Pierre Gillibert

Gabriele Julia Ranieri

Robinson'

Julia Robinson's

NUMBERS

Keywords: Mathematics Subject Classification. 11R04, 11U05, 11R80 Rings of algebraic integers, Undecidability, Totally real fields, Julia Robinson's numbers. 1

Introduction

Let R be the ring of the algebraic integers of a subfield of the field Q tr of all totally real algebraic numbers. For every t ∈ R ∪ {+∞}, denote by R t the set of α ∈ R such that, all conjugates β of α satisfy 0 < β < t. Set A(R) = {t ∈ R ∪ {+∞} | R t is infinite}. Observe that A(R) is either an interval or {+∞}. The Julia Robinson Number of R is defined as JR(R) = inf(A(R)). Moreover R has the Julia Robinson Property (JRP) if JR(R) ∈ A(R).

Robinson, in [START_REF] Robinson | On the decision problem for algebraic rings[END_REF], generalizing her methods of [START_REF] Robinson | The undecidability of algebraic rings and fields[END_REF], proves that if a ring R has the JRP, then the semi-ring (N; 0, 1, +, .) is first-order definable in R (hence the arithmetic of R is as complicated as that of N, and in particular R has an undecidable first order theory). There is a lot of examples of rings with the JRP.

If K is a totally real algebraic number field, then JR(O K ) = +∞ (see [START_REF] Robinson | On the decision problem for algebraic rings[END_REF]). In particular, the first order theory of O K is undecidable. Note that Denef and Lipshitz prove a stronger result in [START_REF] Denef | Hilbert's tenth problem for quadratic rings[END_REF][START_REF] Denef | Diophantine sets over algebraic integer rings[END_REF][START_REF] Denef | Diophantine sets over some rings of algebraic integers[END_REF]: if K is a totally real number field, or an extension of degree 2 of a totally real number field then N is Diophantine in O K . It follows from the Matiyasevich's solution to Hilbert 10th problem [START_REF] Matiyasevich | The Diophantineness of enumerable sets (Russian)[END_REF], that the existential positive first order theory of O K is undecidable. The result is even further extended by Shlapentokh in a serie of papers [START_REF] Shlapentokh | Diophantine classes of holomorphy rings of global fields[END_REF][START_REF] Shlapentokh | Diophantine definability over some rings of algebraic numbers with infinite number of primes allowed in the denominator[END_REF][START_REF] Shlapentokh | Defining integrality at prime sets of high density in number fields[END_REF][START_REF] Shlapentokh | Diophantine definability and decidability in large subrings of totally real number fields and their totally complex extensions of degree 2[END_REF][START_REF] Shlapentokh | A ring version of Mazur's conjecture on topology of rational points[END_REF][START_REF] Shlapentokh | Diophantine definability and decidability in extensions of degree 2 of totally real fields[END_REF] to large subrings of K (see also Poonen [17], Poonen and Shlapentokh [START_REF] Poonen | Diophantine definability of infinite discrete nonarchimedean sets and Diophantine models over large subrings of number fields[END_REF]). Mazur and Rubin prove in [START_REF] Mazur | Ranks of twists of elliptic curves and Hilbert's tenth problem[END_REF] a similar result for the algebraic integer rings of an arbitrary number fields, assuming that the Tate-Shafarevich conjecture holds. See Shlapentokh [START_REF] Shlapentokh | Defining integers[END_REF][START_REF] Shlapentokh | Elliptic curve points and Diophantine models of Z in large subrings of number fields[END_REF][START_REF] Shlapentokh | Hilbert's tenth problem for subrings of Q and number fields, Theory and applications of models of computation[END_REF] for more results. See also Matiyasevich [12] for more on the Hilbert's 10th problem, and Moret-Bailly [START_REF] Moret-Bailly | Elliptic curves and Hilbert's tenth problem for algebraic function fields over real and p-adic fields[END_REF][START_REF] Moret-Bailly | Sur la définissabilité existentielle de la non-nullité dans les anneaux[END_REF] for similar problems in function fields.

On the other hands for infinite algebraic extension of Q far less is known. The Julia Robinson Property seems particularly useful. Vidaux and Videla observe in [START_REF] Vidaux | A note on the Northcott property and undecidability[END_REF] that if a totally real algebraic field K has the Northcott property, then JR(O K ) = +∞. Using the result of Bombieri and Zannier in [START_REF] Bombieri | A note on heights in certain infinite extensions of Q[END_REF], Vidaux and Videla deduce that for every positive integer d, the field generated by all totally real abelian extensions of Q of degree d has its ring of algebraic integers with JR equals to +∞. In particular the ring has undecidable first order theory.

Denote by Z tr the ring of all totally real algebraic integers. As noted by Robinson in [START_REF] Robinson | On the decision problem for algebraic rings[END_REF], it follows from Schur [START_REF] Schur | Über die Verteilung der Wurzeln bei gewissen algebraischen Gleichungen mit ganzzahligen Koeffizienten[END_REF] or Kronecker [START_REF] Kronecker | Zwei Sätze über Gleichungen mit ganzzahligen Coefficienten[END_REF] that JR(Z tr ) = 4, and that Z tr has the JRP (also see Robinson [START_REF] Robinson | Intervals containing infinitely many sets of conjugate algebraic integers[END_REF] or Jarden and Videla [START_REF] Jarden | Undecidability of families of rings of totally real integers[END_REF]). In particular the first order theory of Z tr is undecidable (see [START_REF] Robinson | On the decision problem for algebraic rings[END_REF]). In [START_REF] Vidaux | Definability of the natural numbers in totally real towers of nested square roots[END_REF]Introduction], Vidaux and Videla, starting from a remark of Robinson, ask the following question Question: Can any algebraic number (or even real number) be realized as the JR number of a ring of integers in Q tr ? Then, they find a family of rings O contained in Z tr , such that JR(O) is distinct from 4 and +∞ (see [START_REF] Vidaux | Definability of the natural numbers in totally real towers of nested square roots[END_REF]Theorem 1.4] for the details). On the other hand, they do not know if at least an element of the family is the ring of the algebraic integers of its quotient field. The paper [START_REF] Vidaux | Definability of the natural numbers in totally real towers of nested square roots[END_REF] is also interesting because it contains a very precise bibliography on undecidability problem over algebraic rings. Finally, very recently, Vidaux and Videla (see [START_REF] Vidaux | A note on the Northcott property and undecidability[END_REF]) discovered an interesting relation between the finiteness of the Julia Robinson Number and the Northcott Property. Starting from this, Widmer defined the Northcott Number and gave interesting examples of families of algebraic rings with bounded Northcott Numbers (see [START_REF] Widmer | Northcott Number and Undecidability of certain Algebraic Rings[END_REF] for the details).

In this paper we find infinitely many real numbers t > 4 such that t is the Julia Robinson Number of a ring of the algebraic integers of a subfield of Q tr . This yields the first example of ring of the algebraic integers of a totally real field with JR distinct from 4 and +∞. More precisely, we prove the following result.

Theorem 8.8. Let t ≥ 1 be a square-free odd number. Then the following statement holds.

(1) There are infinitely many fields K such that O K has Julia Robinson's number 2 √ 2t + 2 √ 2t.

(2) There are infinitely many fields K such that O K has Julia Robinson's number 8t.

To prove Theorem 8.8, we choose 0 < a < b coprime integers satisfying certain conditions (see for instance Section 7) and we set α

= a+i √ b 2 -a 2 b
. Then α is a quadratic number with absolute value equal to 1. We recall that a number field L is CM if it is a quadratic imaginary extension of a totally real field (or equivalently L is not contained in R and there exists ι in Aut(L/Q) such that for every embedding of L over C, ι is the complex conjugation, see for instance [START_REF] Washington | Introduction to cyclotomic fields[END_REF]Chapter 4]). Amoroso and Nuccio [START_REF] Amoroso | Algebraic Numbers of Small Weil's Height in CM-fields: on a Theorem of Schinzel[END_REF]Proposition 2.3] prove that a field is CM if and only if it is generated over Q by an element whose all conjugates have absolute value equal to 1. Then for every positive integer n, the field generated by an nth root of α is a CM field, and so it contains a totally real field of index 2. We choose a positive integer r such that, for every prime number dividing b, v p (b) is coprime with r and, for every positive integer k, we set K r k the maximal totally real field contained in the field generated by a r k -th root of α. Let K be the union of all K r k and O K its ring of the algebraic integers. By characterizing the algebraic integers of K r k for every k, we get a very precise description of the algebraic integers of K and we compute the JR Number of K (see Theorem 7.10). Moreover O K has the JRP and so the first order theory of O K is undecidable (see Corollary 7.11).

In Section 8 we use some results of analytic number theory to get the explicit examples of Julia Robinson's number given in Theorem 8.8.

Notations

Let F be an algebraic extension of Q, we denote by O F the ring of all algebraic integers of F . Assume that F is a number field. Let p be a prime number, β be in F * , and P be a prime ideal of O F over p, let e be its ramification index over Q and λ ∈ Z be the exponent of P in the factorization of the fractional ideal (β). We define the P -adic valuation of β as λ e . Then in particular, the P -adic valuation of β should not be an integer. We always denote by v p the p-adic valuation and we always say that the P -adic valuation extends the p-adic valuation.

For every real number γ, we set ⌈γ⌉ the smallest integer ≥ γ, and ⌊γ⌋ the largest integer ≤ γ.

Some preliminaries

Given an integer b ≥ 1, and given p n1 1 p n2 2 . . . p ns s the prime decomposition of b, and u ≥ 0 in Q, we denote

r (b, u) = p ⌈n1u⌉ 1 p ⌈n2u⌉ 2 . . . p ⌈nsu⌉ s .
The next Lemma gives some basic properties of r (b, u). Lemma 3.1. Let b ≥ 1 be an integer, let u ≥ 0 be in Q. Then r (b, u) is an integer, and r (b, u) b -u is an algebraic integer. Also note that r (b, 0) = 1 and r (b, 1) = b.

Let K be a number field and let α be an element of K. Choose an nth root of α and denote it by α 1 n . We need to study some properties of a linear combination with coefficients in K of some powers of α 1 n . We do this in the following Lemmas. Lemma 3.2. Let n ≥ 1 be an integer, let t be an integer relatively prime to n,

-n + 1 ≤ k 1 , k 2 ≤ n -1, and ℓ 1 , ℓ 2 in Z. Assume that ℓ 1 + k1t n = ℓ 2 + k2t n .
Then one of the following holds:

(1)

k 1 = k 2 . (2) k 1 < 0 < k 2 and k 2 = k 1 + n. (3) k 2 < 0 < k 1 and k 1 = k 2 + n. Proof. Assume that ℓ 1 + k1t n = ℓ 2 + k2t n . Hence nℓ 1 +k 1 t = nℓ 2 +k 2 t. Thus (k 1 -k 2 )t is a multiple of n. Then it follows, as t is relatively prime to n, that k 1 -k 2 is a multiple of n. Since -2n + 2 ≤ k 1 -k 2 ≤ 2n -2, we have that k 1 -k 2 ∈ {-n, 0, n}. If k 1 -k 2 = 0 then k 1 = k 2 . If k 1 -k 2 = n, then k 1 = n + k 2 . Moreover, we have k 1 = n + k 2 ≥ n -n + 1 > 0 and k 2 = k 1 -n ≤ n -1 -n < 0. If k 1 -k 2 = -n, then k 2 = n + k 1 and k 1 = -n + k 2 ≤ -n + n -1 < 0. Similarly k 2 = k 1 + n ≥ n -1 -n < 0.
Lemma 3.3. Let K be a number field, n be a natural number, p be a prime number, α be in K. Let ν 1,p , and ν 2,p be valuations over K extending v p to K, let (λ k | -n + 1 ≤ k ≤ n -1) be a family in K. Assume that the following conditions hold:

(1) ν 1,p (α) is a positive integer relatively prime to n.

(2) ν 2,p (α) is a negative integer relatively prime to n.

(3)

γ = n-1 k=-n+1 λ k α k n is an algebraic integer. (4) For all -n + 1 ≤ k ≤ n -1, we have ν 1,p (λ k ) = ν 2,p (λ k ) is an integer. Then for all -n + 1 ≤ k ≤ n -1, we have ν 1,p (λ k ) ≥ -kν2,p(α) n , ν 1,p (λ k ) ≥ -kν1,p(α) n
, and ν 1,p (λ 0 ) ≥ 0.

