Chiara Bianchini

Gisella Croce
email: gisella.croce@univ-lehavre.fr

Antoine Henrot
email: antoine.henrot@univ-lorraine.fr

ON THE QUANTITATIVE ISOPERIMETRIC INEQUALITY IN THE PLANE WITH THE BARYCENTRIC DISTANCE

Keywords: 1991 Mathematics Subject Classification. (2010) 28A75, 49J45, 49J53, 49Q10, 49Q20 Isoperimetric inequality, quantitative isoperimetric inequality, isoperimetric deficit, barycentric asymmetry, shape derivative, optimality conditions

. For that purpose, we consider a shape optimization problem in which we minimize the ratio δ(Ω)/λ 2 0 (Ω) both in the class of compact connected sets and in the class of convex sets.

Introduction

In the last thirty years quantitative isoperimetric inequalities have received much attention in the litterature. The purpose is to quantify the distance that a subset of R n , Ω, has from an n-dimensional ball B of the same measure in terms of the so called isoperimetric deficit δ(Ω):

δ(Ω) = P (Ω) -P (B) P (B) (1)
(here and later | • | indicates the n-dimensional Lebesgue measure). Several kinds of distances have been proposed to establish quantitative isoperimetric inequalities of the type

[dist(Ω, B)] k ≤ C δ(Ω). (2)
In 1989, Fuglede [START_REF] Fuglede | Stability in the isoperimetric problem for convex or nearly spherical domains in R n[END_REF] used the Hausdorff distance of a set Ω from the ball of same volume centered at the barycentre x G of Ω. We recall that the barycenter of a set Ω is defined as x G = 1 |Ω| ˆΩ x dx . He called it the uniform spherical deviation. He proved a series of inequalities for convex sets and nearly spherical sets, that is, star-shaped sets with respect to their barycentre (which may be taken to be 0) written as

{y ∈ R n : y = tx(1 + u(x)), x ∈ S n-1 , t ∈ [0, 1]},
where u : S n-1 → R is a positive Lipschitz function, with u L ∞ ≤ 3 20n and ∇u L ∞ ≤ 1 2 . In [START_REF] Fusco | On a Bonnesen type inequality involving the spherical deviation[END_REF], the authors proved the same kind of inequalities as [START_REF] Ambrosio | Functions of bounded variation and free discontinuity problems[END_REF] for a more general family of sets, and considering as distance the minimum of the Hausdorff distances of a set from all the balls of R n with fixed volume.

[GIRARE la frase] L. E. Fraenkel proposed a different kind of method to count the distance from a ball (now called Fraenkel asymmetry λ(•)), to enlarge the family of sets for which a quantitative isoperimetric inequality hold:

λ(Ω) = inf y∈R n |Ω∆B y | |Ω| (3)
where B y is the ball of center y and such that |B y | = |Ω| and ∆ is the symmetric difference of sets. This distance can be seen as an L 1 distance between Ω and any ball B y , centered at y ∈ R n , with the same measure as Ω. On the contrary, the Hausdorff distance is in some sense an L ∞ distance between sets. Many mathematicians studied quantitative isoperimetric inequalities with the Fraenkel asymmetry, establishing sharp inequalities (see for example [START_REF] Hall | On asymmetry and capacity[END_REF], [START_REF] Hall | A quantitative isoperimetric inequality in n-dimensional space[END_REF], [START_REF] Giorgi | Sulla proprietà isoperimetrica dell'ipersfera, nella classe degli insiemi aventi frontiera orientata di misura finita, (Italian)[END_REF], [START_REF] Campi | Isoperimetric deficit and convex plane sets of maximum translative discrepancy[END_REF], [START_REF] Alvino | A sharp isoperimetric inequality in the plane[END_REF], [START_REF] Figalli | A mass transportation approach to quantitative isoperimetric inequalities[END_REF], [START_REF] Fusco | The sharp quantitative isoperimetric inequality[END_REF], [START_REF] Cicalese | A selection principle for the sharp quantitative isoperimetric inequality[END_REF], [START_REF] Fusco | The quantitative isoperimetric inequality and related topics[END_REF], [START_REF] Dambrine | Stability in shape optimization with second variation[END_REF]) and existence of an optimal set for the related shape optimization problem (see [START_REF] Cicalese | Best constants for the isoperimetric inequality in quantitative form[END_REF] and [START_REF] Bianchini | On the quantitative isoperimetric inequality in the plane[END_REF]).

In the spirit of the Fraenkel asymmetry, Fuglede proposed in [START_REF] Fuglede | Lower estimate of the isoperimetric deficit of convex domains in R n in terms of asymmetry[END_REF] the barycentric asymmetry:

λ 0 (Ω) = |Ω∆B x G | |Ω|
where B x G is the ball centered at the barycentre x G of Ω and such that |Ω| = |B x G |. Notice that λ 0 (Ω) is obviously easier to compute than λ(Ω). Fuglede proved that there exists a positive constant (depending only on the dimension n) such that δ(Ω) ≥ C(n) λ 2 0 (Ω), for any convex subsets Ω of R n .

In this paper we propose two kinds of generalizations of Fuglede's results [START_REF] Fuglede | Lower estimate of the isoperimetric deficit of convex domains in R n in terms of asymmetry[END_REF], in dimension n = 2, by considering the shape optimization problem min G 0 (Ω), where

G 0 (Ω) = δ(Ω) λ 2 0 (Ω) , (5)
in two different families of sets.

(1) We prove that there exists a strictly positive constant C such that the inequality G 0 (Ω) ≥ C holds for compact connected sets (see Section 3). Notice that, as already observed by Fuglede, the connectedness assumption is necessary in order to have a well posed problem (cf. Remark 3.4). [START_REF] Ambrosio | Functions of bounded variation and free discontinuity problems[END_REF] The existence of a minimizer of the shape functional G 0 is proved in the class of convex sets (see Section 4). We also study the regularity of the optimal set in Section 5 and we write different kinds of optimality conditions. Some observations about the existence and the shape of an optimal set for the minimization of G 0 in the plane are here presented.

-In this paper we do not prove the existence of an optimal set for the minimization of G 0 within the class of compact connected sets (see Remark 3.3). However we formulate a conjecture about the shape of a possible optimal set. -We conjecture that the optimal set for G 0 within the class of convex sets is the stadium S, found in [START_REF] Alvino | A sharp isoperimetric inequality in the plane[END_REF] as the optimal set for the minimization of

G(Ω) := δ(Ω) λ 2 (Ω) ; (6)
in Section 5 we give some arguments in this direction. -In [START_REF] Bianchini | On the quantitative isoperimetric inequality in the plane[END_REF] our aim was to compute the infimum of G(Ω). We underline that, since λ(Ω) ≤ λ 0 (Ω), the infimum of G 0 , would entail an estimate from below of the infimum of G(Ω). We note that for a given set Ω, λ 0 (Ω) is much simpler to compute than λ(Ω). As observed by Fuglede [START_REF] Fuglede | Lower estimate of the isoperimetric deficit of convex domains in R n in terms of asymmetry[END_REF], an estimate from below of the infimum of G(Ω) is given in Lemma 2.1 of [START_REF] Hall | On asymmetry and capacity[END_REF]: one has G(Ω) ≥ 0.02 for every Ω ⊂ R 2 ; see also [START_REF] Figalli | A mass transportation approach to quantitative isoperimetric inequalities[END_REF] for an estimate in any dimension. In [START_REF] Li | An analytic proof of the planar quantitative isoperimetric inequality[END_REF] the authors found that G(Ω) ≥ 0.0625 for every Ω in the plane. However, better estimates should be possible. In [START_REF] Cicalese | A selection principle for the sharp quantitative isoperimetric inequality[END_REF], [START_REF] Bianchini | On the quantitative isoperimetric inequality in the plane[END_REF] the conjectured optimal set for G(Ω) is described: a kind of mask with two axes of symmetry and two optimal disks for the Fraenkel asymmetry, whose boundary is composed by arcs of circle with three different radii. If this conjecture is proved, it would provide the following sharp estimate: G(Ω) ≥ 0.3931.

Preliminaries

Given a subset E of R 2 , we denote by E c its complementary set and by co(E) its convex hull. C indicates the set of open convex bounded subsets of R 2 . For ε > 0 we denote by

E ε the ε-enlargement of E, that is, E ε = {x ∈ R 2 : dist(x, E) ≤ ε}
where dist is the Euclidean distance. Diam(E) denotes the diameter of E, that is

Diam(E) = sup x,y∈E |x -y|.
For Ω ⊂ R 2 measurable and bounded, δ(Ω) indicates the isoperimetric deficit as defined in [START_REF] Alvino | A sharp isoperimetric inequality in the plane[END_REF], where P (Ω) is the perimeter in the Minkowski sense, that is:

P (Ω) = lim ε→0 + |Ω ε | -|Ω| ε .
We will explain later in Remark 3.4 why this notion of perimeter is adapted to our problem and why the classical perimeter in the sense of De Giorgi, denoted by P DG (Ω) in the sequel, is not suitable here (we refer to [START_REF] Ambrosio | Functions of bounded variation and free discontinuity problems[END_REF][START_REF] Henrot | Shape variation and optimization. A geometrical analysis[END_REF] for definitions). While considering the Fraenkel asymmetry (3) for a set Ω, we refer to a ball B x as to an optimal ball

if |B x | = Ω and λ(Ω) = |Ω∆B x |/|Ω|.
We are interested in investigating the shape functionals G 0 , G defined in ( 5), [START_REF] Cicalese | A selection principle for the sharp quantitative isoperimetric inequality[END_REF], respectively. Let D be a fixed closed disk, we indicate by K(D) the set of all compact connected subsets of D, while K indicates the class of connected compact subsets of R 2 . We recall that the Hausdorff distance between two sets K 1 and K 2 in K is defined by

d H (K 1 , K 2 ) := max sup x∈K1 dist(x, K 2 ), sup x∈K2 dist(x, K 1 ) .
We recall the classical Blaschke's Theorem (cfr. Theorem 2.2.3 in [START_REF] Henrot | Shape variation and optimization. A geometrical analysis[END_REF]): Theorem 2.1. Let {K n } be a sequence in K(D). Then there exists a subsequence which converges in the Hausdorff metric to a set K ∈ K(D).

Theorem 2.2. Let {K n } be a sequence of compact convex sets converging in the Hausdorff metric to a set K. Then K is compact and convex.

We will also use the following semicontinuity result, analogous to the Golab Theorem for the Minkowski perimeter in the plane, proved by Henrot and Zucco in [START_REF] Henrot | Optimizing the first Dirichlet eigenvalue of the Laplacian with an obstacle[END_REF]:

Theorem 2.3. Let {K n } ⊂ R 2 be a sequence contained in K(D) converging to a set K ∈ K(D) in the Hausdorff metric. Then P (K) ≤ lim inf n→∞ P (K n ).
We will also use the following consequences of the Hausdorff convergence of sets (see Proposition 2.2.21 of [START_REF] Henrot | Shape variation and optimization. A geometrical analysis[END_REF]). Here χ K denotes the characteristic function of a set K.

Proposition 2.4. Let K n , K in K(D). If K n → K in the Hausdorff metric, then (1)
|K n \ K| → 0 (2) χ K ≥ lim sup n→∞ χ Kn a.e. ( 3
) If χ Kn → χ in L 1 (Ω) (or even weak-star in (L 1 , L ∞ )), then χ ≤ χ K .
We also recall a compactness result about the L 1 convergence of sets, that is, the L 1 convergence of characteristic functions of sets. See [START_REF] Henrot | Shape variation and optimization. A geometrical analysis[END_REF] for the proof. Proposition 2.5. Let K n be a sequence of sets contained in an open set with finite measure, such that P DG (K n ) + |K n | is uniformly bounded. Then there exists a set K such that χ Kn → χ K in L 1 , up to a subsequence.

Remark 2.6. The perimeters P DG and P of a set satisfy the inequality P DG (K) ≤ P (K) if K ⊂ R 2 is a compact connected set, as remarked in [START_REF] Henrot | Optimizing the first Dirichlet eigenvalue of the Laplacian with an obstacle[END_REF].

In [START_REF] Bianchini | On the quantitative isoperimetric inequality in the plane[END_REF] we found the infimum value of G for sequences converging to a ball. We used the notion of De Giorgi perimeter to define the isoperimetric deficit. In view of the above relation between the two notions of perimeters P DG , P one has the following result.

Theorem 2.7. Let {Ω ε } ε>0 be a sequence of planar sets converging to a ball B in the sense that

|B∆Ω ε | → 0 as ε → 0. Then inf lim inf ε→0 δ(Ω ε ) λ 2 (Ω ε ) ≥ π 8(4 -π)
.

We will use the following result about the minimization of G within the class convex sets, proved in [START_REF] Alvino | A sharp isoperimetric inequality in the plane[END_REF].

Theorem 2.8. There exists an optimal set for the minimization problem inf K∈C G(K), where C is the family of convex planar sets. The infimum is attained by an explicitely described stadium S and min K∈C G(K) = G(S) ≈ 0.406. Remark 2.9. In the sequel we will use the set D given by two balls, each one of area π 2 , connected by a segment of length 2, whose direction passes through their centers. We will call it dumbbell. We observe that its Minkowski perimeter counts twice the length of the segment and therefore

G 0 (D) = δ(D) λ 2 0 (D) = √ 2 -1 4 + 1 2π ≈ 0.26 < δ(S) λ 2 0 (S) = G 0 (S) ≈ 0.406 ,
where S is the stadium of the above theorem.

In the following we will use nearly spherical sets, studied by Fuglede in [START_REF] Fuglede | Stability in the isoperimetric problem for convex or nearly spherical domains in R n[END_REF], that is star-shaped sets E parametrized as

E = {y ∈ R 2 : y = tx(1 + u(x)), x ∈ S 1 , t ∈ [0, 1]}, (7)
with u : S 1 → (0, +∞) a Lipschitz function. Moreover, we will assume that the barycenter is at the origin and that |E| = π. Let B be the unit ball centered at 0. Then, it is straightforward to check:

|E∆B| = 1 2 ˆ2π 0 |(1 + u) 2 -1| , H 1 (∂E) = ˆ2π 0 (1 + u) 2 + |u | 2 , ˆ2π 0 cos θ(1 + u) 3 = 0 = ˆ2π 0 sin θ(1 + u) 3 , ˆ2π 0 (1 + u) 2 = 2π .
We will use the following result by Fuglede (Lemma 2.2 in [START_REF] Fuglede | Stability in the isoperimetric problem for convex or nearly spherical domains in R n[END_REF]):

Theorem 2.10. Let K n be a sequence of convex compact sets of area π, converging in the Hausdorff metric to the unit ball B. Assume K n to be parametrized in the form

K n = {y ∈ R 2 : y = tx(1 + u n (x)), x ∈ S 1 , t ∈ [0, 1]},
where u n is a Lipschitz function. The following estimate holds:

u n L ∞ ≤ 2 1 + u n L ∞ 1 -u n L ∞ u n 1 2 L ∞ .
In [START_REF] Croce | A Poincaré type inequality with three constraints[END_REF] the following result has been proved. It will be useful in the shape optimization problem within the class of convex sets.

