On Sums of Consecutive Squares. Abstract. In this paper we consider the problem of characterizing those perfect squares

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1 On Sums of Consecutive Squares A. Bremner R.J. Stroeker y N. Tzanakis z Abstract In this paper we consider the problem of characterizing those perfect squares that can be expressed as the sum of consecutive squares where the initial term in this sum is k. This problem is intimately related to that of nding all integral points on elliptic curves belonging to a certain family which can be represented by a Weierstra equation with parameter k. All curves in this family have positive rank, and for those of rank 1 a most likely candidate generator of innite order can be explicitly given in terms of k. We conjecture that this point indeed generates the free part of the Mordell-Weil group, and give some heuristics to back this up. We also show that a point which is modulo torsion equal to a nontrivial multiple of this conjectured generator cannot be integral. For k in the range 1 k 100 the corresponding curves are closely examined, all integral points are determined and all solutions to the original problem are listed. It is worth mentioning that all curves of equal rank in this family can be treated more or less uniformly in terms of the parameter k. The reason for this lies in the fact that in Sinnou David's lower bound of linear forms in elliptic logarithms which is an essential ingredient of our approach the rank is the dominant factor. Also the extra computational eort that is needed for some values of k in order to determine the rank unconditionally and construct a set of generators for the Mordell-Weil group deserves special attention, as there are some unusual features Mathematics subject classication: 11D5, 11G05 Key words and phrases: diophantine equation, elliptic curve, elliptic logarithm Department of Mathematics, Arizona State University, Tempe, AZ , USA, abremner@math.la.asu.edu y Econometric Institute, Erasmus University, P.O.Box 1738, 3000 DR Rotterdam, The Netherlands, stroeker@wis.few.eur.nl z Department of Mathematics, University of Crete, Iraklion, Greece, tzanakis@talos.cc.uch.gr

2 A. Bremner, R.J. Stroeker and N. Tzanakis Contents 1 Preliminaries 1.1 Introduction The family of curves The Mordell-Weil groups 5.1 Rank calculations The exceptional case k = Constructing generators A rank 1 conjecture Determination of integral points Elliptic logarithms LLL-reduction All integral points for 1 k The rank 1 case Introduction 1 Preliminaries Everyone is familiar with the Pythagorean identity = 5 and many with the identity resulting from Lucas' \Square Pyramid" problem, = 70 : The problem of determining those squares equal to the sum of consecutive squares has attracted considerable interest throughout the years: the reader is referred to Guy [5, Problem D3] for a comprehensive list of both historical and contemporary references. We are interested in integer solutions of k + (k + 1) + + (k + n 1) = t ; (1) which equation may be written in the form of an elliptic curve E k : 1 3 n3 + k 1 n + k k + 1 n = t : () 6

3 On sums of consecutive squares 3 Most authors to date have considered n as xed and asked for corresponding pairs of integers k; t if any. It is known that there exist solutions for innitely many n, and in particular all such n < 1000 have been determined. The analysis in this instance depends upon an associated Pellian equation. Alternatively, one can consider k as xed and ask for corresponding integer pairs n; t (when k = 1 this is the Lucas problem mentioned above). The analysis now depends upon the theory of elliptic curves; a few explorations have been made in this direction (Platiel & Rung [8], Rung [9]; see also Kuwata & Top [7]). The present paper oers a systematic investigation of this approach, and all integer solutions n; t of () are found in the range 1 k 100. Stroeker & Tzanakis [14] and Gebel, Petho & Zimmer [4] have studied specic elliptic curves over Q, showing that when the rational Mordell-Weil group of the curve is known, then nding all integer points can be reduced to a practicably ecient process. Both papers employ similar methods, not following the traditional well established path of solving Thue equations, but instead relying on a highly nontrivial lower bound for linear forms in elliptic logarithms recently obtained by Sinnou David [3]. Where the calculations in [4] leading to the computation of the Mordell-Weil group are based on the assumption of the Birch and Swinnerton-Dyer conjectures, the results of [14] are unconditional. This is also one of the objectives of the present paper, and our results for k in the range 1 k 100 do not depend on any of the usual conjectures. However, in practice, this often means that an extensive amount of computational eort is required. In [13] Stroeker takes the elliptic logarithm method one step further and examines the parametrized family of elliptic curves that arises from demanding that the sum of consecutive cubes be a square. He is able systematically to treat the rst 50 curves of the family, showing that certain aspects of the computations can be successfully carried through uniformly in terms of the parameter. The current paper is modelled on this latter, though with some extra features. First, to determine the Mordell-Weil rank unconditionally in eleven cases required an extra argument; in particular for k = 68, it was found necessary to invoke the arithmetic of a number eld with class-group of order A detailed discussion is devoted to this exceptional case, because it is rather surprising that the nontrivial structure of this class-group ultimately clinches the argument. Some of the curves in our range have generators of large height, and an extra descent was necessary in order to compute the corresponding Mordell-Weil groups. Second, the curve () possesses an `obvious' integer solution for each k, namely (n; t) = (1; k). It turns out that the point Q k on the corresponding elliptic curve has innite order, and one might reasonably ask two associated questions in the case that the curve has rank 1: (i) Is Q k always a generator, and (ii) Can any multiple of Q k modulo torsion give rise (on specialization) to a nontrivial integer solution of equation ()? In the range 1 k 100, question (i) can be answered in the armative, and we oer some suggestions as to the reason why the answer should be yes for all suciently large

