On a conjecture of Rudin on squares in Arithmetic Progressions
aa r X i v : . [ m a t h . N T ] J a n ON A CONJECTURE OF RUDINON SQUARES IN ARITHMETIC PROGRESSIONS
ENRIQUE GONZ ´ALEZ-JIM´ENEZ AND XAVIER XARLES
Abstract.
Let Q ( N ; q, a ) denotes the number of squares in the arithmeticprogression qn + a , for n = 0 , , . . . , N −
1, and let Q ( N ) be the maximumof Q ( N ; q, a ) over all non-trivial arithmetic progressions qn + a . Rudin’s con-jecture asserts that Q ( N ) = O ( √ N ), and in its stronger form that Q ( N ) = Q ( N ; 24 ,
1) if N ≥
6. We prove the conjecture above for 6 ≤ N ≤
52. Weeven prove that the arithmetic progression 24 n + 1 is the only one, up toequivalence, that contains Q ( N ) squares for the values of N such that Q ( N )increases, for 7 ≤ N ≤
52 (hence, for N = 8 , , , , , ,
41 and 52). Thisallow us to assert, what we have called Super–Strong Rudin’s Conjecture: letbe N = GP k + 1 ≥ k , where GP k is the k -th generalizedpentagonal number, then Q ( N ) = Q ( N ; q, a ) with gcd( q, a ) squarefree and q > q, a ) = (24 , Introduction
A well–known result by Fermat states that no four squares in arithmetic pro-gression over Z exist. This result may be reformulated in the following form: infour consecutive terms of a non-constant arithmetic progression, there are at mostthree squares. Hence, it is natural to ask how many squares there may be in N consecutive terms of a non-constant arithmetic progression.Following Bombieri, Granville and Pintz [1], given q and a integers, q = 0, wedenote by Q ( N ; q, a ) the number of squares in the arithmetic progression qn + a ,for n = 0 , , . . . , N − i = 0 instead of i = 1). Denoteby Q ( N ) the maximum of Q ( N ; q, a ) over all non-trivial arithmetic progressions.Notice that Fermat’s result is equivalent to Q (4) = 3.As a consequence of Fermat’s result and of its own result on arithmetic pro-gressions, Szemer´edi [25] proved an old Erd¨os [12] conjecture: Q ( N ) = o ( N ). Thisbound was improved by Bombieri, Granville and Pintz [1] to Q ( N ) = O ( N / o (1) ),and by Bombieri and Zannier [2] to Q ( N ) = O ( N / o (1) ). Moreover, the so calledRudin’s conjecture ([21, end of § Q ( N ) = O ( √ N ), and in itsstronger form that (what we called Strong Rudin’s Conjecture): Q ( N ) = Q ( N ; 24 ,
1) = r N + O (1) if N ≥ Mathematics Subject Classification.
Primary: 11G30, 11B25,11D45; Secondary: 14H25.
Key words and phrases.
Arithmetic progressions, squares, covering collections, elliptic curveChabauty.The first author was partially supported by the grant MTM2009–07291. The second authorwas partially supported by the grant MTM2006–11391.
Notice that Q (5; 24 ,
1) = 3, but Q (5; 120 ,
49) = 4 (since 7 = 49 , = 169 , =289 , , = 529) ∗ . Therefore Q (5) = 4 because Q (5) cannot be 5 by Fermat’sresult.We will prove that the arithmetic progression 24 n + 1 is the only one, up toequivalence, that contains Q ( N ) squares for the values of N such that Q ( N ) in-creases, for 7 ≤ N ≤
52 (hence, for N = 8 , , , , , ,
41 and 52). Thisallow us to assert, what we have called Super-Strong Rudin’s Conjecture: let be N = GP k + 1 ≥ k , where GP k is the k -th generalized pentagonalnumber, then Q ( N ) = Q ( N ; q, a ) with gcd( q, a ) squarefree and q > q, a ) = (24 , Theorem 1.
Let N be a positive integer, then: (S) Q ( N ) = Q ( N ; 24 , if ≤ N ≤ . (SS) If ≤ N = GP k + 1 ≤ for some integer k , then Q ( N ) = Q ( N ; q, a ) with gcd( q, a ) squarefree and q ≥ if and only if ( q, a ) = (24 , . Preliminaries: Equivalences, translation to geometry andnotations.
We denote by N the set of non-negative integers.Observe first of all, that in order to compute Q ( N ), there is no lost of generalityto only consider the arithmetic progressions qn + a with gcd( q, a ) squarefree.For any subset I ⊆ N , we denote by Z I = { ( q, a ) ∈ Z | gcd( q, a ) squarefree , q = 0 and qi + a is a square ∀ i ∈ I } , and by z I = Z I its cardinality. We will prove that if I is a finite subset of N of cardinality bigger than 3 then z I < ∞ . Moreover, notice that, if J ⊆ I , then Z I ⊆ Z J , so z I ≤ z J . Since we are interested on the subsets I such that z I = 0, ifthere is some subset J with z J = 0, the same is true for all the subsets I containing J . Observe also that for all I ⊂ { , . . . , N − } with I > Q ( N ) we have bydefinition that z I = 0.Given q and a integers, q = 0, we denote by S ( q, a ) = { i ∈ N | qi + a is a square } the set of values where the arithmetic progression takes its squares, and, given N ≥
2, by S N ( q, a ) the set S ( q, a ) T { , , . . . , N − } . Therefore S N ( q, a ) ≤ Q ( N ).The following lemma is elementary, and its proof is left to the reader. Lemma 2. (i) S (24 ,
1) = {GP k } k ∈ Z the progression † of generalized pentagonal numbers. (ii) S (8 ,
1) = {T k } k ∈ N the progression ‡ of triangular numbers. ∗ It has been proved in [15] that the first quadratic number field where there are five squaresin arithmetic progression is Q ( √ Q ( √ , , , , † GP k = k (3 k − / ‡ T k = k ( k + 1) / N A CONJECTURE OF RUDIN ON SQUARES IN ARITHMETIC PROGRESSIONS 3
Given any subset of the naturals numbers I ⊂ N , we will numerate its elementsby increasing order starting from n , so if n i and n i +1 are elements in I , then n i < n i +1 and there is no element m ∈ I such that n i < m < n i +1 .We define the following three operations on the subsets I ⊂ N : • Let i ∈ Z such that n + i >
0, we denote by I + i the translated subset I + i = { j ∈ N | j − i ∈ I } . • Let r ∈ Q ∗ such that ri ∈ N for all i ∈ I , we denote by rI the subset of N definedby rI = { ri | i ∈ I } . • If I is a finite set, I = { n , . . . , n k } , we denote by I s , the symmetric of I , as I s = { n + n k − i | i ∈ I } . Therefore a finite subset is symmetric if I s = I .We say that two finite subsets I and J of N are equivalent, and denote by I ∼ J ,if there exists I = I , I , . . . , I k = J finite subsets of N such that I i +1 = I i + j or I i +1 = rI i or I i +1 = I si , for all i = 1 , . . . , k − I of N , we will denote by n I the positive integer P i ∈ I i .Then we have a bijection between the set of finite subsets of N and N (the emptyset corresponding to 0). Given two finite subsets I and J of N , we will say that I < J if n I < n J .We say that a finite subset I of N is primitive if 0 ∈ I , the elements of I arecoprime and n I ≤ n I s . Then any finite subset of N is equivalent to a uniqueprimitive subset. Lemma 3.
Let I and J be two finite subsets of N . If I ∼ J , then z I = z J . Proof.