Theorem 2.11. Let m be defined by

m = inf u∈L ˆ2π 0 [(u ) 2 -u 2 ]dθ ˆ2π 0 |u|dθ 2
where L is the space of H 1 (0, 2π) functions satisfying the constraints:

(L1) ˆ2π 0 u dθ = 0 (L2) ˆ2π 0 u cos(θ) dθ = 0 = ˆ2π 0 sin(θ)u dθ (L3) u(0) = u(2π). Then m = 1 2(4 -π) . 3. Minimization of δ(Ω) λ 2 0 (Ω)
within the class of compact connected sets

In this section, we consider compact connected sets K ∈ K of positive measure (in order the shape functionals δ and λ 0 be well-defined). We are going to prove the following result. Theorem 3.1. There exists C > 0 such that for any planar compact connected set K ∈ K it holds

λ 2 0 (K) ≤ Cδ(K).
In the proof we will use the following simple lemma, which involves the notion of d L 1 (E, F ) of sets E, F ∈ K, defined as the L 1 distance of their characteristic functions. Lemma 3.2. Let B 1 and B 2 be two balls such that their area equals π and the distance between their centers equals a ≤ 2. Then

d L 1 (B 1 , B 2 ) = 4 arcsin a 2 + 2a 1 - a 2 4 .
Moreover, for small values of a it holds

d L 1 (B 1 , B 2 ) = 4a + o(a).
Proof. Up to a rotation we can assume that B 1 = B (0,0) and B 2 = B (a,0) , where B (a,0) denote the ball of area π centered at (a, 0), 0 ≤ a ≤ 2. Let τ = arcsin(a/2). The quantity d L 1 (B (0,0) , B (a,0) ) is equal to 4 times the area of the domain E whose boundary is composed by the following three arcs:

(1) (a + cos t, sin t), t ∈ (0, α), α = π 2 + τ ; (2) (cos t, sin t), t ∈ (0, β), β = π 2 -τ ; (3) (t, 0), t ∈ (1, 1 + a).
Hence by Green's theorem

d L 1 (B 1 , B 2 ) = 4|E| = 1 2 ˆπ 2 +τ 0 (a + cos τ ) cos τ + sin 2 t dτ - 1 2 ˆπ 2 -τ 0 1 + 0 dτ = τ + a 2 cos τ,
and we are done. Moreover as a tends to 0 it follows

d L 1 (B 1 , B 2 ) = 4a + o(a)..
We are now going to prove Theorem 3.1, showing that for a minimizing sequence K n , lim inf n→∞ G 0 (K n ) > 0. We will distinguish the cases where K n converges to a ball or to a set different from a ball.

Proof of Theorem 3.1. Let K n be a minimizing sequence in K, that is,

G 0 (K n ) = δ(K n ) λ 2 0 (K n ) → inf E∈K G 0 (E) = inf E∈K δ(E) λ 2 0 (E)
.

Without loss of generality, we can assume that all the sets K n have area π. By Theorem 2.8 one has

G 0 (K n ) ≤ G 0 (S) = G(S) ≈ 0.406 ,
where S denotes the stadium introduced in Theorem 2.8. Since λ 0 (E) ≤ 2 for any set E, we get

P (K n ) ≤ 16.6 . (8)
Therefore the sets K n are all contained in a fixed ball, since they are connected and their perimeter is uniformly bounded. Theorem 2.1 gives us the existence of a connected compact set K towards which K n converges in the Hausdorff metric. There exists a set K such that χ Kn → χ K in L 1 and | K| = π, by Proposition 2.5 and Remark 2.6. We are going to prove that K = K (we note that the only Hausdorff convergence does not guarantee that |K| = π).

By point (3) in Proposition 2.4 applied to K c n et K c , we have χ K ≤ χ K , (9)
since

χ Kn → χ K in L 1 . Therefore |K| ≥ π = | K| . ( 10
) Since K n → K in the Hausdorff metric, K is contained into the ε-enlargement K ε n of K n .
By the definition of the Minkowski perimeter, we have, for every small ε > 0,

|K| ≤ |K ε n | = |K n | + εP (K n ) + o(ε) = π + εP (K n ) + o(ε) . (11)
Since P (K n ) are uniformly bounded, inequality [START_REF] Fuglede | Stability in the isoperimetric problem for convex or nearly spherical domains in R n[END_REF] yields |K| ≤ π. This inequality and (10) imply |K| = π. We deduce that K = K a.e. from [START_REF] Giorgi | Sulla proprietà isoperimetrica dell'ipersfera, nella classe degli insiemi aventi frontiera orientata di misura finita, (Italian)[END_REF].

Since K n → K in L 1 , as n → ∞, we have λ 0 (K n ) → λ 0 (K)
. Indeed, by the triangle inequality,

π|λ 0 (K n ) -λ 0 (K)| ≤ d L 1 (K n , K) + d L 1 (B, B n ) .
The first term in the right hand side tends to 0, as n → ∞ by the L 1 convergence of K n to K. The second one tends to 0 by Lemma 3.2, since

x Gn 1 -x G 1 ≤ 1 π ˆKn\K |x 1 |dx 1 dx 2 ,
and the last term tends to 0, since the diameter of K n is uniformly bounded and |K n \K| → 0, as n → ∞.

The same holds for the second coordinate. Now, only two possibilities may occur: either λ 0 (K) > 0 or λ 0 (K) = 0.

(1) In the first case, since K n is a minimizing sequence and inf

E∈K δ(E) λ 2 0 (E) > 0, one has 0 < P (K) ≤ lim inf n→∞ P (K n ) by Theorem 2.3. Therefore lim inf n→∞ δ(K n ) λ 2 0 (K n ) ≥ δ(K) λ 2 0 (K) > 0 .
(2) In the second case we can assume that λ(

K n ) = 2ε n → 0, with ε n → 0 as n → ∞, since λ 0 (K n ) ≥ λ(K n ). By Theorem 2.7 one has δ(K n ) ≥ 0.45 • 4ε 2 n .
We are now going to prove that there exists an absolute positive constant A such that

|λ(K n ) -λ 0 (K n )| ≤ 4A π ε n . ( 12
) Therefore δ(K n ) λ 2 0 (K n ) ≥ δ(K n ) λ(K n ) + 4A π ε n 2 ≥ 1.8 2 + 4A π 2 , (13)
which implies that

lim inf n→∞ δ(K n ) λ 2 0 (K n ) > 0 .
To prove [START_REF] Fuglede | Lower estimate of the isoperimetric deficit of convex domains in R n in terms of asymmetry[END_REF] it is sufficient to find a positive constant A such that

G n -F n ≤ Aε n ( 14
)
where

G n = (x Gn 1 , x Gn 2 )
is the barycentre of K n and F n is the centre of an optimal ball for λ(K n ). Indeed, by the triangle inequality,

d L 1 (K n , B Gn ) ≤ d L 1 (K n , B Fn ) + d L 1 (B Gn , B Fn ) ,
where B Fn is an optimal ball for K n with respect to the Fraenkel asymmetry. This inequality together with [START_REF] Fusco | On a Bonnesen type inequality involving the spherical deviation[END_REF] and Lemma 3.2 imply [START_REF] Fuglede | Lower estimate of the isoperimetric deficit of convex domains in R n in terms of asymmetry[END_REF]. We want to prove [START_REF] Fusco | On a Bonnesen type inequality involving the spherical deviation[END_REF], which will end the proof of this case. We can always assume that an optimal ball for the Fraenkel asymmetry is centered in 0, that is, F n = 0. We are now going to estimate x Gn

1 = 1 π ˆKn x 1 dx 1 dx 2 . Writing the last integral on (K n \ B) ∪ B \ (B \ K n ) and recalling that 1 π ˆB x 1 dx 1 dx 2 = 0, we get |x Gn 1 | = 1 π ˆKn\B x 1 dx 1 dx 2 - ˆB\Kn x 1 dx 1 dx 2 ≤ 1 π ˆKn\B |x 1 | dx 1 dx 2 + 1 π ˆB\Kn |x 1 | dx 1 dx 2 .
We observe that diam(

K n \ B) ≤ diam(K n ). Now, diam(K n ) → 2. Indeed by definition of Hausdorff convergence, K n ⊂ B ε and B ⊂ K ε n . Therefore diam(K n ) ≤ 2 + 2ε and 2 ≤ diam(K n ) + 2ε.
Using this property to estimate the first of the last two terms, we get

|x Gn 1 | ≤ 2ε n π + ε n π = 3 π ε n ,
where we have used that

|K n \ B| = |B \ K n | = ε n , since λ(K n ) = 2ε n .
The same estimate can be obtained for |x Gn 2 |. Therefore

G n -F n = G n -0 ≤ Aε n = 3 √ 2 π ε n (15)
and ( 14) is proved.

Remark 3.3. We conjecture that the infimum of G 0 in the class of the connected sets is attained by the dumbbell described in Remark 2.9.

In the case where the minimizing sequence K n converges to the ball, we get the following estimate from below of lim inf n→∞ G 0 (K n ) (see [START_REF] Fusco | The quantitative isoperimetric inequality and related topics[END_REF] with

A = 3 √ 2 π given in (15)): lim inf n→∞ G 0 (K n ) ≥ 0.13 .
Notice that this estimate is lower than the value of G 0 computed on the dumbbell (see Remark 2.9); this is the reason why the existence of an optimal set for this problem is still an open problem. We remark that the technique introduced in [START_REF] Bianchini | On the quantitative isoperimetric inequality in the plane[END_REF], cannot be applied to the functional G 0 in the class K of connected compact sets in order to exclude sequences converging to a ball. Remark 3.4. As mentioned by Fuglede (we refer to [START_REF] Fusco | The quantitative isoperimetric inequality and related topics[END_REF]), the assumption that Ω is connected cannot be removed. Indeed one can construct the following sequence of non-connected sets Ω n , given by the union of the disk centered in (2, 0), of radius R n = 1 -1 n , and the disk centered in -2(n-1) 2 2n-1 , 0 , of radius

r n = 2n-1 n 2 . It is easy to check that |Ω n | = π, the barycentre of Ω n is the origin, δ(Ω n ) = R n +r n -1 → 0 as n → ∞ and λ 0 (Ω n ) = 2. Thus lim n→∞ G 0 (Ω n ) = 0.
This example shows why the classical De Giorgi perimeter is not suitable for the barycentric asymmetry λ 0 . Indeed, the set Ωn obtained by connecting the above two balls by a long segment has the same De Giorgi perimeter as the perimeter of Ω n , since the De Giorgi perimeter of the long segment is null. Thus lim n→∞ G 0 ( Ωn ) = 0. On the contrary, for the Minkowski perimeter, δ(Ω n ) → +∞, since the length of the long segment counts twice.

Remark 3.5. The notion of Minkowski perimeter has a central role in the second part of the above proof, in inequality [START_REF] Fuglede | Stability in the isoperimetric problem for convex or nearly spherical domains in R n[END_REF], to prove that |K| = π.

4. Minimisation of δ(Ω) λ 2 0 (Ω)
within the class of compact convex sets

In this section we prove the following result :

Theorem 4.1. There exists an optimal set for inf

Ω∈C G 0 (Ω).
We recall that C is the family of convex planar sets.

Proof. Let K n be a minimizing sequence of convex compact sets so that

lim inf G 0 (K n ) = inf Ω⊂C G 0 (Ω).
It follows a uniform bound on δ(Kn) λ 2 0 (Kn) and hence using the definition of λ 0 we have that δ(K n ) is uniformly bounded. Therefore the sets K n are all contained in a fixed ball, since they are convex and they perimeter is uniformly bounded.

Theorems 2.1 and 2.2 entail the existence of a convex compact set towards which K n converges in the Hausdorff metric. Now, as in the proof of the previous theorem, two possibilities may occur:

(1) K n converges to a ball B in the Hausdorff metric;

(2) K n converges to a set K different from a ball in the Hausdorff metric. In the next theorem we are going to analyse the first case, proving that lim inf G 0 > G 0 (S) = 0.406 where S is the stadium of Theorem 2.8. This means that a minimizing sequence cannot converge to a ball. Therefore the only possibility for a minimizing sequence is the second one. In this case we can prove that K is a minimizer by using an analogous argument to that of the proof of case (2) of Theorem 3.1. Let E be a nearly spherical set with barycenter in 0, and assume E to be written in polar coordinates with respect to 0 as in [START_REF] Cicalese | Best constants for the isoperimetric inequality in quantitative form[END_REF]. Hence if E is a convex set, then the functional G 0 (E) can be written as a function J(u) of the parameter u and, according to the computations in Section 2, we obtain:

J(u) = π 2 ˆ2π 0 (1 + u) 2 + u (θ) 2 -1 dθ 1 2 ˆ2π 0 |(1 + u) 2 -1|dθ 2 . ( 16
)
We are interested in minimizing the functional J(u) with respect to u ∈ N L, that is, u ∈ H 1 (0, 2π), satisfying the constraints of fixed area and barycentre in 0:

(NL1) 1 2π ˆ2π 0 (1 + u) 2 dθ = 1; (NL2) ˆ2π 0 cos(θ)[1 + u(θ)] 3 dθ = 0 = ˆ2π 0 sin(θ)[1 + u(θ)] 3 dθ; (NL3) u(0) = u(2π).
This leads to a complicated problem in the calculus of variations, thus, the strategy consists in replacing this problem by a simpler one which can be seen as a sort of linearization. In order to do this, we define

m = inf u∈L ˆ2π 0 [(u ) 2 -u 2 ]dθ ˆ2π 0 |u|dθ 2 ( 17
)
where L is the space of H 1 (0, 2π) functions satisfying the constraints:

(L1)

ˆ2π 0 u dθ = 0; (L2) ˆ2π 0 u cos(θ) dθ = 0 = ˆ2π 0 sin(θ)u dθ; (L3) u(0) = u(2π).
Proposition 4.3. Let 0 < ε < 1/24. Consider J : N L → R defined in [START_REF] Hall | On asymmetry and capacity[END_REF] and let

m ε := inf{J(u), u L ∞ = ε, u ∈ N L}. (18)
It holds lim inf

ε→0 m ε ≥ π 4 m. (19)
Proof. The proof follows the following steps:

(1) we estimate the infimum value fo the optimization problem [START_REF] Harrell | On the maximization of a class of functionals on convex regions and the characterization of the farthest convex set[END_REF] by considering an auxiliary problem (problem (24)), whose infimum yields a smaller value; (2) we prove that the auxiliary problem has a minimizer u ε ;

(3) we prove that v ε = uε ε (which belongs to the unit sphere of L ∞ ) is bounded in H 1 and converges uniformly to some function v 0 ; (4) passing to the limit as ε → 0, we prove that v 0 is a test function for the optimization problem [START_REF] Hall | A quantitative isoperimetric inequality in n-dimensional space[END_REF] whence the desired inequality holds. In the sequel of the proof C will denote an absolute constant independent of ε.

Step 1. Since ˆ2π 0 (2u + u 2 ) = 0 by (NL1), the minimization of J(u) is equivalent to the minimization (with the same constraints) of

J 1 (u) = π 2 ˆ2π 0 (1 + u) 2 + u 2 -1 dθ - ˆ2π 0 (u + u 2 /2)dθ 1 2 ˆ2π 0 |(1 + u) 2 -1|dθ 2 .
We are going to estimate the numerator of J 1 ; more precisely we are going to prove that ˆ2π

0 1 + 2u + u 2 + u 2 -1 -(u + u 2 /2) dθ ≥ 1 2 ˆ2π 0 u 2 -u 2 - 1 4 u 4 -uu 2 dθ + Cε 3 , (20)
for some constant C. Assume that ε < 1/24 and

u L ∞ ≤ 3 √ ε ( 21
)
(this is possible by the estimate u L ∞ ≤ ε and Theorem 2.10). We first observe that for |ρ| ≤ 1 2 ,

1 + ρ ≥ 1 + ρ 2 - ρ 2 8 + ρ 3 16 - ρ 4 8 . (22)
By [START_REF] Lamboley | Polygons as optimal shapes with convexity constraint[END_REF], one has the estimate |2u

+ u 2 + u 2 | ≤ 2ε + ε 2 + 9ε ≤ 12ε ≤ 1 2 .
We can apply [START_REF] Lamboley | Regularity and singularities of optimal convex shapes in the plane[END_REF]

to 2u + u 2 + u 2 to infer (1 + u) 2 + u 2 -1 ≥ u + 1 2 u 2 - 1 8 u 4 - 1 2 uu 2 + Cε 3 .
Therefore (20) holds. We are going to estimate the denominator of

J 1 . Since |2u + u 2 | ≤ (2 + ε)|u| one has that 1 2 ˆ2π 0 |(1 + u) 2 -1| 2 ≤ 1 + ε 2 2 ˆ2π 0 |u| 2 .
Therefore, under the constraints (NL1), (NL2), (NL3) one has J(u) = J 1 (u) ≥ J 2 (u), where

J 2 (u) = π 2 1 2 ˆ2π 0 u 2 -u 2 - 1 4 u 4 -uu 2 + Cε 3 1 + ε 2 2 ˆ2π 0 |u| 2 . ( 23
)
Defining

m ε := inf{J 2 (u), u L ∞ = ε, u ∈ N L} , (24)
we have m ε ≥ m ε .