4 4 A. Bremner, R.J. Stroeker and N. Tzanakis k. We answer question (ii) in the negative by means of a p-adic approach (p = ; 3), involving straightforward but intricate double induction arguments. 1. The family of curves Under the substitution (x; y) = (1n + 1k 6; 7t) (3) with inverse (n; t) = 1 1 (x + 6) k; 1 7 y the curve E k at () transforms into the following Weierstra form E k : y = x 3 36x 864k(k 1)(k 1): (4) We shall denote by E k (Q) the rational Mordell-Weil group of this curve. rational point T k on E k of order, namely There is a T k = (6(k 1); 0); (5) and with the substitution (X; Y ) = (x 6(k 1); y) (6) (4) transforms to Y = X(X + 18(k 1)X + 7(6k 6k + 1)) (7) with T k transforming to (0; 0). For P E k (Q), the coordinates (x(p ); y(p )) will always be relative to (4), and the coordinates (X(P ); Y (P )) will always be relative to (7). The discriminant k of E k is given by and the j-invariant j k by Some simple facts are easy to establish. k = (1k 1k 1)(6k 6k + 1) j k = =(1k 1k 1)(6k 6k + 1) :

5 On sums of consecutive squares 5 Lemma 1. (i) E 1 (Q) tors ' Z=Z Z=Zand E k (Q) tors ' Z=Zfor k, (ii) The rank r k of E k (Q) satises r k 1 for k 1. Proof. (i) For k = 1, the torsion statement follows from Silverman [10, p. 311]. For k, we use the well-known fact (Silverman [10, p. 176]) that if a prime p does not divide the discriminant k of E k, then E k (Q) tors injects in E k;p (F p ) where E k;p is the reduction mod p of E k. With p = 5, we have E k;5 (F 5 ) ' ( Z=Z Z=4Z for k 0; 1; 3 mod 5; Z=8Z for k ; 4 mod 5: Thus je k (Q) tors j divides 8. Certainly E k (Q) has precisely the one point T k of order, since 6(k 1) is the only real zero of the right-hand side of (4). Furthermore, E k (Q) possesses no point of order 4, for such a point P (x; y) satises P = (6(k 1); 0), implying 6(k 1) = (x + 36) + 691k(k 1)(k 1)x 4y : But this forces x to be exactly divisible by, and hence 6(k 1) should be exactly divisible by an even power of, which is clearly absurd. This shows (i), with the immediate consequence that the point Q k = (1k + 6; 7k) on (4), corresponding to (n; t) = (1; k) on (), cannot be of nite order, which shows r k 1. Our goal is to determine all integer solutions of () in the range 1 k 100. We shall actually do more and determine all integer solutions of (4) in the range 1 k 100; integer solutions of () correspond via the transformation (3) and its inverse to a subset of integer solutions of (4). The attack falls into two distinct parts: determination of the Mordell-Weil groups and subsequent determination of the integer points. The Mordell-Weil groups In this part the Mordell-Weil groups for k in the range 1 k 100 will be computed completely and unconditionally. As the torsion subgroups have been determined in the previous part, that leaves the rank and the generators of innite order..1 Rank calculations The rst step is to compute the rank of each curve in the family. Connell's Apecs program was able to determine rank unconditionally in the range 1 k 100 except in the 11 cases k = 9, 40, 49, 51, 53, 57, 68, 77, 84, 93, 99. To ll in these gaps we used the following descent arguments.

