arXiv:1710.11382v1 [math.NT] 31 Oct 2017 rtodrter) hr salto xmlso ig ihteJRP. the with rings of examples of lot a is There theory). order first of rv togrrsl n[,4 ] if 5]: 4, [3, in result stronger a prove oal elagbacnmes o every For numbers. algebraic real totally n[4 hti oal elagbacfield o algebraic Videla real and totally Vidaux a if useful. JR( that particularly [34] seems in Property Robinson Julia ieaddc htfreeypstv integer positive every for that deduce Videla ui oisnNumber Robinson Julia atclr h rtodrter of theory order first the particular, R R,te h eirn ( semi-ring the then JRP, Property Robinson Julia fdge fattlyra ubrfil then field number real totally a of 2 degree of elaeinetnin of extensions abelian real ei of metic qast + to equals oisnsnumbers. Robinson’s ySlpnohi ei fppr 2,2,2,2,2,2]t ag s large to 29] 28, 27, 26, 25, [24, papers of serie a in Shlapentokh by rmteMtysvc’ ouint ibr 0hpolm[3,ta the that [13], problem of 10th theory order Hilbert first to positive solution Matiyasevich’s the from n oe-aly[4 5 o iia rbesi ucinfields. function in problems Hilbert’s the similar on for num more 15] arbitrary for [14, [12] an Matiyasevich Moret-Bailly of also and Shlap See rings See results. integer Rubin holds. more and algebraic conjecture for Tate-Shafarevich Mazur the the for that [18]). result assuming Shlapentokh similar and a Poonen [11] [17], Poonen also (see { ∪ If oisn n[0,gnrlzn e ehd f[1,poe hti ring a if that proves [21], of methods her generalizing [20], in Robinson, e od n phrases. and words Key Let 2010 Date nteohrhnsfrifiieagbacetninof extension algebraic infinite for hands other the On α O K ∈ + K oebr1 2017. 1, November : R }| ∞} + = ) ahmtc ujc Classification. Subject Mathematics R sattlyra leri ubrfil,te JR( then field, number algebraic real totally a is h e fteJlaRbno’ ubrta ecntuti un is construct we that Number Robinson’s Julia the of set the tutn nifiiefml frnso leri neeso integers algebraic of rings of of family infinite an structing Abstract. R etern fteagbacitgr fasbedo h field the of subfield a of integers algebraic the of ring the be uhta,alconjugates all that, such Q sa opiae sta of that as complicated as is ∞ R hs ui oisnsNme sdsic rm4ad+ and 4 from distinct is Number Robinson’s Julia whose npriua h ighsudcdbefis re theory. order first undecidable has ring the particular In . ∞ t sinfinite is sn h euto obeiadZniri 2,Vdu and Vidaux [2], in Zannier and Bombieri of result the Using . eprilyase oaqeto fVdu n ieab con- by Videla and Vidaux of question a to answer partially We IREGLIETADGBIL RANIERI GABRIELE AND GILLIBERT PIERRE UI OISNSNUMBERS ROBINSON’S JULIA of ig fagbacitgr;Udcdblt;Ttlyreal Totally Undecidability; integers; algebraic of Rings } JP fJR( if (JRP) N bev that Observe . Q R 0 ; O fdegree of sdfie sJR( as defined is , K 1 , 1. sudcdbe h euti vnfrhrextended further even is result The undecidable. is + O . , Introduction K K sfis-re enbein definable first-order is ) β R sudcdbe oeta ee n Lipshitz and Denef that Note undecidable. is sattlyra ubrfil,o nextension an or field, number real totally a is d of 10,1U5 11R80. 11U05, 11R04, ) N 1 a t igo leri neeswt JR with integers algebraic of ring its has ∈ n nparticular in and , A α ( A t R aif 0 satisfy K ( ∈ R sete nitra or interval an either is ) R d N inf( = ) h edgnrtdb l totally all by generated field the , ). R a h otct rpry then property, Northcott the has sDohniein Diophantine is { ∪ t < β < + A ∞} oal elsubfields real totally f ( O Q R K eoeby denote , ) Moreover )). a esi nw.The known. is less far + = ) R bounded. Set . ∞ R a nundecidable an has noh[0 1 32] 31, [30, entokh Moreover . hnetearith- the (hence ∞ O 0hproblem, 10th A big of ubrings se[0) In [20]). (see K { ( tfollows It . R + R existential e fields, ber ed;Julia fields; R Q R = ) ∞} t rv in prove tr a the has a the has h set the bserve The . fall of { t K ∈ 2 P. GILLIBERT AND G. RANIERI
Denote by Ztr the ring of all totally real algebraic integers. As noted by Robinson in [20], it follows from Schur [23] or Kronecker [10] that JR(Ztr) = 4, and that Ztr has the JRP (also see Robinson [22] or Jarden and Videla [9]). In particular the first order theory of Ztr is undecidable (see [20]). In [33, 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 Qtr?
Then, they find a family of rings O contained in Ztr, such that JR(O) is distinct from 4 and + (see [33, 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 [33] is also interesting because it contains a very precise bibliography on undecidability problem over algebraic rings. Finally, very recently, Vidaux and Videla (see [34]) 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 [36] 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 Qtr. 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 OK has Julia Robinson’s num- ber 2√2t +2√2t. (2) There are infinitely many fields K such that OK has Julia Robinson’s num- ber 8t. To prove Theorem 8.8, we choose 0
2. Notations
Let F be an algebraic extension of Q, we denote by OF 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 OF 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 ∈ λ define the P -adic valuation of β as e . Then in particular, the P -adic valuation of β should not be an integer. We always denote by vp 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. ⌈ ⌉ ≥ ⌊ ⌋ ≤ 3. Some preliminaries
n1 n2 ns Given an integer b 1, and given p1 p2 ...ps the prime decomposition of b, and u 0 in Q, we denote≥ ≥ n1u n2u nsu r (b,u)= p1⌈ ⌉p2⌈ ⌉ ...ps⌈ ⌉ . 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, u ≥ ≥ and r (b,u) b− 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 1 α and denote it by α n . We need to study some properties of a linear combination 1 with coefficients in K of some powers of α 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, ≥ k1t k2t n +1 k1, k2 n 1, and ℓ1,ℓ2 in Z. Assume that ℓ1 + n = ℓ2 + n . Then −one of the≤ following≤ holds:−
(1) k1 = k2. (2) k1 < 0 < k2 and k2 = k1 + n. (3) k2 < 0 < k1 and k1 = k2 + n.