Step 2. We prove that problem (24) has a minimizer u ε for every fixed ε > 0. Let u ε n be a minimizing sequence for J 2 . We know that u ε n L ∞ = ε and (u ε n ) L ∞ ≤ 3 √ ε for every n (by ( 21)). Therefore u ε n → u ε weakly in W 1,∞ (0, 2π) and uniformly in (0, 2π), as n → ∞. To pass to the limit, as n → ∞, in J 2 (u ε n ), we need to study the integral in the numerator of J 2 . We will use a standard argument in the calculus of variations. Notice that for small |s| and |ξ| (recall that

|s| ≤ ε ≤ 1 24 and |ξ| ≤ 3 √ ε ), the function j(s, ξ) = ξ 2 (1 -s) -s 2 -1 4 ξ 4 is convex with respect to ξ. This gives j(s, ξ) ≥ j(s, η) + ∇ ξ j(s, η) • (ξ -η) , where ∇ ξ j(s, ξ) = 2ξ(1 -s) -ξ 3 . Therefore ˆ2π 0 j(u ε n , (u ε n ) ) ≥ ˆ2π 0 j(u ε n , u ε ) + ˆ2π 0 ∇ ξ j(u ε n , u ε ) • ((u ε n ) -u ε ). ( 25
)
The uniform convergence of u ε n to u ε implies that j(u ε n , u ε ) converges in L 1 (0, 2π) to j(u ε , u ε ), as n → ∞. Moreover (u ε n ) -u ε converges weakly to 0 in L ∞ (0, 2π) and ∇ ξ j(u ε n , u ε ) L ∞ is bounded uniformly in n and ε. Passing to the lim inf in (25), we get lim inf

n→+∞ ˆ2π 0 j(u ε n , (u ε n ) ) ≥ ˆ2π 0 j(u ε , u ε ) ,
which gives an estimate for the numerator of J 2 (u). We deduce the existence of a minimizer u ε for J 2 .

Step 3. Let us consider the renormalising sequence v ε = uε ε . We are going to prove some estimates on v ε which allows to compute the limit of v ε , as ε → 0. The estimates on v ε will be established thanks to the test function w ε , defined here below.

Let a ε = π/4 -επ/6 and b ε = 3π/4 -επ/6. Let w ε be the function, piecewise affine, π-periodic, defined by

w ε (t) =            ε t a ε t ∈ [0, a ε ] , -2ε t -a ε b ε -a ε + ε t ∈ [a ε , b ε ] , -ε π -t π -b ε t ∈ [b ε , π] .
It is easy to see that w ε satisfies w ε L ∞ = ε and condition (NL3). It also satisfies (NL2), since (1 + w ε ) 3 is π-periodic and therefore orthogonal to sine and cosine. We are going to check (NL1), that is,

2 ˆπ 0 w ε + ˆπ 0 w ε2 = 0.
Elementary calculations provide:

ˆπ 0 w ε = 1 2 a ε ε - 1 2 (π -b ε )ε, ˆπ 0 w ε2 = π ε 2 3 .
Therefore (NL1) is satisfied as soon as a ε + b ε -π = -ε/3 which holds true by the definition of a ε and b ε . We also remark that ˆ2π

0 w ε 2 ≤ Cε 2 , ˆ2π 0 w ε2 ≤ Cε 2 , ˆ2π 0 w ε 4 ≤ Cε 4 , ˆ2π 0 w ε w ε 2 ≤ Cε 3 and ˆ2π 0 |w ε | = επ ≥ ε 2 .
These estimates imply that J 2 (w ε ) ≤ C for every ε.

Therefore J 2 (u ε ) ≤ J 2 (w ε ) ≤ C which yields to ˆ2π 0 (u ε ) 2 ≤ ˆ2π 0 u 2 ε + 1 4 (u ε ) 4 + u ε (u ε ) 2 dθ + Cε 3 + C ˆ2π 0 |u ε | 2 .
We deduce that

ˆ2π 0 (u ε ) 2 ≤ Cε 2 .
From the definition of v ε we have v ε L ∞ = 1, moreover by the above estimate we get ˆ2π

0 v ε 2 ≤ C, (26)
and hence the sequence v ε is bounded in H 1 (0, 2π) and, as ε → 0, v ε converges weakly in H 1 (0, 2π) and uniformly to some function v 0 , up to subsequences. Using [START_REF] Lamboley | Polygons as optimal shapes with convexity constraint[END_REF] we deduce that v ε L ∞ ≤ 3 √ ε , and then using (26), we have ˆ2π

0 v ε 4 ≤ 9 ε ˆ2π 0 v ε 2 ≤ C ε . Moreover ˆ2π 0 v ε v ε 2 ≤ ˆ2π 0 v ε 2 ≤ C by (26) and v ε L ∞ = 1.
Step 4. We now prove that the function v 0 found in Step 3 is a test function for the optimization problem [START_REF] Hall | A quantitative isoperimetric inequality in n-dimensional space[END_REF]. This will allow us to prove the statement of this Proposition. We observe that, by (24) and the definition of v ε , we have

m ε = π 2 1 2 ˆ2π 0 v ε 2 -v 2 ε - ε 2 4 v ε 4 -εv ε v ε 2 + Cε 1 + ε 2 2 ˆ2π 0 |v ε | 2 . ( 27
)
Passing to the limit in (27), we get

lim inf ε→0 m ε ≥ π 2 1 2 ˆ2π 0 v 0 2 -v 2 0 ˆ2π 0 |v 0 | 2 .
On the other hand, passing to the limit in (NL1) and (NL2), we see that v 0 satisfies (L1) and (L2) and therefore is an admissible test function for the optimization problem [START_REF] Hall | A quantitative isoperimetric inequality in n-dimensional space[END_REF]. For example, (NL1) is equivalent to

ˆ2π 0 (u 2 ε + 2u ε ) = 0. This implies that 0 ≤ ˆ2π 0 u 2 ε = -2 ˆ2π 0 u ε ≤ Cε 2 by the estimate u ε L ∞ ≤ ε. Therefore, by the definition of v ε , one has ˆ2π 0 v ε → 0, as ε → 0, which gives ˆ2π 0 v 0 = 0. The inequality lim inf ε→0 m ε ≥ lim inf ε→0 m ε ≥ π 4 m follows.
We can now prove Theorem 4.2.

Proof. As we already observed, we can write K n in polar coordinates (see (??)); this implies that

lim inf n→∞ G 0 (K n ) = lim inf ε→0 m ε ≥ π 4 m
where m, m ε are defined by ( 17), [START_REF] Harrell | On the maximization of a class of functionals on convex regions and the characterization of the farthest convex set[END_REF], respectively. . By Proposition 4.3 it is sufficient to prove the estimate π 4 m > 0.41 .

Notice that by Theorem 2.11, it holds m = 1 2(4 -π)

. This implies that π 4 m ≈ 0.457, and the desired estimate holds.

Remark 4.4. Although Fuglede [START_REF] Fuglede | Stability in the isoperimetric problem for convex or nearly spherical domains in R n[END_REF] was interested in the uniform spherical deviation, that is, the Hausdorff distance of a set E from the ball of same measure centered at the barycenter of E, one can easily deduce from his results the inequality δ(E) ≥ C(n)λ 2 0 (E) for nearly spherical sets (see Theorem 3.1 in [START_REF] Fusco | The quantitative isoperimetric inequality and related topics[END_REF]), where C(n) is a constant depending on the dimension. In particular the following estimate can be proved in the plane:

δ(E) ≥ 1 16 λ 0 (E) 2 .
However this estimate is not sufficient to exclude sequences converging to the ball. Our first attempt to prove Theorem 4.2 was the following. For the denominator of J one has

1 2 ˆ2π 0 |(1 + u(θ)) 2 -1|dθ 2 ≤ 1 2 ˆ2π 0 |u(θ)|(2 + ε)dθ 2 = 1 + ε 2 2 u 2 L 1 . For the numerator ˆ2π 0 (1 + u) 2 + u (θ) 2 dθ -1 dθ ≥ ˆ2π 0 u + u 2 + (u ) 2 2 - [4u 2 + 4u 3 + 4uu + u 4 + u 4 + 2u 2 u 2 ] 2 8 + 8u 3 16 dθ ≥ ˆ2π 0 (u ) 2 2 - u 2 2 dθ + o(2) ≥ c -1 2 ˆ2π 0 u 2 dθ + o(2)
where c = 4 is such that ˆ2π

0 (u ) 2 dθ ≥ c ˆ2π 0 u 2 dθ . Therefore lim inf ε→0 J(u) ≥ 3π 4 ˆ2π 0 u 2 dθ ˆ2π 0 |u|dθ 2 ≥ 3 8 = 0.375
by Hölder inequality. Again, this estimate is not sufficient to exclude sequences converging to the ball.

5.

On the regularity and on the shape of the optimal convex set

In this section we prove that an optimal set for the minimization of G 0 within the class of planar convex sets has C 1,1 boundary. About its shape, we conjecture that the stadium S which minimizes G (see Theorem 2.8) also minimizes the functional G 0 . In particular we show that within the class of stadia, S is the only one satisfying the optimality conditions, presented in Theorem 5.4. Unfortunately we are not able to prove that S is an optimizer within the class all convex sets.

The proof of the regularity of the optimal set makes the use of the first order optimality condition in the same spirit of [START_REF] Lamboley | Regularity and singularities of optimal convex shapes in the plane[END_REF].

Let us first recall how to write these optimality conditions, in the case of convexity constraint, when representing the boundary of the convex set with the gauge function. We recall that the gauge function v is defined through the polar representation of the boundary of the convex set as v(θ) = 1/r(θ), where (r(θ) cos θ, r(θ) sin θ) is a boundary point. Thus we consider the optimization problem

min j(v), v ∈ H 1 (0, 2π), v + v ≥ 0, ˆ2π 0 dθ v 2 (θ) = m 0 (29)
(see Proposition 2.3.3 of [START_REF] Lamboley | Polygons as optimal shapes with convexity constraint[END_REF]), where j(v) represents an analytic functional related to a shape functional.

Proposition 5.1. Assume that v 0 solves (29) where j : H 1 (0, 2π) → R is C 2 . Then there exist ξ 0 non-negative, µ ∈ R such that ξ 0 = 0 on the support of (v 0 + v 0 ) and

dj(v 0 ; ϕ) =< ξ + ξ , ϕ > -µ dm(v 0 ; ϕ)
for every ϕ ∈ H 1 (0, 2π), where dj(v; ϕ) indicates the derivative of j(v) along the deformation ϕ and dm(v 0 ; ϕ) is the corresponding derivative of m(v) := ´2π 0 dθ v 2 (θ) . In this section, we prove the following regularity result: For the proof, we first express the optimality condition of Proposition 5.1 in our context: Proposition 5.3. Let r, θ be the polar coordinates. Let v(θ) = 1 r(θ) be the gauge function used to describe the boundary of a set E. The optimal set satisfies the following condition: there exists ξ ∈ H 1 such that ξ ≥ 0 for every θ ∈ [0, 2π] and ξ(θ) = 0 at every θ corresponding to strictly convex boundary points of E and there exist μ0 , μ1 , μ2 ∈ R such that

- 1 2πλ 2 0 v + v (v 2 + v 2 ) 3 2 - 2δ πλ 3 0 sign(v 2 -1) v 3 = ξ + ξ - μ0 v 3 -3 μ1 cos θ + μ2 sin θ 2v 4 . ( 30
)
Proof. We are going to apply Proposition 5.1. To do that, we need to compute the derivative of the functional G 0 and those of the constraints; we will consider the analytic expressions given by J(v), with constraint v ∈ N L. With abuse of notation we write J(v) to consider the functional J(v) rewritten in terms of the gauge function v, as considered in [START_REF] Lamboley | Polygons as optimal shapes with convexity constraint[END_REF].

(1) The derivative of J(v) along the deformation ϕ is

dJ(v; ϕ) = 1 2πλ 2 0 dP (v; ϕ) -2 δ λ 3 0 dλ 0 (v; ϕ),
where the derivative of P is

dP (v; ϕ) = - ˆv + v (v 2 + v 2 ) 3 2
ϕ .

The derivative of πλ 0 is the derivative of

ˆ{v<1} χ Ω\B + ˆ{v>1} χ B\Ω which gives π dλ 0 (v; ϕ) = 1 2 ˆ{v<1} -2ϕ v 3 + 1 2 ˆ{v>1} 2ϕ v 3 .
(2) The constraints (NL) rewritten in terms of v are

1 2 ˆdθ v 2 (θ) = π, 1 2
ˆ2π 0 cos θ v 3 (θ) = 0 1 2 ˆ2π 0 sin θ v 3 (θ) = 0.
By derivating each left-hand sides of the above conditions, and introducing the corresponding Lagrange multipliers µ 0 , µ 1 , µ 2 , we obtain respectively:

- ˆμ 0 v 3 , -3 ˆμ 1 cos θ 2v 4 , - 3
ˆμ 2 sin θ 2v 4 , so that the desired result follows by Proposition 5.1.

We are now able to prove Theorem 5.2:

Proof. We are going to use the notations of the above proposition. Notice that on the strictly convex parts of the boundary, ξ = 0 and v satisfies a second order ordinary differential equation:

(1) in the exterior of the unit ball v < 1, and so v is continuous by equation (29). By a classical bootstrap argument v is C ∞ ; (2) in the exterior of the unit ball v > 1, and so v is continuous by equation (29). By a bootstrap argument v is C ∞ ; (3) on the boundary of the unit ball v = 1, v is bounded, but not continuous. Thus v is W 2,∞ there. This and the above proposition imply that on strictly convex parts on ∂B, v is C 1,1 . Now, let us prove that Ω is C 1 . If this was not the case, we would have a corner for some θ 0 . This implies that the gauge function is such that v + v contains a Dirac mass, with a positive weight at θ 0 . Thus, the H 1 function ξ appearing in the optimality condition (30) also satisfies that the quantity ξ + ξ contains a Dirac mass at θ 0 . Now, since ξ(θ 0 ) = 0 and ξ ≥ 0, the weight of this Dirac mass must be non-negative, in contradiction with the minus sign appearing in the left-hand side of (30) in front of v + v.

We are left with the conjunctions between a strictly convex part of the boundary and a non strictly convex part. For that, it is sufficient to remark that any C 2 (R + )(C ∞ (R + )) function, which is zero at x = 0, can be extended by 0 on R, getting a C 1,1 (R) function. This ends the proof that an optimal set is C 1,1 .