This is a straightforward computation, just by checking the three possibleelementary equivalences. For example, if J = mI , for m ∈ Z > , we may assign toevery element ( q, a ) ∈ Z J , the element ( qmd , ad ) ∈ Z I , where d is the largest squaredividing gcd( qm, a ). And to every element ( q, a ) ∈ Z I , the element ( qmd , am d ) ∈ Z J ,where d = gcd( q, a, m ). The assertion in the case J = I + i is elementary.In the last case J = I s , we may restrict ourselves to the case 0 ∈ I . Then, we onlyneed to observe that to every element ( q, a ) ∈ Z J , the element ( − q, a + q ( N − Z I . ✷ Let I = { n , n , . . . , n k } ⊂ N be a finite subset such that k > K be a field.We denote by C I the curve in P k ( K ) defined by the system of equations C I : (cid:8) ( n i +2 − n i +1 ) X i − − ( n i +2 − n i ) X i + ( n i +1 − n i ) X i +1 = 0 (cid:9) i =1 ,...,k − . This curve is defined over any field, and, if the characteristic of the field is not 2,it contains 2 k trivial points T I corresponding to the values X i = 1 for all i = 0 , .., k .The following proposition collects some useful facts about the curves C I thatwill be used in the sequel. Proposition 4.
Let I = { n , n , . . . , n k } ⊂ N be a finite subset such that k > and C I be the associated curve. Then (1) If K is a field with characteristic , then the curve C I is a non-singularprojective curve of genus g k = ( k − k − + 1 . ENRIQUE GONZ´ALEZ-JIM´ENEZ AND XAVIER XARLES (2) If k > , then for any i ∈ I , the natural map C I → C I \{ i } is of degree and ramified on the k − points with X i = 0 . (3) If J is another finite set with I ∼ J , then C I ∼ = C J , with the isomorphismbeing the identity or the natural involution in P k given by [ x , . . . , x k ] [ x k , . . . , x ] . (4) Consider the map ι : C I → P k defined by ι ([ x : · · · : x k ]) = [ x : · · · : x k ] .Then there is a natural bijection between ι ( C I ( Q ) \ T I ) and Z I . Proof.
The first three items are well–known facts on the curves C I (cf. [1]). Now,we prove last statement. By (3), we may suppose that I is primitive, and, inparticular, n = 0. On one hand, let [ x : · · · : x k ] ∈ C I ( Q ) \ T I , and without lostof generality assume that x , . . . , x k are coprime integers. Then the correspondingelement of Z I is (( x − x ) /n , x ). On the other hand, let ( q, a ) ∈ Z I . Then thepoint [ a : a + n q : · · · : a + n k q ] belongs to ι ( C I ( Q ) \ T I ). ✷ In fact, in order to compute Z I , we may carry out it modulo the group ofautomorphisms Υ I of the curve C I generated by the automorphisms τ i ( x i ) = − x i and τ i ( x j ) = x j if i = j , for i = 0 , . . . , k . Notice that C I ( Q ) / Υ ∼ = Im( ι ).Finally, observe that, if I ⊂ N only has three elements, the corresponding curve C I is a genus 0 curve. And since it has rational points, C I ( Q ) = P ( Q ) and hence z I = ∞ . 3. First elementary cases
In order to study the first values of N , we will consider the subsets I ⊂ N ofcardinality 4. In this case the curve C I has genus 1. Moreover, C I is an ellipticcurve over any field, since C I always has the rational points T I . Proposition 5.
Given any subset I of { , , . . . , } with four elements, we havethat z I = ∞ unless I is equivalent to one of the following five subsets: { , , , } , { , , , } , { , , , } , { , , , } and { , , , } . In which case, z I = 0 . Proof.
To prove that z I = 0 for the given finite sets I in the proposition, one onlyneeds to show that C I ( Q ) = T I . Or, equivalently, that C I ( Q ) = 8. Using standardtransformations (see section 4), one may put the corresponding elliptic curves inWeierstrass form, and then compute the Cremona reference. We obtain the ellipticcurves , , , and respectively. All of them have rank 0and torsion subgroup isomorphic to Z / Z ⊕ Z / Z . Hence C I ( Q ) = 8 in thesecases. For all the other cases, one easily shows that the rank of the correspondingelliptic curve C I is 1. Therefore, they have an infinite number of points. ✷ Observe that all the subsets in the proposition with z I = 0 are symmetric subsets.We will see that this is true for any subset with four elements. Corollary 6. Q (6) = Q (7) = 4 and Q (8) = 5 . Proof.
First of all, we clearly have that 4 = Q (5) ≤ Q (6) ≤
5. Suppose we have aprimitive subset I of 5 elements inside { , , , , , } such that z I >
0. Notice that
N A CONJECTURE OF RUDIN ON SQUARES IN ARITHMETIC PROGRESSIONS 5 if I does not contain 5, I \ { } will be { , , , , } , which contains J = { , , , } .Since z J = 0 by the proposition 5, z I = 0. So I = { , , , , , } \ { i } for 0 < i < i = 1 or i = 4, then I contains { , , , } ∼ { , , , } or { , , , } respectively, and we have again the same result; and if i = 2 or i = 3,then I contains { , , , } or { , , , } ∼ { , , , } , and applying another case ofproposition 5, we conclude.Now, we are going to prove that Q (7) = 4. We again proceed with the samestrategy. We consider I a primitive subset of { , . . . , } with five elements and weshow that z I = 0 if we find a subset J of I with four elements appearing in the listof the proposition 5, hence with z J = 0. We may suppose that I = { , i, j, k, } for some 0 < i < j < k <
6. We have 10 cases, but only 6 cases to considerbecause the symmetries: the first case { , , , , } is by Fermat; the second case { , , , , } because it contains { , , , } ∼ { , , , } , hence again by Fermat; { , , , , } contains { , , , } , { , , , , } contains { , , , } , { , , , , } con-tains { , , , } and the last subset { , , , , } contains { , , , } ∼ { , , , } .Now, since Q (8) ≤ Q (7) + 1 = 5, to prove Q (8) = 5 we only need to exhibitan arithmetic progression with five squares in the first 8 terms, and the arithmeticprogression 1 + 24 n do the job. Note that S (24 ,
1) = { , , , , } . ✷ So, the first strategy to detect subsets I with cardinality bigger that 3 and z I = 0is to find a subset J of I with four elements such that z J = 0. This will be carryout considering the associated elliptic curve E J and showing that it only containsthe (eight) trivial points T J . In the next section we will study some necessaryconditions where this happens.4. Four squares in arithmetic progressions
Let be I = { n , n , n , n } with n < n < n < n . For any ( q, a ) ∈ Z I , considerthe eight points in the three dimensional projective space [ x , x , x , x ] ∈ P suchthat x j = qn j + a . They all lie in the curve C I given by the sytem of equations C I : ( ( n − n ) X − ( n − n ) X + ( n − n ) X = 0 , ( n − n ) X − ( n − n ) X + ( n − n ) X = 0 . We have that C I is an elliptic curve since it has genus 1 (being the intersection oftwo quadric surfaces in P ) and it has the rational points [1 , ± , ± , ± m = n − n n − n , m = n − n n − n . Note that they are both strictly positive rational numbers. Then we may write theequations of C I as C I : ( X − ( m + 1) X + m X = 0 ,m X − ( m + 1) X + X = 0 . Now, we parametrize the first equation as[ X : X : X ] = [( m + 1) − m + 1) t + t : ( m + 1) − t + t : ( m + 1) − t ] . Next, we substitute X , X , X in the second equation and we obtain a new equationof the curve, depending on the parameter t (note that t = ( X − X ) / ( X + X )): C I : X = t + 4 m t − m + 4 m + 2 m m + 1) t + 4 m ( m + 1) t + ( m + 1) . ENRIQUE GONZ´ALEZ-JIM´ENEZ AND XAVIER XARLES
Observe that the set I is symmetric if and only if m = m . Lemma 7.
Let E I be the elliptic curve defined by the Weierstrass form E I : y = x ( x − m m )( x + m + m + 1) . Then, there exists a Q -isomorphism φ : C I −→ E I such that φ ([1 , , , , , .Furthermore, if we denote by F I = φ ( T I ) , then F I = 8 . Proof.