k1t k2t Proof. Assume that ℓ1 + n = ℓ2 + n . Hence nℓ1 +k1t = nℓ2 +k2t. Thus (k1 k2)t is a multiple of n. Then it follows, as t is relatively prime to n, that k k− is a 1 − 2 multiple of n. Since 2n +2 k1 k2 2n 2, we have that k1 k2 n, 0,n . If k k = 0 then− k = k≤. − ≤ − − ∈ {− } 1 − 2 1 2 If k1 k2 = n, then k1 = n + k2. Moreover, we have k1 = n + k2 n n +1 > 0 and k =− k n n 1 n< 0. ≥ − 2 1 − ≤ − − If k1 k2 = n, then k2 = n + k1 and k1 = n + k2 n + n 1 < 0. Similarly k = k −+ n n− 1 n< 0. − ≤− − 2 1 ≥ − − Lemma 3.3. Let K be a number field, n be a natural number, p be a prime number, α be in K. Let ν1,p, and ν2,p be valuations over K extending vp to K, let (λk n +1 k n 1) be a family in K. Assume that the following conditions hold:| − ≤ ≤ − (1) ν1,p(α) is a positive integer relatively prime to n. (2) ν2,p(α) is a negative integer relatively prime to n. (3) n 1 − k γ = λkα n is an algebraic integer. k= n+1 X− 4 P. GILLIBERT AND G. RANIERI
(4) For all n +1 k n 1, we have ν (λ )= ν (λ ) is an integer. − ≤ ≤ − 1,p k 2,p k kν (α) Then for all n +1 k n 1, we have ν (λ ) − 2,p , ν (λ ) − ≤ ≤ − 1,p k ≥ n 1,p k ≥ kν (α) − 1,p , and ν (λ ) 0. l m n 1,p 0 ≥ l m 1 Proof. By conditions (1) and (2), K(α n )/K is totally ramified over ν1,p and ν2,p. 1 Then, ν and ν extend in a unique way to K(α n ). Denote I = k Z 1,p 2,p { ∈ | n +1 k n 1 . As γ is an algebraic integer, it follows that ν1,p(γ) 0, and −ν (γ) ≤0.≤ Given− k} I, the following equalities hold: ≥ 2,p ≥ ∈ k kν1,p(α) ν (λ α n )= ν (λ )+ . 1,p k 1,p k n
k kν2,p(α) kν2,p(α) ν (λ α n )= ν (λ )+ = ν (λ )+ . 2,p k 2,p k n 1,p k n Denote
kν1,p(α) k S = ν (λ )+ k I = ν (λ α n ) k I 1 1,p k n | ∈ 1,p k | ∈ n o kν2,p(α) k S = ν (λ )+ k I = ν (λ α n ) k I . 2 1,p k n | ∈ 2,p k | ∈ n o Let u be the smallest element of S1 S2. Assume that u< 0 and that u S1. Fix k1ν1,p(α∪) ∈ k1 such that u = ν1,p(λk1 )+ n . k Assume that for all k = k , we have u<ν (λ α n ), then ν (γ) = u < 0; 6 1 1,p k 1,p a contradiction. Therefore, there is k2 I such that k2 = k1 and the following equalities hold: ∈ 6
k1ν1,p(α) k2 k2ν1,p(α) ν (λ )+ = u = ν (λ α n )= ν (λ )+ . 1,p k1 n 2,p k2 1,p k2 n
Up to an exchange of k1 and k2, we can assume that k1 < k2, hence it follows from Lemma 3.2 that k1 < 0 < k2 and k2 = k1 + n. As k2 > 0 and ν1,p(α) > 0 >ν2,p(α), it follows that k ν (α) k ν (α) ν (λ )+ 2 2,p <ν (λ ) <ν (λ )+ 2 1,p = u . 1,p k2 n 1,p k2 1,p k2 n Hence u is not the smallest element of S S ; a contradiction. With a similar 1 ∪ 2 argument if u S2, we can find another contradiction. It follows that u 0. ∈ kν1,p(α) ≥ kν1,p(α) For all k I, we have ν1,p(λk)+ n u 0. Hence ν1,p(λk) n , ∈ ≥ ≥ kν (α)≥ − therefore, as ν (λ ) is an integer it follows that ν (λ ) − 1,p . A similar 1,p k 1,p k ≥ n kν (α) argument yields ν (λ )= ν (λ ) − 2,p . l m 1,p k 2,p k ≥ n l m Lemma 3.4. Let K be a number field, n N, p be a prime number, α K. Let ∈ ∈ νp be a valuation over K extending vp to K. Assume that the following conditions hold:
(1) νp(α) is an integer relatively prime to n. (2) For all λ K∗, νp(λ) is an integer. 1 ∈ Then [K(α n ): K]= n. JULIA ROBINSON’S NUMBERS 5
1 Proof. We extend νp to K(α n ). By abuse of notation, we still denote νp the n k n extension. Let (λk 0 k n 1) be a family of K such that: k=0 λkα = 0. | ≤ ≤ − k k Note that, for all 0 k n, we have ν (λ α n )= ν (λ )+ ν (α). ≤ ≤ p k 1,p k nPp Set I = k N k n 1 and λk = 0 . It follows from Lemma 3.2 that for { ∈ | ≤ − k 6 } ℓ all k = ℓ in I we have νp(λk)+ n νp(α) = νp(λℓ)+ n νp(α). Therefore, if I is not 6 n k 6 k empty, ν ( λ α n ) is the minimum of ν (λ )+ ν (α) for k I. Therefore p k=0 k p k n p ∈ νp(0) = ; a contradiction. 6 ∞P Lemma 3.5. Let K be a field, let α be in K, let n be an integer 2. Assume 1 ≥ that [K(α n ): K] = n. Let k Z be such that k is not a multiple of n. Then k ∈ Tr 1 (α n )=0. K(α n )/K
1 Proof. Set d = (k,n) and let u, v in Z be such that ku + nv = d. Since [K(α n ): n k k K]= n, we have that X d α d is the minimal polynomial of α d over K. It follows k − that Tr 1 (α n ) = 0. K(α n )/K 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. Lemma 3.6. Let a,b be relatively prime in N, with a k 2 2 However, bα n γ is an algebraic integer, therefore nbλ k + nλn k(a + i√b a ) is an algebraic integer. − − − Similarly, k k − −k √ 2 2 √ 2 2 n n k n a + i b a n a i b a n k 2 2 k (bα n ) = b α− = b − = b − − = b − (a i b a ) . b ! b ! − − p −k Hence bα n is an algebraic integer. The following equalities hold n 1 −k − ℓ−k TrL/K (bα n γ) = TrL/K bλℓα n ℓ= n+1 ! X− 1 = TrL/K (bλk)+TrL/K (bλk nα− ) , by (3.1). − 1 1 = nbλk + nbλk nα− , as bλk and bλk nα− are in K. − − 2 2 = nbλk + nλk n(a i b a ). − − − − k 2 2 However, bα n γ is an algebraic integer,p therefore nbλk + nλk n(a i√b a ) 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: x + y x y x + y x y λ cos(x)+µ cos(y) = (λ+µ)cos cos − +(µ λ) sin sin − 2 2 − 2 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. 6 kθ (n k)θ 2 1 λ cos + µ cos − = λ2 + µ2 +2λµ cos θ+ n n 2 2kθ (n 2k)θ + (λ2 µ2)cos + 2(λµ + µ2 cos θ)cos − − n n ! a+i√b2 a2 Lemma 3.9. Let 0 Lemma 3.10. Let 0 u Proof. We can assume that q = v with u, v positive integers. The following equal- ities hold: u i u θ v v iuθ v u iθ u v u 2 2 (√be v ) = b 2 e = b 2 − (be ) = b 2 − a + i b a . − v v u p u 2 √ 2 2 As 2 u it follows that b − is an algebraic integer, moreover (a + i b a ) is ≥ i u θ − an algebraic integer. Therefore √be v is an algebraic integer. Similarly u i u θ v v iuθ v u iθ u v u 2 2 (√be− v ) = b 2 e− = b 2 − (be− ) = b 2 − a i b a . − − i u θ p iqθ iqθ Hence √be− v is algebraic integer. Therefore 2√b cos(qθ)= √b(e + e− ) is an algebraic integer. Similarly 2√b sin(qθ) is an algebraic integer. a Lemma 3.11. Let 0 a+i√b2 a2 Lemma 3.12. Let 0 2r r r r (α)= P1 (p)− I = P1 P2− I, (3.2) with I fractional ideal coprime with (p). In fact if t is the exponent of P1 in the 2 2 factorization of a + i√b a , since P2 = P1 we have that t is the exponent of P in the factorization of −a i√b2 a2. Moreover, we have (a + i√b2 a2)(a 2 − − − − i√b2 a2)= b2. Since p2r divides b2 and p2r+1 does not divide b2, we have t =2r, proving− (3.2). Let ν1,p be respectively ν2,p the valuations associated to P1 respectively P2. Thus by (3.2), r = ν1,p(α)= ν2,p(α)= vp(b). Suppose now that p =− 2. Since 4 divides b and a is coprime with b, a2 b2 1 mod (8). Then by the reciprocity quadratic law (2) splits in the ring of the algebraic− ≡ a+i√b2 a2 2 2 integers of Q(α). Observe that − is a root of the polynomial x ax + b 2 − with integer coefficients. Then (2) divides a+i√b2 a2 and a i√b2 a2. Moreover − − 2 −2 √ 2 2 √ 2 2 a+i√b a (4) does not divide a+i b a and a i b a because 4 − has the trace a − − − equals to 2 , which is not an integer. Write (2) = P1P2 with P1 and P2 prime ideals of the ring of the algebraic integers of Q(α). Since αα = 1, (2) divides a+i√b2 a2 − and (4) does not divide a + i√b2 a2, we get that there exists a positive integer c − and ideal I1, I2 coprime with (2) such that 2c 1 2c c 1 1 c c 1 (α) = (2)P1 I1(b)− = (2)P1 I1(2)− − I2− = P1 I1P2− I2− . (3.3) Let ν1,2 be respectively ν2,2 the valuations associated to P1 respectively P2. Thus by (3.3), ν (α)= c = ν (α) and v (b)= c +1= ν (α) + 1. 