We are going to write the optimality conditions on strictly convex parts in a different way. In particular, this will give the explicit expression of the Lagrange multipliers in (30). We can assume that all the considered sets have area equal to π. Theorem 5.4. Let Ω be an optimal set minimizing the functional G 0 and assume its barycenter is at the origin. Let B be the unit ball centered at the origin. Let

∂Ω IN = ∂Ω ∩ B, ∂Ω OU T = ∂Ω ∩ B c , ∂B IN = ∂B ∩ Ω, ∂B OU T = ∂B ∩ Ω c . Then at every strictly convex boundary point (x, y) of Ω the curvature C(x, y) satisfies: C(x, y) = 1 -3δ + 4δ 2πλ 0 |∂B OU T | -|∂B IN | ± 4δ λ 0 + μ1 x + μ2 y , (31)
(+ at the exterior of B and -in the interior of B) where

μ1 = 4δ πλ 0 ˆ∂B OU T cos tdt - ˆ∂B IN cos tdt , μ2 = 4δ πλ 0 ˆ∂B OU T sin tdt - ˆ∂B IN sin tdt .
Proof. We are going to perform shape variations on the strictly convex parts of ∂Ω. The proof is divided into several steps. Let V be a perturbation, that is a smooth map V : R 2 → R 2 . We denote by V • n the scalar product of V with the outer unit normal vector n to ∂Ω.

(1) Let Ω t = (I + tV )(Ω). Then

|Ω t | = π + t ˆ∂Ω V • n + o(t) .
(2) The barycenter constraint implies that

ˆΩt x dxdy = 0 + t ˆ∂Ω xV • n + o(t). Since by definition x t = 1 |Ω t | ˆΩt
x dxdy, by the above formulas one has

x t = t π ˆ∂Ω x V • n + o(t) .
A similar formula holds for y t :

y t = t π ˆ∂Ω y V • n + o(t) .
(3) Let B t = (I + tW )(B), where

W (x, y) = (a, b) + α(x, y) , with (a, b) = 1 π ˆ∂Ω xV • n, ˆ∂Ω yV • n , α = 1 2π ˆ∂Ω V • n .
(4) The difference between |Ω t ∆B t | and |Ω∆B| is given by two terms:

|Ω t ∆B t | -|Ω∆B| = ± t ˆ∂B W • n ± t ˆ∂Ω V • n :
for the first term of the right hand side + is on ∂B OU T and -is on ∂B IN ; for the last term of the right hand side, + is ∂Ω OU T and -on ∂Ω IN . In the next part of the proof we will write |Ω t ∆B t | = |Ω∆B| + tR. (5) We have

λ 0 (Ω t ) = |Ω t ∆B t | |Ω t | = |Ω∆B| + tR |Ω| + t ´∂Ω V • n = λ 0 (Ω) • 1 + t R |Ω∆B| 1 + t π ´∂Ω V • n = λ 0 (Ω) + t R π - λ 0 π ˆ∂Ω V • n . dλ 0 (Ω, V ) = 1 π ± ˆ∂B W • n ± ˆ∂Ω V • n -λ 0 ˆ∂Ω V • n ,
that is,

dλ 0 (Ω, V ) = 1 π ˆ∂B OU T W • n - ˆ∂B IN W • n + ˆ∂Ω OU T V • n - ˆ∂Ω IN V • n -λ 0 ˆ∂Ω V • n .
(6) If r t is the radius of the ball having the same area as Ω t , then

r t = π + t ´∂Ω V • n π = 1 + t ´∂Ω V • n 2π .
This gives

δ(Ω t ) = P (Ω t ) 2πr t -1 = P (Ω t ) 2π + t ´∂Ω V • n -1 = P (Ω) + t ´∂Ω C V • n 2π + t ´∂Ω V • n -1 ,
where C indicates the curvature of the boundary. With the same computations as for λ 0 δ(Ω t ) = δ(Ω)

+ t ´∂Ω C V • n 2π -t P (Ω) ´∂Ω V • n 4π 2 and so dδ(Ω, V ) = ˆ∂Ω C 2π - (δ + 1)2π 4π 2 V • n = ˆ∂Ω C -δ -1 2π V • n .
The optimality condition for G 0 : dδ λ 2 0 -2δ λ 3 0 dλ 0 = 0, can be written as Remark 5.5. If S denotes the stadium of Theorem 2.8, we can prove that S is the only stadium satisfying the optimality conditions (31).

ˆ∂Ω (C -δ -1)V • n = 4δ λ 0 ˆ∂B OU T W • n - ˆ∂B IN W • n + ˆ∂Ω OU T V • n - ˆ∂Ω IN V • n -λ 0 ˆ∂Ω V • n .
To see that, let us consider a stadium centered at 0, given by the union of a rectangle of dimensions 2r × 2l with r ≤ 1 and two half discs of radius r. Let θ be the angle such that r = sin θ. Assuming without loss of generality that the area is π, one has l = π -π sin 2 θ 4 sin θ .

The perimeter equals 4l + 2πr = π sin θ + π sin θ which implies

δ(θ) = 1 2 sin θ + sin θ 2 -1 ,
while we can compute λ 0 as λ 0 (θ) = 2 π (π -2θ -sin(2θ)) .

On one hand, an optimal stadium is a critical point of the function θ → δ(θ)/λ 2 0 (θ). This leads to solve the nonlinear equation 8 sin θ(1 -sin θ) 2 -cos θ(π -2θ -sin(2θ)) = 0. (32) It is a simple exercise to prove that this equation has a unique solution, providing the stadium S which corresponds to the value θ ∼ 0.5750. On the other hand, writing condition (31) for a stadium yields, with the same notations, to the equation

4 sin θ - 3 2 sin 2 (θ) - 5 2 + 2 (1 -sin θ) 2 (π -2θ) π -2θ -sin(2θ) = 0. ( 33
)
It is easy to check that equation (33) has a unique solution in (0, π/2) and this solves equation (32) too.

Therefore there is only one stadium satisfies (31); this stadium is S, since λ 0 and λ are equal on any symmetric set. Now, to prove that the stadium is indeed the minimizer, it certainly requires to prove first that the Lagrange multilplier μ1 and μ2 are both zero, then a precise analysis should provide the result.

Theorem 4 . 2 .

 42 Let K n be a sequence of convex compact sets converging to a ball in the Hausdorff metric. Then lim inf n→∞ G 0 (K n ) ≥ 0.41.

Theorem 5 . 2 .

 52 A minimizer of G 0 within the class of convex compact sets of the plane has C 1,1 boundary. Moreover the strictly convex parts of the boundary (except the intersection with the circle of the barycentric disk) are of class C ∞ .

  Now, W • n = a cos θ + b sin θ + α (since (x, y) • n = 1 on ∂B), so W • n = cos θ ˆ∂Ω xV • n + sin θ ˆ∂Ω yV • n + 1 2π ˆ∂Ω V • n which gives C = δ + 1 -4δ + 4δ 2πλ 0 |∂B OU T | -|∂B IN | ± 4δ λ 0 + μ1 x + μ2 y .

Acknowledgments. The first author is supported by the GNAMPA group of Istituto Nazionale di Alta Matematica (INdAM).

A part of this work has been realized while the second and the third authors were hosted at IAS at Princeton. The authors thank the Institute for its warm hospitality.

The authors have been supported by the ANR Projects OPTIFORM and SHAPO of the CNRS (Centre National de la Recherche Scientifique) and by the Fir Project 2013 "Geometrical and qualitative aspects of PDE's" of MIUR (Italian Ministry of Education).


Narinder Kumar
email: narinderkw@gmail.com

Priyanka Wadhawan
email: priyanka.wadhawan@gmail.com

REPRESENTATION OF INTEGERS BY SUM OF THREE CUBES -A NEW APPROACH BASED ON SEED EQUATION

Keywords: Cube, Cubic Equation, Integer Root, Linear Equation, Parametrisation, Pythagorean's Triple, Quadratic Equation, Seed Equation. 2010 AMS classification:11D25

We have proved in this paper that numbers can be expressed in algebraic form using one variable and two real rational quantities and thus sum of three cubes can also be expressed in algebraic form as a cubic polynomial. Using skeletal or seed equation, the polynomial can be transformed into a quadratic equation. A seed equation denotes a simple equation that represents a given integer as sum of three cubes including 0, for example, a seed equation for integer 2 is 1 cubed plus 1 cubed plus 0 cubed. Resultant quadratic equation can further be transformed into a linear equation which yields value of the variable and substitution of the value of the variable into the algebraic form of numbers results in the required solution.Notwithstanding, finding a single set of three cubes, we have found, using this approach, multiple sets of cubes. We have also given parametrization for such cubes. Cases, where it is difficult or not possible to determine a seedequation, alternate method has been Narinder Kumar Wadhawan and Priyanka Wadhawan 74 provided. Methods used are unattempted, innovative and easily comprehensible

Introduction

It is an open problem in Number Theory tocharacterisean integer as sum of three cubes. Heath-Brown conjectured that all integers except of the forms 9𝑥 ± 4 or 9𝑥 ± 5, can be expressed as sum of three cubes in multiple ways [START_REF] Heath-Brown | On Solving the Diophantine Equation𝑥 3 + 𝑦 3 + 𝑧 3 = 𝑘 on a Vector Computer[END_REF]. Apart from this statement, number 0 can not be expressed as sum of three distinct cubes on account of Fermat's last theorem that states, three integers 𝑎, 𝑏 and 𝑐 can not be expressed as 𝑎 𝑛 + 𝑏 𝑛 = 𝑐 𝑛 where 𝑛 > 2. Only possible solution is trivial, 𝑎 3 + (-𝑎) 3 + 0 = 0. On the other hand, K. Mahler in the year 1936, found that integer 1 is expressible in infinite ways by parametric (1 -9𝑝 3 ) 3 + (3𝑝 -9𝑝 4 ) 3 + (9𝑝 4 ) 3 = 1, [START_REF] Mahler | Note on Hypothesis K of Hardy and Littlewood[END_REF]. There are other representations and family of parametrisation as well for this integer 1, [START_REF] Avagyan | A New Method in the Problem of Three Cubes[END_REF]. Earlier in the year 1908, A.S. Verebrusov had already discovered that integer 2 is expressible in infinite ways by parametric (1 -6𝑝 3 ) 3 + (1 + 6𝑝 3 ) 3 + (-6𝑝 2 ) 3 = 2, [2], [START_REF] Mordell | On sums of three cubes[END_REF]. Notwithstanding parametric solutions stated above, there is a family of solutions for representing integer 2, [START_REF] Avagyan | A New Method in the Problem of Three Cubes[END_REF], [START_REF] Heath-Brown | On Solving the Diophantine Equation𝑥 3 + 𝑦 3 + 𝑧 3 = 𝑘 on a Vector Computer[END_REF]. For integer 3, Mordell in the year 1953, expressed his opinion that he does not know anything more than to say 3 = 1 3 + 1 3 + 1 3 = 4 3 + 4 3 + (-5) 3 . Heath-Brown, Lioen, and te Riele in the year 1992, found the cubes that sum up to 39 by equation 39 = 134476 3 + 117367 3 + (-159380) 3 [START_REF] Heath-Brown | On Solving the Diophantine Equation𝑥 3 + 𝑦 3 + 𝑧 3 = 𝑘 on a Vector Computer[END_REF]. Thereafter, for finding solutions to 𝑥 3 + 𝑦 3 + 𝑧 3 = 𝑛, where 𝑛 varies from 1 to 1000, many authors implemented computational searches up to (|𝑥|, |𝑦|, |𝑧|) < 10 15 [START_REF] Elsenhans | New sums of three cubes[END_REF]. At present, up to 𝑛 < 100, solutions have been found except two integers 33 and 42 [START_REF] Georgiou | The untracked problem with 33: Mathematicians solves 64-year-old Diophantine puzzle[END_REF]. But now, Andrew Booker and Andrew Southland have settled the cases for these two integers also [START_REF] Booker | On a question of Mordell[END_REF], [START_REF] Pavlus | Sum of Three Cubes Problem Solved forStubborn Number 33[END_REF], [START_REF] Lu | Mathematics cracks centuries-old problem about the number 33[END_REF]. Up to 1000, there are integers 114, 390, 627, 633, 732, 921 and 975 which could not be solved for their representation by sum of three cubes [START_REF] Avagyan | A New Method in the Problem of Three Cubes[END_REF], [7].

Although much work has been done, we, resorting to innovative and unattempted methods, have endeavoured to solve the problem of representations of integers 1, 2, 3, … so on except integers of the forms 9𝑥 ± 4 or 9𝑥 ± 5 by sum of three cubes. For integers of some specific forms, we have derived their parameterisation. We have also proved that a given integer can be expressed by multiple sets of three adding cubes.We have proved the veracity of formulae derived by us, presenting exhaustive examples in Tables. To begin with, we represent an integer 𝑘 as sum of three cubes by the equation 𝑋 3 + 𝑌 3 + 𝑍 3 = 𝑘 (1.1) where 𝑋, 𝑌 and 𝑍 are all integers positive or negative. To determine 𝑋, 𝑌 and 𝑍, we need a seed equation which, in fact, is a skeletal equation for integer 𝑘. This equation, then will be used to determine values of 𝑋, 𝑌 and 𝑍. Seed equations for 𝑘 = 1, can be written easily as

(1) 3 + (0) 3 + (0) 3 = 1 (1.2) or

(1) 3 + (𝑝) 3 + (-𝑝) 3 = 1.

(1.3) where 𝑝 is an integer positive or negative. Some of the seed equations for integers 1, 2, 3, … are given in Table1.1 for better clarity.

Table1.1 Some Seed Equations Integers to Be Represented '𝑘' 𝐴 3 + 𝐵 3 + 𝐶 3

= 𝑘

Integers to Be Represented '𝑘'

𝐴 3 + 𝐵 3 + 𝐶 3 = 𝑘 1 (1) 3 + (𝑝) 3 + (-𝑝) 3 = 1 7 (2) 3 + (-1) 3 + (0) 3 = 7 2 (1) 3 + (1) 3 + (0) 3 = 2 8 (2) 3 + (0) 3 + (0) 3 = 8 3 (1) 3 + (1) 3 + (1) 3 = 3 9 (2) 3 + (1) 3 + (0) 3 = 9 6
(2) 3 + (-1) 3 + (-1) 3 = 6

10

(2) 3 + (1) 3 + (1) 3 = 10 It may happen that seed equation for a given 𝑘 is not easily determinable, such cases will be dealt at the end. We submit, a real rational number, say 𝑛, can be expressed in algebraic form, 𝑛 = 𝑎 • 𝑥 + 𝐴, where 𝑎 and 𝐴 are real rational quantities assigned by us and 𝑥 is a variable. From above said equation, we can write, 𝑥 = 𝑛-𝐴 𝑎 . To illustrate, how a number, say 𝑛, can be expressed, we take the example for 𝑛 = 5. Let 𝑎 and 𝐴 have values

+ 𝑐 3 ) + 3𝑥 2 (𝑎 2 • 𝐴 + 𝑏 2 • 𝐵 + 𝑐 2 • 𝐶) + 3𝑥 (𝑎 • 𝐴 2 + 𝑏 • 𝐵 2 + 𝑐 • 𝐶 2 ) +𝐴 3 + 𝐵 3 + 𝐶 3 -𝑘 = 0.
(1.5) Since 𝑎, 𝑏, 𝑐, 𝐴, 𝐵 and 𝐶 have values as assigned by us, therefore, coefficients of 𝑥 3 , 𝑥 2 , 𝑥 and constant term of above equation are known. Being a cubic equation, finding integer values of 𝑥, satisfying this equation, is tedious, hence determination of values of 𝑋 = 𝑎 • 𝑥 + 𝐴, 𝑌 = 𝑏 • 𝑥 + 𝐵 and 𝑍 = 𝑐 • 𝑥 + 𝐶 will also be tedious and difficult. To tide over this difficulty, we will transform cubic into a linear equation.