6 6 A. Bremner, R.J. Stroeker and N. Tzanakis At a rational point of (7), put X = A =B, squarefree, A; B Z, (A; B) = 1. There results a quartic of type c 0 A 4 + c 1 A B + c B 4 = C (8) on which we seek points. John Cremona's algorithm \mwrank" (see []) will quickly sieve out all quartics (8) locally unsolvable for some prime p (including 1). Therefore we can safely assume that (8) is everywhere locally solvable. Then the associated quadric c 0 X + c 1 X Y + c Y = Z (9) is everywhere locally solvable, and hence globally solvable. Let (; ; ) be a point of (9); then (9) may be rationally parametrized as follows: X : Y : Z = W W V + (c 0 + c 1 )V : (W c 0 V ) : W (c 0 + c 1 )W V + c 0 V : It follows from (8) that ha = W W V + (c 0 + c 1 )V hb = (W c 0 V ); (10) where the squarefree part of h is a divisor of the resultant of the two quadratics in W; V, namely, 4 (c 1 4c 0 c ). That is, the squarefree part of h divides (c 1 4c 0 c ). The possibilities for h can be tested in (10), discarding those for which the pair of quadrics is not everywhere locally solvable. For a remaining value of h, the second quadric at (10) being locally solvable implies it is globally solvable, and so rationally parametrizable. Substituting into the rst quadric at (10) results in a homogeneous quartic in two variables being a square. In ten of the eleven exceptional cases listed above, all the resulting quartics turn out to be locally unsolvable. The rather tedious but straightforward details of these cases are omitted; verication should not pose any serious problems. However, for k = 68 everywhere locally solvable quartics remain, so that we are still uncertain about the expected non-existence of global solutions. We had to do some rethinking at this point, and the proof we found in the end to show that these quartics can possess no global solution is interesting enough in itself to justify a detailed description. Moreover, it clearly shows the power that sophisticated software like Pari/gp puts at one's ngertips. After this the rank will have been determined unconditionally for all k in the range 1 k 100. The rank values are listed in Table 1; here we just indicate their distribution, namely 31 cases of rank 1, 5 cases of rank, 14 cases of rank 3, and 3 cases of rank 4.

7 On sums of consecutive squares 7. The exceptional case k = 68 Details are provided here that show the rank r 68 of (4) for k = 68 E 68 : y = x(x + 430x ) (11) is unconditionally equal to. We refer to [, Chapter III, 3.6] and [10, Chapter III] for background and notational conventions. For the computation of the rank we also need the isogenous curve E 0 68 : y = x(x 4860x ) (1) and the standard -isogenies : E 68! E 0 68 and ^ : E 0 68! E 68. It is well-known that and we will show that je 68 (Q)=^(E 0 68(Q))j je 0 68(Q)=(E 68 (Q))j = r+ ; je 68 (Q)=^(E 0 68 (Q))j = je0 68 (Q)=(E 68(Q))j = : Putting x = a =b, ; a; b Z, squarefree, (; b) = 1, (a; b) = 1 in (11) gives a a b b 4 = c ; j : (13) Moreover, the group E 68 (Q)=^(E 0 68(Q)) is isomorphic to the subgroup of Q =(Q ) generated by the factors for which the diophantine equation (13) has an integer solution. Cremona's \mrank" tells us that global solutions exist at = 1, 3, 7337, , and the remaining four values of, =, 6, 7337, give everywhere locally solvable curves, but the existence of a global solution for these values remains undecided. However it is easy to see that je 68 (Q)=^(E 0 68(Q))j = or 3, depending, respectively, on the existence or non-existence of a solution for =. When =, with parametrization a a b b 4 = c a : b : c = 639u 996uv + 115v : u v : 498u + 16uv + 996v so that for coprime integers U, V, there exist integers h, 0, satisfying 639U 996UV + 115V = h 0 (14) U V = h ;