1,2 − 2,2 2 1,2 The following Corollary immediately follows from Lemmas 3.4, Lemma 3.5, and Lemma 3.12. a+i√b2 a2 Corollary 3.13. Let 0 1 is coprime to n. Then [Q(α n ): Q(α)] = n, and for all integers k not multiple of n k we have Tr 1 (α n )=0. Q(α n )/Q(α) a π Lemma 3.14. With the hypothesis of Corollary 3.13, let θ = arccos( b ) ]0, 2 [. θ 2θ (n 1)θ θ ∈ Then 1, cos n , cos n ,..., cos −n is a Q-basis of Q(cos n ). kθ Moreover, TrQ(cos θ )/Q cos =0 for all 1 k n 1. n n ≤ ≤ − 1 1 1 n n n θ α +α Proof. By Corollary 3.13, [Q(α ): Q]= 2n. Observe that cos n = 2 . 1 2 θ n θ So X 2cos n X + 1 is the minimal polynomial of α over Q(cos n ), and so θ− [Q(cos n ): Q]= n, as illustrated by the following diagram. 1 Q(α n ) ● ss ●● 2 ss ●●n sss ●● sss ●● θ Q(cos( n )) Q(α) ▲▲▲ ✈✈ ▲▲ ✈✈ ▲▲ ✈✈ n ▲▲ ✈✈ 2 ▲▲ ✈✈ Q θ 2θ (n 1)θ θ To show that 1, cos n , cos n ,..., cos −n is a Q-basis of Q(cos n ), it is sufficient to prove that they are linearly independent over Q. Assume that we are given n 1 kθ λ0,...,λn 1 in Q such that − λk cos = 0. The following equalities hold − k=0 n n 1 n 1 k k n 1 k n−k − kθ P − α n + α n − α n + αα n 0= λ cos = λ + λ = λ + λ k n 0 k 2 0 k 2 kX=0 kX=1 Xk=1 Therefore the following equality holds n 1 − λk + λn kα k λ + − α n =0 . 0 2 Xk=1 1 2 n−1 1 Thus, as 1, α n , α n ,...,α n is a basis of Q(α n ) over Q(α), it follows that λk =0 for all k. k k k kθ α n +α n By Corollary 3.13, we have Tr k (α n ) = 0. Since cos = , we Q(α n )/Q n 2 kθ immediately get Tr θ cos = 0. Q(cos n )/Q n Lemma 3.15. With the hypothesis of Corollary 3.13, Let θ = arccos( a ) ]0, π [. b ∈ 2 Let (λk 0 k n 1) be a family in Q. Then the conjugates of the element n| 1 ≤ ≤ kθ − θ − λ0 + k=1 λk cos( n ), counted with multiplicity in Q(cos( n )), are the elements n 1 k(θ+2πt) λ + − λ cos( ) with 0 t n 1. 0 Pk=1 k n ≤ ≤ − a+i√b2 a2 1 P iθ iθ 1 n Proof. Note that α = e = b − , and α = e− = α− . We have [Q(α ): Q(α)] = n, hence for each 0 t n 1 we have a unique morphism, denoted by 1 ≤ ≤ − 1 i 2πt 1 ϕt : Q(α n ) C preserving Q(α), such that ϕt(α n ) = e n α n . The embedding 1 → Q(α n ) C preserving Q(α), are ϕ0,...,ϕn 1. Since α is an algebraic number → − 1 whose Galois conjugates have absolute value equal to 1, then α n has the same 1 property. Then, by [1, Proposition 2.3], Q(α n ) is a CM field. In particular the complex conjugation is well-defined and it is independent of any embedding of 10 P. GILLIBERT AND G. RANIERI 1 1 Q(α n ) over C. The embedding Q(α n ) C are ϕ0,...,ϕn 1, and their complex → − conjugates ϕ0,..., ϕn 1. Let k be in Z, and 0 t n 1. The following equalities hold − ≤ ≤ − k ikθ k 1 i 2πt 1 k i 2πkt ikθ i k(θ+2πt) ϕt e n = ϕt α n = ϕt α n = (e n α n ) = e n e n = e n . (3.4) Therefore − kθ 1 ikθ ikθ i k(θ+2πt) i k(θ+2πt) k(θ +2πt) ϕ (cos( )) = ϕ e n + e n = e n + e− n = cos( ). t n 2 t n (3.5) 1 1 n n θ Set K = Q(α +α )= Q(cos( n )). Note that ϕt↾K = ϕt↾K, for all 0 t n 1. n 1≤ ≤ −kθ − Hence the morphisms K C are ϕ0↾K,...,ϕn 1↾K. Set β = λ0+ k=1 λk cos( n ) Let 0 t n 1. The→ following equalities− hold ≤ ≤ − P n 1 − kθ ϕt (β)= ϕt λ0 + λk cos n ! kX=1 n 1 − kθ = λ + λ ϕ cos 0 k t n kX=1 n 1 − k(θ +2πt) = λ + λ cos . 0 k n kX=1 Lemma 3.16. Let 0 kθ (n k)θ Lemma 3.11 implies that 2b λk cos + λn k cos − is an algebraic in- n − n kθ (n k)θ teger. However 2b and uv are relatively prime, hence λk cos +λn k cos − n − n is an algebraic integer. As λ0 is an integer, it follows that n 1 − kθ λ + λ cos 0 k n Xk=1 is an algebraic integer. n 1 kθ − Reciprocally assume that λ0 + k=1 λk cos n is an algebraic integer. The following equality holds: P n 1 n 1 n 1 − kθ − λk ikθ −ikθ − λk k −k λ + λ cos = λ + e n + e n = λ + α n + α n . 0 k n 0 2 0 2 kX=1 kX=1 Xk=1 By Lemma 3.12 there are valuations ν1,2 and ν2,2 defined over Q(α), extending the 2-adic valuation, such that ν (α) = ν (α) = v (b) 1. However, v (b) 1 1,2 − 2,2 2 − 2 − is relatively prime to n. Moreover, for all x Q(α)∗, we have that ν1,2(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 1 [Q(α n ): Q(α)] = n. As ν1,2(α) = v2(b) 1 is relatively prime to n, it follows from Lemma 3.3 that v (λ ) 0, and the following− inequalities hold 2 0 ≥ λ k(v (b) 1) v k 2 − , for all 1 k n. 2 2 ≥ n ≤ ≤ Therefore (3) holds. Let p be an odd prime dividing b. Thus by Lemma 3.12, there are valuations ν1,p and ν extending v to Q(α) such that ν (α) = ν (α) = v (b). As v (b) is 2,p p 1,p − 2,p p p relatively prime to n, it follows from Lemma 3.3 that vp(λ0) 0, and the following inequalities hold ≥ λ kv (b) v (λ )= v k p , for all 1 k n. p k p 2 ≥ n ≤ ≤ Therefore (2) holds. It follows from Lemma 3.6 that nλ is an integer, however we also have v (λ ) 0 0 p 0 ≥ for all odd primes p dividing n, and v2(λ0) 0, therefore λ0 is an integer. Therefore (1) holds. ≥ λk λn−k √ 2 2 By Lemma 3.6, nb 2 + n 2 (a + i b a ) 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 λn k + n λk (a + i√b2 a2) is an algebraic integer. 2 2 − Therefore, applying TrQ(α)/Q, we obtain that n(bλk + aλn k), n(bλn k + aλk) − − are integers, and nλ i√b2 a2 is an algebraic integer. k − We already proved that v2(λk) 0, and v2(λn k) 0, hence v2(u(λk +λn k)) ≥ − ≥ − ≥ 0, and v2(v(λk λn k)) 0. Similarly if −p be− an odd≥ prime number that divides n, then we have already proved that vp(λk) 0, and vp(λn k) 0, hence vp(u(λk + λn k)) 0, and ≥ − ≥ − ≥ vp(v(λk λn k)) 0. − − ≥ 2 2 2 2 Let p be a prime number that does not divides n(b a ). As nλki√b a is an algebraic integer, it follows that nλ (b2 a2) is an− integer, hence v (λ−) = k − p k 12 P. GILLIBERT AND G. RANIERI 2 2 vp(nλk(b a )) 0. Similarly vp(λn k) 0, therefore vp(u(λk + λn k)) 0, and − ≥ − ≥ − ≥ vp(v(λk λn k)) 0. Let p−be an− odd≥ prime number dividing b2 a2 = (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(bλk + aλn k), n(bλn k + aλk) are integers, it − − − follows that n(bλk + aλn k)+ n(bλn k + aλk)= n(λk + λn k))(a + b) is an integer. − − − As p does not divide n and p does not divide a+b, it follows that vp(u(λk +λn k)= − vp(n(λk + λn k))(a + b)) 0. − ≥ √ 2 2 √ 2 2 1 2 2 Moreover vp(λk) = vp(nλki b a ) vp( b a ) 0 2 vp(b a ) = 1 b a − − − ≥ − 1 − v (b a). However, −2 is a square-free integer, thus v (v) = v (b a) . − 2 p − v p ⌊ 2 p − ⌋ Therefore v (v) > 1 v (b a) 1. It follows that v (vλ ) = v (v)+ v (λ ) > p 2 p − − p k p p k 1 v (b a) 1 1 v (b a)= 1. However v (vλ ) is an integer, thus v (vλ ) 0. 