The proof of the existence of the isomorphism φ is an straightforward com-putation. For example, it may be carried out using the formulae in [8, section 7.2].The set F I is described by the table below i P i Q i = φ ( P i )0 [1 , , , O = [0 , , − , , − ,
1] (0 , − , , , −
1] ( m m , − , − , ,
1] ( − m − m − , , , − ,
1] ( − m , − m ( m + 1))5 [1 , − , ,
1] ( − m , m ( m + 1))6 [ − , , ,
1] ( m ( m + m + 1) , − m ( m + 1)( m + m + 1))7 [1 , , , −
1] ( m ( m + m + 1) , m ( m + 1)( m + m + 1))Therefore, we have F I = 8 since m , m > ✷ Corollary 8.
The set F I is a subgroup of E I if and only if I is symmetric. Fur-thermore, if I is not symmetric, then z I > . Proof.
First, observe that the opposites of the non-Weierstrass points on F I donot belong to F I unless m = m . Since m and m are strictly positive numbers,then, for example, − Q = ( − m , − m ( m + 1)) ∈ F I if and only if it is equalto Q . This shows one implication. Finally, if m = m , one easily checks thatthe non-Weierstrass points are of order 4, and their doubles are equal to the point( m m , F I ∼ = Z / Z ⊕ Z / Z . ✷ Remark . One may use the isomorphism φ in order to find explicitly which arith-metic progression corresponds to the set of points {− P , − P , − P , − P } . If wesuppose I is primitive, in particular with n = 0, so it is of the form { , n , n , n } ,with n , n and n coprime, then the arithmetic progression a + nq given by (cid:26) a = (( n + n − n ) − n n ) ,q = 2 ( n + n − n )( n − n − n )( n − n + n ) , has squares for n = 0 , n , n , n . Using this construction, we show in the tablebelow the arithmetic progression ( q, a ) for all the equivalence classes of 4-tuples I ⊂ { , . . . , N − } , for 5 ≤ N ≤
7, such that C I ( Q ) = T I . Note that in all of thesecases rank E I ( Q ) = 1. N A CONJECTURE OF RUDIN ON SQUARES IN ARITHMETIC PROGRESSIONS 7
N I
An Arithmetic progression ( q, a ) such that I ⊂ S ( q, a )5 { , , , } (120 , { , , , } (24 , { , , , } (168 , { , , , } (840 , { , , , } (8 , { , , , } (280 , { , , , } (24 , Remark . One even show that, if I is not symmetric, the subgroup gener-ated by F I is infinite, unless m + m + 1 is a square and m = − ( m + 2 − p m + m + 1) /
3, or m + m is a square and m = m + 2 p m + m , whichare in fact the same case by interchanging m and m . Or equivalentment, E I is Q -isomorphic to E t : y = x ( x + 1 + 4 t )( x + 16 t ( t + 1)) , for some t ∈ Q . In this case, h P : P ∈ F I i = F I ∪ {− P , − P , − P , − P } ∼ = Z / Z ⊕ Z / Z . Furthermore, the points − P , − P , − P , − P correspond to the arithmetic progres-sion with a = 0 and q = 1. Thus, if we suppose that I is primitive then there exist s , s ∈ Z such that n = s , n = s and n = ( s ± s ) .We have seen that if I ⊂ N has four elements, a necessary condition to have z I = 0 is that I is symmetric. In the sequel, we obtain more necessary conditions,some of them under the Parity Conjecture.Observe that the number of symmetric subsets with 4 elements contained in { , , . . . , N } may be explicitly computed in terms of N (and it is the sequenceA002623 in [22], with n = N − N : N − N N − − N . Since the number of subsets with 4 elements is a polynomial of degree 4 in N , thereare 2 /N + O ( N − ) symmetric subsets among all the subsets with four elements.We do not know how many of the equivalence clases of subsets with four elementsare symmetric, but we suspect is of the same order.In the primitive symmetric case, that is if I = { , n , n , n + n } with 0 < n
0, and even with infinite number of elements. For example,this is the case if the torsion subgroup has more than 8 elements. This case onlymay occur, by Mazur’s theorem, if the torsion subgroup of E t ( Q ) is isomorphicto Z / Z ⊕ Z / Z . Or, equivalently, some of the four 4-torsion points given by thepoints with x -coordinate equal to ± t is the double of some rational point. We use ENRIQUE GONZ´ALEZ-JIM´ENEZ AND XAVIER XARLES the standard formulae (or by a 2-descent argument) to obtain that this happens ifand only if the x -coordinate and the x -coordinate +1 are squares. Lemma 11.
Let t be a positive rational number, then E ′ t ( Q ) tors ∼ = ( Z / Z ⊕ Z / Z if t = (cid:0) s − s (cid:1) , for some s ∈ Q , Z / Z ⊕ Z / Z otherwise. In fact, we may exactly characterize which primitive sets have their correspond-ing elliptic curve with torsion subgroup of order 16, and even which arithmeticprogressions correspond to the torsion points.
Corollary 12.
Let I = { , n , n , n + n } be a primitive symmetric subset of N .Then the elliptic curve E I has torsion isomorphic to Z / Z ⊕ Z / Z if and only if n , n and n + n are squares. The torsion points of E I correspond to the constantarithmetic progression along with the arithmetic progressions with a = 0 and q = 1 and with a = n + n and q = − . Proof.
Since t = n /n is a square, both n and n are squares. Since t + 1 =( n + n ) /n is also a square, also n = n + n is a square. Trivially, the arithmeticprogression with a = 0 and q = 1 verifies that a n i = a + n i q = n i are squares, andthe one with a = n + n and q = − a n i = a + n i q = n − i are alsosquares. And they correspond to the torsion points of order 8. ✷ Concerning the general symmetric case, we find one parametric subfamilies, andeven two parametric subfamilies of E ′ t , with other rational points apart of the trivialones. Example . Let z and z be non-zero rational numbers, and consider t = 14 (cid:18) z + 1 z + z + 1 z (cid:19) and x = − ( z + z ) z z . Then x ( x + 1)( x + t ) is a square. Moreover, if z , z = ± z = ± z and z = 1 /z ,then we obtain a non-trivial point in E ′ t .Recall that the Parity Conjecture asserts that any elliptic curve E defined over Q (or a general number field) that has root number W ( E ) = − Proposition 14.
Let < a < b be coprime integers, and let E a,b be the ellipticcurve defined by E a,b : y = x ( x + a )( x + b ) . Then W ( E ) = − if and only if α ( a, b ) ≡ µ ( a, b ) (mod 2) , where α ( a, b ) = { p odd prime | p divides ab } + { p prime | p divides ( b − a ) and p ≡ } ,µ ( a, b ) = ( if ab ≡ or ab ≡ and ( b − a )8 ≡ , otherwise. Proof.