8 8 A. Bremner, R.J. Stroeker and N. Tzanakis where h is a squarefree divisor of the resultant of the two polynomials in the left-hand side; so h j However, 3jh implies U V mod 3, which is impossible. Thus h j From (14), 3jh 0, so 0 = 3, and 13U 33UV + 384V = 3h (15) U V = h : (16) The quadrics (15) and (16) are locally solvable for precisely the following eight values of h: Write (15), (16) in the form h = 1; ; 3; 46; 377; 377; 3 377; : (17) ( 13 + t)u 33UV + (384 t)v = h(3 + t ) (18) where t is chosen so that the left-hand side is a singular quadratic; this demands t = 0 +, where = 0, and (18) may then be written or, equivalently, (83U + (10 + )V ) = h(10 + )(3 + (0 + ) ) 9 + ( ) = 6(83U + (10 + )V ) =(h(10 + )): Dene the number eld K = Q(') where ' = 606 3, so that ' ' = 0. Further, let L = Q(). Then Norm K=L(3 + ') = 6(83U + (10 + )V ) =h(10 + ): (19) The following arithmetic information about K; L was obtained by use of Pari/gp. In L there are the prime factorizations () = p p 0, (3) = p 3p 0, (83) = p 3 83p 0, (3) = 83 p, 3 (377) = p, (10 + ) = 377 p0 p0. A Z-basis for the ring of integers O 83 L is f1; g; and the following congruences hold: mod p p 0 p 3 p 0 3 p 83 p In K, p = q, p 0 = q 0, p 3 = q 3, p 0 3 = q 0 3, (') = q 0 q 3q 0 3 q 7337, (1 + ') = q q ; a Z-basis for the ring of integers OK is f1; '; 1 3 ' ; 1 3 '3 g; and the class-group is of order 1618 and of type Z 504 Z 4 Z Z Z. The following congruences hold:

9 On sums of consecutive squares 9 mod q q 0 q 3 q 0 3 ' ' ' Now at (19), the greatest ideal common factor of (3 + ') and (3 ') divides (6; '; 3 + '). Since 0() ) U 0() ) V 0() at (16), we have (; ) = 1; and certainly (; 3) = 1. Further, (; ) = 1, for any common prime divisor divides the resultant at (15), (16), so f; 3; 377g, that is, f3; 377g ( odd). But then (15), (16) force U V 0 mod. Thus (6; ) = 1, and the above g.c.d. divides (6; '; 3 + '). Further, 0 mod q 7337 ) 0 mod 7337, and from (15), 13(U V ) V (U + 18V ) 0 mod 7337, giving U V 0 V (U + 18V ) mod 7337, contradicting (U; V ) = 1. So the above gcd is q q 3 q 0 3 ( odd), q 0 q 3q 0 3 ( even). At (19), let the ideal (83U + (10 + )V ) = p 0 83a, so that we obtain as ideals: Norm K=L(3 + ') = p p 3 p 0 3 (h) 1 a : (0) Now, from (17), (h) S [ p p 0 S where S is the set f(1); p 3; p 377; p 3 p 377g; and thus (0) implies an ideal equation ( p p 3 p 0 3 Norm K=L(3 + ') = b p 0 p for an integral ideal 3p 0 b of OL. (1) 3 b From the above remarks on greatest common divisor, (1) implies an ideal equation in K of type ( q q 3 q 0 3 (3 + ') = B q 0 q for an integral ideal 3q 0 B of OK: 3 B But this gives a contradiction in the class-group of K. For in the group Z 504 Z 4 Z Z Z, q [466; 0; 1; 0; 0]; q 0 [38; ; 1; 0; 0]; q 3 [30; 0; 0; 1; 0]; q 0 3 [; ; 0; 0; 0] so that q q 3 q 0 3 [14; ; 1; 1; 0], q 0 q 3q 0 3 [90; 0; 1; 1; 0]. If B [" 0 ; " 1 ; " ; " 3 ; " 4 ], then B [" 0 ; " 1 ; 0; 0; 0], implying (3 + ') [; ; 1; 1; 0] is in the principal class, a contradiction. Consequently there are no global solutions for =, and je 68 (Q)=^(E 0 68(Q))j =. It should be noted here that the ideal classes are given relative to a certain (unspecied) ordered basis. This basis will generally change with each interactive session, resulting in