2 p − − − 2 p − − p k p k ≥ Similarly vp(vλn k) 0, and so vp(v(λk λn k)) 0. − ≥ − − ≥ With a similar argument, we can prove that if p divides b + a then vp(u(λk + λn k)) 0 and vp(v(λk λn k)) 0. − ≥ − − ≥ Hence, for all prime p we have vp(u(λk + λn k)) 0 and vp(v(λk λn k)) 0. − ≥ − − ≥ Therefore u(λk + λn k) and v(λk λn k) are integers. That is (4) holds. − − − 4. Trace in general TrK/Q(β) Given an algebraic number β in a number field K, we denote T (β) = [K:Q] . TrQ(β)/Q(β) Note that T does not depends on the choice of K. In particular T (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. ∈ Lemma 4.1. 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). ≥ ≤ 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). ≥ Lemma 4.3. Let β be a totally real algebraic number. Then β has a conjugate T (β2), or a conjugate T (β2). ≥ ≤− Proof.p Lemma 4.2 implies thatpβ2 has a conjugate T (β2). Hence we have a ≥ conjugate γ of β such that γ2 T (β2), therefore γ T (β2) or γ T (β2). ≥ ≥ ≤− Lemma 4.4. Let β be a totally real algebraic numberp such that T (βp)=0. Then δ(β) 2 T (β2). ≥ Proof. Wep can assume that β = 0. We see from Lemma 4.2 that β2 has a conjugate 6 T (β2). Therefore β has a conjugate T (β2) or a conjugate T (β2). In the≥ second case, we can change β to ≥β, and so assume that β ≤has − a conjugate − p p T (β2). ≥ p JULIAROBINSON’SNUMBERS 13 Let γ1 be the largest conjugate of β, and γ2 be the smallest conjugate of β. From Lemma 4.2 it follows that γ1 > 0, and γ2 < 0. Let u Q be such that u>γ1. Set 2 2 u T (β ) 2 ∈ v = −2u . As u>γ1 T (β ), it follows that v > 0. The following equalities≥ hold: p T ((β v)2)= T (β2) 2vT (β)+ v2 − − = T (β2)+ v2 as T (β) = 0. u2 T (β2) 2 = T (β2)+ − 2u u4 2u2T (β2)+ T (β2)2 = T (β2)+ − 4u2 u4 +2u2T (β2)+ T (β2)2 = 4u2 u2 + T (β2) 2 = . 2u u2+T (β2) It follows from Lemma 4.3 that β v has a conjugate 2u or a conjugate u2+T (β2) − ≥ 2u . ≤− u2+T (β2) If β v has a conjugate 2u , it follows that γ1 the largest conjugate of β satisfies− ≥ u2 + T (β2) u2 T (β2) u2 + T (β2) γ v + = − + = u , 1 ≥ 2u 2u 2u 2 2 which contradicts u>γ . Therefore β v has a conjugate u +T (β ) . It follows 1 − ≤− 2u that γ2 the smallest conjugate of β satisfies u2 + T (β2) u2 T (β2) u2 + T (β2) T (β2) γ v = − = . 2 ≤ − 2u 2u − 2u − u 2 This inequality is true for all u Q larger than γ . It follows that γ T (β ) . 1 2 γ1 2 ∈ ≤ − T (β ) 2 Thus γ1 γ2 γ1 + 2 T (β ). − ≥ γ1 ≥ Lemma 4.5. Let β be a totallyp real algebraic number, set r = T (β2 T (β2))2 . − If T (β)=0 and T (β3)=0 then δ(β) 2 2√r. ≥ Proof. Denote by γ1 the largest conjugatep of β and by γ2 the smallest conjugate of β. If γ2 γ1 , then we can change β to β, and so assume that γ1 = γ1 γ2 , in particular| | ≥ | γ2| is the largest conjugate of−β2. Note that T (β2 T (β2))| = 0,| ≥ hence | | 1 − it follows from Lemma 4.4 that δ(β2 T (β2)) 2 T ((β2 T (β2))2)=2√r. − ≥ − However all conjugates of β2 are positives, hence γ2 δ(β2)= δ(β2 T (β2)) 2√r, 1 ≥ p − ≥ it follows that γ1 2√r. Consider the following≥ function p f : R R → x x γ + 2 r +4x2T (β2). 7→ − 1 q Note that f(0) = γ1 + 2√r 0 and thatp limx f(x)=+ . Moreover f is continuous, therefore− we can pick≤c 0, such that →∞f(c)=0. Set ∞s = r +4c2T (β2). p ≥ 14 P. GILLIBERT AND G. RANIERI Hence c = γ 2 r +4c2T (β2)= γ 2√s. (4.1) 1 − 1 − Let d Q be such that d>cq. Setpρ = (β d)2 T ((β qd)2). Note that T (ρ) = 0, moreover∈ the following equalities hold − − − ρ = β2 2dβ + d2 T (β2 2dβ + d2) − − − = β2 2dβ + d2 (T (β2)+2dT (β)+ d2) − − = β2 T (β2) 2dβ. − − Thus T (ρ2)= T (β2 T (β2) 2dβ)2 − − = T (β2 T (β2))2 +4d2T (β2) 4dT (β3)+4dT (β2)T (β) − − 2 2 = r +4 d T (β ) 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 of β − 3 such that (γ d)2 2√s. Assume that γ d 2√s, then, as d>c it follows 3 − ≥ 3 − ≥ from (4.1) that γ d+ 2√s>c+ 2√s = γ ; a contradiction, as γ is the largest 3 ≥ 1 p 1 conjugate of β. Therefore γ d 2√s, hence γ γ d 2√s. Therefore, p 3 − ≤−p 2 ≤ 3 ≤ − for all d Q such that d>c, we have γ d 2√s. Hence γ c 2√s = ∈ p 2 ≤ − p2 ≤ − γ 2 2√s, thus γ γ γ (γ 2 2√s)=2 2√s 2 2√r. 1 − 1 − 2 ≥ 1 − 1 − p ≥ p p p p p 5. Trace in the fields for the examples In this section we fix 0 iθ iθ Proof. Set α = e . Set K = Q(e n ), note that for every 1 k,ℓ n 1, the following equalities hold: ≤ ≤ − kθ ℓθ 1 kiθ −kiθ ℓiθ −ℓiθ cos cos = e n + e n e n + e n n n 4 1 (k+ℓ)iθ (k−ℓ)iθ (−k+ℓ)iθ (−k−ℓ)iθ = e n + e n + e n + e n . 4 Therefore, n na n 2 + 2b if k = ℓ = 2 n if k = ℓ = n kθ ℓθ 2 2 Tr 1 cos cos = na 6 n Q(α n )/Q(α) n n if k = n ℓ = 2b − 6 2 0 otherwise. Hence: n na n 2 + 2b if k = ℓ = 2 n n kθ ℓθ 2 if k = ℓ = 2 TrQ(cos( θ ))/Q cos cos = na 6 n n n n if k = n ℓ = 2b − 6 2 0 otherwise. kθ Moreover TrQ(cos( θ ))/Q(cos( )) = 0 for all 1 k n 1. The Lemma follows n n ≤ ≤ − directly. n Remark 5.2. Given λ, µ Q, n R, and 1 k < 2 . It follows from Lemma 5.1 that ∈ ∈ ≤ kθ (n k)θ 2 N(λ, µ)=2T λ cos + µ cos − . n n ! Moreover if 2 R, then ∈ θ 2 N(λ, λ)=2T λ cos . 2 ! 6. 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 λ2 + µ2. The minimum value is λ2 + µ2. − Proof. A rotation ofp the angle x, of the vector (λ, µ), hasp first coordinate λ cos x + µ sin x. Hence the maximum possible value is the length of the vector (λ, µ), which is λ2 + µ2. Lemmap 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 λ2 + µ2 +2λµ cos(θ). − − Proof. For all x R, denote f(x)= λ cos(pqθ + x)+ µ cos((1 q)θ x). It follows, from Lemma 3.7,∈ that − − θ 1 2q θ 1 2q f(x) = (λ + µ)cos cos − θ + x + (µ λ) sin sin − θ + x 2 2 − 2 2 16 P. GILLIBERT AND G. RANIERI Therefore, by Lemma 6.1, the maximum value of f is θ θ V = (λ + µ)2 cos2 + (µ λ)2 sin2 . 