Recall that the root number W ( E ) of an elliptic curve E over Q is equal tothe product of local root numbers W p ( E ), where p runs over all the prime numbersand infinity. One always has that W ∞ ( E ) = − W p ( E ) = 1 if the curve has good N A CONJECTURE OF RUDIN ON SQUARES IN ARITHMETIC PROGRESSIONS 9 or non-split multiplicative reduction at p , and W p ( E ) = − p . Since a and b are coprime integers, E a,b is minimalat an odd prime. The reduction is good if p does not divide ab ( b − a ), and it issplit multiplicative if p divides ab , or if p divides b − a and − p . Hence we obtain W ( E ) = ( − β W ( E ), where β = α ( a, b ) + 1.Now, to compute the root number at 2, we need to carry out a more detailedanalysis, since the reduction can be additive in this case. First, one shows by achange of variables that the curve E a,b is isomorphic to the curve given by theequation y + xy + ry = x + rx (this is the so called Tate normal form), where r = ab a + b ) . Given s ∈ Q , we denote by v ( s ) the 2-adic valuation of s . The curvehas good reduction at 2 (hence W ( E ) = 1) if and only if v ( r ) = 0, so if v ( a ) = 2or v ( b ) = 2. And it has split multiplicative reduction (hence W ( E ) = −
1) if v ( r ) >
0, so if v ( a ) > v ( b ) >
2, and it has additive reduction otherwise.When the valuation of r is negative, we will consider the original equation ofthe curve E a,b , which is an integral model. We will need the following standardinvariants of the equation: j = j ( E a,b ) = 2 ( a + b − a b ) a b ( a + b ) ( a − b ) , c = 2 ( a + b − a b ) , ∆ = 2 a b ( a + b ) ( a − b ) , c = 2 ( a + b )(2 b − a )(2 a − b ) . Now, if v ( r ) = −
1, hence v ( a ) = 1 or v ( b ) = 1. Then the curve has potentiallygood reduction ( since v ( j ) = 4), and we can look at the tables in [17]. We obtain v (∆) = 8, v ( c ) = 4 and v ( c ) = 6, hence this information is not enough to obtainthe sign. We consider c ′ = c / and c ′ = c / , and we compute 2 c ′ + c ′ (mod 16).After a case by case computation one obtain that 2 c ′ + c ′ ≡ I ∗ . Next, one computes that 2 c ′ + c ′ ≡ W ( E ) is − v ( a ) = v ( b ) = 0, we need to take in account the valuations v ( a + b )and v ( a − b ). First we need to determine when the curve E a,b has potentiallymultiplicative reduction in order to apply the formulae by Rohrlich [20]. This isequivalent to the case v ( j ) <
0. Since in our case v ( j ) = 8 − v ( a − b ) − v ( a + b ),we have potentially multiplicative reduction if and only if v ( a − b ) + v ( a + b ) ≥ s ∈ Q , we denote by s = s − v ( s ) . Then, we obtain that W ( E ) ≡ − c (mod 4). But observe that c = ( a + b )(2 b − a )(2 a − b ) ≡ ( a + b ) ≡ , hence W ( E ) = − v ( a ) = v ( b ) = 0. In this case, we have v ( a − b ) + v ( a + b ) = 3. Therefore, we obtainthat v (∆) = 10, v ( c ) = 4 and v ( c ) = 6. One easily shows that the necessarycondition c ≡ I ∗ in the tables of [17] is always satisfied, sowe obtain that W ( E ) is 1 in this case.In summary, we obtain W ( E ) = 1 if and only if v ( ab ) = 0 and v ( a − b ) = 3,or v ( ab ) = 2. ✷ Corollary 15.
Let < n < n be coprime integers and I = { , n , n , n + n } .Assume that the Parity Conjecture holds for the curve E n ,n , and that α ( n , n ) ≡ µ ( n , n ) (mod 2) . Then z I = ∞ . Remark . Cohn [9] studied the special symmetric case { , , n, n + 2 } when n ≤ { , , n, n + 2 } forany positive integer n . This condition is that the number of odd primes dividing n has the same parity than the number of primes congruent to 1 mod 4 dividing n −
4. The disadvantage is that this condition is under the Parity Conjecture forthe elliptic curve E ,n . Note that we may suppose that n is odd since in the evencase we can reduce { , , n, n + 2 } to { , , n/ , ( n + 1) / } .5. Five squares in arithmetic progressions: the technique
We will see in section 6 that the results of section 4 are not enough to show theRudin’s Conjecture even for small values of N. In this section we will study how toprove, for some subsets I with 5 elements, that z I = 0 even if it is not zero for anysubset J of I with 4 elements. Moreover, we will be able to determine Z I in somecases.In order to prove these type of results, we need to be able to compute the rationalpoints of some genus 5 curves whose Jacobians are product of elliptic curves, allof them of rank greater than 0. Hence it is not possible to apply the classicalChabauty method (see [7], [10], [13], [23], [24], [19]). We will instead apply thecovering collections technique, as developed by Coombes and Grant [11], Wetherell[26] and others, and specifically a modification of what is now called the ellipticcurve Chabauty method developed by Flynn and Wetherell in [14] and by Bruin in[4]. In fact, we will follow the same technique we applied in [16], though we couldmay be use also a similar technique as the one we use in [15] to study 5 squares inarithmetic progression over quadratic fields.First, we fix the notations. We consider a primitive subset I ⊂ N with 5 elements,and C I the associated curve as in proposition 4. Then if we want to prove that z I = 0, this will be equivalent to prove that C I ( Q ) only contains the trivial points T I . In fact, since the genus of C I is greater than one, its set of rational pointsalways is finite, hence we may even try to compute them.Observe that C I has 5 different maps to the elliptic curves corresponding to C J ,for J a subset of I with four elements. As we have already seen in the previoussection, the corresponding elliptic curves have all its 2-torsion points defined over Q , a fact that we will use to build unramified coverings of C I .The method has two parts. Suppose we have a curve C over a number field K ,and an unramified map χ : C ′ → C of degree greater than one, may be defined overa finite extension L of K , along with a nice quotient C ′ → H , for example a genus1 quotient. We consider the different unramified coverings χ ( s ) : C ′ ( s ) → C buildby all the twists of the given one. We obtain that C ( K ) = [ s χ ( s ) ( { P ∈ C ′ ( s ) ( L ) : χ ( s ) ( P ) ∈ C ( K ) } ) , the union being disjoint. Moreover, only a finite number of twists have rationalpoints, and the finite set of twists with points locally everywhere can be explicitlydescribed. The method depends first on being able to compute the set of twists,and second, on being able to compute the points P ∈ C ′ ( s ) ( L ) such that χ ( s ) ( P ) ∈ N A CONJECTURE OF RUDIN ON SQUARES IN ARITHMETIC PROGRESSIONS 11 C ( K ), by computing their images in H ( s ) ( L ). C ′ χ π C H s C ′ ( s ) χ ( s ) π ( s ) C H ( s ) In our case, the coverings we are searching for will be defined over Q , but thegenus 1 quotients of such coverings are, in general, not defined over Q , but in aquadratic or in a biquadratic extension. The way we will construct the coverings(factorizing quartic polynomials) will also give us the genus 1 quotients and thefield where they are all defined.In order now to construct the coverings of the curve C I , we first rewrite the curveas the projectivization (and desigularization) of a curve in A given by equationsof the form y = p ( x ) and y = p ( x ), where p ( x ) and p ( x ) are separable degree4 polynomials with coefficients in Q . This is possible because of the special formof the curve (essentially, because it has two degree 2 maps to elliptic curves thatcorrespond to involutions that commute with each other), and in our case we willsee it can be done in 10 different ways.Next, we will consider a factorization of the polynomials p i ( x ) as product of twodegree two polynomials p i, ( x ) and p i, ( x ), may be defined over a larger field K (in our case, a quadratic field). This factorization p i ( x ) = p i, ( x ) p i, ( x ) determinesan unramified degree two covering χ : F ′ i → F i of the genus 1 curve F i given by y i = p i ( x ), as we describe in the next proposition. Proposition 17.
Let F be a genus curve over a number field K given by a quarticmodel of the form y = q ( x ) , where q ( x ) is a degree four monic polynomial in K [ x ] .Thus, the curve F has two rational points at infinity, and we fix an isomorphismfrom F to its Jacobian E = Jac( F ) defined by sending one of these points at infinityto the zero point of E . Then (1) Any -torsion point of the curve E defined over K corresponds to a fac-torization of the polynomial q ( x ) as a product of two quadratic polynomials q ( x ) , q ( x ) ∈ L [ x ] , where L/K is an algebraic extension of degree at most . (2) Given such a -torsion point P , the degree two unramified covering χ : F ′ → F corresponding to the degree two isogeny φ : E ′ → E determinedby P can be described as the map from the curve F ′ defined over L , withaffine part in A given by the equations y = q ( x ) and y = q ( x ) and themap given by χ (( x, y , y )) = ( x, y y ) . (3) Given any degree two isogeny φ : E ′ → E , consider the Selmer group Sel ( φ ) as a subgroup of K ∗ / ( K ∗ ) . Let S L ( φ ) be a set of representatives in L ofthe image of Sel ( φ ) in L ∗ / ( L ∗ ) via the natural map. For any δ ∈ S L ( φ ) ,define the curve F ′ ( δ ) given by the equations δy = q ( x ) and δy = q ( x ) ,and the map to F defined by χ ( δ ) ( x, y , y ) = ( x, y y /δ ) . Then F ( K ) ⊆ [ δ ∈S L ( φ ) χ ( δ ) ( { ( x, y , y ) ∈ F ′ ( δ ) ( L ) : x ∈ K or x = ∞} ) . Proof.