Proof. By conditions (1) and (2), K(α 1 n )/K is totally ramified over ν 1,p and ν 2,p . Then, ν 1,p and ν 2,p extend in a unique way to K(α

1 n ). Denote I = {k ∈ Z | -n + 1 ≤ k ≤ n -1}.
As γ is an algebraic integer, it follows that ν 1,p (γ) ≥ 0, and ν 2,p (γ) ≥ 0. Given k ∈ I, the following equalities hold:

ν 1,p (λ k α k n ) = ν 1,p (λ k ) + kν 1,p (α) n . ν 2,p (λ k α k n ) = ν 2,p (λ k ) + kν 2,p (α) n = ν 1,p (λ k ) + kν 2,p (α) n .
Denote

S 1 = ν 1,p (λ k ) + kν 1,p (α) n | k ∈ I = ν 1,p (λ k α k n ) | k ∈ I S 2 = ν 1,p (λ k ) + kν 2,p (α) n | k ∈ I = ν 2,p (λ k α k n ) | k ∈ I .
Let u be the smallest element of S 1 ∪ S 2 . Assume that u < 0 and that u ∈ S 1 . Fix

k 1 such that u = ν 1,p (λ k1 ) + k1ν1,p (α) n
. Assume that for all k = k 1 , we have u < ν 1,p (λ k α k n ), then ν 1,p (γ) = u < 0; a contradiction. Therefore, there is k 2 ∈ I such that k 2 = k 1 and the following equalities hold:

ν 1,p (λ k1 ) + k 1 ν 1,p (α) n = u = ν 2,p (λ k2 α k 2 n ) = ν 1,p (λ k2 ) + k 2 ν 1,p (α) n .
Up to an exchange of k 1 and k 2 , we can assume that

k 1 < k 2 , hence it follows from Lemma 3.2 that k 1 < 0 < k 2 and k 2 = k 1 + n. As k 2 > 0 and ν 1,p (α) > 0 > ν 2,p (α), it follows that ν 1,p (λ k2 ) + k 2 ν 2,p (α) n < ν 1,p (λ k2 ) < ν 1,p (λ k2 ) + k 2 ν 1,p (α) n = u .
Hence u is not the smallest element of S 1 ∪ S 2 ; a contradiction. With a similar argument if u ∈ S 2 , we can find another contradiction. It follows that u ≥ 0. For all k ∈ I, we have

ν 1,p (λ k ) + kν1,p(α) n ≥ u ≥ 0. Hence ν 1,p (λ k ) ≥ - kν1,p(α) n , therefore, as ν 1,p (λ k ) is an integer it follows that ν 1,p (λ k ) ≥ -kν1,p(α) n . A similar argument yields ν 1,p (λ k ) = ν 2,p (λ k ) ≥ -kν2,p(α) n . Lemma 3.4.
Let K be a number field, n ∈ N, p be a prime number, α ∈ K. Let ν p be a valuation over K extending v p to K. Assume that the following conditions hold:

(1) ν p (α) is an integer relatively prime to n.

(2) For all λ ∈ K * , ν p (λ) is an integer.

Then [K(α 1 n ) : K] = n.
Proof. We extend ν p to K(α 1 n ). By abuse of notation, we still denote ν p the extension. Let (λ k | 0 ≤ k ≤ n -1) be a family of K such that:

n k=0 λ k α k n = 0. Note that, for all 0 ≤ k ≤ n, we have ν p (λ k α k n ) = ν 1,p (λ k ) + k n ν p (α). Set I = {k ∈ N | k ≤ n -1 and λ k = 0}. It follows from Lemma 3.2 that for all k = ℓ in I we have ν p (λ k ) + k n ν p (α) = ν p (λ ℓ ) + ℓ n ν p (α). Therefore, if I is not empty, ν p ( n k=0 λ k α k n ) is the minimum of ν p (λ k ) + k n ν p (α) for k ∈ I. Therefore ν p (0) = ∞; a contradiction. Lemma 3.5. Let K be a field, let α be in K, let n be an integer ≥ 2. Assume that [K(α 1 n ) : K] = n. Let k ∈ Z be such that k is not a multiple of n. Then Tr K(α 1 n )/K (α k n ) = 0. Proof. Set d = (k, n) and let u, v in Z be such that ku + nv = d. Since [K(α 1 n ) : K] = n, we have that X n d -α k d is the minimal polynomial of α k d over K. It follows that Tr K(α 1 n )/K (α k n ) = 0.
In the sequel, we are interested in the case when K is a quadratic imaginary field and α ∈ K such that α has absolute value equal to 1. In the following Lemma we apply some of the previous results in this particular case. . Let n ≥ 1 be an integer. Assume that α

1 n generates an extension of degree n over Q(i √ b 2 -a 2 ). Let (λ k | -n + 1 ≤ k ≤ n -1) be a family in Q. Assume that n-1 k=-n+1 λ k α k n is an algebraic integer. Then, for all 1 ≤ k ≤ n -1, both nbλ -k + nλ n-k (a + i √ b 2 -a 2 ) and nbλ k + nλ k-n (a -i √ b 2 -a 2
) are algebraic integers, and nλ 0 is an integer.

Proof. Set γ = n-1 k=-n+1 λ k α k n . Set K = Q(i √ b 2 -a 2 ) and set L = K(α 1 n ). It follows from Lemma 3.5 that Tr L/K (α k n ) = 0 , for all k ∈ Z which is not a multiple of n. (3.1)
Therefore Tr L/K (γ) = Tr L/K (λ 0 ) = nλ 0 . However, γ is an algebraic integer, therefore nλ 0 is an algebraic integer. As λ 0 ∈ Q, it follows that nλ 0 is an integer. Let 1 ≤ k ≤ n -1 be an integer. The following equalities hold:

(bα k n ) n = b n α k = b n a + i √ b 2 -a 2 b k = b n-k (a + i b 2 -a 2 ) k .
Therefore bα k n is an algebraic integer. The following equalities hold

Tr L/K (bα k n γ) = Tr L/K n-1 ℓ=-n+1 bλ ℓ α ℓ+k n = Tr L/K (bλ -k ) + Tr L/K (bλ n-k α) , by (3.1). = nbλ -k + nbλ n-k α , as bλ -k and bλ n-k α are in K. = nbλ -k + nλ n-k (a + i b 2 -a 2 ).
However, bα

k n γ is an algebraic integer, therefore nbλ -k + nλ n-k (a + i √ b 2 -a 2
) is an algebraic integer.

Similarly, (bα

-k n ) n = b n α -k = b n a + i √ b 2 -a 2 b -k = b n a -i √ b 2 -a 2 b k = b n-k (a-i b 2 -a 2 ) k .
Hence bα

-k
n is an algebraic integer. The following equalities hold

Tr L/K (bα -k n γ) = Tr L/K n-1 ℓ=-n+1 bλ ℓ α ℓ-k n = Tr L/K (bλ k ) + Tr L/K (bλ k-n α -1 ) , by (3.1). = nbλ k + nbλ k-n α -1 , as bλ k and bλ k-n α -1 are in K. = nbλ k + nλ k-n (a -i b 2 -a 2 ).
However, bα

-k n γ is an algebraic integer, therefore nbλ k + nλ k-n (a -i √ b 2 -a 2
) is an algebraic integer.

Let K and α be as in the previous Lemma. Since α has absolute value equal to 1, there exists θ ∈ R such that α = cos(θ) + i sin(θ). In various cases, we shall need some basic trigonometric formulas, as Euler's formula, from which it follows the following result. Lemma 3.7. Let λ, µ, x and y in R. Then:

λ cos(x)+µ cos(y) = (λ+µ) cos x + y 2 cos x -y 2 +(µ-λ) sin x + y 2 sin x -y 2
Similarly the following lemma is a consequence of Euler's formula.

Lemma 3.8. For all λ, µ, θ, n = 0, and k in R the following equality holds.

λ cos kθ n + µ cos (n -k)θ n 2 = 1 2 λ 2 + µ 2 + 2λµ cos θ+ + (λ 2 -µ 2 ) cos 2kθ n + 2(λµ + µ 2 cos θ) cos (n -2k)θ n Lemma 3.9. Let 0 < a < b be relatively prime integers, set α = a+i √ b 2 -a 2 b
. Assume that b is even. Let q ≥ 0 be in Q. Then r b 2 , q α q is an algebraic integer.

Proof. As b is even, it follows that a+i

√ b 2 -a 2 2
is an algebraic integer. The following equalities hold

r b 2 , q α q = r b 2 , q a + i √ b 2 -a 2 b q = r 2 b , q 2 b -q a + i √ b 2 -a 2 2 q .
From Lemma 3.1 we have r 2 b , q 2 b -q is an algebraic integer, moreover a+i

√ b 2 -a 2 2
is an algebraic integer. Therefore r b 2 , q α q is an algebraic integer.

Lemma 3.10. Let 0 < a < b be relatively prime integers, set θ = arccos( a b ) ∈]0, π 2 [. Let n ≥ 2 be an integer and

1 ≤ k ≤ n -1. Let -1 2 ≤ q ≤ 1 2 be in Q. Then 2 √
b cos(qθ) and 2 √ b sin(qθ) are algebraic integers.

Proof. We can assume that q = u v with u, v positive integers. The following equalities hold:

( √ be i u v θ ) v = b v 2 e iuθ = b v 2 -u (be iθ ) u = b v 2 -u a + i b 2 -a 2 u . As v 2 ≥ u it follows that b v 2 -u is an algebraic integer, moreover (a + i √ b 2 -a 2 ) u is an algebraic integer. Therefore √ be i u v θ is an algebraic integer. Similarly ( √ be -i u v θ ) v = b v 2 e -iuθ = b v 2 -u (be -iθ ) u = b v 2 -u a -i b 2 -a 2 u . Hence √ be -i u v θ is algebraic integer. Therefore 2 √ b cos(qθ) = √ b(e iqθ + e -iqθ
) is an algebraic integer. Similarly 2 √ b sin(qθ) is an algebraic integer.

Lemma 3.11. Let 0 < a < b be relatively prime integers, let θ = arccos( a b ) ∈ ]0, π 2 [. Let n ≥ 2 be an integer and

1 ≤ k ≤ n -1. Let λ and µ in Q be such that (λ + µ) √ b + a and (µ -λ) √ b + a are algebraic integers. Then 4b(λ cos( kθ n ) + µ cos( (n-k)θ n )) is an algebraic integer.
Proof. Set:

γ = λ cos kθ n + µ cos (n -k)θ n .
It follows from Lemma 3.7 that . Let p be a prime number dividing b and, if p = 2 suppose that 4 divides b. Then (p) splits in Q(α). Let P 1 and P 2 be the prime ideals over p and ν 1,p , respectively ν 2,p the associated valuations. Then, if p is odd ν 1,p (α) = -ν 2,p (α) = v p (b), where, as before, v p is the p-adic valuation. On the other hand

γ = (λ + µ) cos θ 2 cos (2k -n)θ 2n + (µ -λ) sin θ 2 sin (2k -n)θ 2n . Note that, since θ = arccos( a b ) ∈]0, π 2 [, cos θ 2 = 1 2 (1 + cos(θ)) = 1 2 (1 + a b ) = b + a 2b . Similarly sin θ 2 = b -a 2b . Therefore 2 √ 2bγ = 2 √ 2b (λ + µ) b + a 2b cos (2k -n)θ 2n + (µ -λ) b -a 2b sin (2k -n)θ 2n = (λ + µ) √ b + a2 √ b cos (2k -n)θ 2n + (µ -λ) √ b -a2 √ b sin (2k -n)θ 2n . Note that -1 2 ≤ (2k-n) 2n ≤ 1 2 . It follows, from Lemma 3.10, that 2 √ b cos (2k-n)θ
, if p = 2, ν 1,2 (α) = -ν 2,2 (α) = v 2 (b) -1.
Proof. First observe that αα = 1. Let R be the ring of the algebraic integers of

Q(α). Suppose first that p is odd. Since p does not divide b 2 -a 2 , (p) is not ramified in R. Suppose that (p) divides a + i √ b 2 -a 2 in R.
Then, since (p) is fixed by the complex conjugation, p divides b and αα = 1, we should have

p divides a + i √ b 2 -a 2 and a -i √ b 2 -a 2 in R.
In this case, (p) should divide 2a and this is not possible because p is odd and p does not divide a. This proves that pR = P 1 P 2 , with P 1 and P 2 distinct prime ideals of R and that, for every i, j in {1, 2}, i = j, if

P i divides a + i √ b 2 -a 2 , then P j divides a -i √ b 2 -a 2 . Suppose that P 1 divides a + i √ b 2 -a 2 and P 2 divides a -i √ b 2 -a 2
and let r be the exponent of p in the prime factorization of b. Then, since αα = 1, by the unique prime decomposition of fractional ideal of R with integer exponents, we get that the fractional ideal (α) generated by α can be written as

(α) = P 2r 1 (p) -r I = P r 1 P -r 2 I, (3.2)
with I fractional ideal coprime with (p). In fact if t is the exponent of

P 1 in the factorization of a + i √ b 2 -a 2 , since P 2 = P 1 we have that t is the exponent of P 2 in the factorization of a -i √ b 2 -a 2 . Moreover, we have (a + i √ b 2 -a 2 )(a - i √ b 2 -a 2 ) = b 2 .
Since p 2r divides b 2 and p 2r+1 does not divide b 2 , we have t = 2r, proving (3.2).

Let ν 1,p be respectively ν 2,p the valuations associated to

P 1 respectively P 2 . Thus by (3.2), r = ν 1,p (α) = -ν 2,p (α) = v p (b).
Suppose now that p = 2. Since 4 divides b and a is coprime with b, a 2 -b 2 ≡ 1 mod [START_REF] Marcus | Number Fields[END_REF]. Then by the reciprocity quadratic law (2) splits in the ring of the algebraic integers of Q(α). Observe that a+i

√ b 2 -a 2 2 is a root of the polynomial x 2 -ax + b 2 with integer coefficients. Then (2) divides a+i √ b 2 -a 2 and a-i √ b 2 -a 2 . Moreover (4) does not divide a+i √ b 2 -a 2 and a-i √ b 2 -a 2 because a+i √ b 2 -a 2 4
has the trace equals to a 2 , which is not an integer. Write (2) = P 1 P 2 with P 1 and P 2 prime ideals of the ring of the algebraic integers of Q(α).