10 10 A. Bremner, R.J. Stroeker and N. Tzanakis dierent representations for the ideal classes. Also, the algorithms used in Pari for constructing the class-group are correct under GRH. However, in most cases it should be comparativelyeasy to verify the results Pari produces. Next we consider the isogenous curve (1). In a completely analogous way it can be seen that, starting with A A B B4 = C ; j ; it suces to show that there are no global solutions for =. Now, the class-group of the relevant quartic number eld generated by a zero of x x is isomorphic to Z 4 Z 4 Z 4 Z. An argument similar to that used before applies and ultimately we nd je 0 68 (Q)=(E 68(Q))j =, thus proving that r 68 =..3 Constructing generators The next step towards the construction of Mordell-Weil bases is to nd on each curve the maximal number of linearly independent points. Apecs was used extensively but failed to nd the right number of points in six instances. It was necessary to perform the extra descent described above and search the resulting quartics for global solutions, which, when found, could be pulled back to the corresponding points of (4). This descent nds the points of large height in Table 1; and it is clear why the Apecs search failed to nd them. Further we remark that any determination of generators of a Mordell-Weil group will depend on estimation of height functions on the curve, in particular the relation between the logarithmic height h(p ) and the canonical height ^h(p ) of a point P on the curve. Silverman [11] gives general estimates for the dierence h(p ) ^h(p ), but it turns out that these are not precise enough for our purposes, and it was necessary to tailor his arguments specically to the curve (4). Lemma. (i) For P E 1 (Q), (ii) For k and P E k (Q), 3 log 3 log 6 4:076 h(p ) ^h(p ) log 6 + 4:504 1 (k 1) h(p ) ^h(p ) log log C k where C k = (6k 6k + 1) Q p pep=, the product running over all primes p for which p ep exactly divides 1k 1k 1 and e p.

11 On sums of consecutive squares 11 Proof. (i) Example. of Silverman [11]. (ii) This is a careful book-keeping exercise using the methods and formulae of Silverman [11], [1]. Brief details of the proof are given below. Theorem 4.1 of Silverman [11] gives 1 4 log+ jjj p p (P ) 1 log+ jx(p )j p 1 1 log jj p () for all nite primes p (where p is the local component of the height); and that for primes p dividing the denominator of j to the rst power, () may be replaced by 1 1 log+ jjj p p (P ) 1 log+ jx(p )j p 1 1 log jj p: (3) Summing over the nite primes results in where F (k) ^h(p ) 1 (P ) 1 1 ln(den(x(p ))) log jj; (4) 1 F (k) = 1 4 ln(jden(j)j) ln(squarefree part of j1 + 1k 1k j): (5) To compute the component 1 (P ) of the height of P at innity, refer to Silverman [1], noting it is the height there termed 0 that is used in the inequalities (), (3). See also [1, Remark, p. 341]. The cubic in x, x 3 36x 864k(k 1)(k 1), has precisely one real zero at x = 1k 6; so for P (x; y) E(R) then necessarily x 1k 6, and h(p ) = ln(num(x)) = ln(x) + ln(den(x)). Dene a sequence of reals fx n g by x 0 = x; x n+1 = x4 n + 7x n + 691k(k 1)(k 1)x n (x 3 n 36x n 864k(k 1)(k 1)) and a sequence fz n g by z n = Z(x n ), where Z(x) = x4 + 7x + 691k(k 1)(k 1)x x 4 : Then 1 (P ) = 1 1 ln jj ln(z(x)x4 ) P 1 n=1 4 n ln jz n j so that (4) implies F (k) 1 1 ln jj ^h(p ) 1 h(p ) 1 8 ln Z(x) 1 8 1X n=1 4 n ln jz n j 0: (6)