First we prove (1) and (2). Suppose we have such a factorization q ( x ) = q ( x ) q ( x ) over some extension L/K , with q ( x ) and q ( x ) monic quadratic poly-nomials. Then the covering χ : F ′ → F from the curve F ′ defined over L , with affine part in A given by the equations y = q ( x ) and y = q ( x ) and the mapgiven by χ (( x, y , y )) = ( x, y y ), is an unramified degree two covering. So F ′ isa genus 1 curve, and clearly it contains the preimage of the two points at infinity,which are rational over L , hence it is isomorphic to an elliptic curve E ′ . Choosingsuch isomorphism by sending one of the preimages of the fixed point at infinity to O , we obtain a degree two isogeny E ′ → E , which corresponds to a choice of a twotorsion point.So, if the polynomial q ( x ) decomposes completely in K , the assertions (1) and(2) are clear since the number of decompositions q ( x ) = q ( x ) q ( x ) as above isequal to the number of points of exact order 2. Now the general case is proved byGalois descent: a two torsion point P of E is defined over K if and only if the degreetwo isogeny E ′ → E is defined over K , so if and only if the corresponding curve F ′ is defined over K . Hence the polynomials q ( x ) and q ( x ) should be definedover an extension of L of degree ≤
2, and in case they are not defined over K , thepolynomials q ( x ) and q ( x ) should be Galois conjugate over K .Now we show the last assertion. First, notice that the curves F ′ ( δ ) are twistedforms (or principal homogeneous spaces) of F ′ , and it becomes isomorphic to F ′ over the quadratic extension of L adjoining the square root of δ .Consider the case where L = K . So F ′ is defined over K . For any δ ∈ Sel( φ ),consider the associated homogeneous space D ( δ ) ; it is a curve of genus 1 along witha degree 2 map φ ( δ ) to E , without points in any local completion, and isomorphicto E ′ (and compatible with φ ) over the quadratic extension K ( √ d ). Moreover, itis determined by such properties (see section 8.2 in [8]). So, by this uniqueness, itmust be isomorphic to F ′ ( δ ) along with χ ( δ ) . The last assertion also is clear fromthe definition of the Selmer group.Now, the case L = K . The assertion is proved just observing that the commu-tativity of the diagram Sel( φ ) Sel( φ L ) K ∗ / ( K ∗ ) L ∗ / ( L ∗ ) where the map Sel( φ ) → Sel( φ L ) is the one sending the corresponding homogeneousspace to its base change to L . ✷ In order to apply the method to a 5-tuple I ⊂ N , we first explain how to constructmodels of C I as the ones described above. We need first to choose a subset J = { n , n , n } ⊂ I with three elements, which determines a partition I = J ⊔ { n , n } of I , the n i not necessarily ordered, and everything will depend of that choice.Second, we write the equations of C I of the form C I : X = ( m + 1) X − m X ,X = − m X + ( m + 1) X ,X = − m X + ( m + 1) X , where m = n − n n − n , m = n − n n − n , m = n − n n − n . N A CONJECTURE OF RUDIN ON SQUARES IN ARITHMETIC PROGRESSIONS 13
Now, we parametrize the first equation as it has been done on section 4:[ X : X : X ] = [( m + 1) − m + 1) t + t : ( m + 1) − t + t : ( m + 1) − t ] , and substituting in the next two equations, we obtain the new equations of thecurve, depending on the parameter t : C I : (cid:8) y = p ( t ) , y = p ( t ) (cid:9) , where, for i = 1 , y i = X i and p i ( t ) = t + 4 m i t − m + 4 m i + 2 m i m + 1) t + 4 m i ( m + 1) t + ( m + 1) . For i = 1 ,
2, by lemma 7, we obtain that the genus 1 curve F i : y i = p i ( t ) is Q -isomorphic to the elliptic curve E i : y = x ( x − m m i )( x + m + m i + 1) . Next, we need to choose factorizations of the polynomials p i ( t ) as product oftwo quadratic polynomials over some quadratic extension K/ Q . We describe in thenext elementary lemma all these factorizations, relating them to the corresponding2-torsion points in the corresponding elliptic curve E i . Lemma 18.
For i = 1 , , denote by D i, = m i (1 + m i ) , D i, = (1 + m i )( m i + m + 1) , D i, = m i ( m i + m + 1) , and choose an square root α i,j = p D i,j . Then the polynomial p i ( t ) factorizes over Q ( α i,j ) as a product of two quadratic polynomials p i,j, + ( t ) and p i,j, − ( t ) , dependingon j , where p i, , ± ( t ) = t + 2( m i ± α i, ) t ∓ α i, m − m i m − m − − m i ∓ α i, ,p i, , ± ( t ) = t + 2( m i ± α i, ) t + m + 1 ,p i, , ± ( t ) = t + 2( m i ∓ α i, ) t − m − − m i ± α i, . These factorizations correspond, by the proposition 17, to the -torsion points in E i ( Q ) with x -coordinate equal to r i, = m m i , r i, = − m − m i − and r i, = 0 . By the previous lemma and the proposition 17, one can construct Galois coversof C I with Galois group ( Z / Z ) , depending on the choice of the subset J ⊂ I above and the choice of j , j ∈ { , , } . The coverings can be described as theprojectivization (and desingularization) of the curve in A given by C ′ : { y , + = p ,j , + ( t ) , y , − = p ,j , − ( t ) , y , + = p ,j , + ( t ) , y , − = p ,j , − ( t ) } , which is a curve of genus 17, along with the map χ : C ′ → C I defined as χ ( t, y , + , y , − , y , + , y , − ) = ( t, y , + y , − , y , + y , − ) . These coverings can be defined over Q , although we choose to show them in thisform defined over the field Q ( α ,j , α ,j ), which is at most a biquadratic extensionof Q , in order to consider appropriate genus 1 quotients of them.Next, we choose one genus 1 quotient of the form H ± , ± : z = p ,j , ± ( t ) p ,j , ± ( t ) . There are four such quotients, but depending on the degree of the field Q ( α ,j , α ,j )there can be all of them conjugates over Q , or to have two conjugacy classes if thedegree is 2, or all independent if the degree is 1. For any element δ = ( δ , δ ) ∈ ( Q ∗ ) , we consider the twist C ′ ( δ ,δ ) of the cover χ , given by C ′ ( δ ,δ ) : (cid:26) δ y , + = p ,j , + ( t ) , δ y , − = p ,j , − ( t ) δ y , + = p ,j , + ( t ) , δ y , − = p ,j , − ( t ) (cid:27) , along with the map χ ( δ ,δ ) ( t, y , + , y , − , y , + , y , − ) = ( t, ( y , + y , − ) /δ , ( y , + y , − ) /δ ) . We obtain C ( Q ) ⊆ [ δ ∈ D χ ( δ ) ( { ( t, y , + , y , − , y , + , y , − ) ∈ C ′ ( δ ) ( Q ( α ,j , α ,j )) : t ∈ P ( Q ) } ) , for some finite subset D ⊂ ( Q ∗ ) . The Proposition 17 allows us to describe theset D in terms of the Selmer groups of some isogenies. For any such δ = ( δ , δ ),consider the quotients H ( δ δ ) ± , ± : δ δ z = p ,j , ± ( t ) p ,j , ± ( t )which, in fact, only depend on the product δ δ . We obtain { t ∈ Q | ∃ Y ∈ Q ( α ,j , α ,j ) such that ( t, Y ) ∈ C ′ ( δ ) ( Q ( α ,j , α ,j )) }⊆ { t ∈ Q | ∃ w ∈ Q ( α ,j , α ,j ) such that ( t, w ) ∈ H δ ± , ± ( Q ( α ,j , α ,j )) } . The following diagram illustrates, for a choice of ( j , j ), all the curves and mor-phisms involved in our problem: C ′ ( δ ,δ ) C H ( δ δ ) ± , ± π F ′ ( δ )1 F ′ ( δ )2 F ′ F F F ′ P E ′ E E E ′ Note that the previous construction also depends on the choice of a subset J = { n , n , n } ⊂ I .In the next lemma we describe a finite set S ⊂ ( Q ∗ ) enough to cover all thepossible values t giving points of C I , modulo the group of automorphisms Υ. Lemma 19.