Since αα = 1, (2) divides a+i √ b 2 -a 2 and (4) does not divide a + i √ b 2 -a 2
, we get that there exists a positive integer c and ideal

I 1 , I 2 coprime with (2) such that (α) = (2)P 2c 1 I 1 (b) -1 = (2)P 2c 1 I 1 (2) -c-1 I -1 2 = P c 1 I 1 P -c 2 I -1 2 . (3.3)
Let ν 1,2 be respectively ν 2,2 the valuations associated to P 1 respectively P 2 . Thus by

(3.3), ν 1,2 (α) = c = -ν 2,2 (α) and v 2 (b) = c + 1 = ν 1,2 (α) + 1.
The following Corollary immediately follows from Lemmas 3.4, Lemma 3.5, and Lemma 3.12.

Corollary 3.13. Let 0 < a < b be coprime integers. Set α = a+i √ b 2 -a 2 b
. Let p be a prime number dividing b. Let n > 0 be an integer. If p is odd we assume that v p (b) is coprime to n. If p is even we assume that 4 divides b, and that v 2 (b) -

1 is coprime to n. Then [Q(α 1 n ) : Q(α)] = n,
and for all integers k not multiple of n we have Tr

Q(α 1 n )/Q(α) (α k n ) = 0. Lemma 3.14. With the hypothesis of Corollary 3.13, let θ = arccos( a b ) ∈]0, π 2 [. Then 1, cos θ n , cos 2θ n , . . . , cos (n-1)θ n is a Q-basis of Q(cos θ n ). Moreover, Tr Q(cos θ n )/Q cos kθ n = 0 for all 1 ≤ k ≤ n -1. Proof. By Corollary 3.13, [Q(α 1 n ) : Q] = 2n. Observe that cos θ n = α 1 n +α 1 n 2 . So X 2 -2 cos θ n X + 1 is the minimal polynomial of α 1 n over Q(cos θ n ), and so [Q(cos θ n ) : Q] = n, as illustrated by the following diagram. Q(α 1 n ) Q(cos( θ n )) 2 s s s s s s s s s s Q(α) n • • • • • • • • • Q 2 ✈ ✈ ✈ ✈ ✈ ✈ ✈ ✈ ✈ ✈ n ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ ▲ To show that 1, cos θ n , cos 2θ n , . . . , cos (n-1)θ n is a Q-basis of Q(cos θ n )
, it is sufficient to prove that they are linearly independent over Q. Assume that we are given λ

0 , . . . , λ n-1 in Q such that n-1 k=0 λ k cos kθ n = 0. The following equalities hold 0 = n-1 k=0 λ k cos kθ n = λ 0 + n-1 k=1 λ k α k n + α k n 2 = λ 0 + n-1 k=1 λ k α k n + αα n-k n 2
Therefore the following equality holds

λ 0 + n-1 k=1 λ k + λ n-k α 2 α k n = 0 .
Thus, as 1, α

1 n , α 2 n , . . . , α n-1 n is a basis of Q(α 1 n ) over Q(α), it follows that λ k = 0 for all k.
By Corollary 3.13, we have Tr

Q(α k n )/Q (α k n ) = 0. Since cos kθ n = α k n +α k n 2 , we immediately get Tr Q(cos θ n )/Q cos kθ n = 0. Lemma 3.15. With the hypothesis of Corollary 3.13, Let θ = arccos( a b ) ∈]0, π 2 [. Let (λ k | 0 ≤ k ≤ n -1) be a family in Q. Then the conjugates of the element λ 0 + n-1 k=1 λ k cos( kθ n ), counted with multiplicity in Q(cos( θ n )), are the elements λ 0 + n-1 k=1 λ k cos( k(θ+2πt) n ) with 0 ≤ t ≤ n -1. Proof. Note that α = e iθ = a+i √ b 2 -a 2 b
, and

α = e -iθ = α -1 . We have [Q(α 1 n ) : Q(α)] = n, hence for each 0 ≤ t ≤ n -1 we have a unique morphism, denoted by ϕ t : Q(α 1 n ) → C preserving Q(α), such that ϕ t (α 1 n ) = e i 2πt n α 1 n . The embedding Q(α 1 n ) → C preserving Q(α), are ϕ 0 , . . . , ϕ n-1 .
Since α is an algebraic number whose Galois conjugates have absolute value equal to 1, then α ) is a CM field. In particular the complex conjugation is well-defined and it is independent of any embedding of

Q(α 1 n ) over C. The embedding Q(α 1 n ) → C are ϕ 0 , .
. . , ϕ n-1 , and their complex conjugates ϕ 0 , . . . , ϕ n-1 . Let k be in Z, and 0 ≤ t ≤ n -1. The following equalities hold

ϕ t e ikθ n = ϕ t α k n = ϕ t α 1 n k = (e i 2πt n α 1 n ) k = e i 2πkt n e ikθ n = e i k(θ+2πt) n . (3.4) Therefore ϕ t (cos( kθ n )) = 1 2 ϕ t e ikθ n + e -ikθ n = e i k(θ+2πt) n + e -i k(θ+2πt) n = cos( k(θ + 2πt) n ). (3.5) Set K = Q(α 1 n +α 1 n ) = Q(cos( θ n )). Note that ϕ t ↾K = ϕ t ↾K, for all 0 ≤ t ≤ n-1. Hence the morphisms K → C are ϕ 0 ↾K, . . . , ϕ n-1 ↾K. Set β = λ 0 + n-1 k=1 λ k cos( kθ n ) Let 0 ≤ t ≤ n -1.
The following equalities hold

ϕ t (β) = ϕ t λ 0 + n-1 k=1 λ k cos kθ n = λ 0 + n-1 k=1 λ k ϕ t cos kθ n = λ 0 + n-1 k=1 λ k cos k(θ + 2πt) n . Lemma 3.16. Let 0 < a < b be relatively prime integers, let θ = arccos( a b ) ∈]0, π[. Let n ≥ 2. Let (λ k | 0 ≤ k ≤ n -1) be a family in Q. Let u, v in N * be such that
Then λ 0 + n-1 k=1 λ k cos( kθ n
) is an algebraic integer if and only if for all 1 ≤ k ≤ n-1 the following statements hold

(1) λ 0 is an integer.

(2) For all odd prime number p such that p divides b we have n is an algebraic integer. As λ 0 is an integer, it follows that

v p (λ k ) ≥ kvp(b) n . (3) ν 2 (λ k ) ≥ 1 + k(v2(b)-1) n . (4) u(λ k + λ n-k ) and v(λ k -λ n-k ) are integers.
λ 0 + n-1 k=1 λ k cos kθ n is an algebraic integer. Reciprocally assume that λ 0 + n-1 k=1 λ k cos kθ n
is an algebraic integer. The following equality holds:

λ 0 + n-1 k=1 λ k cos kθ n = λ 0 + n-1 k=1 λ k 2 e ikθ n + e -ikθ n = λ 0 + n-1 k=1 λ k 2 α k n + α -k n .
By Lemma 3.12 there are valuations ν 1,2 and ν 2,2 defined over Q(α), extending the 2-adic valuation, such that

ν 1,2 (α) = -ν 2,2 (α) = v 2 (b) -1. However, v 2 (b) -1
is relatively prime to n. Moreover, for all x ∈ Q(α) * , we have that ν 1,2 (x) is an integer. In fact since 4 divides b and a and b are coprime, by the reciprocity quadratic law, (2) is not ramified over Q(α). Hence it follows from Lemma 3.4 that [Q(α

1 n ) : Q(α)] = n. As ν 1,2 (α) = v 2 (b) -1 is relatively prime to n, it follows from Lemma 3.3 that v 2 (λ 0 ) ≥ 0,
and the following inequalities hold

v 2 λ k 2 ≥ k(v 2 (b) -1) n , for all 1 ≤ k ≤ n.
Therefore (3) holds.

Let p be an odd prime dividing b. Thus by Lemma 3.12, there are valuations ν 1,p and ν 2,p extending v p to Q(α) such that ν 1,p (α) = -ν 2,p (α) = v p (b). As v p (b) is relatively prime to n, it follows from Lemma 3.3 that v p (λ 0 ) ≥ 0, and the following inequalities hold

v p (λ k ) = v p λ k 2 ≥ kv p (b) n , for all 1 ≤ k ≤ n.
Therefore (2) holds. It follows from Lemma 3.6 that nλ 0 is an integer, however we also have v p (λ 0 ) ≥ 0 for all odd primes p dividing n, and v 2 (λ 0 ) ≥ 0, therefore λ 0 is an integer. Therefore (1) holds.

By Lemma 3.6,

nb λ k 2 + n λ n-k 2 (a + i √ b 2 -a 2 ) is an algebraic integer (for all 1 ≤ k ≤ n -1). In particular, if 1 ≤ k ≤ n -1, then also 1 ≤ n -k ≤ n -1, and so nb λ n-k 2 + n λ k 2 (a + i √ b 2 -a 2
) is an algebraic integer. Therefore, applying Tr Q(α)/Q , we obtain that n(bλ k + aλ n-k ), n(bλ n-k + aλ k ) are integers, and nλ k i √ b 2 -a 2 is an algebraic integer. We already proved that v 2 (λ k ) ≥ 0, and

v 2 (λ n-k ) ≥ 0, hence v 2 (u(λ k +λ n-k )) ≥ 0, and v 2 (v(λ k -λ n-k )) ≥ 0.
Similarly if p be an odd prime number that divides n, then we have already proved that v p (λ k ) ≥ 0, and

v p (λ n-k ) ≥ 0, hence v p (u(λ k + λ n-k )) ≥ 0, and v p (v(λ k -λ n-k )) ≥ 0.
Let p be a prime number that does not divides n(b 2 -a 2 ). As

nλ k i √ b 2 -a 2 is an algebraic integer, it follows that nλ k (b 2 -a 2 ) is an integer, hence v p (λ k ) = v p (nλ k (b 2 -a 2 )) ≥ 0. Similarly v p (λ n-k ) ≥ 0, therefore v p (u(λ k + λ n-k )) ≥ 0, and v p (v(λ k -λ n-k )) ≥ 0.
Let p be an odd prime number dividing b 2 -a 2 = (b -a)(b + a). As b and a are relatively prime, it follows that either p divides b -a or p divides b + a. Moreover p does not divide b, and so p does not divides n.

Assume that p divides b -a. As n(bλ

k + aλ n-k ), n(bλ n-k + aλ k ) are integers, it follows that n(bλ k + aλ n-k ) + n(bλ n-k + aλ k ) = n(λ k + λ n-k) )(a + b) is an integer. As p does not divide n and p does not divide a+b, it follows that v p (u(λ k +λ n-k ) = v p (n(λ k + λ n-k) )(a + b)) ≥ 0. Moreover v p (λ k ) = v p (nλ k i √ b 2 -a 2 ) -v p ( √ b 2 -a 2 ) ≥ 0 -1 2 v p (b 2 -a 2 ) = -1 2 v p (b -a). However, b-a v 2 is a square-free integer, thus v p (v) = ⌊ 1 2 v p (b -a)⌋. Therefore v p (v) > 1 2 v p (b -a) -1. It follows that v p (vλ k ) = v p (v) + v p (λ k ) > 1 2 v p (b -a) -1 -1 2 v p (b -a) = -1. However v p (vλ k ) is an integer, thus v p (vλ k ) ≥ 0. Similarly v p (vλ n-k ) ≥ 0, and so v p (v(λ k -λ n-k )) ≥ 0.
With a similar argument, we can prove that if

p divides b + a then v p (u(λ k + λ n-k )) ≥ 0 and v p (v(λ k -λ n-k )) ≥ 0.
Hence, for all prime p we have v p (u(

λ k + λ n-k )) ≥ 0 and v p (v(λ k -λ n-k )) ≥ 0. Therefore u(λ k + λ n-k ) and v(λ k -λ n-k ) are integers. That is (4) holds.

Trace in general

Given an algebraic number β in a number field K, we denote T (β) =

Tr K/Q (β) [K:Q] .
Note that T does not depends on the choice of K. In particular T (β) =

Tr Q(β)/Q (β) [Q(β):Q] .
Also note that T (q) = q for all q ∈ Q. We also denote by δ(β) the largest difference of two conjugates of β. Note that δ(β + q) = δ(β) for all q ∈ Q. The following Lemma is evident. Proof. Set γ = β -T (β). As T (γ) = 0 it follows from Lemma 4.1 that there is ϕ :

Q(β) → R such that ϕ(γ) ≥ 0, hence ϕ(β) = ϕ(γ +T (β)) = ϕ(γ)+T (β) ≥ T (β).
Similarly β has a conjugate smaller than T (β). Proof. Lemma 4.2 implies that β 2 has a conjugate ≥ T (β 2 ). Hence we have a conjugate γ of β such that γ 2 ≥ T (β 2 ), therefore γ ≥ T (β 2 ) or γ ≤ -T (β 2 ). Proof. We can assume that β = 0. We see from Lemma 4.2 that β 2 has a conjugate ≥ T (β 2 ). Therefore β has a conjugate ≥ T (β 2 ) or a conjugate ≤ -T (β 2 ). In the second case, we can change β to -β, and so assume that β has a conjugate ≥ T (β 2 ).

