-1 , -p δ-3/2 , p δ-3/2 , p δ-2 , p δ-2 …) (p 2δ , -p 2δ-1 , -p 2δ-1 , p 2δ-2 , p 2δ-2 , -p 2δ-3 , -p 2δ-3 , p 2δ-4 , p 2δ-4 …) p 3δ , p 3δ-3/2 , -p 3δ-3/2 , p 3δ-3 , p 3δ-3 , p 3δ-9/2 , p 3δ-9/2 , p 3δ-6 , -p 3δ-6 … 7 mod 12 (p, -i.p 1/2 , i.p 1/2 ) (p 2 , p, p) (p δ , -i.p δ-1/2 , i.p δ-1/2 , p δ-1 , p δ-1 , -i.p δ-3/2 , i.p δ-3/2 , p δ-2 , p δ-2 …) (p 2δ , p 2δ-1 , p 2δ-1 , p 2δ-2 , p 2δ-2 , p 2δ-3 , p 2δ-3 , p 2δ-4 , p 2δ-4 …) p 3δ , -i.p 3δ-3/2 , i.p 3δ-3/2 , p 3δ-3 , p 3δ-3 , -ip 3δ-9/2 , ip 3δ-9/2 , p 3δ-6 , p 3δ-6 … 11 mod 12 (p, -i.p 1/2 , i.p 1/2 ) (p 2 , -p, -p) (p δ , -i.p δ-1/2 , i.p δ-1/2 , p δ-1 , p δ-1 , -i.p δ-3/2 , i.p δ-3/2 , p δ-2 , p δ-2 …) (p 2δ , -p 2δ-1 , -p 2δ-1 , p 2δ-2 , p 2δ-2 , -p 2δ-3 , -p 2δ-3 , p 2δ-4 , p 2δ-4 …) p 3δ , i.p 3δ-3/2 , -i.p 3δ-3/2 , p 3δ-3 , p 3δ-3 , i.p 3δ-9/2 , -i.p 3δ- 9/2 , p 3δ-6 , p 3δ-6 …

Environment matrices

We must also distinguish cases modulo 12. For the d = 2, environment matrices have been given at page 129. Look at it more closely, for example, if δ = 3 and p = 1 mod 4. The eigenvalues are p 9 , -p 15/2 , p 15/2 , p 6 , p 6 , -p 9/2 , p 9/2 . We have :

P 376/390

Generation of prime numbers by quadratic sums ans analogues That is also : Then : #(0) p 9 -(p-1).p 15/2 (p-1).p 15/2 (p-1).p 6 (p-1).p 6 -(p-1).p 9/2 (p-1).p 9/2 1/p 3 #(p 2 .g 2u ) p 9 -(p-1).p 15/2 (p-1).p 15/2 (p-1).p 6 (p-1).p 6 -(-1-p 1/2 ).p 9/2 (-1+p 1/2 ).p 9/2 1/2p 3 #(p 2 .g.g 2u ) p 9 -(p-1).p 15/2 (p-1).p 15/2 (p-1).p 6 (p-1).p 6 -(-1+p 1/2 ).p 9/2 (-1-p 1/2 ).p 9/2 1/2p We will give some brief hints.

Useful mathematical tools

We give here only the bare minimum which is useful to our purpose.

Riemann zeta function

Euler work suffices here. The function ζ,, defined only on real numbers greater than 1, is written :

n p

According to [5] p 33, Euler gave several proofs of : ).e γ .ln(x) (6) p ≤ x, x → +∞

Corollary

We have : П 1-a/p ≡ r/ln a (x) (7) p ≤ x, x → +∞ Here a and r are constants. We are interested only in the limit value of the product and thus, in any manner, with the evolution of r when x increases. The point to observe is that constant a, appearing in the left member, is found also in the right member in exponent of the logarithm. Indeed, on the basis of Mertens theorem, let us write for a positive integer :

П (1-1/p) a = П (1-a/p+c 2 /p 2 +…+c a /p a ) ≡ e -aγ /ln a (x) p ≤ x, x → +∞ p ≤ x, x → +∞ Then П (1-a/p)) П(1+(c 2 /p 2 +…+c a /p a )/(1-a/p)) ≡ e -aγ /ln a (x) p ≤ x, x → +∞ p ≤ x, x → +∞

We know that ∑ n 1/n s diverge when s ≤ 1 and converge when s > 1. It is the same for П p (1-1/p s ). Thus, П(1+(c 2 /p 2 +…+c a /p a )/(1-a/p)) is a constant. Hence :