Let I ⊂ N be a -tuple. Fix a subset J = { n , n , n } ⊂ I and j , j ∈{ , , } . For any i = 1 , , denote by φ i : E ′ i → E i the -isogeny corresponding tothe -torsion point ( r i,j i , ∈ E i ( Q ) , by L = Q ( α ,j , α ,j ) and by S L ( φ i ) a setof representatives in L of the image of Sel ( φ i ) in L ∗ / ( L ∗ ) via the natural map.Finally, denote by f S L ( φ ) a set of representatives of Sel ( φ ) modulo the subgroupgenerated by the image of the trivial points T I in this Selmer group. Consider thesubset S ⊂ Q ∗ defined by S = { δ δ : δ ∈ f S L ( φ ) , δ ∈ S L ( φ ) } . Then, for any point P = ( t, y , y ) ∈ C I ( Q ) , τ ∈ Υ and δ ∈ S exist such that τ ( P ) = ( t ′ , y ′ , y ′ ) and t ′ ∈ P ( Q ) such that ( t ′ , w ) ∈ H δ ± , ± ( L ) for any sign ( ± , ± ) . N A CONJECTURE OF RUDIN ON SQUARES IN ARITHMETIC PROGRESSIONS 15
Proof.
We have described at the previous paragraph that any point P = ( t, y , y ) ∈ C I ( Q ) lift to a point in C ′ ( δ ,δ ) ( L ) for some δ i ∈ Sel( φ i ), for i = 1 ,
2. Hence deter-mine a point in H δ δ ± , ± ( L ) with the first coordinate in P ( Q ).If P ∈ C I ( Q ) has image δ in the Selmer group Sel( φ ), then τ ( P ) has image δ δ τ ,if δ τ is the image of τ ( T ) in Sel( φ ) for some trivial point T with corresponding δ = 1. The previous sentence is true because the automorphisms belongs to Υcorrespond to translation by trivial points in the corresponding elliptic curve, if wefix (as we did) the zero point to be a trivial point (see Lemma 11 in [27] for a proofin a special case). But the action of Υ in C I ( Q ) is transitive on the set of trivialpoints T I . ✷ Now, the method allows us to conclude if we are able to compute, for some choiceof a subset J = { n , n , n } ⊂ I and j , j ∈ { , , } , and for any δ ∈ S , all thepoints ( t, w ) ∈ H δ ± , ± ( Q ( α ,j , α ,j )) with t ∈ Q for some choice of the signs ( ± , ± ).This last computation can be done in two steps as follows:(1) We first need to determine if there is some point in H δ ± , ± ( Q ( α ,j , α ,j )).In the special case δ = (1 , δ ’s we choose) may have no rational points if it represents an element ofthe Tate-Shafarevich group of its Jacobian. We use the method describedby Bruin and Stoll in [5]. In particular, we have used their implementationin Magma [3] to determine if this happens.(2) Secondly, we will choose an isomorphism with its Jacobian Jac( H δ ± , ± ) andthen we may use the elliptic curve Chabauty technique as it was developedby Bruin at [4] to compute this set if the rank of its group of Q ( α ,j , α ,j )-rational points is less than the degree of Q ( α ,j , α ,j ) over Q . We alsoneed to determine a subgroup of finite index of this group to carry out theelliptic curve Chabauty method. Here, we have used the implementationin Magma too.Hence we have 90 possible choices of J , j and j , and we need to find one ofthem where we can carry out all these computations for all the elements δ ∈ S .In practice, we only consider the case where the field Q ( α ,j , α ,j ) is at mosta quadratic extension of Q , essentially because of the computation of the rankand/or a subgroup of finite index in Jac( H δ ± , ± )( Q ( α ,j , α ,j )) is too expensivecomputationally for number fields of higher degree.5.1. The algorithm at work.
We have implemented in
Magma V2.18-8 the al-gorithm developed above. In the following we describe this algorithm in a fewexamples. For these 5-tuples I ⊂ N we show how it works. In the case that theoutput of the algorithm is true then we obtain Z I , otherwise we give detailedinformation about the reasons why the algorithm does not work. • I = { , , , , } : this is the first case having no rang zero elliptic quotients.First, we need to choose a subset J ⊂ I , and two values j , j ∈ { , , } such thatthe field L = Q ( α ,j , α ,j ) is of degree less or equal to 2. The subset J = { , , } and the pair ( j , j ) = (2 ,
1) do the job. In this case L = Q ( √
10) and we have the following factorizations: p , , + ( t ) = t − / t + 2 , p , , + ( t ) = t + 1 / − √ − t + 1 / √
10 + 14) ,p , , − ( t ) = t − t + 2 , p , , − ( t ) = t + 1 / √ − t + 1 / − √
10 + 14) . Note that in fact in this case we have Q ( α , ) = Q . Next step is to compute theset S (see Lemma 19). We have S = { , , , } . Now for any δ ∈ S , we mustcompute all the points ( t, w ) ∈ H δ ± , ± ( Q ( √ t ∈ P ( Q ) for some choice ofthe signs ( ± , ± ) where H δ ± , ± : δw = p , , ± ( t ) p , , ± ( t ) . For δ = 1 ,
6, we have that rank Z H δ + , + ( Q ( √ t . For δ = 1 (resp. δ = 6) weobtain t = ∞ (resp. t = 0). For the values t = ∞ and t = 0 we obtain the trivialpoints [1 : ± ± ± ± ∈ C I ( Q ). For δ = 2 ,
3, using the method describedby Bruin and Stoll in [5] we obtain H δ − , + ( Q ( √ ∅ .The following table shows all the previous data, where at the last column appearsthe arithmetic progression attached to the corresponding t : δ signs H δ signs ( L ) = ∅ ? rank Z H δ signs ( L ) t ( q, a )1 (+ , +) no 1 ∞ (0 , − , +) yes − − − − , +) yes − − − , +) no 1 0 (0 , I = { , , , , } , J = { , , } , ( j , j ) = (2 , , L = Q ( √ C I ( Q ) = { [1 : ± ± ± ± } ; andtherefore z I = 0 if I = { , , , , } . • I = { , , , , } : this is the Rudin sequence. Let be J = { , , } and ( j , j ) =(3 , L = Q ( √
14) and S = {± , ± , ± , ± } . The following tablesummarises all the computations made in this case: δ signs H δ signs ( L ) = ∅ ? rank Z H δ signs ( L ) t ( q, a )1 (+ , +) no 1 ∞ (0 , − , +) no 1 − − , − ) no 1 3 (24 , − , − ) no 1 − − , − ) no 1 5 / , − , − ) no 1 − −
10 (+ , +) no 1 0 (0 , −
10 (+ , +) no 1 − − I = { , , , , } , J = { , , } , ( j , j ) = (3 , , L = Q ( √ C I ( Q ) = { [1 : ± ± ± ± , [1 : ± ± ±
11 : ± } . That is, Z I = { (24 , } for I = { , , , , } . N A CONJECTURE OF RUDIN ON SQUARES IN ARITHMETIC PROGRESSIONS 17 • I = { , , , , } : this is an interesting example where there appear manyvalues for t . Looking at the table below we obtain Z I = { (120 , } . δ signs H δ signs ( L ) = ∅ ? rank Z H δ signs ( L ) t ( q, a )1 (+ , +) no 1 ∞ , , / , , − , +) no 1 0 , / / , / , , , +) yes − − −− , +) yes − − − I = { , , , , } , J = { , , } , { j , j } = { , } , L = Q ( √ C I ( Q ) = { [1 : ± ± ± ± , [1 : ±
11 : ±
19 : ±
29 : ± } . • I = { , , , , } : in this case we have at less two possible choices of J and( j , j ) where the algorithm works obtaining z I = 0. In the first case L = Q ( √ δ signs H δ signs ( L ) = ∅ ? rank Z H δ signs ( L ) t ( q, a )1 (+ , +) no 1 0 , ∞ (0 , , +) no 1 − − , +) no 1 − −
21 ( − , +) no 1 − − I = { , , , , } , J = { , , } , { j , j } = { , } , L = Q ( √ L = Q as the table below shows: δ signs H δ signs ( L ) = ∅ ? rank Z H δ signs ( L ) t ( q, a )1 (+ , − ) no 0 1 , ∞ (0 , , +) yes − − −− , +) no 0 0 , , − , +) yes − − − I = { , , , , } , J = { , , } , { j , j } = { , } , L = Q • I = { , , , , } : this is the second 5-tuple where the algorithm does notwork. The first one is I = { , , , , } and the reason is that 10 CPU hours wasnot enough to finish the computations for I . In the following table appear all thesubsets J ⊂ I and pairs ( j , j ) such that L = Q ( √ D ) for some D ∈ Z . Note that inall the previous cases, p ( t ) and p ( t ) do not factorize over Q . Therefore is enoughto check the signs (+ , +) and ( − , +). For δ = 1 we have computed an upper boundof the rank (denoted by rank ∗ ) of the Mordell-Weil group of the Jacobians of thecurves H , + ( L ) and H − , + ( L ), where in all those cases is greater than 1. Thereforewe can not apply elliptic curve Chabauty and the algorithm outputs false . J { j , j } D rank ∗ Z H , + ( Q ( √ D )) rank ∗ Z H − , + ( Q ( √ D )) { , , } { , } − { , , } { , }