2 − 2 s The following equalities hold θ θ θ θ V 2 = (λ2 + µ2) cos2 + sin2 +2λµ cos2 sin2 2 2 2 − 2 = λ2 + µ2 +2λµ cos(θ) . Therefore the maximum value is V = λ2 + µ2 +2λµ cos(θ). Corollary 6.3. Let λ, µ be real numbers,p and ε> 0. Set λ + µ A = max 1, 2π | | | | . ε Let θ be a real number. Set s = λ2 + µ2 +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 ≥ ≤ ≤ − ∈ k(θ +2πt) (n k)(θ +2πt) λ cos + µ cos − √s ε . n n ≥ − Proof. For each t Z, we set ∈ k(θ +2πt) (n k)(θ +2πt) α = λ cos + µ cos − t n n kθ 2πkt (n k)θ 2πkt = λ cos + + µ cos − . n n n − n Let r be an integer such that rk 1 mod n. By Lemma 6.2, we can pick x R such that ≡ ∈ kθ (n k)θ √s = λ cos + x + µ cos − x . (6.1) n n − nx Set q = 2π . Set t = rq, hence kt = krq q mod n. The following equalities hold ⌊ ⌋ ≡ kθ 2πq (n k)θ 2πq α = λ cos + + µ cos − . (6.2) t n n n − n The following inequalities hold 2πq 2π nx 2π ε 0 x = x . (6.3) ≤ − n − n 2π ≤ n ≤ λ + µ j k | | | | 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 ε α √s ( λ + µ ) = ε . | t − |≤ | | | | λ + µ | | | | It follows that α √s ε. t ≥ − Lemma 6.4. Let γ be a real number, ℓ =0 be an integer, and d> 1 be an integer. d 1 6 tℓ − Assume that d ∤ ℓ. Then t=0 cos γ + d 2π =0. P JULIAROBINSON’SNUMBERS 17 i ℓ 2π d Proof. Set x = e d . Note that x = 1. The following equalities hold: d 1 d 1 − t − tℓ cos γ + 2π = exp iγ + i 2π d ℜ d t=0 t=0 X X d 1 − = eiγ xt ℜ t=0 ! X xd 1 = eiγ − ℜ x 1 − =0 Notation 6.5. For a R and b R∗, we denote Rb(a) the only real number b b ∈ ∈ c [ , [ such that a c bZ. Note that for all t R∗ we have R (ta)= tR (a). ∈ − 2 2 − ∈ ∈ tb b a Rb(a) We will use the particular case R2π( b 2π)= b 2π Note that if a,b are integers, than Rb(a) is an integer. Also observe that if a,a′,b are integers, then R (aa′) R (a) R (a′) . | b | ≤ | b | × | b | Lemma 6.6. Let λ,µ,θ1,θ2,ρ be real numbers, and u be an integer. If R2π(uρ) = π 6 uR2π(ρ) and R2π(uρ) < 2 , then there is r max(1, u 1) such that λ cos(θ1 + rρ)+ µ cos(θ | rρ) |0. | |≤ | |− 2 − ≥ Proof. We cannot have u = 0 or u = 1. Assume that u = 1. It follows, from R (uρ) = uR (ρ), that ρ = π. Hence λ cos(θ +ρ)+µ cos(θ − ρ)= (λ cos(θ )+ 2π 6 2π 1 2 − − 1 µ cos(θ2)). Hence r = 0, or r = 1 satisfies the required condition. We now assume that u 2. We can assume that R (ρ)= ρ, that is ρ [ π, π[. ≥ 2π ∈ − Set λ′ = λ cos(θ1)+ µ cos(θ2), and µ′ = µ sin(θ2) λ sin(θ1). Given r Z, denote f(r)= λ cos(θ + rρ)+ µ cos(θ rρ), hence the following− equalities hold∈ 1 2 − f(r)= λ(cos(θ ) cos(rρ) sin(θ ) sin(rρ)) + µ(cos(θ ) cos(rρ) + sin(θ ) sin(rρ)) 1 − 1 2 2 = λ′ cos(rρ)+ µ′ sin(rρ). If λ′ 0, then f(0) = λ′ 0, as required. Assume that λ′ < 0. We first want to ≥ ≥ π π find s such that R2π(sρ)= sρ and sρ [ 2 , 2 [. π π 6∈ − If ρ [ 2 , 2 [, then s = 1 satisfies the desired conditions. Assume that ρ π π 6∈ − π π∈ [ 2 , 2 [. As uρ [ π, π[, we can take the smallest s 1 such that sρ [ 2 , 2 [. From− the minimality6∈ − we have (s 1)ρ [ π , π [, however≥ ρ [ π ,6∈π [,− hence − ∈ − 2 2 ∈ − 2 2 sρ [ π, π[. Thus R2π(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. − − ≥ ≤− Lemma 6.7. Let a Z and b > 0 an integer, then there exists u Z 0 such ∈ ∈ \{ } that u √b and R (u) √b. | |≤ | 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. (| |≤ 18 P. GILLIBERT AND G. RANIERI √ √b The set of solutions is a parallelogram of height 2 b, and base 2 b . Hence the surface is 4. Therefore, by Minkowski’s Theorem (see for instance [8, Lemma, p. 137]), there is a solution (u, v) Z2 (0, 0) . ∈ \{ } Thus we have u √b and R (ua) ua + vb √b. Moreover if u = 0, | | ≤ | b | ≤ | | ≤ then v = 0 and b vb = ua + vb √b; a contradiction. Therefore u = 0 as required.6 ≤ | | | | ≤ 6 Lemma 6.8. 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. 6 6 − (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(Rn(rℓ)) = Rn(urℓ), and Rn(urℓ) √n. (3) There are6 non-zero integers r, u such≤ that u √n, rℓ 1 mod n, u(R (rk)) = R (urk), and R (urk) √n. | | ≤ ≡ n 6 n 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 < √n. If gcd(ℓ,n) > 1 ≤ gcd(k,n) 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 , so u = u √n and R (urk)= R ( rk n)= ≡ gcd(k,n) | | ≤ n n gcd(k,n) 0. As rℓ 1 mod n and k Theorem 6.9. Let λ1,µ1, λ2,µ2 be real numbers and ε> 0. Set λ + λ + µ + µ 2 A = max 17, 2π | 1| | 2| | 1| | 2| . ε ! Let θ1,θ2 be real numbers. Set 2 2 2 2 s = min(λ1 + µ1 +2λ1µ1 cos(θ1), λ2 + µ2 +2λ2µ2 cos(θ2)) . JULIAROBINSON’SNUMBERS 19 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 ≤ − 6 6 − ∈ k(θ +2πt) (n k)(θ +2πt) ℓ(θ +2πt) (n ℓ)(θ +2πt) λ cos 1 +µ cos − 1 +λ cos 2 +µ cos − 2 > √s ε . 1 n 1 n 2 n 2 n − Proof. Let x be as in Lemma 6.2, that is kθ (n k)θ λ cos 1 + x + µ cos − 1 x = λ2 + µ2 +2λ µ cos(θ) √s . 1 n 1 n − 1 1 1 1 ≥ q (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≤ ≤ − 6 6 − ∈ k(θ +2πj) (n k)(θ +2πj) α = λ cos 1 + µ cos − 1 j 1 n 1 n ℓ(θ +2πj) (n ℓ)(θ +2πj) β = λ cos 2 + µ cos − 2 . j 2 n 2 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 . The following≤ equalities hold: ∈ ≡ ⌊ 2πd ⌋ k(θ +2πrq) (n k)(θ +2πrq) α = λ cos 1 + µ cos − 1 rq 1 n 1 n kθ 2πkrq (n k)θ 2πkrq = λ cos 1 + + µ cos − 1 1 n n 1 n − n kθ 2πdq (n k)θ 2πdq = λ cos 1 + + µ cos − 1 . 1 n n 1 n − n Note that the following inequalities hold: 2πdq 2πd nx 2πd 2π 2π ε 0 x = x . ≤ − n − n 2πd ≤ n ≤ √n ≤ √A ≤ λ1 + µ1 j k | | | | However, cos(a) cos(b) a b for all a,b R. It follows that | − | ≤ | − | ∈ kθ (n k)θ ε α λ cos 1 + x +µ cos − 1 x ( λ + µ ) = √s ε . rq ≥ 1 n 1 n − − | 1| | 1| λ + µ − | 1| | 1| Set r = n and r = k . So kr = nr . For all t Z the following equalities hold: 1 d 2 d 1 2 ∈ k(θ + 2(rq + tr )π) (n k)(θ + 2(rq + tr )π) α = λ cos 1 1 + µ cos − 1 1 rq+tr1 1 n 1 n k(θ +2πpq) 2πktr (n k)(θ +2πrq) 2πktr = λ cos 1 + 1 + µ cos − 1 1 1 n n 1 n − n k(θ +2πrq) (n k)(θ +2prπ) = λ cos 1 +2πtr + µ cos − 1 2πtr 1 n 2 1 n − 2 = αrq. 20 P. GILLIBERT AND G. RANIERI ℓ(θ2+2πrq) (n ℓ)(θ2+2πrq) Set γ1 = n and γ2 = − n . For all t Z the following equalities hold: ∈ ℓ(θ +2πrq) 2πr ℓ (n ℓ)(θ +2πrq) 2πr ℓ β = λ cos 2 + t 1 + µ cos − 2 t 1 rq+tr1 2 n n 2 n − n ℓ ℓ = λ cos γ + t 2π + µ cos γ t 2π . 