Let γ 1 be the largest conjugate of β, and γ 2 be the smallest conjugate of β. From Lemma 4.2 it follows that γ 1 > 0, and

γ 2 < 0. Let u ∈ Q be such that u > γ 1 . Set v = u 2 -T (β 2 ) 2u . As u > γ 1 ≥ T (β 2 ), it follows that v > 0.
The following equalities hold:

T ((β -v) 2 ) = T (β 2 ) -2vT (β) + v 2 = T (β 2 ) + v 2 as T (β) = 0. = T (β 2 ) + u 2 -T (β 2 ) 2u 2 = T (β 2 ) + u 4 -2u 2 T (β 2 ) + T (β 2 ) 2 4u 2 = u 4 + 2u 2 T (β 2 ) + T (β 2 ) 2 4u 2 = u 2 + T (β 2 ) 2u 2 . It follows from Lemma 4.3 that β -v has a conjugate ≥ u 2 +T (β 2 ) 2u or a conjugate ≤ -u 2 +T (β 2 ) 2u . If β -v has a conjugate ≥ u 2 +T (β 2 ) 2u
, it follows that γ 1 the largest conjugate of β satisfies

γ 1 ≥ v + u 2 + T (β 2 ) 2u = u 2 -T (β 2 ) 2u + u 2 + T (β 2 ) 2u = u , which contradicts u > γ 1 . Therefore β -v has a conjugate ≤ -u 2 +T (β 2 ) 2u
. It follows that γ 2 the smallest conjugate of β satisfies

γ 2 ≤ v - u 2 + T (β 2 ) 2u = u 2 -T (β 2 ) 2u - u 2 + T (β 2 ) 2u = - T (β 2 ) u .
This inequality is true for all u ∈ Q larger than γ 1 . It follows that

γ 2 ≤ -T (β 2 ) γ1 . Thus γ 1 -γ 2 ≥ γ 1 + T (β 2 ) γ1 ≥ 2 T (β 2 ).
γ 1 = |γ 1 | ≥ |γ 2 |, in particular γ 2 1 is the largest conjugate of β 2 . Note that T (β 2 -T (β 2 )) = 0, hence it follows from Lemma 4.4 that δ(β 2 -T (β 2 )) ≥ 2 T ((β 2 -T (β 2 )) 2 ) = 2 √ r. However all conjugates of β 2 are positives, hence γ 2 1 ≥ δ(β 2 ) = δ(β 2 -T (β 2 )) ≥ 2 √ r, it follows that γ 1 ≥ 2 √ r. Consider the following function f : R → R x → x -γ 1 + 2 r + 4x 2 T (β 2 ). Note that f (0) = -γ 1 + 2 √ r ≤ 0 and that lim x→∞ f (x) = +∞. Moreover f is continuous, therefore we can pick c ≥ 0, such that f (c) = 0. Set s = r + 4c 2 T (β 2 ). Hence c = γ 1 -2 r + 4c 2 T (β 2 ) = γ 1 -2 √ s. (4.1) Let d ∈ Q be such that d > c. Set ρ = (β -d) 2 -T ((β -d) 2
). Note that T (ρ) = 0, moreover the following equalities hold

ρ = β 2 -2dβ + d 2 -T (β 2 -2dβ + d 2 ) = β 2 -2dβ + d 2 -(T (β 2 ) + 2dT (β) + d 2 ) = β 2 -T (β 2 ) -2dβ.
Thus

T (ρ 2 ) = T (β 2 -T (β 2 ) -2dβ) 2 = T (β 2 -T (β 2 )) 2 + 4d 2 T (β 2 ) -4dT (β 3 ) + 4dT (β 2 )T (β) = r + 4d 2 T (β 2 ) ≥ s.
Therefore, by Lemma 4.4, we have

δ(ρ) ≥ 2 √ s, thus δ((β -d) 2 ) = δ(ρ) ≥ 2 √ s. However all conjugates of (β -d) 2 are positives, hence there is a conjugate γ 3 of β such that (γ 3 -d) 2 ≥ 2 √ s. Assume that γ 3 -d ≥ 2 √ s, then, as d > c it follows from (4.1) that γ 3 ≥ d+ 2 √ s > c+ 2 √ s = γ 1 ; a contradiction, as γ 1 is the largest conjugate of β. Therefore γ 3 -d ≤ -2 √ s, hence γ 2 ≤ γ 3 ≤ d-2 √ s. Therefore, for all d ∈ Q such that d > c, we have γ 2 ≤ d -2 √ s. Hence γ 2 ≤ c -2 √ s = γ 1 -2 2 √ s, thus γ 1 -γ 2 ≥ γ 1 -(γ 1 -2 2 √ s) = 2 2 √ s ≥ 2 2 √ r.

Trace in the fields for the examples

In this section we fix 0 < a < b relatively prime integers. We assume that if 2 divides b, then 4 divides b. Denote θ = arccos( a b ). We consider R a set of positive integers such that for all n, m ∈ R the following statements hold

• gcd(m, n) ∈ R.
• For all s that divides n we have s ∈ R.

• Let p be an odd prime that divides b, then v p (b) is relatively prime to n.

• If b is even then v 2 (b) -1 is relatively prime to n. Denote Q = k n | n ∈ R and 1 ≤ k ≤ n .
Note that it follows from Corollary 3.13 that T cos kθ n = 0 , for all n ∈ R, and all 1 ≤ k ≤ n. Therefore,

For all x, y ∈ Q we set N (x, y) = x 2 + y 2 + 2xy a b . Lemma 5.1. Let n be in R. Let (λ k | 0 ≤ k ≤ n -1) be a family in Q. Then T   λ 0 + n-1 k=1 λ k cos kθ n 2   = λ 2 0 + 1 2 n-1 k=1 λ 2 k + λ k λ n-k a b = λ 2 0 + 1 2 ⌊ n 2 ⌋ k=1 N (λ k , λ n-k ) .
Tr Q(α 1 n )/Q(α) cos kθ n cos ℓθ n =          n 2 + na 2b if k = ℓ = n 2 n 2 if k = ℓ = n 2 na 2b if k = n -ℓ = n 2 0
otherwise.

Hence:

Tr Q(cos( θ n ))/Q cos kθ n cos ℓθ n =          n 2 + na 2b if k = ℓ = n 2 n 2 if k = ℓ = n 2 na 2b if k = n -ℓ = n 2 0 otherwise. Moreover Tr Q(cos( θ n ))/Q (cos( kθ n )) = 0 for all 1 ≤ k ≤ n -1. The Lemma follows directly. Remark 5.2. Given λ, µ ∈ Q, n ∈ R, and 1 ≤ k < n 2 . It follows from Lemma 5.1 that N (λ, µ) = 2T λ cos kθ n + µ cos (n -k)θ n 2 . Moreover if 2 ∈ R, then N (λ, λ) = 2T λ cos θ 2 2 .

Some analysis

In this section we give various estimation of maximal value of functions that will be necessary to the calculation of the Julia Robinson Number in the next section. Lemma 6.1. Let λ, µ be real numbers. Then the maximum value, with x ∈ R, of

λ cos x + µ sin x is λ 2 + µ 2 . The minimum value is -λ 2 + µ 2 .
Proof. A rotation of the angle x, of the vector (λ, µ), has first coordinate λ cos x + µ sin x. Hence the maximum possible value is the length of the vector (λ, µ), which is λ 2 + µ 2 . Lemma 6.2. Let λ, µ, θ, q be real numbers. Then the maximum value, for x ∈ R, of λ cos(qθ + x) + µ cos((1 -q)θ -x) is λ 2 + µ 2 + 2λµ cos(θ).

Proof. For all x ∈ R, denote f (x) = λ cos(qθ + x) + µ cos((1 -q)θ -x). It follows, from Lemma 3.7, that

f (x) = (λ + µ) cos θ 2 cos 1 -2q 2 θ + x + (µ -λ) sin θ 2 sin 1 -2q 2 θ + x
Therefore, by Lemma 6.1, the maximum value of f is

V = (λ + µ) 2 cos 2 θ 2 + (µ -λ) 2 sin 2 θ 2 .
The following equalities hold

V 2 = (λ 2 + µ 2 ) cos 2 θ 2 + sin 2 θ 2 + 2λµ cos 2 θ 2 -sin 2 θ 2 = λ 2 + µ 2 + 2λµ cos(θ) .
Therefore the maximum value is V = λ 2 + µ 2 + 2λµ cos(θ).

Corollary 6.3. Let λ, µ be real numbers, and ε > 0. Set

A = max 1, 2π |λ| + |µ| ε .
Let θ be a real number. Set

s = λ 2 + µ 2 + 2λµ cos(θ) .
Then for all n ≥ A, and all

1 ≤ k ≤ n -1 with gcd(k, n) = 1, there is t ∈ Z such that λ cos k(θ + 2πt) n + µ cos (n -k)(θ + 2πt) n ≥ √ s -ε .
Proof. For each t ∈ Z, we set

α t = λ cos k(θ + 2πt) n + µ cos (n -k)(θ + 2πt) n = λ cos kθ n + 2πkt n + µ cos (n -k)θ n - 2πkt n .
Let r be an integer such that rk ≡ 1 mod n. By Lemma 6.2, we can pick

x ∈ R such that √ s = λ cos kθ n + x + µ cos (n -k)θ n -x . (6.1)
Set q = ⌊ nx 2π ⌋. Set t = rq, hence kt = krq ≡ q mod n. The following equalities hold

α t = λ cos kθ n + 2πq n + µ cos (n -k)θ n - 2πq n . (6.2)
The following inequalities hold Lemma 6.6. Let λ, µ, θ 1 , θ 2 , ρ be real numbers, and u be an integer.

0 ≤ x - 2πq n = x - 2π n nx 2π ≤ 2π n ≤ ε |λ| + |µ| . ( 6
If R 2π (uρ) = uR 2π (ρ) and |R 2π (uρ)| < π 2 , then there is |r| ≤ max(1, |u| -1) such that λ cos(θ 1 + rρ) + µ cos(θ 2 -rρ) ≥ 0. Proof. We cannot have u = 0 or u = 1. Assume that u = -1. It follows, from R 2π (uρ) = uR 2π (ρ), that ρ = π. Hence λ cos(θ 1 + ρ) + µ cos(θ 2 -ρ) = -(λ cos(θ 1 ) + µ cos(θ 2 )
). Hence r = 0, or r = 1 satisfies the required condition.

We now assume that u ≥ 2. We can assume that R 2π (ρ) = ρ, that is ρ ∈ [-π, π[. Set λ ′ = λ cos(θ 1 ) + µ cos(θ 2 ), and µ ′ = µ sin(θ 2 ) -λ sin(θ 1 ). Given r ∈ Z, denote f (r) = λ cos(θ 1 + rρ) + µ cos(θ 2 -rρ), hence the following equalities hold f (r) = λ(cos(θ 1 ) cos(rρ) -sin(θ 1 ) sin(rρ)) + µ(cos(θ 2 ) cos(rρ) + sin(θ 2 ) sin(rρ)) = λ ′ cos(rρ) + µ ′ sin(rρ).

If λ ′ ≥ 0, then f (0) = λ ′ ≥ 0, as required. Assume that λ ′ < 0. We first want to find s such that R 2π (sρ) = sρ and sρ ∈

[-π 2 , π 2 [. If ρ ∈ [-π 2 , π 2 [, then s = 1 satisfies the desired conditions. Assume that ρ ∈ [-π 2 , π 2 [. As uρ ∈ [-π, π[, we can take the smallest s ≥ 1 such that sρ ∈ [-π 2 , π 2 [. From the minimality we have (s -1)ρ ∈ [-π 2 , π 2 [, however ρ ∈ [-π 2 , π 2 [, hence sρ ∈ [-π, π[. Thus R 2π (sρ) = sρ, in particular |s| ≤ |u| -1.
Note that cos(sρ) = cos(-sρ) ≤ 0. Moreover sin(sρ) = -sin(-sρ). Hence

f (s) = λ ′ cos(sρ) + µ ′ sin(sρ) ≥ 0 or f (-s) = λ ′ cos(sρ) -µ ′ sin(sρ) ≥ 0.
The proof is similarly if u ≤ -2.

(1) 1 < gcd(k, n) ≤ √ n or 1 < gcd(ℓ, n) ≤ √ n. (2) There are non-zero integers r, u such that |u| ≤ √ n, rk ≡ 1 mod n, u(R n (rℓ)) = R n (urℓ), and R n (urℓ) ≤ √ n. (3) There are non-zero integers r, u such that |u| ≤ √ n, rℓ ≡ 1 mod n, u(R n (rk)) = R n (urk), and R n (urk) ≤ √ n. Proof. Note that k, ℓ, n -k, n -ℓ are all distinct in {1, . . . , n}, thus n ≥ 5. As gcd(k, ℓ, n) = 1, it follows that gcd(ℓ, n) × gcd(k, n) divides n Assume that gcd(k, n) > √ n. Then gcd(ℓ, n) ≤ n gcd(k,n) < √ n. If gcd(ℓ, n) > 1 then (1) holds. Assume that gcd(ℓ, n) = 1. Let r be an inverse of ℓ modulo u, thus rℓ ≡ 1 mod n. Set u = n gcd(k,n) , so |u| = u ≤ √ n and R n (urk) = R n ( rk gcd(k,n) n) = 0. As rℓ ≡ 1 mod n and k < n it follows that n ∤ rk, thus R n (rk) = 0. So u(R n (rk)) = R n (urk), therefore (3) holds.
Similarly if gcd(ℓ, n) > √ n, then (1) or ( 3) holds. Hence we can assume that gcd(k, n) ≤ √ n and gcd(ℓ, n) ≤ √ n. If gcd(k, n) > 1, or gcd(ℓ, n) > 1 then (1) holds. So we can assume that gcd(k, n) = 1 = gcd(ℓ, n).