10 2 2 { , , } { , } − { , , } { , } I = { , , , , } • I = { , , , , } : this example shows one case where for all subsets J ⊂ I of three elements and for all j , j ∈ { , , } we have that L = Q ( α ,j , α ,j ) is abiquadratic extension of Q . • Note that for all the 5-tuples I ⊂ N such that our algorithm has worked outwe have obtained z I = 0 or z I = 1, except in the case I = { , , , , } . Thetable below shows that Z I = { (24 , , ( − , } , that is z I = 2. δ signs H δ signs ( L ) = ∅ ? rank Z H δ signs ( L ) t ( q, a )1 (+ , +) no 1 ∞ , − , − /
11 (0 , − , , +) no 1 −
12 ( − , , − ) no 0 2 , /
11 ( − , , − ) no 0 2 , /
11 ( − , , − ) no 1 12 /
11 ( − , , +) no 1 16 / −
12 (24 , − , , − ) no 0 2 , /
11 ( − , , − ) no 0 2 , /
11 ( − , I = { , , , , } , J = { , , } , { j , j } = { , } , L = Q ( √ C I ( Q ) = { [1 : ± ± ± ± , [49 : ±
36 : ±
25 : ±
16 : ± , [49 : ±
361 : ±
625 : ±
841 : ± } .6. Summary of the computations
One of the main objectives of this article is to prove Rudin’s conjecture until N = 52. For this purpose, we have developed a method based on the computation ofthe rational points of the curves C I attached to finite subsets I ⊂ N . Furthermore,at section 2, we have defined an equivalence relation on the finite subsets of N such that for any pair of finite subsets I, J such that I ∼ J we have that C I is Q -isomorphic to C J , in particular z I = z J ; and such that in any given equivalenceclass we have a primitive representant. Therefore, we restrict our computations toprimitive subsets. Remember that we have attached to any finite subset I ⊂ N a positive integer n I , then we can introduce an ordering on equivalence classes offinite subsets of N .First, let us consider the case of finite subsets I ⊂ N ∩ { , . . . , } of cardinality4. There are 270725 of those subsets. But only 9077 equivalence classes. We haveproved at section 4 that the corresponding curves are elliptic curve over Q . Thefollowing table shows the number of curves for a given rank:rank 0 1 2 3 4 { , . . . , } , 117449 equivalence classes. Then we remove all the subsets I in the previous list with a subset J such that C J ( Q ) is an elliptic curve of rank 0 N A CONJECTURE OF RUDIN ON SQUARES IN ARITHMETIC PROGRESSIONS 19 and it only has 8 torsion points, since in that case z I = 0. After this sieve, 111338subsets remain. Now, at section 5, given a subset I ⊂ N with 5 elements we havedeveloped a method that allows in some case to determine C I ( Q ). The methodconsists, first to choose a subset J ⊂ I of three elements and j , j ∈ { , , } .There are 90 possible choices. Secondly, to compute the finite set S . Afterwards,compute for any δ ∈ S , all the points ( t, w ) ∈ H δ ± , ± ( Q ( α ,j , α ,j )) with t ∈ P ( Q )for some choice of the signs ( ± , ± ). This method has worked out in 26589 genus 5curves C I . For those, there are 26165 cases such that C I ( Q ) = T I and 424 casessuch that C I ( Q ) = T I . For the remaining cases, 84749, our method does not workfor different reasons. Notice that for a fixed δ ∈ S and a choice of ( ± , ± ) there aredifferent reason making that our method will not work. First, we have boundedour computations for the case where the field Q ( α ,j , α ,j ) is at most a quadraticextension of Q , since the algorithms on Magma that we are going to use are betterimplemented than in general number fields. There are 34548 cases where all the90 possible choices give biquadratic fields. For the remaining cases, there are 1033such that
Magma crashed for some unknown reason or there had not been enoughtime (maximum of 10 CPU hours); and there are 49168 cases where we know thereasons because our method has not worked for them. For a given case, we needto decide if H δ ± , ± ( Q ( α ,j , α ,j )) is empty or not. Then the first reason such thatour method does not work is:(BS) Magma does not determine if H δ ± , ± ( Q ( α ,j , α ,j )) is empty or not.Now assuming that we have computed a rational point on H δ ± , ± ( Q ( α ,j , α ,j )).Then two more reason may occur:(Rank) An upper bound of rank Z Jac( H δ ± , ± )( Q ( α ,j , α ,j )) is greater than one.Then, in principle, we can not use the elliptic curve Chabauty method.(noMW) Magma does not determine a subgroup of finite index on the elliptic curveJac( H δ ± , ± )( Q ( α ,j , α ,j )).Notice that more than one reason could happen for a given I making that any ofthe 90 possible choices do not compute C I ( Q ) by our method. The next table showsthe number of cases for the corresponding reasons:(Rank) : 37394 (Rank)+(noMW) : 988(BS) : 630 (BS)+(Rank) : 8526 (Rank)+(noMW)+(BS) : 1523(noMW) : 11 (noMW)+(BS) : 96All these computations (110305 5-tuples such that the algorithm has finished inless than 10 CPU hours) took around 68 days of CPU time on a MacPro4.1 with 2x 2.26 GHz Quad-Core Intel Xeon.The first case where we have not been able to determine C I ( Q ) is I = { , , , , } since 10 CPU hours was not enough. The second one is I = { , , , , } . In thiscase, our method does not work since for all the elliptic quotients defined overquadratic fields the upper bound for the rank is greater than 1.7. Consequences and comments
The main goal of this article is to give new evidences for the Rudin’s Conjectures.First, given a positive integer N ≥ Q ( N ) = Q ( N ; 24 , know Q ( N ) for some N then we attempt to compute Q ( N + 1). We have that Q ( N ) ≤ Q ( N + 1) ≤ Q ( N ) + 1. Therefore we must compute Z I for any I ⊂{ , . . . , N } such that I = Q ( N ) + 1. Note that if z I = 0 for any such tuples I ,then Q ( N + 1) = Q ( N ). Otherwise Q ( N + 1) = Q ( N ) + 1.In section 3 we have proved Q (6) = Q (7) = 4, and Q (8) = 5 since Q (8; 24 ,
1) = 5.Following the same strategy, that is with the computations of subsets of 4 integers,we even prove that Q (9) = Q (10) = Q (11) = 5. But it is not enough to showthat Q (12) = 5, since for I = { , , , , , } all the genus 1 quotients attached tosubsets of I of four elements have positive rank. However, by using the methods insection 5, we prove that for J = { , , , , } we have z J = 0, therefore z I = 0.Now, in the general case, the strategy we followed was to consider all the primi-tive subsets I of 5 elements in { , . . . , } where we are not able to compute C I ( Q ),either using the genus 1 quotients or by the methods in section 5, as we have de-scribed in the section 6. Using this list we recursively compute the list N C ( k )of all the primitive subsets I of k elements, k ≥
6, such that we are not able tocompute C I ( Q ), by finding all the primitive subsets I of k elements whose subsetsof k − N C ( k −
1) (See table 1). Note thatwe determined C I ( Q ) for all the subsets of { , . . . , } with more than 10 elements. k I ⊂ { , . . . , } number of I {
0, 1, 2, 6, 10 } {
0, 1, 2, 7, 12, 15 } ¶ {
0, 1, 6, 8, 11, 19, 23 } {
0, 1, 3, 11, 17, 22, 23, 30 } {
0, 2, 4, 13, 14, 19, 30, 33, 41 } {
0, 2, 7, 14, 17, 24, 37, 40, 43, 48 } Table 1.