2 1 d 2 2 − d d 1 − Note that gcd(d, ℓ) = 1. Then, we can deduce from Lemma 6.4 that t=0 βpr+tr1 = 0, hence there is 1 t d 1 such that βpr+tr 0. Therefore ≤ ≤ − 1 ≥ P α + β α √s ε . rq+tr1 rq+tr1 ≥ rq ≥ − Now assume that Lemma 6.8(2) holds. Let r, u be non-zero integers such that nx u √n, rk 1 mod n, uRn(rℓ) = Rn(urℓ), and Rn(urℓ) √n. Set q = 2π . For| |≤ each t Z≡we have 6 ≤ ⌊ ⌋ ∈ kθ 2πq 2πt (n k)(θ ) 2πq 2πt α = λ cos 1 + + + µ cos − 1 r(q+t) 1 n n n 1 n − n − n If t u 1 then the following inequalities hold | | ≤ | |− 2πq 2πt 2π 2π t 2π u 2π ε + x + | | | | . | n n − |≤ n n ≤ n ≤ √A ≤ λ1 + µ1 | | | | As before it follows that kθ (n k)θ α λ cos 1 + x + µ cos − 1 x ε . r(q+t) ≥ 1 n 1 n − − We obtain the following inequality α √s ε , for all t u 1. (6.5) r(q+t) ≥ − | | ≤ | |− Note that ℓ(θ +2πrq) 2πrℓ (n ℓ)(θ +2πrq) 2πrℓ β = λ cos 2 + t + µ cos − 2 t . r(q+t) 2 n n 2 n − n ℓ(θ2+2πrq) (n ℓ)(θ2+2πrq) 2πrℓ Set γ1 = n , γ2 = − n , and ρ = n . Therefore the following equality holds. β = λ cos(γ + tρ)+ µ cos(γ tρ) . (6.6) r(q+t) 2 1 2 2 − As uR (rℓ) = R (urℓ), we have uR ( rℓ 2π) = R (u rℓ 2π), therefore uR (ρ) = n 6 n 2π n 6 2π n 2π 6 R (uρ). As R (urℓ) √n we have R (uρ)= R (u rℓ 2π)= Rn(urℓ) 2π 2π < 2π n ≤ 2π 2π n n ≤ √n π . Hence by Lemma 6.6 there is t < u such that 2 | | | | β = λ cos(θ + tρ)+ µ cos(θ tρ) 0 . r(q+t) 2 1 2 2 − ≥ Therefore by (6.5) we have α + β √s ε. r(q+t) r(q+t) ≥ − With a similar and simpler proof, we obtain the following variation of Theo- rem 6.9. Theorem 6.10. Let λ1,µ1, λ2 be real numbers, and ε> 0. Set λ + µ A = max 1, 2 2π | 1| | 1| . ε Let θ1,θ2 be in R. Set 2 2 s = λ1 + µ1 +2λ1µ1 cos(θ1) . JULIAROBINSON’SNUMBERS 21 n Then for all n A even, and all 1 k,ℓ n 1 with gcd(k, 2 )=1 there is t Z such that ≥ ≤ ≤ − ∈ k(θ +2πt) (n k)(θ +2πt) θ +2πt λ cos 1 + µ cos − 1 + λ cos 2 > √s ε . 1 n 1 n 2 2 − 7. Determination of the Julia Robinson Number a π iθ Let 0 L = Lrk and K = Krk . k N k N [∈ [∈ We say that K is the totally real field associated to (a,b,r). Denote by N the smallest common multiple of v2(b) 1, and all vp(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 b a − u2 (respectively v−2 ) are square-free. Note that the conditions in the following notation are similar to (2)-(4) in Lemma 3.16 2 Notation 7.1. Let 0 k N 1. Denote by Sk the set of all pairs (x, y) Q that satisfies the following≤ conditions:≤ − ∈ k+1 (1) For all odd prime number p such that p divides b we have vp(x) N vp(b) N k ≥ and vp(y) N− vp(b); ≥ k+1 N k (2) We have v2(x) 1+ N (v2(b) 1) and v2(y) 1+ N− (v2(b) 1); (3) u(x + y) and v(≥x y) are integers.− ≥ − − We denote by U = S1 ...SN 1. ∪ − Notation 7.2. As before we consider the following map N : R2 R → a (x, y) N(x, y)= x2 + y2 +2xy . 7→ b Set s = inf N(x, y) 0 k N 1 and (x, y) S . The goal of this section | ≤ ≤ − ∈ k is to prove that the Julia Robinson Number of OK is s + s. p ⌈ ⌉ 2 Note that (x, y) Sk if and only if (y, x) SN k 1. We say that (θ,K,s,U,N : R R) are the objects associated∈ to (a,b,r). ∈ − − → Lemma 7.3. Let 1 k k k+1 Lemma 7.4. Let 0 k N 1. Let x, y Q.Then for all q ] N , N [ Q, we have (x, y) S if and≤ only≤ if the− following three∈ conditions hold:∈ ∩ ∈ k (1) For all odd prime number p such that p divides b we have vp(x) qvp(b) and v (y) (1 q)v (b); ≥ p ≥ − p (2) We have v2(x) 1+ q(v2(b) 1) and v2(y) 1+(1 q)(v2(b) 1); (3) u(x + y) and v(≥x y) are integers.− ≥ − − − Proof. Let q ] k , k+1 [ Q. Let p be an odd prime number dividing b and let (x, y) ∈ N N ∩ be in Sk. kvp(b) (k+1)vp(b) Assume that there is an integer ℓ such that N <ℓ< N , then k < N N ℓ < k+1. However, vp(b) N, hence ℓ is an integer; a contradiction. There- vp(b) | vp(b) kvp(b) (k+1)vp(b) kvp (b) (k+1)vp(b) fore there is no integer in ] N , N [. Note that qvp(b) ] N , N [. ∈ k+1 As vp(x) is an integer it follows that vp(x) qvp(b) if and only if vp(x) N vp(b). ≥ N k ≥ Similarly vp(y) (1 q)vp(b) if and only if vp(y) N− vp(b). 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 λ0, λ1,...,λn 1 be in Q. The following≥ statement are equivalent: − n 1 ℓθ − O (1) λ0 + ℓ=1 λℓ cos( n ) is in K . (2) λ0 Z, and for all 1 ℓ n 1 we have (λℓ, λn ℓ) Sk, where 0 k ∈ ≤ ≤ − − ∈ ≤ ≤ N 1Psuch that ℓ ] k , k+1 [. − n ∈ N N Proof. Assume (1). Observe that N is the product of v2(b) 1 and all vp(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 = k . ≤ ≤ − ≤ ≤ n 6 N By Lemma 3.16 we have λ0 Z, andfor all1 ℓ n 1 the following statement holds. ∈ ≤ ≤ − (i) For all odd prime number p dividing b, v (λ ) ℓvp(b) ; p ℓ ≥ n ℓ(v (b) 1) (ii) v (λ ) 1+ 2 − ; l m 2 ℓ ≥ n (iii) u(λℓ + λn ℓ)l and v(λℓ +m λn ℓ) are integers. − − ℓ k k+1 ℓ k Let 1 ℓ n 1. Let0 k N 1 be such that n [ N , N [. As n = N , we have ℓ k≤ k+1≤ − ≤ ≤ − ∈ 6 n ] N , N [. Hence it follows from Lemma 7.4 that (λℓ, λn ℓ) Sk. Therefore (2) holds.∈ − ∈ Assume (2). Let 1 ℓ n 1. Let 0 k N 1 such that ℓ ] k , k+1 [. ≤ ≤ − ≤ ≤ − n ∈ N N As (λℓ, λn ℓ) Sk it follows from Lemma 7.4 that (i), (ii), and (iii) holds. By − ∈ n 1 ℓθ − Lemma 3.16, we have that λ0 + ℓ=1 λℓ cos( n ) is an algebraic integer, moreover it belongs to K, therefore (1) holds. P Lemma 7.6. Let M be a discrete Z-submodule of Q2. Then JULIAROBINSON’SNUMBERS 23 (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 X2 + 2 a 2 Y +2XY b = C. Since 0 Then the set B given by EC2 minus the interior part of EC1 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 ∈ 6 ≤ (x′,y′) M (0, 0) , because if (x′,y′) B, then N(x′,y′) C N(x, y). ∈ \{ } 6∈ ≥ 2 ≥ Lemma 7.7. Let (λ, µ) be in S . Then there are infinitely many q Q [0, 1] k ∈ ∩ such that λ cos(qθ)+ µ cos((1 q)θ) OK . Moreover we can choose q such that the denominator of q is any large− enough∈ divisor of a power of r. e 2k+1 t 1 Proof. Take coprime integers t and n = r such that 2N n < 2N , Note that | − | t there are infinitely many t,n that satisfy this condition. In particular n is in the k k+1 n t N k 1 N k interval ] N , N [. It follows that n− ] −N− , N− [, moreover (µ, λ) SN k 1. ∈ t n t ∈ − − Therefore by Lemma 7.5 we have λ cos( θ)+ µ cos(( − )θ) O . n n ∈ K Corollary 7.8. Denote by t the product of all odd primes (without counting mul- tiplicity) dividing b. The following statements hold. (1) s2 is an integer, and 8t s2. | (2) s 2√2. (3) s ≥ 4t. 2≤ b a 2 b+a 2 (4) s 8t v−2 and s 8t u2 . In particular if b a is a square then s =8t. (5) 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 k N 1 and ≤ ≤ − (λ, µ) S (0, 0) such that s = N(λ, µ). So s2 = λ2 + µ2 +2λµ a . ∈ k \{ } b Let q = 1 be as in Lemma 7.7, that is λ cos(qθ)+ µ cos((1 q)θ) O . So 6 2 p − ∈ K T ((λ cos(qθ)+ µ cos((1 q)θ))2)= 1 N(λ, µ)= 1 s2 is an integer. − 2 2 Let p be an odd prime number dividing b. It follows from the definition of Sk that v (b) v (b) v (λ) (k + 1) p 1 and v (µ) (N k) p 1 . (7.1) p ≥ N ≥ p ≥ − N ≥ Therefore the following statement holds vp(λµ)= vp(λ)+ vp(µ) (7.2) v (b) v (b) (k + 1) p + (N k) p (7.3) ≥ N − N v (b) (N + 1) p (7.4) ≥ N v (b)+1 . (7.5) ≥ p 24 P. GILLIBERT AND G. RANIERI Hence v (2λµ a ) 1, therefore p b ≥ a a v (s2)= v (λ2 + µ2 +2λµ ) min(v (λ2), v (µ2), v (2λµ )) 1= v (8t) . (7.6) p p b ≥ p p p b ≥ p Similarly the following statement hold v (b) 1 v (b) 1 v (λ) 1 + (k + 1) 2 − 2 and v (µ) 1 + (N k) 2 − 2 . 