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Euler products П (1-a/p) ≡ c.e -aγ /ln a (x) p ≤ x, x → +∞

We can proceed the same way with either a positive or negative (using П (1+1/p) ≡ (6/π 2 ).e γ .ln(x)). By interpolation, the relation extends to all real number a. Indeed, for 0 < a < 1, we can still write, b p > 0 being related to p and all the more small as p is large : П (1+1/p) a ≡ П (1+a/p-b p /p 2 ) p ≤ x, x → +∞ p ≤ x, x → +∞ Representation of 1/r = ΓP(a)

A rough evaluation of the evolution of 1/r pending to exponent a reveals a graph with the form of a Gamma function. However, the function diverges not at negative integers x-ordinates (and on zero), but at negative prime numbers xordinates. By analogy, we thus note this function ΓP. In spite of the initial resemblance, the behaviours of the two functions remain quite distinct. We will note in addition that there is no divergence of f(a) at a = -2, although this point is initially a singularity. Still the special nature of 2 !

Case modulo 4

We exploit the results of Bombieri asymptotic sieve and Friedlander and Iwaniec enumeration. According to [10], the method of Bombieri asymptotic sieve allows to write : Results ( 9) and ( 10) are also in agreement with the theory of the numbers of classes of quadratic forms which implies (cf for example [2]) : П 1-(-1) (p-1)/2 /p = 2. П 1+(-1) Then, using Mertens second theorem : П 1-1/p ≡ 3 1/2 .u 1/2 .e -γ/2 /ln 1/2 (x) (11) p ≤ x, p = 1 mod 4 П 1-1/p ≡ (2/3 1/2 ).u 1/2 .e -γ/2 /ln 1/2 (x) (12) p ≤ x, p = 3 mod 4

In addition, as П 1-1/p ≡ 2.e -γ /ln(x) p > 2, p ≤ x and with П (1+1/p).(1-1/p) = 8/π 2 p>2 it follows П 1+1/p ≡ (4/π 2 ).e γ .ln(x) p > 2, p ≤ x then П 1+1/p ≡ (4/3 1/2 /π).u 1/2 .e γ/2 .ln 1/2 (x) (13) p ≤ x, p = 1 mod 4 and П 1+1/p ≡ (3 1/2 /π).u 1/2 .e γ/2 .ln 1/2 (x) (14) p ≤ x, p = 3 mod 4

The Landau-Ramanujan constant appears in a result of Number Theory which states that the proportion of the positive integers lower than x and sum of two squares is, for large x, proportional to x/ln 1/2 (x). According to [2], the Landau-Ramanujan constant is expressed in the form : L = (1/2 1/2 ) П (1-1/p 2 ) -1/2 p ≤ x, p = 3 mod 4

That is also L = (3π) 1/2 .u ≈ 0,767495030959866 u 4

Constant L is given (according to Wikipédia) for 0.76422365358922… Value of u is close to the unit (variation of about +0.43 %).

Case modulo 6

We are interested here by the quadratic abundance factors. Earlier, we observed the fractional ratio (=2/1) concerning П p = 1 mod 4 (1-1/p).П p = 3 mod 4 (1+1/p) and П p = 1 mod 4 (1+1/p).П p = 3 mod 4 (1-1/p). On the same model, we can "guess" П p = 1 mod 6 (1-1/p).П p = 5 mod 6 (1+1/p) and П p = 1 mod 6 (1+1/p).П p = 5 mod 6 (1-1/p) in a 4/3 ratio. Hence, we would have immediately using П p>3 (1+1/p).( 1 It follows lim #{x 1 2 +x 1 .x 2 +x 2 2 = p} = q/Ln(q) (17) p → +∞

The multiplicative coefficient in front of q/Ln(q) is reduced to 1, which would be quite satisfactory to validate (15).

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Euler products

In fact, this last relation is easily verified when one uses the conjecture given at page 84. Indeed, it results from that conjectures that x 1 2 +x 1 .x 2 +x 2 2 = p has two symmetrical solutions when p = 1 mod 6. In addition, we have one solution for p = 3 (direct verification) and we have no solution for p = 5 mod 6. The last point is immediately proved by the showing that x 1 2 +x 1 .x 2 +x 2 2 is always equal to 0, 1, 3 or 4 mod 6 and thus 5 modulo 6 is banned (like 2 mod 6). As prime numbers 1 mod 6 and 5 mod 6 densities are the same (according to Dirichlet), the density of the solutions of x 1 2 +x 1 .x 2 +x 2 2 = p is that of the prime numbers p/ln(p).