We list the first primitive subsets (in the natural orderexplained in section 2) with k elements that we are not able todetermine C I ( Q ), together with the number of such subsets.Furthermore, using now the subsets I with 5 elements where we explicitly deter-mine C I ( Q ) and containing other points apart from the trivial ones, we explicitlycompute, for N ≥
8, all the arithmetic progressions ( q, a ) such that S N ( q, a ) = Q ( N ) except § for N = 11 ,
12 . In the table 2 we summarize these results.The computations from the table 2 allow us to prove what we have called Super–Strong Rudin’s Conjecture up to level 52: let be 8 ≤ N = GP k + 1 ≤
52 for someinteger k , then Q ( N ) = Q ( N ; q, a ) with gcd( q, a ) squarefree and q > q, a ) = (24 , I ⊂ { , . . . , N } ¶ Note that I = { , , , , , } ⊂ S (24 , , ∈ Z I , but we are not able tocompute the exact value of z I . § For the 5-tuples { , , , , } , { , , , , } , { , , , , } , { , , , , } , { , , , , } and { , , , , } we have not been able to compute the rational points of the corresponding genus 5curve. N A CONJECTURE OF RUDIN ON SQUARES IN ARITHMETIC PROGRESSIONS 21 k GP k N Q ( N ) Arithmetic Progressions ( q, a ) − , −
10 (24 , , § (24 , , , § (24 , , , , , , , −
14 6 (24 , , , , − −
18 7 (24 , −
20 (24 , , , , , , , , , , , −
25 (24 , , , , , − −
31 9 (24 , −
34 (24 , , , , , −
39 10 (24 , , , − −
49 11 (24 , , , , , , , Table 2.
In the first column k is an integer, in the second thecorresponding generalized pentagonal number GP k , in the thirdan integer N , in the fourth Q ( N ) and in the last one appear thearithmetic progressions qn + a with gcd( q, a ) squarefree and q > Q ( N ) squares for n ∈ { , . . . , N − } .with I ≥
5. One consequence of our computations is that, for the subsets I of { , . . . , } with I ≥ z I , we have obtained that z I ≤
1, except for one case where z I ≤
2. But it is easy to see that this is not truein general.
Lemma 20.
Consider a i , q i ∈ Z . Then: (1) If q q is not a square, then the set S ( q , a ) ∩ S ( q , a ) is infinite. (2) S ( q , a ) ∩ S ( q , a ) ∩ S ( q , a ) is finite. (3) If the Bombieri-Lang conjecture is true, there exists an r such that, forany set of r pairs ( q i , a i ) of coprime integers, T ri =1 S ( q i , a i ) has at most 4elements. Proof.
The set S ( q , a ) ∩ S ( q , a ) can be described also by the set of integersolutions of the equation x − q q x = q a − q q a , which is a Pell type equation with a solution. Hence it has an infinite number ofsolutions. In the case we have three pairs, we look for integer solutions of an equationgiving a genus 1 curve, so it has a finite number of them by Siegel’s Theorem.If we have more than three pairs, the resulting curve will be of genus biggerthan 1. So, suppose we have r pairs such that J = T ri =1 S ( q i , a i ) has more than4 elements, so there is a subset I ⊂ J with 5 elements in it. This means thatthe corresponding curve C I will have genus 5, and with C I ( Q ) ≥ r + 8 (and,if q i = 0 for all i ∈ I , in fact ≥ r + 1)). But thanks to the results from [6],the Bombieri-Lang conjecture implies there is an absolute bound for the number ofrational points of genus 5 curves over Q . Hence such r is bounded by above. ✷ Example . Using the ideas of the previous lemma, it is easy to construct oneparametric families of subsets I ⊂ N with 5 elements along with two different andnon-constant arithmetic progressions taking squares in I . For example, for anyinteger s >
1, we have that S ( s − , ∩ S ( s +1 , ⊃ { , s, s (4 s − , s (8 s − s +1) , s (32 s − s +14 s − } . As a consequence, we have built a one-parametric family of genus 5 non-hyperellip-tic curves (of the form C I ) having at least 3 ·
16 = 48 points.
Remark . If the Bombieri-Lang conjecture is true then thanks to the results from[6] we have that there exists a bound B ( g, Q ) such that any curve of genus g definedover Q satisfying C ( Q ) ≤ B ( g, Q ). For the special case g = 5, Kulesz [18] founda biparametric family of hyperelliptic curves of genus 5 with 24 automorphismsover Q with at least 96 points such that specializing he is able to found a genus5 hyperelliptic curve C defined over Q such that C ( Q ) = 120. For the non-hyperelliptic case we have that the curve C I attached to a 5-tuple I ⊂ N is of genus5 and has 16 automorphisms over Q . The example 21 shows a one-parametric familyof genus 5 non-hyperelliptic curves with at least 3 ·
16 = 48 points. Furthermore,we have found the following curves attached to 5-tuples I ⊂ N such that C I ( Q ) ≥ ·
16 = 80. For this search, we have looked for 5-tuples such that have pointscorresponding to S (24 b, a ) with a = 1 + 24 k square for some b, k ∈ N . The table 3shows the results we have obtained. I Arithmetic progression ( q, a ) such that I ⊂ S ( q, a ) { , , , , } (240 , ,
25) (120 , , { , , , , } (72 ,
1) (120 , , , { , , , , } (1344 , , , , { , , , , } (120 , ,
49) (24 , , Table 3.
Some 5-tuples I with z I ≥ k ≥
5, a constant c ( k )should exist such that z I ≤ c ( k ) for all I ⊂ N with I = k . In particular, z I ≤ c (5)for all I ⊂ N . The previous examples show that c ( k ) ≥ c (5) ≥ c (5) > Acknowledgements.
We would like to thank Noan Elkies for useful discussionabout the remark 22 and to Nils Bruin for fixing a bug in the elliptic curve Chabauty magma routine that appeared on Magma v2.18-7.
N A CONJECTURE OF RUDIN ON SQUARES IN ARITHMETIC PROGRESSIONS 23
Data:
All the
Magma and
Sage sources are available on the first author’s web-page.
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Universidad Aut´onoma de Madrid, Departamento de Matem´aticas and Instituto deCiencias Matem´aticas (ICMat), Madrid, Spain
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