2 ≥ N ≥ 2 ≥ − N ≥ (7.7) Therefore v (b) 1 v (b) 1 v (λµ) 1 + (k + 1) 2 − + 1 + (N k) 2 − (7.8) 2 ≥ N − N v (b) 1 2 + (N + 1) 2 − (7.9) ≥ N 2+ v (b) . (7.10) ≥ 2 So v (2λµ a ) 3, therefore 2 b ≥ a v (s2)= v (λ2 + µ2 +2λµ ) 3= v (8t) . (7.11) 2 2 b ≥ 2 It follows from (7.6) and (7.11) that 8t s2, so (1) holds. In particular 8 s2, so | ≤ s 2√2, that is (2) holds. ≥ vp(b) Let p be an odd prime number dividing b. Note that vp(4t)=1= N and v2(b) 1 l m v2(4t)=2= 1+ N− , and 4tu and 4tv are integers (as 4t is an integer) thus (4t, 0) S0. Thereforel s m N(4t, 0)=4t, so (3) holds. ∈ N 1 ≤ b 1 v2(b) 1 Set k = − . Set x =2r , . Note that v (x)=1+ − . As v (b) 1 is 2 p2 2 2 ⌈ 2 ⌉ 2 − relatively prime to b, it follows that v (b) 1 is odd, thus v (x)=1+ v2(b) . Note 2 2 2 that N v (b) 1 thus the following inequality− holds ≥ 2 − N +1 v (b) 1 v (b) 1 v (b) 1 1 (v (b) 1)+1 = 2 − + 2 − +1 2 − + +1 = v (x) . (7.12) 2N 2 − 2 2N ≤ 2 2 2 vp(b) vp(b)+1 Let p be an odd prime number dividing b. Then vp(x)= 2 = 2 , as vp(b) is odd. As N v (b), the following inequality holds ⌈ ⌉ ≥ p N +1 v (b) v (b) v (b) 1 v (b)= p + p p + = v (x) . (7.13) 2N p 2 2N ≤ 2 2 p k+1 N k N+1 As v is relatively prime to b and N = N− = 2N , it follows from (7.12) and x x (7.13) that ( v , v ) satisfies the condition (1) and (2) of Notation 7.1. Moreover x x − x x x x u( v v ) = 0 and v( v + v )=2x are integers, hence ( v , v ) Sk. Therefore the following− inequality holds − ∈ x x x2 a s N , =2 1 . (7.14) ≤ v − v v2 − b v2(b) However, as v2(x)=1+ 2 and for all odd prime p dividing b we have vp(x) = vp(b)+1 2 2 , and for all other prime p we have vp(x) = 0, it follows that x = 4tb. It 4tb b a b a follows from (7.14) that s 2 v2 −b =8t v−2 . ≤ x x b+a A similar proof, considering ( u , u ) Sk, yelds s 8t u2 . 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 (λ) 1, and v (µ) 1. − 2 ≥ 2 ≥ JULIAROBINSON’SNUMBERS 25 Therefore λ and µ are integers. Moreover it follows from (7.1) and (7.7) that 4t divides λ and µ. 2 2 2 a 2 2 If λµ 0 then s = N(λ, µ)= λ + µ +2λµ b max(λ ,µ ) however 4t divides both λ and≥ µ, and (λ, µ) = (0, 0), thus s 4t. ≥ Assume that λµ < 0. The6 following statement≥ holds a a N(λ, µ)= λ2 + µ2 +2λµ = (λ + µ)2 2λµ 1 (λ + µ)2 . b − − b ≥ However 4t λ + µ, so if λ + µ = 0 we have s = N(λ, µ) 4t. Now assume| that λ + µ = 0,6 that is λ = µ. The following≥ holds − p a a N(λ, µ)= λ2 + λ2 2λ2 =2λ2 1 . − b − b 2 It follows from (7.5) that vp(λ ) = vp(λµ) vp(b)+1= vp(4bt), similarly (7.10) 2 ≥ 2 implies v2(λ ) = v2(λµ) = 2+ v2(b) = v2(4tb). Therefore 4tb divides λ , so a 2 2 N(λ, µ) 8tb 1 b =8t(b a) 16t = (4t) , and so s = N(λ, µ) 4t. However,≥ from− (2) we also− have≥s 4t, thus s =4t. Therefore (5) holds.≥ ≤ p Corollary 7.9. There are infinitely many β OK such that all conjugates of β are in [0, s + s]. In particular the Julia Robinson∈ Number of O is at most s + s. ⌈ ⌉ K ⌈ ⌉ Proof. By Lemma 7.3, for every k between 0 and N 1, S is a discrete Z-module. − k Then, by Lemma 7.6, there exists k and (λ, µ) Sk such that N(λ, µ) = s. By Lemma 7.7, there exist infinitely many q [0, 1]∈ Q such that γ = λ cos(qθ)+ ∈ ∩ q µ cos((1 q)θ) OK . From Lemma 3.15 we see that the conjugates of γq are the λ cos(qθ −+2πqt)+∈ µ cos((1 q)θ 2πqt), with t Z. Hence by Lemma 6.2, every − − ∈ conjugate γq′ of γ satisfies 2 2 γ′ λ + µ +2λµ cos(θ)= N(λ, µ)= s. | q|≤ Hence every conjugate ofp s + γq is positive, andp it is smaller than s + s. This proves that the Julia Robinson⌈ ⌉ Number of O is at most s + s. ⌈ ⌉ K ⌈ ⌉ Theorem 7.10. The Julia Robinson Number of OK is s + s. Moreover OK has the Julia Robinson Property. ⌈ ⌉ Proof. From Corollary 7.9 we already know that the Julia Robinson Number of OK is at most s + s. Moreover there are infinitely many β OK such that all conjugates of β⌈are⌉ in [0, s + s], thus to prove that O has∈ the Julia Robinson ⌈ ⌉ K Property, we only have to prove that the Julia Robinson Number of OK is s + s. 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 Corollary 7.11. The first order theory of OK 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 ∈ \{ } ≥ 2 kθ (n k)θ r, let 1 k n 1. Then λ cos + µ cos − Q if and only if 2k = n. ≤ ≤ − n n ∈ kθ (n k)θ θ 2 1+cos θ b+a Proof. Set β = λ cos n + µ cos −n . Note that cos 2 = 2 = 2b belongs to Q. Thus if n =2k, then the following statement holds θ 2 β2 = (λ + µ)cos Q . 2 ∈ Assume that n =2k, hence k = n k. We can assume that k 8. Example of Julia Robinson’s Number In this section we construct particular a and b such that we can explicitly com- pute 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. ≥ ≤ ≡ 28 P. GILLIBERT AND G. RANIERI The following theorem was proved by Prachar [19] (see also Hooley [7]). Theorem 8.2. Let k 1 be an integer and i relatively prime to k. Then we have the following equivalence≥ 6 p2 Q(x; i, k) x x . ∼ →∞ π2k p2 1 p k Y| − 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.≥ ≤ ≤ ≡ − 6 p2 3 Proof. Set l = π2k p k p2 1 . Note that kl 5 . From Theorem 8.2 we have | − ≥ Q Q(x;1, k) Q(x; 1, k) lim = l and lim − = l . x x →∞ x →∞ x 3 Set 3η = ε(l 5k ). Pick A 0 such that for all x A and i = 1 we have Q(x;i,k) − ≥ ≥ ± x l η. | − |≤A+1 Set C = 1 ε . Let b C, thus b εb 1 A. Therefore we have − ≥ − − ≥ Q(b εb; i, k) − l η . | b εb − |≤ − It follows that Q(b εb; i, k) (b εb)l (b εb)η . (8.1) | − − − |≤ − 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; i, k) Q(b εb; i, k)+ εbl 2b εbη 3bη . | − − |≤ − ≤ 3 3 Q(b; i, k) Q(b εb; i, k) εbl 3bη = εbl εb(l )= εb . (8.4) − − ≥ − − − 5k 5k Similarly from (8.2) and (8.3) we deduce 3 Q(b + εb; i, k) Q(b; i, k) εbl 3bη = εb . (8.5) − ≥ − 5k εb 1 Consider the set Xi = b + ijk + i j integer and 0 j k− . Hence X [b,b + εb], moreover X{ is the set of| all integers in [b,b≤ + εb≤] congruent} to 1 1 ⊆ 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(b εb; 1, k) εb 3 square-free integers congruent− to 1 modulo k−. However− all− − ≥ 5k − integer in [b εb,b] congruent to 1 modulo k are in X 1. Therefore there are 3− − − εb 1 at least εb 5k square-free integers in X 1. However X 1 has k− elements, − − 3 k 3 therefore the proportion of square-free elements in X 1 is at least εb 5k εb = 5 . − 3 Similarly, by (8.5), the proportion of square-free elements in X1 is at least 5 . Note that X1 X 1, r 2b r is a bijection. Therefore there is a square-free r X such that→ 2b −r is square-free.7→ − Set a = r b, thus b + a = r is square-free, ∈ 1 − − JULIAROBINSON’SNUMBERS 29 and b a =2b r is square-free. Moreover as r X1, we have 0 < r b εb. Also note that− a r− 1 mod k. ∈ − ≤ ≡ ≡ 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 Robin- son’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. Lemma 8.4. Let (a1,b1, r1) be good; let (a2,b2, r2) be good. Denote by K1 (resp., K2) the totally real field associated to (a1,b1, r1) (resp., (a2,b2, r2)). Assume that r , r are even. If (b + a )(b + a )b b is not a square then K = K . 