In addition, we can observe that : П 1-1/p p = 1 mod 6 = (7/6).u П 1-1/p p = 5 mod 6 and П 1+1/p p = 1 mod 6 = (7/8).u П 1+1/p p = 5 mod 6

where u is again a factor close to 1 (positive variation of some per thousand). This gives then using relations ( 5) and ( 6) :

П 1-1/p ≡ (7/3) 1/2 .(u -1/2 ).e -γ/2 /ln 1/2 (x) (18) p ≤ x, p = 1 mod 6 П 1-1/p ≡ 2.(3/7) 1/2 .(u -1/2 ).e -γ/2 /ln 1/2 (x) (19) P ≤ x, p = 5 mod 6 П 1+1/p ≡ ((21) 1/2 /2π).(u 1/2 ).e γ/2 .ln 1/2 (x) (20) p ≤ x, p = 1 mod 6 П 1+1/p ≡ (4.(3/7) 1/2 /π).(u 1/2 ).e γ/2 .ln 1/2 (x) (21) p ≤ x, p = 5 mod 6

These few examples open a window on the actual evaluation of more complex abundance factors (Euler products of unspecified arithmetic series of p). Quite a lot remains to be made to start with the explicit determinations of u in the two preceding cases (modulo 4 and modulo 6).

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Conclusions and major research axis

CONCLUSIONS AND MAJOR RESEARCH AXIS

Asymptotic ecology and species

The mainstay of asymptotic diophantine evaluation rests on Euler products. At each sequence, our building blocks, multiple entries tables method results in a characteristic matrix of each independent diophantine species called cardinal matrix. The gathering of these species (by addition) is expressed in a convolution (multiplication) of these matrices in an environment which can be made common (with common dimension to the matrices). The so-called matrix of environment of the eigenvectors is the binding agent of the eigenvalues of the cardinal matrices.

Sustainable developments

We do not present a chapter in continuity with exercise 11 for overlapping variables diophantine equations. The reader has already understood it, the main problem is not the approximate evaluation of the law of surface but well the determination of the (boundless) volume imposed by selected complex equations. Thus, beyond the quadratic case n = 2, for an equation like (z 1 n+1 -z 2 n+1 )/(z 1 -z 2 ) < p+c (where z 1 ≠z 2 ), the reader may try to find the constant of correction of volume compared to z 1 n +z 2 n < p+c (this last volume being easy to evaluate while referring to our exercises). Of course, the evaluation of the abundance factors of (z 1 n+1 -z 2 n+1 )/(z 1 -z 2 ) = p+c in a literal way will not be self-evident either.

The numerical evaluation of the cardinal matrices eigenvalues is not in general a difficulty. It is negotiated with the size of the computer and the choice of a good algorithm. The literal evaluation of these eigenvalues is another business. The study undertaken here shows that, a priori, each environment is a peculiar case based on sequences p primitive roots, primitives whose behaviour is considered difficult to encircle. A landscape comparison is to say that environments 1 and 2 are, from now on, largely under control and that environment 3, 4, 6 and 12 include here or there wild growths. But the infinity of the other cases present only weeds. Thus, a considerable work remains in front of us.

The geometer will find a fascinating problem in multiple factors environments : locus of eigenvalues focus, locus of angle rotation.

The exercise on Fermat-Catalan cardinal matrices clarified the fundamental interest of constructive of sequence matrices.

The framework is conjectural and the suggested fitting thus remains to be demonstrated. Who will control their evaluation is likely to control the field of the asymptotic evaluations.

Other topics remain to be cleared up and the reader will have only the embarrassment of choice here : detail of environment 3 case, conjectures concerning environment 6 and environments of 2d type, fundamental environments 2 r and p r cases in their general feature, study of the inclusive and exclusive frequencies in the polynomial cases, proof of the reciprocity in general case, evaluation of Euler product with conditions of congruence carrying on polynomials of prime numbers, fine resolution of overlapping polynomial problems, etc.

Opening on another space

We relied on the notion of equiprobability to build the asymptomatic representatives of variables. Thus, in principle, the method outlined here applies only to equations with an infinite number of solutions. However, it is natural to wonder if its scope may eventually extend to equations with a finite number of solutions. The reader will find an article in this respect on our online sites [18] [19]. Against all odds, there is an application continuity via appropriate treatment : the number of solutions is linearly dependent of the singular series affected by the inverse power of the degree of the polynomials of the chosen diophantine equation.