Transformation of Cubic Equation into a Linear Equation

For solving cubic equation in an easy way, it is transformed into a quadratic equation and, then to linear equation.Transformation into a quadratic equation will be achieved by use of seed equation that has already been explained in the introduction section. Let the seed equation for integer 𝑘 be 𝐴 3 + 𝐵 3 + 𝐶 3 = 𝑘. Substituting this value of 𝑘 in Equation (1.5) and on simplifying, it gets transformed into 𝑥{𝑥 2 (𝑎 3 + 𝑏 3

+ 𝑐 3 ) + 3𝑥 (𝑎 2 • 𝐴 + 𝑏 2 • 𝐵 + 𝑐 2 • 𝐶) + 3(𝑎 • 𝐴 2 + 𝑏 • 𝐵 2 + 𝑐 • 𝐶 2 )} = 0.
This equation has one root at 𝑥 = 0 and this root is ignored on account of the fact, it will yield the given seed equation. Ignoring root at 𝑥 = 0, the equation, then reduces to quadratic form 𝑥 2 (𝑎 3 + 𝑏 3

+ 𝑐 3 ) + 3𝑥 (𝑎 2 • 𝐴 + 𝑏 2 • 𝐵 + 𝑐 2 • 𝐶) + 3(𝑎 • 𝐴 2 + 𝑏 • 𝐵 2 + 𝑐 • 𝐶 2 ) = 0.
(2.1) If Equation (2.1) is solvable for real rational roots, say 𝑥 equal to 𝑝 and 𝑞 such that substitution of value of the roots in (𝑎 • 𝑥 + 𝐴), (𝑏 • 𝑥 + 𝐵) and (𝑐 • 𝑥 + 𝐶) yields integer values on account of our choice of 𝑎, 𝑏 and 𝑐, then required cubes are given by relations

(𝑎 • 𝑝 + 𝐴) 3 + (𝑏 • 𝑝 + 𝐵) 3 + (𝑐 • 𝑝 + 𝐶) 3 = 𝑘, (2.

2) and

(𝑎 • 𝑞 + 𝐴) 3 + (𝑏 • 𝑞 + 𝐵) 3 + (𝑐 • 𝑞 + 𝐶) 3 = 𝑘.

(2.3) If the quadratic Equation (2.1) can not be solved or difficult to solve for real rational root or roots, it will be further transformed into a linear equation by equating constant term to zero.That is

(𝑎 • 𝐴 2 + 𝑏 • 𝐵 2 + 𝑐 • 𝐶 2 ) = 0.
(2.4) The transformed equation, then will be 𝑥 2 (𝑎 3 + 𝑏 3 + 𝑐 3 ) + 3𝑥 (𝑎 2 • 𝐴 + 𝑏 2 • 𝐵 + 𝑐 2 • 𝐶) = 0. Again this equation has one root, 𝑥 = 0 and this root will yield seed equation, hence ignored. Transformed linear equation will, then be

𝑥 = - 3(𝑎 2 •𝐴+𝑏 2 •𝐵+𝑐 2 •𝐶) (𝑎 3 +𝑏 3 +𝑐 3 )
= 𝑟 (say).

(2. Assignment of values to 𝑎, 𝑏 and 𝑐 is very crucial. These are so chosen that either cubic equation (1.5) or quadratic equation (2.1) or linear equation (2.5) yields at least one real and rational root such that substitution of that root in (𝑎 • 𝑥 + 𝐴), (𝑏 • 𝑥 + 𝐵) and (𝑐 • 𝑥 + 𝐶) gives integer values. To achieve this result, certain algebraic identities will be used. With this concept of transformation of cubic equation, we proceed to determine three cubes that sum up to a given integer.

Use of a Seed Equation

We will take up relatively easier cases first, then harder and ultimately the hardest, where it is difficult to find a seed equation.

Representation of Integer 2 and Other Integers Expressible as 𝟐𝒑 𝟑 , Where 𝒑 Is an Integer

Seed equation for integer 2 is (1) 3 + (1) 3 + (0) 3 = 2. That makes 𝑋 = 𝑎 • 𝑥 + 1, 𝑌 = 𝑏 • 𝑥 + 1 and 𝑍 = 𝑐 • 𝑥. We have chosen, 𝐴 = 1, 𝐵 = 1 and 𝐶 = 0 intentionally so as to get rid of constant term in Equation (3.1). Equation, then can be written,

(𝑎 • 𝑥 + 1) 3 + (𝑏 • 𝑥 + 1) 3 + (𝑐 • 𝑥) 3 = 2. (3.1) On simplification, 𝑥 2 (𝑎 3 + 𝑏 3 + 𝑐 3 ) + 3𝑥 (𝑎 2 + 𝑏 2 ) + 3 (𝑎 + 𝑏 ) = 0.
For transformation into linear equation, we put 𝑎 + 𝑏 = 0. Transformed linear equation is 𝑥 (𝑎 3 + 𝑏 3 + 𝑐 3 ) + 3(𝑎 2 + 𝑏 2 ) = 0. On putting, 𝑏 = -𝑎,

𝑥 = - 6 𝑐 ( 𝑎 2 𝑐 2 ) = - 6 𝑐 (𝑝 2 ),
where 𝑝 = 𝑎/𝑐. Substituting this value of 𝑥 in Equation (2.6)

(1 -6𝑝 3 ) 3 + (1 + 6𝑝 3 ) 3 + (-6𝑝 2 ) 3 = 2. (3.
2) This parameterisation proves, there are infinite values of integer 𝑝 that satisfy Equation (3.2), hence there are infinite ways to represent 2 as sum of three cubes. This relation was discovered by A.S. Verebrusov who found this parametrisation [2]. But we have given the simplest method of determining family of sets of three cubes that sum up to 2. On multiplying identity (3.2) with 𝑑 3

, {𝑑(1 -6𝑝 3 )} 3 + {𝑑(1 + 6𝑝 3 )} 3 + {-𝑑(6𝑝 2 )} 3 = 2𝑑 3 . (3.
3) This identity is a parametrisation for the integers 2𝑑 3 expressed as sum of three cubes.

Representation of Integers 3 and 𝟑𝒑

𝟑 Seed equation for integer 3 is (1) 3 + (1) 3 + (1) 3 = 3. That makes 𝑋 = (𝑎 • 𝑥 + 1), 𝑌 = (𝑏 • 𝑥 + 1) and 𝑍 = (𝑐 • 𝑥 + 1). Corresponding equation is (𝑎 • 𝑥 + 1) 3 + (𝑏 • 𝑥 + 1) 3 + (𝑐 • 𝑥 + 1) 3 = 3.
(3.4) On simplification, 𝑥 3 (𝑎 3 + 𝑏 3 + 𝑐 3 ) + 3𝑥 2 (𝑎 2 + 𝑏 2 + 𝑐 2 ) + 3𝑥(𝑎 + 𝑏 + 𝑐) = 0 . For transformation into linear equation, we put 𝑎 + 𝑏 + 𝑐 = 0 and simplify it, then

𝑥 = 2 ( 1 𝑐 - 𝑐 𝑎𝑏
).

Substituting this value of 𝑥 in Equation (3.4), we get

{1 + 2𝑎 ( 1 𝑐 - 𝑐 𝑎𝑏 )} 3 + {1 + 2𝑏 ( 1 𝑐 - 𝑐 𝑎𝑏 )} 3 + {1 + 2𝑐 ( 1 𝑐 - 𝑐 𝑎𝑏 )} 3 = 3. (3.5) Let 𝑏 = 𝑐 • 𝑑, then 𝑎 = -(𝑐 • 𝑑 + 𝑐) since 𝑎 = -(𝑏 + 𝑐),
where 𝑑 is real rational quantity. On putting these values of 𝑎 and 𝑏 in Equation (3.5) and simplifying,

{-1 -2 (𝑑 + 1 𝑑 )} 3 + {1 + 2 (𝑑 + 1 𝑑+1 )} 3 + {1 + 2 ( 1 𝑑 + 𝑑 𝑑+1 )} 3 = 3. (3.6) For integer solutions, 2 (𝑑 + 1 𝑑 ) , 2 (𝑑 + 1 𝑑+1
) and 2 (

1 𝑑 + 𝑑 𝑑+1
) must be integers. To achieve this objective, 𝑑 and 𝑑 + 1 must be factors of 2 and either of the two, must not be zero. Due to reciprocity of 𝑑 and 𝑑 + 1 with 1/𝑑 and 1/(𝑑 + 1) respectively as is apparent from Equation (3.6), values of 1/𝑑 and 1/(𝑑 + 1) will also satisfy Equation (3.6) if values of 𝑑 and (𝑑 + 1) satisfy this equation. For example, 𝑑 = -2 or 1/𝑑 = -2 or 𝑑 = 1 will satisfy the Equation (3.6) and yield (4) 3 + (-5) 3 + (4) 3 = 3 , (4) 3 + (4) 3 + (-5) 3 = 3, (-5) 3 + (4) 3 + (4) 3 = 3.

(3.7) These equations have same cubes but latter two equations have cubes displaced from their original position and, in this way, these three equations amount to one set of cubes. Multiplication of Equation (3.7) with 𝑝 3 , where 𝑝 is an integer, gives parametrisation for 3𝑝 3 (4𝑝) 3 + (4𝑝) 3 + (-5𝑝) 3 = 3𝑝 3 .

(3.8) 3.2a Multiple Sets of Three Cubes That Represent Integer 𝟑𝒑 𝟑 Equation (3.6), when multiplied with 𝑝 3 , where 𝑝 is an integer, takes the form

{-𝑝 -2𝑝 (𝑑 + 1 𝑑 )} 3 + {𝑝 + 2𝑝 (𝑑 + 1 𝑑+1 )} 3 + {𝑝 + 2𝑝 ( 1 𝑑 + 𝑑 𝑑+1 )} 3 = 3𝑝 3 . (3.9)
For integer values of cubes, values of 𝑑 should be such that 2𝑝(𝑑 + 1/𝑑), 2𝑝{𝑑 + 1/(𝑑 + 1)} and 2𝑝{1/𝑑 + 𝑑/(𝑑 + 1)} must be integers. For integer 2, we have already found out factors in preceding paragraph and these will also be applicable to 2𝑝 as 2 is a factor of 2𝑝 but to have more factors, 2𝑝 must be of the form 2𝑝 = 𝑞(𝑞 + 1), where 𝑞 is an integer positive or negative such that either 𝑞 or (𝑞 + 1) is not zero. Let us take the case, where 𝑞 = 2. In that case, factors of 2𝑝 (or 6) of the form 𝑑(𝑑 + 1), are (2)(3) and (1)(2) ignoring factors (-3)(-2) and (-2)(-1) as these yield same cubes but displaced from their original positions. Therefore, putting 𝑑 = 1 and 𝑝 = 3, we get (-15) 3 + (12) 3 + (12) 3 = 3(3) 3 and on putting 𝑑 = 2 and 𝑝 = 3, we get (-18) 3 + (17) 3 + (10) 3 = 3(3) 3 . This proves proceeding Lemmas3.1, 3.2 and 3.3. Lemma 3.1: Integer 3 is represented by sum of cubes

{-1 -2 (𝑑 + 1 𝑑 )} 3 + {1 + 2 (𝑑 + 1 𝑑 + 1 )} 3 + {1 + 2 ( 1 𝑑 + 𝑑 𝑑 + 1 )} 3 = 3,
where 𝑑 has such values that 𝑑 and 𝑑 + 1 are factors of 2. Also 2 (𝑑 +

{-𝑝 -2𝑝(1 + 1)} 3 + {𝑝 + 2𝑝 (1 + 1 2 )} 3 + {𝑝 + 2𝑝 (1 + 1 2 )} 3 = 3𝑝 3 and {-𝑝 -2𝑝 (𝑞 + 1 𝑞 )} 3 + {𝑝 + 2𝑝 (𝑞 + 1 𝑞 + 1 )} 3 + {𝑝 + 2𝑝 ( 1 𝑞 + 𝑞 𝑞 + 1 )} 3 = 3𝑝 3
provided 𝑞 is not further factorable as 𝑞 1 (𝑞 1 + 1). From above, this can be concluded, when 𝑘 is of form form 3(𝑃!/2!) 3 , where 𝑃 ≥ 3, then 𝑘 can be represented by multiple sets of three cubes. Number of such sets of three cubes equals to the number of factors of the form 𝑑(𝑑 + 1) of 𝑃!. Also, when 𝑃 ≥ 5, then number of sets of cubes is more than 𝑃. This proves Lemma3.4. Lemma3.4: If integer 𝑝 = (𝑃!/2!) and 𝑃 ≥ 3, then integer 𝑘 = 3(𝑃!/2!) 3 is represented by sum of cubes

3.2b Multiple Sets or

{-𝑝 -2𝑝 (𝑑 + 1 𝑑 )} 3 + {𝑝 + 2𝑝 (𝑑 + 1 𝑑 + 1 )} 3 + {𝑝 + 2𝑝 ( 1 𝑑 + 𝑑 𝑑 + 1 )} 3 = 3𝑝 3 = 3 ( 𝑃! 2! ) 3
in as many sets as𝑃! has distinct factors of the form 𝑑(𝑑 + 1). When 𝑃 ≥ 5, then 3𝑝 3 can be represented by sets of cubes in as many numbers as equal to or more than 𝑃. Also, If 𝑃! has 𝑛 sets of distinct factors, then using this method, there will be 𝑛 distinct sets of three cubes to represent 3𝑝 3 . Let 𝑎 = -9𝑝 2 , 𝑏 = 3(1 -3𝑝 3 ), 𝑐 = 9𝑝 3 , then quadratic equation gets transformed into (1 -9𝑝 3 )𝑥 2 + 9𝑥𝑝 4 -𝑝 2 = 0 which has roots, 𝑥 = 𝑝 and 𝑥 = -𝑝/(1 -𝑝 3 ). On putting these values, 𝑥 = 𝑝 or 𝑥 = -𝑝/(1 -𝑝 3 ) in Equation (3.10), we get the identity (1 -9𝑝 3 ) 3 + (3𝑝 -9𝑝 4 ) 3 + (9𝑝 4 ) 3 = 1.(3.11) This is a parameterisation for representation of integer 1 and proves, there are infinite values of 𝑝, hence infinite ways to represent 1 as sum of three cubes. This was first given by K. Mahler in 1936 [START_REF] Mahler | Note on Hypothesis K of Hardy and Littlewood[END_REF]. But we have given the simplest method of determining family of sets of three cubes that sum up to 1.Referring to identity (3.11), if this is multiplied by 𝑑 3 , where 𝑑 is an integer, it gets transformed into {𝑑(1 -9𝑝 3 ) 3 } 3 + {𝑑(3 -9𝑝 3 )} 3 + {𝑑(9𝑝 4 )} 3 = 𝑑 3 (3.12) which is a parametrisation for sum of three cubes that equals 𝑑 3 .