Let 1 ≤ r < n be the inverse of k modulo n, and 1 ≤ q < n be the inverse of ℓ modulo n.

(qk)| ≤ √ n. Assume that |R n (rℓ)| ≥ n - √ n. Set u = R n (rℓ), note that |u| ≤ √ n, more- over n ≥ u(R n (rℓ)) = (R n (rℓ)) 2 ≥ n - √ n 2 = n - √ n, and so |R n (urℓ)| ≤ √ n. Moreover u(R n (rℓ)) ≥ n - √ n > √ n, hence u(R n (rℓ)) = R n (urℓ), therefore (2) holds. Similarly if |R n (rℓ)| ≥ n - √ n then (3) holds. We can assume that |R n (rℓ)| < n - √ n and |R n (qk)| < n - √ n. Thus |R n (rℓ)| × |R n (qk)| < n - √ n. However, rkqℓ ≡ 1 mod n, hence |R n (rℓ)| × |R n (qk)| ≡ ±1 mod n. Therefore |R n (rℓ)|×|R n (qk)| = 1. It follows that |R n (rℓ)| = |R n (qk)| = 1, so that rℓ ≡ ±1 mod n, thus k ≡ krℓ ≡ ±ℓ mod n. Therefore k = ℓ or k = n -ℓ; a contradiction.
Theorem 6.9. Let λ 1 , µ 1 , λ 2 , µ 2 be real numbers and ε > 0. Set

A = max 17, 2π |λ 1 | + |λ 2 | + |µ 1 | + |µ 2 | ε 2 .
Let θ 1 , θ 2 be real numbers. Set

s = min(λ 2 1 + µ 2 1 + 2λ 1 µ 1 cos(θ 1 ), λ 2 2 + µ 2 2 + 2λ 2 µ 2 cos(θ 2 )) .
Then for all n ≥ A, and all 1 ≤ k, ℓ ≤ n -1 with k = ℓ, k = n -ℓ, and gcd(k, ℓ, n) = 1, there is t ∈ Z such that

λ 1 cos k(θ 1 + 2πt) n +µ 1 cos (n -k)(θ 1 + 2πt) n +λ 2 cos ℓ(θ 2 + 2πt) n +µ 2 cos (n -ℓ)(θ 2 + 2πt) n > √ s-ε .
Proof. Let x be as in Lemma 6.2, that is

λ 1 cos kθ 1 n + x + µ 1 cos (n -k)θ 1 n -x = λ 2 1 + µ 2 1 + 2λ 1 µ 1 cos(θ) ≥ √ s . (6.4) Let n be ≥ A, let 1 ≤ k, ℓ ≤ n -1 be with k = ℓ, k = n -ℓ, and gcd(k, ℓ, n) = 1.
For all j ∈ Z, we denote

α j = λ 1 cos k(θ 1 + 2πj) n + µ 1 cos (n -k)(θ 1 + 2πj) n β j = λ 2 cos ℓ(θ 2 + 2πj) n + µ 2 cos (n -ℓ)(θ 2 + 2πj) n .
One of the statement of Lemma 6.8( 1), ( 2), or (3) holds. Assume that Lemma 6.8(1) holds. By permuting k and ℓ, we can assume 1

< gcd(k, n) ≤ √ n. Set d = gcd(k, n). Let r ∈ Z such that kr ≡ d mod n. Set q = ⌊ nx 2πd ⌋.
The following equalities hold:

α rq = λ 1 cos k(θ 1 + 2πrq) n + µ 1 cos (n -k)(θ 1 + 2πrq) n = λ 1 cos kθ 1 n + 2πkrq n + µ 1 cos (n -1 n - 2πkrq n = λ 1 cos kθ 1 n + 2πdq n + µ 1 cos (n -k)θ 1 n - 2πdq n .
Note that the following inequalities hold:

0 ≤ x - 2πdq n = x - 2πd n nx 2πd ≤ 2πd n ≤ 2π √ n ≤ 2π √ A ≤ ε |λ 1 | + |µ 1 | . However, |cos(a) -cos(b)| ≤ |a -b| for all a, b ∈ R. It follows that α rq ≥ λ 1 cos kθ 1 n + x +µ 1 cos (n -k)θ 1 n -x -(|λ 1 |+|µ 1 |) ε |λ 1 | + |µ 1 | = √ s-ε .
Set r 1 = n d and r 2 = k d . So kr 1 = nr 2 . For all t ∈ Z the following equalities hold:

α rq+tr1 = λ 1 cos k(θ 1 + 2(rq + tr 1 )π) n + µ 1 cos (n -k)(θ 1 + 2(rq + tr 1 )π) n = λ 1 cos k(θ 1 + 2πpq) n + 2πktr 1 n + µ 1 cos (n -k)(θ 1 + 2πrq) n - 2πktr 1 n = λ 1 cos k(θ 1 + 2πrq) n + 2πtr 2 + µ 1 cos (n -k)(θ 1 + 2prπ) n -2πtr 2 = α rq . Set γ 1 = ℓ(θ2+2πrq) n and γ 2 = (n-ℓ)(θ2+2πrq) n
. For all t ∈ Z the following equalities hold:

β rq+tr1 = λ 2 cos ℓ(θ 2 + 2πrq) n + t 2πr 1 ℓ n + µ 2 cos (n -ℓ)(θ 2 + 2πrq) n -t 2πr 1 ℓ n = λ 2 cos γ 1 + t ℓ d 2π + µ 2 cos γ 2 -t ℓ d 2π .
Note that gcd(d, ℓ) = 1. Then, we can deduce from Lemma 6.4 that

d-1 t=0 β pr+tr1 = 0, hence there is 1 ≤ t ≤ d -1 such that β pr+tr1 ≥ 0. Therefore α rq+tr1 + β rq+tr1 ≥ α rq ≥ √ s -ε .
Now assume that Lemma 6.8(2) holds. Let r, u be non-zero integers such that |u| ≤ √ n, rk ≡ 1 mod n, uR n (rℓ) = R n (urℓ), and

R n (urℓ) ≤ √ n. Set q = ⌊ nx 2π ⌋. For each t ∈ Z we have α r(q+t) = λ 1 cos kθ 1 n + 2πq n + 2πt n + µ 1 cos (n -k)(θ 1 ) n - 2πq n - 2πt n
If |t| ≤ |u| -1 then the following inequalities hold

| 2πq n + 2πt n -x| ≤ 2π n + 2π|t| n ≤ 2π|u| n ≤ 2π √ A ≤ ε |λ 1 | + |µ 1 | .
As before it follows that

α r(q+t) ≥ λ 1 cos kθ 1 n + x + µ 1 cos (n -k)θ 1 n -x -ε .
We obtain the following inequality α r(q+t) ≥ √ s -ε , for all |t| ≤ |u| -1. (6.5)

Note that

β r(q+t) = λ 2 cos ℓ(θ 2 + 2πrq) n + t 2πrℓ n + µ 2 cos (n -ℓ)(θ 2 + 2πrq) n -t 2πrℓ n . Set γ 1 = ℓ(θ2+2πrq) n , γ 2 = (n-ℓ)(θ2+2πrq) n
, and ρ = 2πrℓ n . Therefore the following equality holds.

β r(q+t) = λ 2 cos(γ 1 + tρ) + µ 2 cos(γ 2 -tρ) . (6.6)
As uR n (rℓ) = R n (urℓ), we have uR 2π

( rℓ n 2π) = R 2π (u rℓ n 2π), therefore uR 2π (ρ) = R 2π (uρ). As R n (urℓ) ≤ √ n we have R 2π (uρ) = R 2π (u rℓ n 2π) = Rn(urℓ) n 2π ≤ 2π √ n < π 2 .
Hence by Lemma 6.6 there is |t|

< |u| such that β r(q+t) = λ 2 cos(θ 1 + tρ) + µ 2 cos(θ 2 -tρ) ≥ 0 .
Therefore by (6.5) we have α r(q+t

) + β r(q+t) ≥ √ s -ε.
With a similar and simpler proof, we obtain the following variation of Theorem 6.9. Theorem 6.10. Let λ 1 , µ 1 , λ 2 be real numbers, and ε > 0. Set

A = max 1, 2 2π |λ 1 | + |µ 1 | ε . Let θ 1 , θ 2 be in R. Set s = λ 2 1 + µ 2 1 + 2λ 1 µ 1 cos(θ 1 ) .
Then for all n ≥ A even, and all

1 ≤ k, ℓ ≤ n -1 with gcd(k, n 2 ) = 1 there is t ∈ Z such that λ 1 cos k(θ 1 + 2πt) n + µ 1 cos (n -k)(θ 1 + 2πt) n + λ 2 cos θ 2 + 2πt 2 > √ s -ε .

Determination of the Julia Robinson Number

Let 0 < a < b be coprime integers, let θ = arccos

( a b ) ∈]0, π 2 [. Denote α = e iθ = a+i √ b 2 -a 2 b
. Let r ≥ 2 be an integer. Assume the following:

• 4 divides b; • v 2 (b) -1 is coprime to r; • r divides b;
• For all odd prime p dividing b, we have v p (b) coprime to r. We fix such a, b, r in the whole section. Subsequentely if (a, b, r) satisfies those conditions we say that (a, b, r) is good.

Given n ≥ 1, denote

L n = Q(e iθ n
), and

K n = Q(cos( θ n )). Moreover, set L = k∈N L r k and K = k∈N K r k .
We say that K is the totally real field associated to (a, b, r). Denote by N the smallest common multiple of v 2 (b) -1, and all v p (b) with p odd prime.

Let u be the square factor of b + a, and v be the square factor of b -a, that is b+a u 2 (respectively b-a v 2 ) are square-free. Note that the conditions in the following notation are similar to (2)-( 4) in Lemma 3.16 Notation 7.1. Let 0 ≤ k ≤ N -1. Denote by S k the set of all pairs (x, y) ∈ Q 2 that satisfies the following conditions:

(1) For all odd prime number p such that p divides b we have

v p (x) ≥ k+1 N v p (b) and v p (y) ≥ N -k N v p (b); (2) We have v 2 (x) ≥ 1 + k+1 N (v 2 (b) -1) and v 2 (y) ≥ 1 + N -k N (v 2 (b) -1); (3)
u(x + y) and v(x -y) are integers. We denote by U = S 1 ∪ . . . S N -1 . Notation 7.2. As before we consider the following map

N : R 2 → R (x, y) → N (x, y) = x 2 + y 2 + 2xy a b . Set s = inf N (x, y) | 0 ≤ k ≤ N -1 and (x, y) ∈ S k . The goal of this section is to prove that the Julia Robinson Number of O K is ⌈s⌉ + s.
Note that (x, y) ∈ S k if and only if (y, x) ∈ S N -k-1 . We say that (θ, K, s, U, N : R 2 → R) are the objects associated to (a, b, r).

Lemma 7.3. Let 1 ≤ k < N . Then S k is a discrete Z-submodule of Q 2 .
Proof. By the well-known property that for every valuation v over a field F and α, β ∈ F , v(α + β) ≥ min(v(α), v(β)), we immediately have that if (x 1 , y 1 ) and (x 2 , y 2 ) satisfy the conditions (1) and (2) of Notation 7.1, then (x 1 + x 2 , y 1 + y 2 ) also satisfies those conditions. Finally, it is clear that if (x 1 , y 1 ) and (x 2 , y 2 ) satisfy (1) For all C ∈ R there are only finitely many (x, y)

∈ M such that N (x, y) ≤ C. (2) If M is not trivial, then there is (x, y) ∈ M \ {(0, 0)} such that, for all (x ′ , y ′ ) ∈ M \ {(0, 0)} we have N (x, y) ≤ N (x ′ , y ′ ).
Proof. Let C be a real number and consider the affine conic with equation X 2 + Y 2 + 2XY a b = C. Since 0 < a < b by hypothesis, the locus of its zeros in R 2 is an ellipse. Then the set

E C = {(x, y) ∈ R 2 N (x, y) ≤ C} is compact. Since M is discrete, there are only finitely many (x, y) ∈ M such that N (x, y) ≤ C.
By hypothesis M is discrete, then every point of M is isolated. Then take C 1 ∈ R such that E C1 only contains (0, 0) and C 2 ∈ R such that E C2 contains at least a point of M distinct from (0, 0) (observe that C 2 exists because M is not trivial). Then the set B given by E C2 minus the interior part of E C1 is a compact, and so it contains a finite number of elements of M . Take (x, y) ∈ B such that N (x, y) ≤ N (w, z) for every (w, z) ∈ B. Then (x, y) = (0, 0) and N (x, y) ≤ N (x ′ , y ′ ) for every

(x ′ , y ′ ) ∈ M \ {(0, 0)}, because if (x ′ , y ′ ) ∈ B, then N (x ′ , y ′ ) ≥ C 2 ≥ N (x, y).
Lemma 7.7. Let (λ, µ) be in S k . Then there are infinitely many q ∈ Q ∩ [0, 1] such that λ cos(qθ) + µ cos((1 -q)θ) ∈ O K . Moreover we can choose q such that the denominator of q is any large enough divisor of a power of r.