12 1 A. Bremner, R.J. Stroeker and N. Tzanakis Now using x 6(k 1), Similarly, so that Z(x) 1 + 3k(k 1) + (k 1) (k 1) + 1 (k 1) = 1 3 : 4 (k 1) ln jz n j ln 3 (k 1) 1X n=1 Substituting into (6) results in where the left-hand side is 4 n ln jz n j 3 ln 3 1 : (k 1) F (k) 1 1 ln jj ^h(p ) 1 h(p ) 1 3 ln 1 3 ; (k 1) F (k) 1 1 ln(1 3 6 ) 1 1 ln(den(j)) = 1 ln(1) 1 4 ln C k; by (5). This completes the proof of Lemma (ii). At the nal step, determination of bases for the Mordell-Weil groups, we choose to consider four cases, according to rank. Rank 1 (31 instances) Suppose the known point P is not a generator. It is easy to verify in each case that neither P nor P + T k lies in E k (Q), so there exists m Z, Q E k (Q) with P = mq, m 3, and where, without loss of generality, we may suppose m prime. By looking at E k (F p ) for a suitable prime p, it can be shown in each of the cases that m 6= 3. So m 5, and ^h(q) = 1 ^h(mq) = 1 ^h(p ) m m 5^h(P 1 ). Using the bounds of Lemma, a simple search shows no such Q can exist in any of the cases. Ranks 3, 4 (14, 3 instances, respectively) In each case the known independent points do not have particularly large height, and the following idea of Silverman is applicable. Let P 1 ; : : : ; P r be a maximal set of independent points in E k (Q) corresponding to a set of generators for E k (Q)=E k (Q) tors. Select a complete set S of r representatives for E k (Q)=E k (Q) from the set r X i=1 " i P i "i f0; 1g

13 On sums of consecutive squares 13 and set B = max P S ^h(p ). Then Ek (Q) is generated by all points X of E k (Q) satisfying ^h(x) B. The most intransigent of these cases occurred at k = 74 with B = 13:186, leading to h(p ) 0:8066 and a run-time of several hours on a SUN workstation. Rank (5 instances) The above method of Silverman works in many instances, but cannot deal with the cases where one of the two known independent points has large height, for example k = 75, where the known points have heights.6069 and We need to introduce a further idea. Suppose the known points P 1 ; P generate a subgroup of index m in E k (Q)=E k (Q) tors. It is straightforward to verify in all the cases that m is odd, by showing that if " 1 P 1 + " P + " 3 T k E k (Q) for " i f0; 1g, then " i = 0 for i = 1; ; 3. Suppose now m > 1, and let q be a prime dividing m. Then either P qe k (Q) or at least one of the points P 1 rp with r f0; 1; : : : ; q 1 g lies in qe k(q). In the latter case, let P 1 + rp = qq, jrj q 1. Then so that ^h(p 1 + rp ) + ^h(p 1 rp ) = ^h(p 1 ) + ^h(rp ) = ^h(p 1 ) + r ^h(p ) ^h(p 1 ) + 1 (q 1)^h(P ); q ^h(q) = ^h(qq) = ^h(p1 + rp ) ^h(p 1 ) + 1 (q 1)^h(P ); from which ^h(q) q ^h(p1 ) ^h(p ) q q ^h(p1 ) + 1 ^h(p ): (7) Thus either there exists Q E k (Q) satisfying ^h(q) = 1 q ^h(p ) or there exists Q E k (Q) satisfying the inequality (7). In each numerical case we eliminate the possibilities q = 3; 5; 7 by showing that none of P ; P 1 rp with r f0; 1; : : : ; q 1 g lies in qe k(q). As an example, when k = 4, the structure of the groups E 4 (F 19 ) and E 4 (F 137 ) was used to eliminate q = 7. It follows that q 11, with consequently a point Q E k (Q) satisfying either ^h(q) 1 11^h(P ) or ^h(q) 11^h(P 1 ) + 1 ^h(p ). By choosing P 1 to be the point with larger height than P, only the second inequality matters. It implies a manageable bound for h(q), and by search there are no such Q. It only remains to indicate how in practice the search for points of bounded height was carried out. We are searching on the curve (4) for points P with h(p ) < B, for some known bound B. Let X(P ) at (7) be given by r=s, so that using (6), 0 r + 6(k 1)s e B ;