Representation of

Representation of Integer 𝟒 or 𝟓

Using Lemma 1.1, a real rational integer 𝑋 can be expressed as 𝑋 = 9𝑥 + 𝑎 , where 𝑥 is an integer positive or negative and 𝑎 is also an integer positive 0, 1, 2, 3, 4, 5 or a negative integer -1, -2, -3, -4. For examination of divisibility of 𝑋 3 by 9, we expand 𝑋 3 ,

𝑋 3 = (9𝑥 + 𝑎) 3 = (9𝑥) 3 + 27𝑥 • 𝑎(9𝑥 + 𝑎) + 𝑎 3 .
In the above equation, (9𝑥) 3 and 27𝑥 • 𝑎(9𝑥 + 𝑎) are obviously divisible by 9, therefore, we will now discuss divisibility of 𝑎 3 by 9. Taking different values of 𝑎 as stated above, 𝑎 3 is found to have remainders either 0 or 1 or -1 or 0 or 1 or -1 or 0. Therefore, algebraic sum of three cubes i.e. 𝑋 3 + 𝑌 3 + 𝑍 3 will have a remainder which is either algebraic sum of any combination of three integers 0, 1 or -1 or single integer 0 or 1 or -1 repeated thrice. All combinations of the sum of remainders yield 0 or ±1 or ±2 or ±3. Therefore, sum of three cubes when divided by 9 can not have remainders other than as mentioned above. Conversely, it can not have remainder ±4. When division is by 9, remainder -4 is equivalent to remainder 5 and remainder -5 equivalent to 4, therefore, it can also be stated that sum of three cubes, when divided by 9, can not have remainder ±5. Therefore, 𝑘 = 9𝑥 ± 4 or 𝑘 = 9𝑥 ± 5 can not be represented by sum of three cubes. In 1992, Roger Heath-Brown conjectured that every 𝑘 unequal to 4 or 5 modulo 9 has infinitely many representations as sums of three cubes [START_REF] Heath-Brown | On Solving the Diophantine Equation𝑥 3 + 𝑦 3 + 𝑧 3 = 𝑘 on a Vector Computer[END_REF].

Representation of Integer 7 and Other Integers Expressible as 𝟕𝒅 𝟑

Seed equation for integer 7 is (2) 3 + (-1) 3 + (0) 3 At 𝑦 = 4, 44 3 -169 3 + 168 3 = 7. On multiplying above said identity with 𝑑 3 , it gets transforms into

(44𝑑 ) 3 + (-169𝑑 ) 3 + (168𝑑) 3 = 7𝑑 3
which is a parametrisation for integer expressible as 7𝑑 3 , where 𝑑 is an integer. On putting, 𝑦 = 𝑝 2 in above equation, {𝑝 + 3𝑝(𝑝 3 -1)} 3 + {-1 -3𝑝 3 (𝑝 3 -1)} 3 + {3𝑝 3 (𝑝 3 -1)} 3 = 𝑝 3 -1,(3.17)

Representation of an Integers

which is parametrisation for integer 𝑝 3 -1, where 𝑝 is an integer positive or negative.

Table3.1 Sum of Three Cubes That Equals to 𝑝 3 -1 𝑝 𝑋 3 + 𝑌 3 + 𝑍 3 = 𝑝 3 16) reveals that if 𝑝 is assumed as -𝑝, this equation gets transformed into {𝑝 -3𝑝(𝑝 3 + 1)} 3 + {1 + 3(𝑝 3 )(𝑝 3 + 1)} 3 + {-3(𝑝 3 )(𝑝 3 + 1)} 3 = 𝑝 3 + 1. (3.18) At 𝑝 = 2, (-52) 3 + (217) 3 + (-216) 3 = 9. On multiplying above said identity with 𝑑 3 , it gets transformed into

(-52𝑑 ) 3 + (217𝑑 ) 3 + (-216𝑑) 3 = 9𝑑 3
which is a parametrisation for integers expressible as 9𝑝 3 , where 𝑝 is an integer. (3.22) Since (𝑦 2 -1), (2𝑦) and (𝑦 2 + 1) are also Pythagorean's triple for all integer values of 𝑦, therefore, substituting these for 𝐴, 𝐵 and 𝐶 respectively, identity (3.22) takes the form, (-5𝑦 2 + 6𝑦 -1) 3 + (-6𝑦 2 + 4𝑦) 3 + (7𝑦 2 -6𝑦 + 1) 3 = 2𝑦 2 (𝑦 4 -4𝑦 + 3). (3.23) At 𝑦 = -1, 2 and 3, we obtain following sum of cubes (-12) 3 + (-10) 3 + (14) 3 = 2(8), (-9) 3 + (-16) 3 + (17) 3 = 8( 11), (-28) 3 + (-42) 3 + (46) 3 = 18(72). respectively. On putting, 𝑝 = 1/𝑦 in Equation (3.23), (-𝑝 2 + 6𝑝 -5) 3 + (4𝑝 -6) 3 + (𝑝 2 -6𝑝 + 7) = 2(3𝑝 4 -4𝑝 3 + 1).

(3.24) At 𝑝 = 2, 7 and 4, we obtain following sum of cubes

(3) 3 + (2) 3 + (-1) 3 = 34, (-12) 3 + (22) 3 + (14) 3 = 11664, (3) 3
+ (10) 3 + (-1) 3 = 1026. respectively. Also, since 𝐴 = 𝑝 2 -𝑞 2 , 𝐵 = 2𝑝 • 𝑞, 𝐶 = 𝑝 2 + 𝑞 2 are Pythagorean's triple for all real rational values of 𝑝 and 𝑞, therefore

{-(5𝑝 -𝑞)(𝑝 -𝑞)} 3 + {2𝑝(2𝑞 -3𝑝)} 3 + {7𝑝 2 + 𝑞 2 -6𝑝 • 𝑞} 3 = 2𝑝 2 (𝑝 4 + 3𝑞 4 -4𝑝 • 𝑞 3 ).
(3.25) At 𝑝, 𝑞 = (1, 0), 𝑝, 𝑞 = (1, 2) and 𝑝, 𝑞 = (2, 1), we obtain following sum of cubes (-5) 3 + (-6) 3 + (7) 3 = 2, (3) 3 + (2) 3 + (-1) 3 = 2(17), (-9) 3 (-16) 3 + (17) 3 = 8 [START_REF] Pavlus | Sum of Three Cubes Problem Solved forStubborn Number 33[END_REF]. respectively. Table3.3 depicts sum of cubes for 𝑘 = 𝐴 3 -𝐵 3 + 𝐶 3 where 𝐴, 𝐵 and 𝐶 are Pythagorean's triple Table3. [START_REF] Booker | On a question of Mordell[END_REF] Sum of Cubes for 𝑘 = 𝐴 3 -𝐵 3 + 𝐶 3 𝐴 3 -𝐵 3 + 𝐶 3 = 𝑘 (3𝑆 + 𝐴) 3 + (3𝑆 -𝐵) 3 -(3𝑆 -𝐶) 3 𝐴 3 -𝐵 3 + 𝐶 3 = 𝑘 (3𝑆 + 𝐴) 3 + (3𝑆 -𝐵) 3 -(3𝑆 -𝐶) 3 1 3 -0 3 + 1 3 = 2 -5 3 -6 3 + 7 3 5 3 -12 3 -13 3 = -3800 65 3 + 48 3 -73 3

𝑥 = 3 ( 2𝑎 2 -𝑐 2 𝑐 3 ) = 3 𝑐 ( 2𝑎 2 𝑐 2 -1) = 3 𝑐 (2𝑝 2 -1).
Substitution of this value of 𝑥 in Equation (4.10) yields ). Substituting this value of 𝑥 and putting,

{6𝑝 3 -3𝑝 + 1} 3 + {-6𝑝 3 + 3𝑝 + 1} 3 + {-6𝑝 2 + 2} 3 = 2{5 -9𝑝 2 }.
𝑎 𝑐 = 𝑝 in Equation (4.12), {3(2𝑝 3 -𝐶 • 𝑝) + 1} 3 -{3(2𝑝 3 -𝐶 • 𝑝) -1} 3 -{ 3 𝑝 (2𝑝 3 -𝐶 • 𝑝) + 𝐶} 3 = {-( 9𝐶 2 𝑝 ) (2𝑝 3 -𝐶𝑝) + 2 -𝐶 3 } . (4.13)
When 𝐶 = 𝑝 2 , Equation (4.13) gets transformed into

{3𝑝 3 + 1} 3 -{3𝑝 3 -1} 3 -{4𝑝 2 } 3 = {-10𝑝 6 + 2} . ( 4

Method When Seed Equation Is Not Helpful

There may be cases, where use of seed equation may not be fruitful in determining cubes by transformation of cubic into linear equation. For example, it is per se difficult to find three cubes that sum up to 6 except its seed equation (2) 3 + (-1) 3 + (-1) 3 = 6. For finding another set of three cubes, we consider 𝑋 = (𝑎 • 𝑥 + 2), 𝑌 = (𝑏 • 𝑥 -1), 𝑍 = 𝑐 • 𝑥 -1, therefore, (𝑎 • 𝑥 + 2) 3 + (𝑏 • 𝑥 -1) 3 + (𝑐 • 𝑥 -1) 3 = 6. (5.1) On expansion and simplification, it gets transforms into quadratic equation 𝑥 2 (𝑎 3 + 𝑏 3 + 𝑐 3 ) + 3𝑥(2𝑎 2 -𝑏 2 -𝑐 2 ) + 3(4𝑎 + 𝑏 + 𝑐) = 0.

(5.2) On transforming it into linear equation, we put 𝑏 + 𝑐 = -4𝑎 and 𝑐/𝑎 = 𝑦. Resultant equation is

𝑥 = -( 2 𝑎 ) 𝑦 2 + 4𝑦 + 7 4𝑦 2 + 16𝑦 + 21 = -( 2 𝑎 ) 𝑧.
where 𝑧 = 𝑦 2 +4𝑦+7 4𝑦 2 +16𝑦+21 . On putting this value of 𝑧 in Equation (5.1), we get {-4𝑧 + 2} 3 + {2𝑧(4 + 𝑦) -1} 3 + {-2𝑦 • 𝑧 -1} 3 = 6.(5.3) For ensuring adding cubes to be integers, value of 𝑦 should be such that 𝑧 must have either integer value or a value of the form 𝑝/2 , where 𝑝 is an integer. This value is not easily found out. To determine this value of 𝑦, different integer values are assigned one by one to 𝑦 and at each value , it is checked whether resultant value of 𝑧 satisfies our requirement. Consider 𝑝 1 as that value of 𝑧 say at 𝑦 = 𝑦 1 , then {-4𝑝 1 + 2} 3 + {2𝑝 1 (4 + 𝑦 1 ) -1} 3 + {-2𝑦 1 • 𝑧 1 -1} 3 = 6. Alternatively, in stead of integer values, we can either assign values to 𝑦 as 𝑦 1 /2 or 𝑦 1 where 𝑦 1 is an integer so that 𝑧 must, then have integer value. Purpose of such assignments is that {-4𝑧 + 2} , {2𝑧(4 + 𝑦) -1} and {-2𝑦 • 𝑧 -1} must have integer values. In above method, we have put a condition that 𝑏 + 𝑐 = -4𝑎 so that quadratic (5.2) transforms into a linear equation. However, a general solution can also be found out by solving the quadratic equation for its real roots. Let 𝑎 = 𝑎 1 • 𝑐 and 𝑏 = 𝑏 1 • 𝑐 where 𝑎 1 and 𝑏 1 are real rational quantities, then quadratic equation takes the form 𝑐 2 • 𝑥 2 (𝑎 1 3 + 𝑏 1 3 + 1) + 3𝑐 • 𝑥(2𝑎 1 2 -𝑏 1 2 -1) + 3(4𝑎 1 + 𝑏 1 + 1) = 0(5.4)

which has roots

It is reiterated, method of seed equation employs use of a seed or skeletal equation for populating cubes that sum up to a given integer 𝑘. To illustrate it, some seed equations and populated equations are given in the Table 7.1. It has been proved in the paper that every seed equation populates one more equation, that itself proves there is always more than one set of cubes that sum up to given integer. Using the populated equation as seed equation, we can further populate another equation and so on. Notwithstanding, the list given in Table 7.1, there are many other integers that have been described in this paper as expressible by more than one set of three adding cubes. It is prudent to draw attention to sub paragraph 3.2a dealing with integer of the type 3(P!/2!) 3 and Lemma3.4 that states, if integer Table 7.1 Some Seed Equations And Populated Equations 𝑘 Seed Equation Populated Equation 1(1) 3 + (0) 3 + (0) 3 (1 -9𝑝 3 ) 3 + (3𝑝 -9𝑝 4 ) 3 + (9𝑝 4 ) 3 2

(1) 3 + (1) 3 + (0) 3 (1 -6𝑝 3 ) 3 + (1 + 6𝑝 3 ) 3 + (-6𝑝 2 ) 3 3

(1) 3 + (1) 3 + (1) 3 (4) 3 + (4) 3 + (-5) 3 7

(2) 3 + (-1) 3 + (0) 3 (44) 3 + (-169) 3 + (168) 3 8

(2) 3 + (0) 3 + (0) 3 (-16) 3 + (-12) 3 + (18) 3 9

(2) 3 + (1) 3 + (0) 3 (-52) 3 + (217) 3 + (-216) 3 10

(2) 3 + (1) 3 + (1) 3 (4) 3 + (-3) 3 + (-3) 3 p = (P!/2!) 3 and P ≥ 3, then integer 𝑘 = 3(P!/2!) 3 is representable by sum of cubes in as many ways as P! has distinct factors of the form d(d + 1). When P ≥ 5, then 3p 3 can be represented by adding cubes in as many sets as equal to or more than P. Also If P! has n sets of distinct factors, then there will be n distinct sets of three cubes to represent integer 3(P!/2!) 3 . We have also given an example for integer 3(6!/2) 3 which can be represented by as many as eight sets of three cubes by putting different values of 𝑑 using the method adopted by us.We have also proved in sub paragraph 3.7 that if 𝐴, 𝐵 and 𝐶 are Pythagorean's triple and 𝑘 = 𝐴 3 -𝐵 3 + 𝐶 3 , then 𝑘 can also be expressed by relation 𝑘 = (3𝑆 + 𝐴) 3 + (3𝑆 -𝐵) 3 + (-3𝑆 + 𝐶) 3 where 𝑆 = 𝐵 -(𝐴 + 𝐶). That also proves multiplicity of representation of 𝑘. Thus a given integer 𝑘 has multiple representation by three adding cubes.