Proof. Take coprime integers t and n = r e such that | 2k+1 2N -t n | < 1 2N , Note that there are infinitely many t, n that satisfy this condition. In

particular t n is in the interval ] k N , k+1 N [. It follows that n-t n ∈] N -k-1 N , N -k N [, moreover (µ, λ) ∈ S N -k-1 .
Therefore by Lemma 7.5 we have λ cos( t n θ) + µ cos(( n-t n )θ) ∈ O K . Corollary 7.8. Denote by t the product of all odd primes (without counting multiplicity) dividing b. The following statements hold.

(1) s 2 is an integer, and

8t|s 2 . (2) s ≥ 2 √ 2. (3) s ≤ 4t. (4) s 2 ≤ 8t b-a v 2 and s 2 ≤ 8t b+a u 2 .
In particular if b -a is a square then s 2 = 8t. ( 5

≤ k ≤ N -1 and (λ, µ) ∈ S k \ {(0, 0)} such that s = N (λ, µ). So s 2 = λ 2 + µ 2 + 2λµ a b . Let q = 1 2 be as in Lemma 7.7, that is λ cos(qθ) + µ cos((1 -q)θ) ∈ O K . So T ((λ cos(qθ) + µ cos((1 -q)θ)) 2 ) = 1 2 N (λ, µ) = 1 2 s 2 is an integer. Let p be an odd prime number dividing b. It follows from the definition of S k that v p (λ) ≥ (k + 1) v p (b) N ≥ 1 and v p (µ) ≥ (N -k) v p (b) N ≥ 1 . (7.1)
Therefore the following statement holds

v p (λµ) = v p (λ) + v p (µ) (7.2)
≥ (k + 1) v p (b) N + (N -k) v p (b) N (7.3) ≥ (N + 1) v p (b) N (7.4) ≥ v p (b) + 1 . (7.5) Hence v p (2λµ a b ) ≥ 1, therefore v p (s 2 ) = v p (λ 2 + µ 2 + 2λµ a b ) ≥ min(v p (λ 2 ), v p (µ 2 ), v p (2λµ a b )) ≥ 1 = v p (8t) . (7.6)
Similarly the following statement hold

v 2 (λ) ≥ 1 + (k + 1) v 2 (b) -1 N ≥ 2 and v 2 (µ) ≥ 1 + (N -k) v 2 (b) -1 N ≥ 2 . (7.7) Therefore v 2 (λµ) ≥ 1 + (k + 1) v 2 (b) -1 N + 1 + (N -k) v 2 (b) -1 N (7.8) ≥ 2 + (N + 1) v 2 (b) -1 N (7.9) ≥ 2 + v 2 (b) . (7.10) So v 2 (2λµ a b ) ≥ 3, therefore v 2 (s 2 ) = v 2 (λ 2 + µ 2 + 2λµ a b ) ≥ 3 = v 2 (8t) . (7.11)
It follows from (7.6) and (7.11) that 8t|s 2 , so (1) holds. In particular 8 ≤ s 2 , so s ≥ 2 √ 2, that is (2) holds.

Let p be an odd prime number dividing b. Note that v p (4t

) = 1 = vp(b) N and v 2 (4t) = 2 = 1 + v2(b)-1 N
, and 4tu and 4tv are integers (as 4t is an integer) thus (4t, 0) ∈ S 0 . Therefore s ≤ N (4t, 0) = 4t, so (3) holds.

Set k = N -1 2 . Set x = 2r b 2 , 1 2 . Note that v 2 (x) = 1 + ⌈ v2(b)-1 2 ⌉. As v 2 (b) -1 is relatively prime to b, it follows that v 2 (b) -1 is odd, thus v 2 (x) = 1 + v2(b)
2 . Note that N ≥ v 2 (b) -1 thus the following inequality holds

N + 1 2N (v 2 (b)-1)+1 = v 2 (b) -1 2 + v 2 (b) -1 2N +1 ≤ v 2 (b) -1 2 + 1 2 +1 = v 2 (x) . (7.12)
Let p be an odd prime number dividing b.

Then v p (x) = ⌈ vp(b) 2 ⌉ = vp(b)+1 2
, as v p (b) is odd. As N ≥ v p (b), the following inequality holds

N + 1 2N v p (b) = v p (b) 2 + v p (b) 2N ≤ v p (b) 2 + 1 2 = v p (x) . (7.13)
As v is relatively prime to b and k+1 N = N -k N = N +1 2N , it follows from (7.12) and (7.13) that ( x v , -x v ) satisfies the condition (1) and ( 2) of Notation 7.1. Moreover , and for all other prime p we have v p (x) = 0, it follows that x 2 = 4tb. It follows from (7.14) that s ≤ 2 4tb Therefore λ and µ are integers. Moreover it follows from (7.1) and (7.7) that 4t divides λ and µ.

u( x v -x v ) = 0 and v( x v + x v ) = 2x are integers, hence ( x v , -x v ) ∈ S k . Therefore the following inequality holds s ≤ N x v , - x v = 2 x 2 v 2 1 - a b . ( 7
If λµ ≥ 0 then s 2 = N (λ, µ) = λ 2 + µ 2 + 2λµ a b ≥ max(λ 2 , µ 2 ) however 4t divides both λ and µ, and (λ, µ) = (0, 0), thus s ≥ 4t.

Assume that λµ < 0. The following statement holds

N (λ, µ) = λ 2 + µ 2 + 2λµ a b = (λ + µ) 2 -2λµ 1 - a b ≥ (λ + µ) 2 .
However 4t|λ + µ, so if λ + µ = 0 we have s = N (λ, µ) ≥ 4t. Now assume that λ + µ = 0, that is λ = -µ. The following holds

N (λ, µ) = λ 2 + λ 2 -2λ 2 a b = 2λ 2 1 - a b .
It follows from (7.5)

that v p (λ 2 ) = v p (λµ) ≥ v p (b) + 1 = v p (4bt), similarly (7.10) implies v 2 (λ 2 ) = v 2 (λµ) = 2 + v 2 (b) = v 2 (4tb). Therefore 4tb divides λ 2 , so N (λ, µ) ≥ 8tb 1 -a b = 8t(b -a) ≥ 16t 2 = (4t) 2
, and so s = N (λ, µ) ≥ 4t. However, from (2) we also have s ≤ 4t, thus s = 4t. Therefore (5) holds. Proof. By Lemma 7.3, for every k between 0 and N -1, S k is a discrete Z-module. Then, by Lemma 7.6, there exists k and (λ, µ) ∈ S k such that N (λ, µ) = s. By Lemma 7.7, there exist infinitely many q ∈ [0, 1] ∩ Q such that γ q = λ cos(qθ) + µ cos((1 -q)θ) ∈ O K . From Lemma 3.15 we see that the conjugates of γ q are the λ cos(qθ + 2πqt) + µ cos((1 -q)θ -2πqt), with t ∈ Z. Hence by Lemma 6.2, every conjugate γ ′ q of γ satisfies |γ ′ q | ≤ λ 2 + µ 2 + 2λµ cos(θ) = N (λ, µ) = s. Hence every conjugate of ⌈s⌉ + γ q is positive, and it is smaller than ⌈s⌉ + s. This proves that the Julia Let ε > 0 be a real number. We can assume that ε ≤ 1, and if s is not an integer we also assume that ε is smaller than the fractional part of s. In particular q -s + ε < 0 , for all integer q < s. (7.15) Denote by S the set of all pairs (λ, µ) ∈ N -1 k=0 S k such that N (λ, µ) ≤ 2s 2 +s+1. We see from Lemma 7.6 that S is finite. Let B be a power of r such that B ≥ 17 and

B ≥ 4π |λ| + |µ| ε 2
, for all (λ, µ) ∈ S.

Set X = λ cos tθ d + µ cos (d -t)θ d | d ≤ B, 1 ≤ t ≤ d -1
, and (λ, µ) ∈ S and

X ′ = λ cos θ 2 | (λ, λ) ∈ S .
In particular 0 ∈ X, and so X ⊆ X +X. Set Y = {1, 2, . . . , 2⌈s⌉-1}+X +(X ∪X ′ ). Let γ ∈ O K be such that all conjugates of γ are in [0, ⌈s⌉+s]. There is n ∈ N such that β ∈ K n = Q(cos( θ n )). By Lemma 3.14 we can write γ = λ 0 + n-1 t=1 λ t cos tθ n , with λ 0 , λ 1 , . . . , λ n-1 ∈ Q. From Lemma 3.16 we see that λ 0 ∈ Z, and for all 1 ≤ t ≤ n-1 we have (λ t , λ n-t ) in some S k , with 0 ≤ k ≤ N -1. So, by construction of s, either (λ t , λ n-t ) = (0, 0) or N (λ t , λ n-t ) ≥ s 2 . Set β = γ-λ 0 = n-1 t=1 λ k cos tθ n . It follows from Lemma 3.14 that T (β) = 0.

As γ = λ 0 + β has no negative conjugates, it follows from Lemma 4.1 that λ 0 > 0, similarly we have λ 0 < 2⌈s⌉. Moreover the maximal difference of two conjugates of

β is δ(β) = δ(γ) ≤ ⌈s⌉+s. The trace of β 2 is computed in Lemma 5.1, that is T (β 2 ) = 1 2 ⌊ n 2 ⌋
k=1 N (λ k , λ n-k ). Moreover from Lemma 4.4 we have δ(β) ≥ 2 T (β 2 ). Therefore we obtain the following inequality 2 1 2 If C = 0, then β = 0, and so γ = λ ∈ Y . If C = 1 then we can write β = λ cos θ 2 with (λ, λ) ∈ S, or β = λ cos kθ n + µ cos kθ n , with 1 ≤ k ≤ n -1, and (λ, µ) ∈ S. In the first case we see that γ = λ 0 + β ∈ Y . We now treat the second case. We can assume that gcd(k, n) = 1.

⌊ n 2 ⌋ k=1 N (λ k , λ n-k ) ≤ ⌈s⌉ + s ≤ 2s + 1 . Thus 2 ⌊ n 2 ⌋ k=1 N (λ k , λ n-k ) ≤ (2s + 1) 2 . (7.16) Let 1 ≤ k ≤ n -1. If N (λ k , λ n-k ) > 2s 2 +
Assume that n ≥ B. Thus n ≥ A, so from Corollary 6.3 and Lemma 3.15 we obtain a conjugate β ′ of β such that

β ′ ≥ λ 2 + µ 2 + 2λµ cos θ -ε = N (λ, µ) -ε ≥ s -ε. Similarly we have a conjugate β ′′ of β such that β ′′ ≤ -s + ε.
If λ 0 ≥ ⌈s⌉, then a conjugate of γ is λ 0 + β ′ ≥ ⌈s⌉ + s -ε; a contradiction. If λ 0 < ⌈s⌉, then λ 0 < s, so a conjugate of γ is λ 0 + γ ′′ ≤ λ 0 -s + ε which is negative by (7.15); a contradiction. Therefore n ≤ B, and so γ ∈ Y .

If C = 2, once again simplifying fractions, then we can write

β = λ 1 cos kθ n + µ 1 cos (n-k)θ n + λ 2 cos θ 2 with (λ 2 , λ 2 ) ∈ S, (λ 1 , µ 1 ) ∈ S, and gcd(k, n 2 ) = 1, or β = λ 1 cos kθ n + µ 1 cos (n-k)θ n + λ 2 cos ℓθ n + µ 2 cos (n-ℓ)θ n , with 1 ≤ k < ℓ < n -ℓ ≤ n -1, (λ 1 , µ 1 ) ∈ S, (λ 2 , µ 2 ) ∈ S,
and gcd(k, ℓ, n) = 1. In both case, assuming n ≥ B, using respectively Theorem 6.10 and Theorem 6.9 we have conjugates β ′ and β ′′ of β such that β ′ ≥ s -ε, and β ′′ ≤ -s + ε. As before we reach a contradiction. Therefore n ≤ B and so γ = λ 0 + β ∈ Y .

As O K has the Julia Robinson Property, from the result of Robinson in [START_REF] Robinson | On the decision problem for algebraic rings[END_REF] we obtain the following Corollary of Theorem 7.10.

Corollary 7.11. The first order theory of O K is undecidable.

The following lemma express that there are not many "simple" square roots of rationals in K. It will be used to prove that many of the constructed fields are distincts.

Lemma 7.12. Let (λ, µ) ∈ Q \ {(0, 0)}, let n ≥ 2 be an integer dividing a power of

r, let 1 ≤ k ≤ n -1. Then λ cos kθ n + µ cos (n-k)θ n 2 ∈ Q if and only if 2k = n. Proof. Set β = λ cos kθ n + µ cos (n-k)θ n . Note that cos θ 2 2 = 1+cos θ 2
= b+a 2b belongs to Q. Thus if n = 2k, then the following statement holds

β 2 = (λ + µ) cos θ 2 2 ∈ Q .
Assume that n = 2k, hence k = n -k. We can assume that k < n -k. It follows from Lemma 3.8 that

β 2 = 1 2 λ 2 + µ 2 + 2λµ a b + (λ 2 -µ 2 ) cos 2kθ n + 2(λµ + µ 2 a b ) cos (n -2k)θ n .
(7.17) Note that if (λ 2 -µ 2 ) = 0 then λ = ±µ = 0, hence (λµ + µ 2 a b ) = µ 2 ( a b ± 1) = 0. Assume that 2k = n -2k, it follows from Lemma 3.14 that 1, cos 2kθ n , cos n-2kθ n are Q-linearly independent, thus β 2 ∈ Q.