14 14 A. Bremner, R.J. Stroeker and N. Tzanakis whence 0 s e B =(6(k 1)) 1=; 0 r e B 6(k 1)s : (8) In practice, write r = where is a squarefree divisor of 7(6k 6k + 1). Then for each, (8) becomes 1=; 0 s e B eb 6(k 1)s 1= =(6(k 1)) 0 : Of course one can also restrict to those known to correspond to everywhere locally solvable curves, but in our cases the time of running was so short that this minor renement was unnecessary. Table 1: The Mordell-Weil groups E k (Q), k = 1; : : : ; 100 Rank and generators of E k (Q), k = 1; : : : ; 100 rank k r k generators P 1 ; : : : ; P rk on (4) 1 1 (18, 7) 1 (30, 144) 3 (4, 16), (54, 360) 4 (54, 88), (46, 15) 5 1 (66, 360) 6 1 (78, 43) 7 3 (144, 1584), (90, 504), (14, 1196) 8 (10, 576), (86, 4760) 9 (114, 648), (390, 763) 10 (16, 70), (189, 95) 11 (138, 79), (13706/5, /15) 1 (150, 864), (864, 5344) 13 (16, 936), (300, 4860) 14 (174, 1008), (166, 568) 15 (5, 376), (186, 1080) 16 (198, 115), (134, 49096) 17 (474, 9936), (405, 7659) 18 (34, 5544), (, 196) 19 1 (34, 1368) 0 4 (46, 1440), (58, 088), (5, 11376), (396, 700) 1 (58, 151), (1398, 518) continued on next page

15 On sums of consecutive squares 15 continued from previous page (Table 1) Rank and generators of E k (Q), k = 1; : : : ; 100 rank k r k generators P 1 ; : : : ; P rk on (4) 3 (540, 11844), (634, 15416), (14076, ) 3 1 (8, 1656) 4 (94, 178), ( /101761, / ) 5 4 (606, 14040), (333, 3393), (306, 1800), (300, 160) 6 1 (318, 187) 7 3 (714, 1816), (330, 1944), (406, 5896) 8 3 (34, 016), (480, 8640), (930, 770) 9 1 (354, 088) 30 (366, 160), (1777/4, 5649/8) 31 (378, 3), (1784/89, /4913) 3 3 (390, 304), (396, 844), (3886/9, 13895/7) 33 (40, 376), ( / , / ) 34 1 (414, 448) 35 (46, 50), (139164, ) 36 (438, 59), (578/9, 30111/7) 37 1 (450, 664) 38 3 (1158, 383), (58, 1096), (58, 7488) 39 (474, 808), (176/5, 714/15) 40 1 (486, 880) 41 (498, 95), (398896/305, /166375) 4 1 (510, 304) 43 1 (5, 3096) 44 4 (534, 3168), (810, 1978), (11, 35640), ( /5, /15) 45 (546, 340), (590713/16, /64) 46 (558, 331), (909, 455) 47 (570, 3384), ( /465, / ) 48 3 (58, 3456), (3093, ), (530/9, 8970/7) 49 1 (594, 358) 50 (1317, 45549), (1558, 59768) 51 1 (618, 367) 5 3 (780, 15444), (630, 3744), (814, 17416) 53 1 (64, 3816) 54 1 (654, 3888) continued on next page

16 16 A. Bremner, R.J. Stroeker and N. Tzanakis continued from previous page (Table 1) Rank and generators of E k (Q), k = 1; : : : ; 100 rank k r k generators P 1 ; : : : ; P rk on (4) 55 (666, 3960), (440, ) 56 (678, 403), ( /1089, /35937) 57 1 (690, 4104) 58 (70, 4176), (5590, ) 59 1 (714, 448) 60 3 (1110, 31680), (813, 13167), (76, 430) 61 1 (738, 439) 6 1 (750, 4464) 63 (957, 131), (76, 4536) 64 (774, 4608), (1644, 635) 65 3 (1461, 5155), (786, 4680), (1068, 7468) 66 1 (798, 475) 67 3 (810, 484), (1398, 47160), (10198, ) 68 (8, 4896), (53374/5, /15) 69 1 (834, 4968) 70 1 (846, 5040) 71 (858, 511), (144448, ) 7 (870, 5184), (1584, 57816) 73 3 (334, ), (146, 3570), (88, 556) 74 3 (894, 538), (469, ), (35339/400, /8000) 75 (906, 5400), ( / , / ) 76 (918, 547), (1194, 30960) 77 (930, 5544), (164364/5, /15) 78 (94, 5616), (51878/539, /389017) 79 (954, 5688), ( /883041, / ) 80 1 (966, 5760) 81 (978, 583), ( /4096, /6144) 8 1 (990, 5904) 83 3 (53, 10303), (674, 13470), (78768/49, /343) 84 (1014, 6048), (6494/5, /15) 85 1 (106, 610) 86 (1038, 619), (81416/65, /1565) 87 (1050, 664), (1638, 5740) 88 (106, 6336), (13093/11, /1331) continued on next page