Results AndConclusions

The To sum up, let us have a relook what we have proved in the paper. An integer, say 𝑋, can be expressed by the relation, 𝑋 = 𝑎 • 𝑥 + 𝐴, similarly 𝑌 as 𝑏 • 𝑥 + 𝐵 and 𝑍 as 𝑐 • 𝑥 + 𝐶, where 𝑎, 𝑏, 𝑐, 𝐴, 𝐵 and 𝐶 are real rational quantities and 𝑥 is a variable. If 𝑋 3 + 𝑌 3 + 𝑍 3 = 𝑘, then this relation can be written in cubic equation with variable 𝑥. To transform this equation, first into quadratic equation, 𝑘 must be equal to 𝐴 3 + 𝐵 3 + 𝐶 3 . We will, therefore, assign such values to 𝐴, 𝐵 and 𝐶 that

7 and 3 respectively 7 . 1 . 1 . 1 :

 37111 If 𝑎 and 𝐴 have values, 3 and -1 respectively, then 𝑥 = 2. Integer 5 can, therefore, be written in infinite ways by assigning different values to 𝑎 and 𝐴. For example, 5 = 7 • (2/7) + 3 = 3 • 2 -1 = ⋯. That proves proceeding Lemma 1.Lemma A real rational number 𝑛 can always be expressed in the form 𝑛 = 𝑎 • 𝑥 + 𝐴, where 𝑎 and 𝐴 are real rational numbers and 𝑥 variable. 1.1 Algebraic Equation for Sum of Cubes Applying Lemma 1.1,𝑋, 𝑌 and 𝑍 can be written in algebraic form, 𝑋 = 𝑎 • 𝑥 + 𝐴, 𝑌 = 𝑏 • 𝑥 + 𝐵 and 𝑍 = 𝑐 • 𝑥 + 𝐶. Equation (1.1), then can be written (𝑎 • 𝑥 + 𝐴) 3 + (𝑏 • 𝑥 + 𝐵) 3 + (𝑐 • 𝑥 + 𝐶) 3 = 𝑘.(1.4) On expansion and rearrangement, this equation can be written, 𝑥 3 (𝑎 3 + 𝑏 3

  Family of Sets of Three Cubes Representing Integer𝟑(𝑷!/𝟐!) 𝟑 for 𝑷 ≥ 𝟑 Let integer 𝑝 be such that 𝑝 = 𝑃!/2! and if 𝑃 = 3, then 𝑝 = 3!/2! = 3 and 2𝑝 = 6. Six is factorable as (1)(2)(3). Therefore, (1)(2) and (2)(3) are its two pairs of factors. That means 3(3!/2!) 3 can be represented by two sets of three cubes. Taking the case, when 𝑃 = 4, then 𝑝 = 4!/2! = 12 or 2𝑝 = 24 which is factorable as(1)(2)(3)(4), then(1)(2), (2)(3) and (3)(4) are three pairs of factors of the form 𝑑(𝑑 + 1). That means 3(4!/2!) 3 can be represented by three sets of three cubes. Takings another case, when 𝑃 = 5 or 𝑝 = 5!/2! = 60 or 2𝑝 = 120 which is factorable as (1)(2)(3)(4)(5), then (1)(2), (2)(3), (3)(4), (4)(5) and (5)(6) are five pairs of factors of the form 𝑑(𝑑 + 1). That means 3(5!/2!) 3 can be represented by five sets of three cubes. Similarly, it can be proved, when 𝑃 = 6 or 𝑝 = 6!/2! = 360 or 2𝑝 = 720, it is factorable by eight pairs of factors of the form 𝑑(𝑑 + 1). That means 3(6!/2) 3 can be represented by eight sets of three cubes by putting different values of 𝑑. These eight sets are given below. (-1800) 3 + (1440) 3 + (1440) 3 = 3(6!/2!) 3 = 139968000 (-2160) 3 + (2040) 3 + (1200) 3 = 3(6!/2!) 3 = 139968000 (-2760) 3 + (2700) 3 + (1140) 3 = 3(6!/2!) 3 = 139968000 (-3420) 3 + (3384) 3 + (1116) 3 = 3(6!/2!) 3 = 139968000 (-4104) 3 + (4080) 3 + (1104) 3 = 3(6!/2!) 3 = 139968000 (-6210) 3 + (6200) 3 + (1090) 3 = 3(6!/2!) 3 = 139968000 (-6920) 3 + (6912) 3 + (1088) 3 = 3(6!/2!) 3 = 139968000 (-11208) 3 + (11205) 3 + (1083) 3 = 3(6!/2!) 3 = 139968000

  Integers 1 and Other Integers Expressible as 𝒑 𝟑 , Where 𝒑 Is an Integer Seed equation for integer 1 is (1) 3 + (0) 3 + (0) 3 = 1. That makes 𝑋 = 𝑎 • 𝑥 + 1, 𝑌 = 𝑏 • 𝑥 and 𝑍 = 𝑐 • 𝑥. Therefore, (𝑎 • 𝑥 + 1) 3 + (𝑏 • 𝑥) 3 + (𝑐 • 𝑥) 3 = 1. (3.10) On simplification, 𝑥 2 (𝑎 3 + 𝑏 3 + 𝑐 3 ) + 3𝑥 (𝑎 2 ) + 3(𝑎 ) = 0.

= 7 .

 7 That makes 𝑋 = (𝑎 • 𝑥 + 2), 𝑌 = (𝑏 • 𝑥 -1), 𝑍 = 𝑐 • 𝑥 and (𝑎 • 𝑥 + 2) 3 + (𝑏 • 𝑥 -1) 3 + (𝑐 • 𝑥) 3 = 7. (3.13) On putting 4𝑎 + 𝑏 = 0, (𝑐/𝑎) = 𝑦 and simplifying, transformed linear equation is said value of 𝑥, 𝑏 = -4𝑎 and (𝑐/𝑎) = 𝑦 in Equation (3.13), we get

  Expressible as 𝒑 𝟑 -𝟏 and 𝒅 𝟑 (𝒑 𝟑 -𝟏) Where 𝒅 and 𝒑 Are Integers. Seed equation for integer 𝑝 3 -1 is (𝑝) 3 + (-1) 3 + (0) 3 = 𝑝 3 -1. That makes 𝑋 = (𝑎 • 𝑥 + 𝑝), 𝑌 = (𝑏 • 𝑥 -1), 𝑍 = 𝑐 • 𝑥 and (𝑎 • 𝑥 + 𝑝) 3 + (𝑏 • 𝑥 -1) 3 + (𝑐 • 𝑥) 3 = 𝑝 3 -1. (3.15) On expansion, simplification and putting 𝑝 2 • 𝑎 + 𝑏 = 0, (𝑐/𝑎) = 𝑦, Equation 3.15 gets transformed into a linear equation, said value of 𝑥, 𝑏 = -𝑝 2 𝑎 and (𝑐/𝑎) = 𝑦 in Equation (3.15),

1

 1 depicts sum of three cubes that equals to 𝑝 3 -1 when integer 𝑝 varies from 3 to 12.3.7 Representation of Integer𝟗, 𝟗𝒅 𝟑 and Other Integers Expressible as 𝒑 𝟑 + 𝟏 Examination of Equation (

Table3. 2

 2 Sum of Three Cubes That Equals to 𝑝 3 depicts sum of three cubes that equals to 𝑝 3 + 1 where 𝑝 is an integer from 3 to 12. 3.8 Representation of Integer 10 and Other Integers Expressible as 𝟏𝟎𝒑 𝟑 Seed equation for integer 10 is (2) 3 + (1) 3 + (1) 3 = 10. That makes 𝑋 = (𝑎 • 𝑥 + 2), 𝑌 = (𝑏 • 𝑥 + 1), 𝑍 = (𝑐 • 𝑥 + 1) and (𝑎 • 𝑥 + 2) 3 + (𝑏 • 𝑥 + 1) 3 + (𝑐 • 𝑥 + 1) 3 = 10. (3.19) On putting, 4𝑎 + 𝑏 + 𝑐 = 0, 𝑏 = 𝑐 = -2𝑎 and simplifying, 𝑥 = 2/𝑎 (3.20) and on putting this value of 𝑥 in Equation (3.19), we get (4) 3 + (-3) 3 + (-3) 3 = 10. Therefore, parametrisation for 10𝑝 3 is (4𝑝) 3 + (-3𝑝) 3 + (-3𝑝) 3 = 10𝑝 3 . 3.9 Representation of Integers Using Pythagorean's Triples Consider 𝐴, 𝐵 and 𝐶 as Pythagorean's Triple such that 𝐴 2 + 𝐵 2 = 𝐶 2 and its seed equation as 𝑘 = 𝐴 3 -𝐵 3 + 𝐶 3 . Therefore, 𝑋 = (𝑎 • 𝑥 + 𝐴) , 𝑌 = (𝑎 • 𝑥 -𝐵) and 𝑍 = (-𝑎 • 𝑥 + 𝐶) and (𝑎 • 𝑥 + 𝐴) 3 + (𝑎 • 𝑥 -𝐵) 3 + (-𝑎 • 𝑥 + 𝐶) 3 = 𝐴 3 -𝐵 3 + 𝐶 3 .(3.21) On simplification by putting 𝐴 2 + 𝐵 2 -𝐶 2 = 0, being Pythagorean's triple, 𝑥 = -(3/𝑎)(𝐴 -𝐵 + 𝐶). Substituting this value of 𝑥 in Equation (3.21) {-3(𝐴 -𝐵 + 𝐶) + 𝐴} 3 + {-3(𝐴 -𝐵 + 𝐶) -𝐵} 3 + {3(𝐴 -𝐵 + 𝐶) + 𝐶} 3 = 𝐴 3 -𝐵 3 + 𝐶 3 . Let 𝑆 = 𝐵 -(𝐴 + 𝐶), then the Equation takes the form, (3𝑆 + 𝐴) 3 + (3𝑆 -𝐵) 3 + (-3𝑆 + 𝐶) 3 = 𝐴 3 -𝐵 3 + 𝐶 3 .

  (4.11) At 𝑝 = 2, 3 and 4, we obtain (43) 3 + (-41) 3 + (-22) 3 = (-72 + 10) = -62, (154) 3 + (-152) 3 + (-52) 3 = (-162 + 10) = -152, (373) 3 + (-371) 3 + (-94) 3 = (-288 + 10) = -278. 4.5. Miscellaneous Parametrisation Consider the equation (𝑎 • 𝑥 + 1) 3 + (-𝑎 • 𝑥 + 1) 3 + (-𝑐 • 𝑥 -𝐶) 3 = (𝑑 • 𝑥 + 2 -𝐶 3 ) .(4.12) On expansion, simplification, and putting 𝑑 = -3𝑐 • 𝐶 2 , it gets transformed into linear equation 𝑥 = 3 ( 2𝑎 2 -𝐶•𝑐 2 𝑐 3

  .14) At 𝑝 = 2, 3 and 4, we obtain (25) 3 + (-23) 3 + (-16) 3 = -640 + 2 = -638, (82) 3 + (-80) 3 + (-36) 3 = -7290 + 2 = -7288, (193) 3 + (-191) 3 + (-64) 3 = -40960 + 2 = -40958.When 𝐶 = 𝑝, Equation (4.13) gets transformed into {3𝑝 2 (2𝑝 -1) + 1} 3 -{3𝑝 2 (2𝑝 -1) -1} 3 -{2𝑝(3𝑝 -1}3 = 2(1 + 4𝑝 3 -9𝑝 4 ).

( 4 .

 4 15)At 𝑝 = -1, 2 and -2, we obtain (-8) 3 + (10) 3 + (-8)3 = 2(-12) = -24, (37)3 + (-35)3 + (-20)3 = 2(-111) = -222, (-59)3 + (61)3 + (-28)3 = 2(-175) = -350. When 𝐶 = -1,Equation (4.13) gets transformed into {1 + 3𝑝(2𝑝 2 + 1)} 3 -{3𝑝(2𝑝 2 + 1) -1} 3 -{3(2𝑝 2 + 1) -1} 3 = -{6 + 18𝑝 2 }. (4.16) At 𝑝 = 0, 2 and 3, we obtain (1) 3 + (1) 3 + (-2) 3 = -6, (55) 3 + (-53) 3 + (-26) 3 = -78, (172) 3 -(170) 3 -(56) 3 = -168, When 𝐶 = -2,Equation (4.13) gets transformed into {1 + 6𝑝(𝑝 2 + 1)} 3 -{6𝑝(𝑝 2 + 1) -1} 3 -{6(𝑝 2 + 1) -2} 3 = -{62 + 72𝑝 2 }. (4.17) At 𝑝 = 0, 1 and 2, we obtain (1) 3 + (1) 3 + (-4) 3 = -62, (13) 3 + (-11) 3 + (-10) 3 = -(62 + 72) = -134, (61) 3 + (-59) 3 + (-28) 3 = -(62 + 288) = -350. When 𝐶 = 2,Equation (4.13) gets transformed into {1 + 6𝑝(𝑝 2 -1)} 3 -{6𝑝(𝑝 2 -1) -1} 3 -{6(𝑝 2 -1) + 2} 3 = 6{11 -12𝑝 2 }, (4.18) At 𝑝 = 2, 3 and 4, we obtain (37) 3 + (-35) 3 + (-20) 3 = 6(11 -48) = -222, (145) 3 + (-143) 3 + (-50) 3 = 6(11 -108) = -582, (361) 3 + (-359) 3 + (-92) 3 = 6(11 -192) = -1086. When 𝐶 = 1/𝑝,Equation (4.13) gets transformed into{𝑝(3𝑝 3 -1)} 3 -{𝑝(3𝑝 3 -2)} 3 -{(3𝑝 3 = 1 -2𝑝 3 .(4.19) At 𝑝 = -1,2 and -2, we obtain (4) 3 + (-5) 3 + (4) 3 = 3, (46) 3 + (-44) 3 + (-23) 3 = -15, (50) 3 + (-52) 3 + (25) 3 = 17. When 𝐶 = 1/2𝑝,Equation (4.13) gets transformed into {𝑝(12𝑝 3 -1)} 3 -{𝑝(12𝑝 3 -5)} 3 -{2(6𝑝 3 -1)} 3 = 4(2 -5𝑝 3 ).(4.20) At 𝑝 = 1, -1 and 2, we obtain (11) 3 + (-7) 3 + (-10) 3 = -12, (13) 3 + (-17) 3 + (14) 3 = 28, (190) 3 + (-182) 3 + (-94) 3 = -152. On putting 𝑝 = 2𝑡 in Equation (4.13), we obtain {𝑡(96𝑡 3 -1)} 3 -{𝑡(96𝑡 3 -5)} 3 -{(48𝑡 3 -1)} 3 = (1 -20𝑡 3 ). (4.21) At 𝑡 = 1, -1 and 2, 95 3 -91 3 -47 3 = -19, 97 3 -101 3 + 49 3 = 21, 1534 3 -1526 3 -383 3 = -159. In general, consider equation (𝑎 • 𝑥 + 𝑞) 3 -(𝑎 • 𝑥 -𝑞) 3 -(𝑐 • 𝑥 + 𝐶) 3 = (𝑑 • 𝑥 + 2𝑞 3 -𝐶 3 ). (4.22) (31) 3 + (-30) 3 -(14) 3 = 47, (91) 3 + (-90) 3 + (-29) 3 = 182, (205) 3 + (-204) 3 + (-50) 3 = 461 respectively. In addition to the above parametric solution, we can have another parametric solution by putting 𝑟 = 7, 𝑞 = -5 and 𝐶 = 6 in Equation (4.25) which gets transformed into [3𝑝{2𝑝 2 -6} + 7] 3 -[3𝑝{2𝑝 2 -6} + 5] 3 -[{6𝑝 2 -12}] 3 = 216{𝑝 2 -3}{2𝑝 -3} + 2. (4.27) At 𝑝 = 2, 1, 3, we get (19) 3 + (-17) 3 + (-12) 3 = 218, (-5) 3 + (7) 3 + (6) 3 = 434 (115) 3 + (-113) 3 + (42) 3 = 3890 respectively. Thus Equations (4.26) and (4.27) yield parametric solution to represent integer 𝑘 = 9𝑥 + 2 by sum of three cubes. Assuming 𝑝 to be -𝑝, Equation (4.27) gets transformed into[3𝑝{2𝑝 2 -6} -7] 3 -[3𝑝{2𝑝 2 -6} -5] 3 + [{6𝑝 2 -12}] 3 = 216{𝑝 2 -3}{2𝑝 + 3} -2 (4.28) At 𝑝 = 2, 3, 4, we get(5) 3 +(-7) 3 + (12) 3 = 1510, (101) 3 + (-103) 3 + (42) 3 = 11662, (305) 3 + (-307) 3 + (84) 3 = 30884 respectively. Thus Equations (4.28) yields parametric solution to represent integer 𝑘 = 9𝑥 -2 by sum of three cubes. b. When Integers are of the form ±3 modulo 9. Using Equation (4.25) and putting 𝑟 = 1, 𝑞 = 1, 𝐶 = -1, we get -[3𝑝{2𝑝 2 + 1} + 1] 3 + [3𝑝{2𝑝 2 + 1} -1] 3 + [6𝑝 2 + 2] 3 = 9{2𝑝 2 + 1} -3.(4.29) At 𝑝 = 1, 2, 3, we get (-10) 3 + (8) 3 + (8) 3 = 24, (-55) 3 + (53) 3 + (26) 3 = 78, (-172) 3 + (170) 3 + (56) 3 = 168 respectively. Thus Equations (4.29) yields parametric solution to represent integer 𝑘 = 9𝑥 -3 by sum of three cubes. Substituting 𝑝 with -(3𝑞 + 1) in Equation (4.19), we get [3(3𝑞 + 1) 4 + 3𝑞 + 1] 3 -[3(3𝑞 + 1) 4 + 6𝑞 + 2] 3 + [3(3𝑞 + 1) 3 + 1] 3 = 2(3𝑞 + 1) 3 + 1. (4.30) This is a parametrisation for representing integer of the form 9𝑥 + 3. At 𝑞 = 0, 1, 2, we get (4) 3 + (-5) 3 + (4) 3 = 3, (772) 3 + (-776) 3 + (193) 3 = 129, (7210) 3 + (-7217) 3 + (1030) 3 = 687. respectively.