1 2 1 1 2 2 1 2 1 6 2 Proof. Let (θi,Ki,Ui,si,Ni) be the objects associated to (ai,bi, ri), for i =1, 2. As- O O O O sume that K1 = K2, then K1 = K2 , thus JR( K1 ) = JR( K2 ), so by Lemma 7.10 we have s + s = s + s , and so s = s . ⌈ 1⌉ 1 ⌈ 2⌉ 2 1 2 n1 Claim 1. There are n1 2 a power of two, n2 2 an integer, 1 k1 < 2 , 1 k n 1 integers, ≥λ ,µ , λ ,µ in Q such that≥ ≤ ≤ 2 ≤ 2 − 1 1 2 2 k1θ1 (n1 k1)θ1 k2θ2 (n2 k2)θ2 λ1 cos + µ1 cos − = λ2 cos + µ2 cos − =0 . (8.6) n1 n1 n2 n2 6 Proof of Claim. As the infinum in the definition of s1 (cf Notation 7.2) is a mini- mum (by Lemma 7.6), there is (λ ,µ ) U such that s2 = N (λ ,µ ). It follows 1 1 ∈ 1 1 1 1 1 from Lemma 7.7 that there are an odd integer k1 and a power of two n1 4 such k1θ1 (n1 k1)θ1 ≥ that 1 k1 < n1 and β = λ1 cos + µ1 cos − OK . Up to permuting n1 n1 1 ≤ n1 ∈ λ1 and µ1, we can assume that k1 < 2 . As β OK = OK , there is m 1, and η0, η1 ...ηm 1 in Q such that ∈ 1 2 ≥ − n2 k1θ1 (n k1)θ1 ℓθ2 λ1 cos + µ1 cos − = η0 + ηℓ cos . n1 n1 n2 Xℓ=1 Therefore applying T to both side of the equality, and multiplying by two, we obtain the following equality by Lemma 5.1. n2 ⌊ 2 ⌋ 2 N1(λ1,µ1)=2η0 + N2(ηℓ, ηn2 ℓ) − Xℓ=0 n2 However for each 1 ℓ we have either ηℓ =0= ηn ℓ or N2(ηℓ, ηn ℓ) ≤ ≤ 2 2− 2− ≥ s2 = s2 = N (λ ,µ ). As β Q, it follows that there is 1 p n 1 such that 2 1 1 1 1 6∈ ≤ ≤ 2 − ηp = 0. Therefore for all 0 ℓ n2 1, if ℓ = p and ℓ = n2 p we have λℓ = 0, thus6 we obtain ≤ ≤ − 6 6 − k1θ1 (n1 k1)θ1 pθ2 (n2 p)θ2 λ1 cos + µ1 cos − = ηp cos + ηn2 p cos − . n1 n1 n2 − n2 Claim 1. Claim 2. There are x, y in Q such that θ θ x cos 1 = y cos 2 =0 . 2 2 6 30 P. GILLIBERT AND G. RANIERI Proof of Claim. Let n1,n2, k1, k2, λ1,µ1, λ2,µ2 as in Claim 1 with n1 minimal. We n2 can assume that k2 2 . From Lemma 7.12 and (8.6), we have 2k1 = n1 if and only if 2k = n . Assume≤ that n 4. Squaring both side we obtain from Lemma 3.8 2 2 1 ≥ 1 2 2 a1 2 2 2k1θ1 2 a1 (n1 2k1)θ1 (λ1 + µ1 +2λ1µ1 ) + (λ1 + µ1)cos + 2(λ1µ1 + µ1 )cos − 2 b1 n1 b1 n1 1 2 2 a2 2 2 2k2θ2 2 a2 (n2 2k2)θ2 = (λ2 +µ2 +2λ2µ2 )+(λ2 +µ2)cos +2(λ2µ2 +µ2 )cos − 2 b2 n2 b2 n2 (8.7) From Lemma 3.14, taking the trace we obtain 1 2 2 a1 1 2 2 a2 (λ1 + µ1 +2λ1µ1 )= (λ2 + µ2 +2λ2µ2 ) . 2 b1 2 b2 Hence 2 2 2k1θ1 2 a1 (n1 2k1)θ1 (λ1 + µ1)cos + 2(λ1µ1 + µ1 )cos − n1 b1 n1 2 2 2k2θ2 2 a2 (n2 2k2)θ2 = (λ2 + µ2)cos + 2(λ2µ2 + µ2 )cos − (8.8) n2 b2 n2 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 n1. Therefore n1 = 2 and k1 = 1, so 2k1 = n1, thus 2k2 = n2, the result follows for x = λ1 + µ1, and y = λ2 + µ2. Claim 2. As cos θ1 = b1+a1 and cos θ2 = b2+a2 , it follows from Claim 2 that (b + 2 2b1 2 2b2 1 a1)(b2 +a2)b1b2 qis a square; a contradiction.q Therefore K1 and K2 are distincts. The following lemma is an immediate consequence of Hensel’s Lemma (see for instance [16, 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≥pn. 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 mod4, and 2t is a square modulo p. ≡ ≡ p Proof. Given p, q we denote by q the Legendre symbol. Set q−1 2 ε = ( 1) 2 . − q q t q primeY| 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 mod8. 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. [6]), and in both case we have p 3 mod 4. 2 ≡ Moreover by Gauss’s quadratic reciprocity we have p = ε. JULIAROBINSON’SNUMBERS 31 The following equalities hold 2t 2 q = , by multiplicativity. p p p q t q primeY| 2 p−1 q−1 p = ( 1) 2 2 , by Gauss’s quadratic reciprocity. p − q q t q primeY| 2 q−1 2 p 1 = ( 1) 2 , as − is odd and p 2 mod t. p − q 2 ≡ q t q primeY| = ε2 =1 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 ,n ) such that, for all i, m 2 and n are both i i i − i relatively prime to 2t and, for all i = j, (2mi tni 1)(2mj tnj 1) is not a square. 6 − − Proof. Assume that we have (m ,n ),..., (m ,n ) (for 1 i k), such that, for 1 1 k k ≤ ≤ every i = j, (2mi (2t)ni 1)(2mj (2t)nj 1) is not a square. It follows6 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 2mi tni 1. In particular p is relatively prime to 2t. ≤ ≤ − − − p 1 u+1 p 1 u+1 Pick u 1 odd such that (2t) 2 < p , hence (2t) 2 1 mod p . Note that Lemma≥ 8.5 implies that 2t is a square modulo pu+1. Let x6≡be such that 2t x2 − − u+1 p 1 2 p 1 p 1 ≡u+1 mod p , and so x is prime to p. Therefore x − (x ) 2 (2t) 2 mod p , p 1 u+1 ≡ ≡ u+1 so x − 1 mod p . However the multiplicative group (Z/p Z)∗ is cyclic of 6≡ u (p 1)pu u+1 order (p 1)p , hence x − 1 mod p . − u p ≡ u+1 u+1 Note that (p + 1) 1 mod p . As the subgroup of (Z/p Z)∗ generated p 1 ≡ u by x − is not trivial and of order dividing p , it follows that this subgroup contains u (p 1)c u u+1 (p 1)pu p +1. Therefore there is c 1 such that x − p +1 mod p . As x − u+1 (p 1)(≥ c+pu) (p 1)c u≡ u+1 ≡ 1 mod p we have x − x − p + 1 mod p . Therefore as p is odd, we can assume that c is odd.≡ ≡ p 1 p−1 v v 2 c (p 1)c u u+1 Set v = −2 c. Note that 2 t = (2t) x − p + 1 mod p . As u ≡ ≡ (p 1)p is even and relatively prime to t we can pick w0 and w1 odds such that w0 2 is− prime to t, w is prime to t, w v mod (p 1)pu, and w v mod (p 1)p−u. 1 0 ≡ − 1 ≡ − Therefore 2w0 tw1 2vtv pu + 1 mod pu+1, so pu divides 2w0 tw1 1 and pu+1 ≡ ≡ − does not divides 2w0 tw1 1. − Note that v (2w0 tw1 1) = u is odd, moreover p does not appear in any prime p − decomposition of the 2mi tni 1. Therefore for all 1 i k, (2mi tni 1)(2w0 tw1 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 OK has Julia Robinson’s num- ber 2√2t +2√2t. 32 P. GILLIBERT AND G. RANIERI (2) There are infinitely many fields K such that OK has Julia Robinson’s num- ber 8t. Proof. By Lemma 8.4 and Lemma 8.7, we can find infinitely many good (a,b,r), m 1 n such that b =2 − t , 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 7.2. Therefore by Corollary 7.8(4) we have s2 = 8t, thus s = 2√2t. Therefore by Theorem 7.10 the Julia Robinson’s number of OK is 2√2t +2√2t. So (1) holds. Set r = k = 2t. Let C as in Lemma 8.3, for ε = 1 . Consider n 1 an odd 2 ≥ integer such that n is relatively prime to t and 2n+1t max(C, 4t). Set b =2n+1t. Note that there are infinitely many possibilities for such≥ b. By Lemma 8.3 there is 1 1 a 2 b such that a 1 mod k, and b a, b + a are square-free. As k b, it follows≤ ≤ that a is relatively≡ prime to b. − | Consider K the field constructed in Section 7. Let s as in Notation 7.2. 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