Assume that 2k = n -2k, thus 4k = n, so n is even, and so r is even. Moreover from (7.17) we have

β 2 = 1 2 λ 2 + µ 2 + 2λµ a b + (λ 2 -µ 2 + 2λµ + 2µ 2 a b ) cos θ 2 .
Assume that β 2 ∈ Q. However from Lemma 3.14 we see that 1, cos θ 2 are linearly independent, thus λ 2 -µ 2 + 2λµ + 2µ 2 a b = 0. Hence λ is a root of the polynomial Λ

2 + 2µΛ + 2µ 2 a b -µ 2 . Therefore the discriminant ∆ = 4µ 2 -4(2µ 2 a b -µ 2 ) = 8µ 2 -8µ 2 a b = 8µ 2 a+b b is a square. Therefore 2 a+b b is a square, thus v 2 (2 a+b b ) = 1 -v 2 (b)
is even. However r and v 2 (b) -1 are relatively prime, and r is even; a contradiction.

Example of Julia Robinson's Number

In this section we construct particular a and b such that we can explicitly compute the number s (see beginning of Section 7 and Notation 7.2). We obtain the examples of Julia Robinson Numbers in Theorem 8.8. We basically consider two family of cases. For one we choose a, b such that b -a = 1 (which is a square). The other family is such that b -a, b + a are large and square-free, the existence of such pairs (a, b) relies on the fact that the density of square-free integers is large. Notation 8.1. Given x ≥ 0, k ≥ 1, and i relatively prime to k, denote by Q(x; i, k) the number of integer j ≥ 0, such that j ≤ x and j ≡ i mod k.

The following theorem was proved by Prachar [START_REF] Prachar | Über die kleinster quadratfrie Zahl einer arithmetischen Reihe[END_REF] (see also Hooley [START_REF] Hooley | A note on square-free numbers in arithmetic progressions[END_REF]). Theorem 8.2. Let k ≥ 1 be an integer and i relatively prime to k. Then we have the following equivalence

Q(x; i, k) ∼ x→∞ x 6 π 2 k p|k p 2 p 2 -
Q(x; -1, k) x = l .
Set 3η = ε(l -3 5k ). Pick A ≥ 0 such that for all x ≥ A and i = ±1 we have 5k square-free integers in X -1 . However X -1 has εb-1 k elements, therefore the proportion of square-free elements in X -1 is at least εb 3 5k k εb = 3 5 . Similarly, by (8.5), the proportion of square-free elements in X 1 is at least 3 5 . Note that X 1 → X -1 , r → 2b -r is a bijection. Therefore there is a square-free r ∈ X 1 such that 2b -r is square-free. Set a = r -b, thus b + a = r is square-free, and b -a = 2b -r is square-free. Moreover as r ∈ X 1 , we have 0 < r -b ≤ εb. Also note that a ≡ r ≡ 1 mod k.

| Q(x;i,k) x -l| ≤ η. Set C = A+1 1-ε . Let b ≥ C, thus b -εb -1 ≥ A. Therefore we have | Q(b -εb; i, k) b -εb -l| ≤ η . It follows that |Q(b -εb; i, k) -(b -εb)l| ≤ (b -εb)η . ( 8
; i, k) -Q(b -εb; i, k) + εbl| ≤ 2b -εbη ≤ 3bη . Q(b; i, k) -Q(b -εb; i, k) ≥ εbl -3bη = εbl -εb(l - 3 5k ) = εb 3 5k . ( 8
We shall give some explicite examples of finite Julia Robinson's Number distinct from 4. On the other hand we also want to show, for every example of Julia Robinson's Number j obtained, that there exist infinitely many rings of the algebraic integers of totally real fields having Julia Robinson's Number j. We will use the following lemma. Proof. Let (θ i , K i , U i , s i , N i ) be the objects associated to (a i , b i , r i ), for i = 1, 2. Assume that K 1 = K 2 , then O K1 = O K2 , thus JR(O K1 ) = JR(O K2 ), so by Lemma 7.10 we have ⌈s 1 ⌉ + s 1 = ⌈s 2 ⌉ + s 2 , and so s 1 = s 2 .

Claim 1. There are n 1 ≥ 2 a power of two, n 2 ≥ 2 an integer,

1 ≤ k 1 < n1 2 , 1 ≤ k 2 ≤ n 2 -1 integers, λ 1 , µ 1 , λ 2 , µ 2 in Q such that λ 1 cos k 1 θ 1 n 1 + µ 1 cos (n 1 -k 1 )θ 1 n 1 = λ 2 cos k 2 θ 2 n 2 + µ 2 cos (n 2 -k 2 )θ 2 n 2 = 0 . (8.6)
Proof of Claim. As the infinum in the definition of s 1 (cf Notation 7.2) is a minimum (by Lemma 7.6), there is (λ 1 , µ 1 ) ∈ U 1 such that s 2 1 = N 1 (λ 1 , µ 1 ). It follows from Lemma 7.7 that there are an odd integer k 1 and a power of two n 1 ≥ 4 such that 1 ≤ k 1 < n 1 and β = λ 1 cos k1θ1 n1 + µ 1 cos (n1-k1)θ1 n1 ∈ O K1 . Up to permuting λ 1 and µ 1 , we can assume that k 1 < n1 2 . As β ∈ O K1 = O K2 , there is m ≥ 1, and η 0 , η 1 . . . η m-1 in Q such that

λ 1 cos k 1 θ 1 n 1 + µ 1 cos (n -k 1 )θ 1 n 1 = η 0 + n2 ℓ=1 η ℓ cos ℓθ 2 n 2 .
Therefore applying T to both side of the equality, and multiplying by two, we obtain the following equality by Lemma 5.1.

N 1 (λ 1 , µ 1 ) = 2η 2 0 + ⌊ n 2 2 ⌋ ℓ=0 N 2 (η ℓ , η n2-ℓ )
However for each 1 ≤ ℓ ≤ n2 2 we have either η ℓ = 0 = η n2-ℓ or N 2 (η ℓ , η n2-ℓ ) ≥ s 2 2 = s 2 1 = N 1 (λ 1 , µ 1 ). As β ∈ Q, it follows that there is 1 ≤ p ≤ n 2 -1 such that η p = 0. Therefore for all 0 ≤ ℓ ≤ n 2 -1, if ℓ = p and ℓ = n 2 -p we have λ ℓ = 0, thus we obtain

λ 1 cos k 1 θ 1 n 1 + µ 1 cos (n 1 -k 1 )
However, by Lemma 7.12, the number (8.7) is irrational, hence the number in (8.8) is not 0; a contradiction with the minamility of n 1 . Therefore n 1 = 2 and k 1 = 1, so 2k 1 = n 1 , thus 2k 2 = n 2 , the result follows for x = λ 1 + µ 1 , and y = λ 2 + µ 2 . Claim 2.

As cos θ1 2 = b1+a1 2b1 and cos θ2 2 = b2+a2 2b2 , it follows from Claim 2 that (b 1 + a 1 )(b 2 +a 2 )b 1 b 2 is a square; a contradiction. Therefore K 1 and K 2 are distincts.

The following lemma is an immediate consequence of Hensel's Lemma (see for instance [START_REF] Narkiewicz | Elementary and analytic theory of algebraic numbers[END_REF]Theorem 5.6]). Lemma 8.5. Let p ≥ 3 be a prime, let q be relatively prime to p. Let n ≥ 1 be an integer. Assume that q is a square modulo p. Then q is a square modulo p n . Lemma 8.6. Let t be an odd number. Then there are infinitely many prime number p such that p ≡ 2 mod t, p ≡ 3 mod 4, and 2t is a square modulo p.

Proof. Given p, q we denote by p q the Legendre symbol. Set ε = q|t q prime (-1) q-1 2 2 q .

As t is odd, it follows that ε = ±1. If ε = 1 pick p a prime number such that p ≡ 2 mod t and p ≡ 7 mod 8. If ε = -1 pick p a prime number such that p ≡ 2 mod t and p ≡ 3 mod 8. In both case there are infinitely many primes satisfying those conditions by Dirichlet's theorem (cf. [START_REF] Lejeune-Dirichlet | Beweis des Satzes, dass jede unbegrenzte arithmetische Progression, deren erstes Glied und Differenz ganze Zahlen ohne gemeinschaftlichen Factor sind, unendlich viele Primzahlen enthält[END_REF]), and in both case we have p ≡ 3 mod 4.

Moreover by Gauss's quadratic reciprocity we have 2 p = ε.

Lemma 3 . 6 .

 36 Let a, b be relatively prime in N, with a < b. Let α be equal to a+i √ b 2 -a 2 b

2n and 2 √Lemma 3 . 12 .

 2312 b sin (2k-n)θ 2n are algebraic integers. Moreover, we assumed that (λ + µ) √ b + a and (µ -λ) √ b -a are algebraic integers, therefore 4bγ is an algebraic integer. Let 0 < a < b be coprime integers. Set α = a+i √ b 2 -a 2 b

1 n

 1 has the same property. Then, by [1, Proposition 2.3], Q(α 1 n

b+a u 2

 2 and b-a v 2 are square-free integers. Assume the following statements • 4 divides b. • v 2 (b) -1 is relatively prime to n. • If p is an odd prime number such that p divides n then p divides b. • If p is an odd prime number dividing b then v p (b) is relatively prime to n.

= uv λ k 2 e

 2 Proof. As b is even and a is relatively prime to b, it follows that b -a and b + a are odd, thus u and v are odd. Set α = e iθ = a+i √ b 2 -a 2 b . Assume that (1) -(4) hold. Let 1 ≤ k ≤ n -1. It follows from (4) that 2uvλ k is an integer. However by (3) we have v 2 (λ k ) ≥ 1, therefore uv λ k 2 is an integer. It follows from (2) and (3) that r b 2 , k n divides uv λ k 2 , and so by Lemma 3.9 we have uv λ k 2 α k n = uv λ k 2 e kiθ n is an algebraic integer. Therefore uvλ k cos kθ n kiθ n + e -kiθ n is an algebraic integer. Similarly uvλ n-k cos (n-k)θ n is an algebraic integer. Therefore uv λ k cos kθ n + λ n-k cos (n-k)θ n is an algebraic integer. Lemma 3.11 implies that 2b λ k cos kθ n + λ n-k cos (n-k)θ n is an algebraic integer. However 2b and uv are relatively prime, hence λ k cos kθ n +λ n-k cos (n-k)θ

Lemma 4 . 1 .

 41 Let γ be a totally real algebraic number. If T (γ) = 0, then γ has a conjugate ≥ 0 and a conjugate (maybe the same) ≤ 0. Lemma 4.2. Let β be a totally real algebraic number. Then β has a conjugate ≥ T (β) and a conjugate (maybe the same) ≤ T (β).

Lemma 4 . 3 .

 43 Let β be a totally real algebraic number. Then β has a conjugate ≥ T (β 2 ), or a conjugate ≤ -T (β 2 ).

Lemma 4 . 4 .

 44 Let β be a totally real algebraic number such that T (β) = 0. Then δ(β) ≥ 2 T (β 2 ).

Lemma 4 . 5 .

 45 Let β be a totally real algebraic number, set r = T (β 2 -T (β 2 )) 2 . If T (β) = 0 and T (β 3 ) = 0 then δ(β) ≥ 2 2 √ r. Proof. Denote by γ 1 the largest conjugate of β and by γ 2 the smallest conjugate of β. If |γ 2 | ≥ |γ 1 |, then we can change β to -β, and so assume that

Proof.

  Set α = e iθ . Set K = Q(e iθ n), note that for every 1 ≤ k, ℓ ≤ n -

. 3 )= ℜ e iγ x d -1 x - 1 = 0 Notation 6 . 5 .

 31065 However, |cos(a) -cos(b)| ≤ |a -b| for all a, b ∈ R. Therefore, subtracting (6.1) and (6.2), and comparing the argument of the cosine function in (6.3), we obtain|α t -√ s| ≤ (|λ| + |µ|) ε |λ| + |µ| = ε . It follows that α t ≥ √ s -ε.Lemma 6.4. Let γ be a real number, ℓ = 0 be an integer, and d > 1 be an integer. Assume that d ∤ ℓ. Then d-1 t=0 cos γ + tℓ d 2π = 0. Proof. Set x = e i ℓ d 2π . Note that x d = 1. The following equalities hold: For a ∈ R and b ∈ R * , we denote R b (a) the only real number c ∈ [-b 2 , b 2 [ such that a -c ∈ bZ. Note that for all t ∈ R * we have R tb (ta) = tR b (a). We will use the particular case R 2π ( a b 2π) = R b (a) b 2π Note that if a, b are integers, than R b (a) is an integer. Also observe that if a, a ′ , b are integers, then |R b (aa ′ )| ≤ |R b (a)| × |R b (a ′ )|.

Lemma 6 . 7 .Lemma 6 . 8 .

 6768 Let a ∈ Z and b > 0 an integer, then there exists u ∈ Z \ {0} such that |u| ≤ √ b and |R b (u)| ≤ √ b. Proof. If b = 1 the it is easy. Assume that b ≥ 2. Consider the following system with indeterminate u, v in R. |u| ≤ √ b |ua + vb| ≤ √ b. The set of solutions is a parallelogram of height 2 √ b, and base 2 √ b b . Hence the surface is 4. Therefore, by Minkowski's Theorem (see for instance [8, Lemma, p. 137]), there is a solution (u, v) ∈ Z 2 \ {(0, 0)}. Thus we have |u| ≤ √ b and |R b (ua)| ≤ |ua + vb| ≤ √ b. Moreover if u = 0, then v = 0 and b ≤ |vb| = |ua + vb| ≤ √ b; a contradiction. Therefore u = 0 as required. Let 1 ≤ k, ℓ < n be integers such that gcd(k, ℓ, n) = 1, k = ℓ, and k = n -ℓ. Then one of the following statement holds.