.1 Choice of 𝒂

  , 𝒃, 𝒄, 𝑨, 𝑩 and 𝑪 What values should be assigned to 𝐴, 𝐵 and 𝐶 depend upon the value of 𝑘 and that has already been explained while dealing with seed equations. Kindly refer to

	5)
	Depending upon our selection of 𝑎, 𝑏 and 𝑐 if 𝑟 has a real rational value such that
	(𝑎 • 𝑥 + 𝐴), (𝑏 • 𝑥 + 𝐵) and (𝑐 • 𝑥 + 𝐶) are integers, then solution will be
	(𝑎 • 𝑟 + 𝐴) 3 + (𝑏 • 𝑟 + 𝐵) 3 + (𝑐 • 𝑟 + 𝐶) 3 = 𝑘.	(2.6)
	2

Table 1 .1.

 1

  where 𝑑 has such values that 𝑑 and 𝑑 + 1 are factors of 2𝑝. Also 2𝑝 (𝑑 + If integer 2𝑝 has only factors 𝑞 and (𝑞 + 1) and neither 𝑞 nor (𝑞 + 1) is zero, then integer 3𝑝 3 is represented by two sets of sum of cubes as given below

														1 𝑑 ) ,
	2𝑝 (𝑑 +	1 𝑑+1 ) and				2𝑝 ( 𝑑 + 1	𝑑 𝑑+1 ) have integer values.
	Lemma 3.3:
														1 𝑑 ) ,
	2 (𝑑 +	1 𝑑+1 ) and 2( 𝑑 + 1	𝑑 𝑑+1 ) have integer values.
	Lemma3.2: Integer 3𝑝 3 is represented by sum of cubes
	{-𝑝 -2𝑝 (𝑑 +	1 𝑑	)}	3	+ {𝑝 + 2𝑝 (𝑑 +	1 𝑑 + 1	)} 3	+ {𝑝 + 2𝑝 (	1 𝑑	+	𝑑 𝑑 + 1	3 )}	= 3𝑝 3 ,

Acknowledgement:We acknowledge the help provided by the scientific calculators at websites https://www.desmos.com and https://keisan.casio.com/calculator in calculating the values of tedious and large exponential and factorial terms. We further

Parametrisation When Given Integer Is of Form 𝒌 = 𝟗𝒑 𝟑 + 𝟏

When 𝑘 = 9𝑝 3 + 1, we will consider seed equation (1) 3 + (0) 3 + (0) 3 = 1 and for 9𝑝 3 , we will find this value from coefficient of 𝑥 after proper assignment of values to 𝑎, 𝑏 and 𝑐. For that, consider 𝑋 = (𝑎 • 𝑥 + 1), 𝑌 = (-𝑎 • 𝑥), 𝑍 = (𝑐 • 𝑥) and 𝑋 3 + 𝑌 3 + 𝑍 3 = (𝑑 • 𝑥 + 1), then (𝑎 • 𝑥 + 1) 3 + (-𝑎 • 𝑥) 3 + (𝑐 • 𝑥) 3 = 𝑑 • 𝑥 + 1. Substituting this value of 𝑥 in Equation (4.1), {3𝑝 3 -1] 3 + {-3𝑝 3 } 3 + {3𝑝 2 } 3 = 9𝑝 3 -1. (4.2) At 𝑝 = 2, 3 and 4, following are sum of cubes 23 3 -24 3 + 12 3 = 9(2 3 ) -1 = 71, 80 3 -81 3 + 27 3 = 9(3 3 ) -1 = 242, 191 3 -192 3 + 48 3 = 9(4 3 ) -1 = 575. Substituting -𝑝 for 𝑝 {(3𝑝 3 + 1)] 3 + {-3𝑝 3 } 3 + {-3𝑝 2 } 3 = 9𝑝 3 + 1. (4.3) At 𝑝 = 1, 2 and 3, following are sum of cubes 4 3 -3 3 -3 3 = 9(1 3 ) + 1 = 10, 25 3 -24 3 -12 3 = 9(2 3 ) + 1 = 73, 82 3 -81 3 -27 3 = 9(4 3 ) -1 = 244.

Parametrisation, When Given Integer Is of Form 𝒌 = 𝟗(𝒑 + 𝟏)(𝒑 𝟐 -𝟏)

We will consider seed equation for integer 0 as (1) 3 + (-1) 3 + (0) 3 = 0 and for 𝑘 = 9(𝑝 + 1)(𝑝 2 -1), we will assign values to 𝑎, 𝑏 and 𝑐 so as to obtain this value of 𝑘 from coefficient of 𝑥 while transforming it to linear equation. Consider 𝑋 = (𝑎 • 𝑥 + 1), 𝑌 = (𝑐 • 𝑥 -1), 𝑍 = (-𝑎 • 𝑥) and 𝑋 3 + 𝑌 3 + 𝑍 3 = 𝑑𝑥, then

On expanding and rearranging , it takes the form

On putting, 𝑑 = 3(𝑎 + 𝑐) and 𝑎 𝑐 = 𝑝 for reducing it to linear equation, it takes the form

On putting this value of 𝑥 in Equation (4.4), it takes the form {3𝑝(𝑝 2 -1) -1] 3 + {3(𝑝 2 -1) + 1] 3 + {-3𝑝(𝑝 2 -1)] 3 = 9(𝑝 + 1)(𝑝 2 -1).(4.5)

At 𝑝 = 2, 3 and 4, we obtain (17) 3 + (10) 3 + (-18) 3 = 81, 71 3 + 25 3 -72 3 = 9(4)(8) = 288, 179 3 + 46 3 -180 3 = 9(5)(15) = 675. Substituting -𝑝 for 𝑝 and rearranging , Equation (4.5) is into {3𝑝(𝑝 2 -1) + 1] 3 -{3(𝑝 2 -1) + 1] 3 -{3𝑝(𝑝 2 -1)] 3 = 9(𝑝 -1)(𝑝 2 -1). (4.7) On expansion and simplification,

For transforming it linear equation, we put = 2 = 4/3 , then 𝑥 = -4/(3𝑐 3 ). On substituting this value of 𝑥 in Equation (4.7) and putting 𝑝 = 2/3𝑐, (-6𝑝 3 + 1) 3 + (6𝑝 3 ) 3 + (-3𝑝 2 ) 3 = (9𝑝 3 -1) 2 .(4.8) At 𝑝 = 1, 2 and 3, we obtain (-5) 3 + (6) 3 + (-3) 3 = (9 -1) 2 = 64, (-47) 3 + (48) 3 + (-12) 3 = (71) 2 = 5041, (-161) 3 + (162) 3 + (-27) 3 = (242) 2 = 58564. Substituting -𝑝 for 𝑝 and rearranging (6𝑝 3 + 1) 3 + (-6𝑝 3 ) 3 + (-3𝑝 2 ) 3 = (9𝑝 3 + 1) 2 . (4.9) At 𝑝 = 1, 2 and 3, we obtain (7) 3 + (-6) 3 + (-3) 3 = (10) 2 = 100, (49) 3 + (-48) 3 + (-12) 3 = (73) 2 = 5329, (163) 3 + (-162) 3 + (-27) 3 = (244) 2 = 59536. Substituting this value of 𝑥 in Equation (4.24), We have already taken up the cases, when integer 𝑘 = 9𝑥 ± 1 in paragraphs 4.1 to 4.6. Although a few cases of integer 𝑘 = 9𝑥 ± 2 and 𝑘 = 9𝑥 ± 3 have also been considered in aforementioned paragraphs, we, in order, to further elaborate, take up these cases in this sub paragraph.

Parametrisation, When Given Integer Is of Form

a. When Integers are of the form ±2 modulo 9. Let (𝑎 • 𝑥 + 1) 3 + (-𝑎 • 𝑥) 3 + (𝑐 • 𝑥 + 1) 3 = 𝑑 • 𝑥 + 2. On simplifying, putting 𝑑 = 3(𝑎 + 𝑐) and 𝑝 = 𝑎/𝑐, we get, 𝑥 = -(3/𝑐)(1 + 𝑝 2 ). On substituting this value of 𝑥 in the equation, (𝑎 • 𝑥 + 1) 3 + (-𝑎 • 𝑥) 3 + (𝑐 • 𝑥 + 1) 3 = 𝑑 • 𝑥 + 2 and, then rearranging, we get parametric solution, where 𝑃 = (𝑎 1 3 + 𝑏 1 3 + 1), 𝑄 = 3(2𝑎 1 2 -𝑏 1 2 -1) and 𝑅 = 3(4𝑎 1 + 𝑏 1 + 1).

Substituting the value of 𝑥 given by Equation (5.5) in Equation (5.1), we get (𝑎 1 • 𝑧 + 2) 3 + (𝑏 1 • 𝑧 -1) 3 + (𝑧 -1) 3 = 6. That requires 𝑎 1 and 𝑏 1 should have such real rational values that 𝑧, (𝑎 1 • 𝑧) and (𝑏 1 • 𝑧) given by Equation (5.5) must have integer values. Finding such values of 𝑎 1 and 𝑏 1 requires assumption of different values for these till the condition is satisfied.

Method When Seed Equation Is Not Determinable

So far, we dealt with those cases where seed equation was easily determinable. But there are cases, where it is not possible to find or difficult to find a seed equation for a given integer 𝑘. In such cases, an equation will be found for arbitrary integer 𝑘 1 which is different from 𝑘. Let the equation for 𝑘 1 is 𝑎 1 3 + 𝑏

2) where 𝑘 -𝑘 1 = 𝑑. Equation (6.2) is a cubic equation which can have roots either (1) two complex and one real or (2) two irrational and one rational root or (3) three real rational roots. Nature of roots depends upon values of coefficients of 𝑧 3 , 𝑧 2 , 𝑧 and constant term. Assignment of different integer values to 𝑎 2 , 𝑏 2 and 𝑐 will give different roots and the root (or roots) that has (or have) integer value(s) will be the required root(s). Let that integer root be 𝑐 • 𝑥 = 𝑧 or 𝑥 = 𝑧/𝑐, then required cubes will be (𝑎 2 • 𝑧 + 𝑎 1 ) 3 + (𝑏 2 • 𝑧 + 𝑏 1 ) 3 + (𝑧 + 𝑐 1 ) 3 = 𝑘 (6.3) Different procedures and algorithms have been devised by mathematicians to find the desired roots but we are not going into details of these as the purpose of this paper is to represent a given integer by sum of three cubes using novel approach of seed equation.

Multiple Representations as Sums of Three

Cubes for Integer, Where 𝒌 ≠ 𝟒 or 𝒌 ≠ 𝟓 Modulo 𝟗 𝐴 3 + 𝐵 3 + 𝐶 3 = 𝑘. For example, integer 1 can be written as (1) 3 + (0) 3 + (0) 3 , integer 2 can be written as (1) 3 + (1) 3 + (0) 3 and so on. Such equation for 𝑘 is called seed equation as this will populate other three cubes that sum up to 𝑘. To illustrate it, some seed equations are given in Table1.1. When cubic equation is transformed into quadratic, then quadratic can be transformed into linear equation by equating constant term to 0. Linear equation gives value of 𝑥 which can be substituted in 𝑋 = 𝑎 • 𝑥 + 𝐴, 𝑌 = 𝑏 • 𝑥 + 𝐵 and 𝑍 = 𝑐 • 𝑥 + 𝐶. Once 𝑋, 𝑌 and 𝑍 are known, three cubes that sum up to 𝑘 are known. Values of 𝐴, 𝐵 and 𝐶 depend upon the value of 𝑘 but values of 𝑎, 𝑏 and 𝑐 are so chosen that 𝑋 = 𝑎 • 𝑥 + 𝐴, 𝑌 = 𝑏 • 𝑥 + 𝐵 and 𝑍 = 𝑐 • 𝑥 + 𝐶 are integers.

We have also proved when 𝑘 = 3(𝑃!/2!) 3 , then 𝑘 can be represented by sum of three cubes in as ways as number of factor 𝑑(𝑑 + 1). For example when 𝑘 = 3(6!/2!) 3 then 𝑘 can be represented by as many as eight sets of three cubes. We also proved if 𝐴, 𝐵 and 𝐶 are Pythagorean's triple, then 𝑘 = 𝐴 3 -𝐵 3 + 𝐶 3 , can also be represented by relation 𝑘 = (3𝑆 + 𝐴) 3 + (3𝑆 -𝐵) 3 + (-3𝑆 + 𝐶) 3 , where 𝑆 = 𝐵 -(𝐴 + 𝐶). Under such circumstances, an integer 𝑘 can be represented in multiple ways by sum of three cubes provided 𝑘 ≠ 9𝑥 ± 4 or 𝑘 ≠ 9𝑥 ± 5 thanks to Roger-Brown conjecture [START_REF] Heath-Brown | On Solving the Diophantine Equation𝑥 3 + 𝑦 3 + 𝑧 3 = 𝑘 on a Vector Computer[END_REF]. We have proved in Sub-paragraph3.4 why it is not possible to have three cubes that sum up to 𝑘 where 𝑘 = 9𝑥 ± 4 or 𝑘 = 9𝑥 ± 5. Parametrisation for representing integer 𝑘 of the form 𝑘 = 9𝑝 3 + 1, 𝑘 = 9(𝑝 + 1)(𝑝 2 -1), 𝑘 = (9𝑝 3 -1) 2 , 𝑘 = 2(5 -9𝑝 2 ), 𝑘 = {-10𝑝 6 + 2} , 𝑘 = 2(1 + 4𝑝 3 -9𝑝 4 ), 𝑘 = -{6 + 18𝑝 2 } , 𝑘 = -{62 + 72𝑝 2 } , 𝑘 = 6{11 -12𝑝 2 } , 𝑘 = 1 -2𝑝 3 , 𝑘 = 4(2 -5𝑝 3 ), 𝑘 = (1 -20𝑝 3 ), 𝑘 = 9(𝑞 -1)(1 + 𝑞 2 ) + 2, 𝑘 = 216{𝑝 2 -3}{2𝑝 -3} + 2, 9{2𝑝 2 + 1} -3, and 𝑘 = 2(3𝑞 + 1) 3 + 1 as sum of three cubes, has also been given in the paper. Notwithstanding these, general parametrisation for 𝑘 = 2𝐶 2 (4𝐶 -9𝑝 2 • 𝑞) + 2𝑞 3 and 𝑘 = 9{𝑝 2 (𝑟 + 𝑞) -𝐶}{𝑝(𝑟 2 -𝑞 2 ) -𝐶 2 } + 𝑟 3 + 𝑞 3 -𝐶 3 have also been determined. These can facilitate determining three cubes that sum up to a given integer 𝑘 of the above said forms where 𝑝, 𝑞, 𝑟 and 𝐶 are integers. We have also given a method to find three cubes when seed equation is not helpful or is difficult to find out. In addition to these, we have given exhaustive examples to illustrate the novel method of finding three cubes that sum up to a given number 𝑘.

acknowledge the guidance and help provided by the learned Reviewer and learned Editor in bringing the paper to its present form. are thankful to the unknown reviewer for constructive as well as creative suggestions.