  ) If (b -a) and (b + a) are square-free and b ≥ 2t + a, then s = 4t. Proof. It follows from Lemma 7.3 and Lemma 7.6 that there is 0

. 14 )

 14 However, as v 2 (x) = 1 + v2(b)2 and for all odd prime p dividing b we have v p (x) = vp(b)+1 2

  b-a v 2 . A similar proof, considering ( x u , x u ) ∈ S k , yelds s ≤ 8t b+a u 2 . Hence (4) holds. If b ≥ 2t + a, b -a and b + a are square-free then u = v = 1. Thus λ + µ and λ -µ are integers. However as seen in (7.7), we have v 2 (λ) ≥ 1, and v 2 (µ) ≥ 1.

Corollary 7 . 9 .

 79 There are infinitely many β ∈ O K such that all conjugates of β are in [0, ⌈s⌉ + s]. In particular the Julia Robinson Number of O K is at most ⌈s⌉ + s.

  .1) Similarly |Q(b; i, k) -bl| ≤ bη (8.2) and |Q(b + εb; i, k) -(b + εb)l| ≤ (b + εb)η . (8.3) From (8.1) and (8.2) we have |Q(b

. 4 ). ( 8 . 5 )

 485 Similarly from (8.2) and (8.3) we deduceQ(b + εb; i, k) -Q(b; i, k) ≥ εbl -3bη = εb 3 5kConsider the setX i = {b + ijk + i | j integer and 0 ≤ j ≤ εb-1 k }. Hence X 1 ⊆ [b, b + εb], moreover X 1 is the set of all integers in [b, b + εb] congruent to 1 modulo k. Similarly X -1 ⊆ [b -εb, b], moreover X -1 is the set of all integers in [b -εb, b] congruent to -1 modulo k. From (8.4) we have, in the interval [b -εb, b], at least Q(b; -1, k) -Q(bεb; -1, k) ≥ εb 35k square-free integers congruent to -1 modulo k. However all integer in [b -εb, b] congruent to -1 modulo k are in X -1 . Therefore there are at least εb3

Lemma 8 . 4 .

 84 Let (a 1 , b 1 , r 1 ) be good; let (a 2 , b 2 , r 2 ) be good. Denote by K 1 (resp., K 2 ) the totally real field associated to (a1 , b 1 , r 1 ) (resp., (a 2 , b 2 , r 2 )). Assume that r 1 , r 2 are even. If (b 1 + a 1 )(b 2 + a 2 )b 1 b 2 is not a square then K 1 = K 2 .

  First assume that |R

n (rℓ)| > √ n. It follows from Lemma 6.7 that there is u ∈ Z \ {0} such that |u| ≤ √ n and |R n (urℓ)| ≤ √ n. So |u(R n (rℓ))| = |u||R n (rℓ)| > √ n ≥ |R n (urℓ)|, thus (2) holds. Similarly |R n (qk)| > √ n then (3) holds. Therefore we can assume that |R n (rℓ)| ≤ √ n and |R n

  Robinson Number of O K is at most ⌈s⌉ + s. Theorem 7.10. The Julia Robinson Number of O K is ⌈s⌉ + s. Moreover O K has the Julia Robinson Property. Proof. From Corollary 7.9 we already know that the Julia Robinson Number of O K is at most ⌈s⌉ + s. Moreover there are infinitely many β ∈ O K such that all conjugates of β are in [0, ⌈s⌉ + s], thus to prove that O K has the Julia Robinson Property, we only have to prove that the Julia Robinson Number of O K is ⌈s⌉ + s.

  For each such k we have N (λ k , λ n-k ) ≥ s 2 . Thus by(7.16) we have 2Cs 2 ≤ (2s + 1) 2 , so

	C ≤ 2 + 1 s + 1 2s 2 . However, from Lemma 7.8 we have s ≥ 2 it follows that C ≤ 2.	√ 2. As C is an integer

s + 1 then from

(7.16)

we have

4s 2 + 2s + 2 = 2(2s 2 + s + 1) < 2N (λ, µ) ≤ (2s + 1) 2 = 4s 2 + 2s + 1 ; a contradiction. Therefore N (λ k , λ n-k ) ≤ 2s 2 + s + 1. Hence we have (λ k , λ n-k ) ∈ S, for all 1 ≤ k ≤ n -1.

Let C be the number of k ∈ {1, . . . , ⌊ n 2 ⌋} such that (λ k , λ n-k ) = (0, 0).

1 .

 1 Lemma 8.3. Let k ≥ 1 be an integer, let 1 > ε > 0 be a real number. Then there is an integer C such that for all b multiple of k if b ≥ C then there is 1 ≤ a ≤ εb such that a ≡ 1 mod k, and b -a and b + a are square-free.

	Proof. Set l = 6 π 2 k	p|k	p 2 p 2 -1 . Note that kl ≥ 3 5 . From Theorem 8.2 we have
	lim x→∞	Q(x; 1, k) x	= l and lim x→∞

  Proof of Claim. Let n 1 , n 2 , k 1 , k 2 , λ 1 , µ 1 , λ 2 , µ 2 as in Claim 1 with n 1 minimal. We can assume that k 2 ≤ n2 2 . From Lemma 7.12 and (8.6), we have 2k 1 = n 1 if and only if 2k 2 = n 2 . Assume that n 1 ≥ 4. Squaring both side we obtain from Lemma 3.8

		1 2	(λ 2 1 + µ 2 1 + 2λ 1 µ 1	a 1 b 1	) + (λ 2 1 + µ 2 1 ) cos	2k 1 θ 1 n 1	+ 2(λ 1 µ 1 + µ 2 1	a 1 b 1	) cos	(n 1 -2k 1 )θ 1 n 1
	=	1 2	(λ 2 2 + µ 2 2 + 2λ 2 µ 2	a 2 b 2	) + (λ 2 2 + µ 2 2 ) cos	2k 2 θ 2 n 2	+ 2(λ 2 µ 2 + µ 2 2	a 2 b 2	) cos	(n 2 -2k 2 )θ 2 n 2 (8.7)
	From Lemma 3.14, taking the trace we obtain
					1 2	(λ 2 1 + µ 2 1 + 2λ 1 µ 1	a 1 b 1	) =	1 2	(λ 2 2 + µ 2 2 + 2λ 2 µ 2	a 2 b 2	) .
	Hence
	(λ 2 1 + µ 2 1 ) cos	2k 1 θ 1 n 1	+ 2(λ 1 µ 1 + µ 2 1	a 1 b 1	) cos	(n 1 -2k 1 )θ 1 n 1
						= (λ 2 2 + µ 2 2 ) cos	2k 2 θ 2 n 2	+ 2(λ 2 µ 2 + µ 2 2	a 2 b 2	) cos	(n 2 -2k 2 )θ 2 n 2
								n 1			θ 1	= η p cos	pθ 2 n 2	+ η n2-p cos	(n 2 -p)θ 2 n 2 Claim 1. .
								x cos	θ 1 2	= y cos	θ 2 2	= 0 .

Claim 2. There are x, y in Q such that

the conditions (3) of Notation 7.1, then (x 1 +x 2 , y 1 +y 2 ) also. Thus S k is a subgroup of Q 2 .

Observe that condition (3) in Notations 7.1 implies that if (x, y) ∈ S k , then 2uvx ∈ Z and 2uvy ∈ Z. Then S k is contained in the discrete sub-Z-module generated by (1/2uv, 0) and (0, 1/2uv). In particular it is a discrete Z-module. Lemma 7.4. Let 0 ≤ k ≤ N -1. Let x, y ∈ Q.Then for all q ∈] k N , k+1 N [∩Q, we have (x, y) ∈ S k if and only if the following three conditions hold:

(1) For all odd prime number p such that p divides b we have v p (x) ≥ qv p (b) and v p (y)

Let p be an odd prime number dividing b and let (x, y) be in S k .

Assume that there is an integer ℓ such that

ℓ is an integer; a contradiction. Therefore there is no integer in ]

. Therefore the condition Lemma 7.4(1) and Notation 7.1(1) are equivalent.

Similarly Lemma 7.4(2) and Notation 7.1(2) are equivalent. The last condition is the same. Lemma 7.5. Let n ≥ 1 be an integer dividing a power of r. Let λ 0 , λ 1 , . . . , λ n-1 be in Q. The following statement are equivalent:

(1) [START_REF] Amoroso | Algebraic Numbers of Small Weil's Height in CM-fields: on a Theorem of Schinzel[END_REF]. Observe that N is the product of v 2 (b) -1 and all v p (b) for p|b odd primes, which are each coprime to r, thus N and r are coprime. As n divides a power of r it follows that n and N are coprime, therefore for all 1 ≤ ℓ ≤ n -1 and all 0 ≤ k ≤ N we have ℓ n = k N . By Lemma 3.16 we have λ 0 ∈ Z, and for all 1 ≤ ℓ ≤ n-1 the following statement holds.

it follows from Lemma 7.4 that (i), (ii), and (iii) holds. By Lemma 3.16, we have that λ 0 + n-1 ℓ=1 λ ℓ cos( ℓθ n ) is an algebraic integer, moreover it belongs to K, therefore (1) holds.

Then

The following equalities hold 2t p = 2 p q|t q prime q p , by multiplicativity.

, by Gauss's quadratic reciprocity.

is odd and p ≡ 2 mod t.

Therefore 2t is a square modulo p.

Lemma 8.7. Let t ≥ 1 be a square-free odd number. There are infinitely many couples of positive integers (m i , n i ) such that, for all i, m i -2 and n i are both relatively prime to 2t and, for all i = j, (2 mi t ni -1)(2 mj t nj -1) is not a square.

Proof. Assume that we have (m

It follows from Lemma 8.6 that there is a prime p such that p ≡ 2 mod t, p ≡ 3 mod 4, 2t is a square modulo p and, for all 1 ≤ i ≤ k, p does not divide 2 mi t ni -1. In particular p is relatively prime to 2t.

Pick u ≥ 1 odd such that (2t)

p-1 2 ≡ 1 mod p u+1 . Note that Lemma 8.5 implies that 2t is a square modulo p u+1 . Let x be such that 2t ≡ x 2 mod p u+1 , and so x is prime to p. Therefore x p-1 ≡ (x 2 ) p-1 2

≡ (2t)

p-1 2 mod p u+1 , so x p-1 ≡ 1 mod p u+1 . However the multiplicative group (Z/p u+1 Z) * is cyclic of order (p -1)p u , hence x (p-1)p u ≡ 1 mod p u+1 .

Note that (p u + 1) p ≡ 1 mod p u+1 . As the subgroup of (Z/p u+1 Z) * generated by x p-1 is not trivial and of order dividing p u , it follows that this subgroup contains p u +1. Therefore there is c ≥ 1 such that x (p-1)c ≡ p u +1 mod p u+1 . As x (p-1)p u ≡ 1 mod p u+1 we have x (p-1)(c+p u ) ≡ x (p-1)c ≡ p u + 1 mod p u+1 . Therefore as p is odd, we can assume that c is odd.

p-1 2 c ≡ x (p-1)c ≡ p u + 1 mod p u+1 . As (p-1)p u is even and relatively prime to t we can pick w 0 and w 1 odds such that w 0 -2 is prime to t, w 1 is prime to t, w 0 ≡ v mod (p -1)p u , and w 1 ≡ v mod (p -1)p u . Therefore 2 w0 t w1 ≡ 2 v t v ≡ p u + 1 mod p u+1 , so p u divides 2 w0 t w1 -1 and p u+1 does not divides 2 w0 t w1 -1.

Note that v p (2 w0 t w1 -1) = u is odd, moreover p does not appear in any prime decomposition of the 2 mi t ni -1. Therefore for all 1 ≤ i ≤ k, (2 mi t ni -1)(2 w0 t w1 -1) is not a square. Theorem 8.8. Let t ≥ 1 be a square-free odd number. Then the following statement holds.

(1) There are infinitely many fields K such that O K has Julia Robinson's number 2 √ 2t + 2 √ 2t.

(2) There are infinitely many fields K such that O K has Julia Robinson's number 8t.

Proof. By Lemma 8.4 and Lemma 8.7, we can find infinitely many good (a, b, r), such that b = 2 m-1 t n , with m -2 and n relatively prime to 2t, a = b -1 and r = 2t, and all the totally real associated fields are distinct. Fix such (a, b, r). Let K be the totally real field associated to (a, b, r), let s as in Notation 7.

Acknowledgement

The authors thank Francesco Amoroso for the invitation to the Université de Caen. The friendly ambiance and a discussion, with Francesco Amoroso, yields to the explicit example in Theorem 8.8 [START_REF] Bombieri | A note on heights in certain infinite extensions of Q[END_REF]. The authors also thank Florence and Jean Gillibert and Philippe Satgé for helpful discussions. Finally, the authors thank Denis Simon, for his very useful calculations using Paris, allowing us to find our examples.

Funding

The first author was supported by CONICYT, Proyectos Regulares FONDECYT no. 1150595, and by project number P27600 of the Austrian Science Fund (FWF). The second author was supported by CONICYT, Proyectos Regulares FONDECYT no. 1140946.