17 On sums of consecutive squares 17 continued from previous page (Table 1) Rank and generators of E k (Q), k = 1; : : : ; 100 rank k r k generators P 1 ; : : : ; P rk on (4) 89 (1074, 6408), ( / , / ) 90 1 (1086, 6480) 91 (1098, 655), (1968, 7963) 9 1 (1110, 664) 93 1 (11, 6696) 94 (1134, 6768), (38713/4, /8) 95 (1146, 6840), ( /5, /15) 96 (1158, 691), (967461/59, /1167) 97 (1170, 6984), (5777/9, /7) 98 (118, 7056), (1677, 55809) 99 1 (1194, 718) 100 (106, 700), (18481, ).4 A rank 1 conjecture From Table 1 it can be seen that for all 31 curves of rank 1 the point Q k = (1k + 6; 7k) on (4) serves as a generator for E k (Q)=E k (Q) tors. Can this be a coincidence? We think not, but we have no more hard evidence than these 31 examples. Nevertheless, we wish to formulate the Conjecture. If the curve given by (4) has rank 1, then E k (Q)=E k (Q) tors ' hq k i: Support of a heuristic nature may be found in the following remark which contains ideas due to Samir Siksek. We are grateful to him for allowing us to use them. Remark. The Szpiro ratio of an elliptic curve E over Q is the ratio E = log (discriminant E)= log (conductor E); and is conjectured to be bounded. Hindry and Silverman [6] show that all non-torsion points P E(Q) satisfy ^h(p ) (0 E ) :1 4 E log(discriminant E), so that applied to the curves E k, we obtain an estimate ^h(p ) > c log(k) where c is an absolute and

18 18 A. Bremner, R.J. Stroeker and N. Tzanakis eective constant, provided E is bounded. On the other hand, our estimates indicate that ^h(q k ) is asymptotic to 1 log(kp 6). So if Q k generates a subgroup of index m in E k (Q), then Q k = mq 0 k + T, for some Q0 k E(Q) and torsion point T, and m c log(k) < m ^h(q 0 k ) = ^h(q k ) 1 log(kp 6) implying a uniform bound on the index m for suciently large k. It now seems plausible that for a rank 1 curve, there can only be nitely many k where the index m exceeds 1. For otherwise, there exists m 0 N such that Q k = m 0 (x; y) as an equation in E k (Q) is solvable for x; y Q for innitely many k. Equating rst components, there results an equation F (x; k) = 0 of degree m 0 in x, which is known to have innitely many rational solutions x; k. Further, Q k = m 0 (x; y) forces x to be an integer, so F (x; k) = 0 has innitely many integer solutions x; k. Consequently, F (x; k) must represent a curve of genus 0, which seems unlikely in general. 3 Determination of integral points Now that the rank r k and a complete set of generators for E k (Q) are known, set E k (Q)=E k (Q) tors = hp 1 ; : : : ; P rk i: For P E k (Q), there exist integers m 1 ; : : : ; m rk such that P = m 1 P m rk P rk + P 0 ; (9) where P 0 is a torsion point, satisfying (from Lemma 1) P 0 = 0 in E k (Q). For integral P = (x(p ); y(p )) we intend to estimate the integral vector m = (m 1 ; : : : ; m rk ). Once (small) upper bounds for its coordinates m i are known, an attempt can be made to recover all integral points by direct search. 3.1 Elliptic logarithms Considering m as a column vector, then ^h(p ) = m T H k m where H k is the r k r k heightpairing matrix H k = 1 hp i; P j i with hr; Si = ^h(r + S) ^h(r) ^h(s) the Neron-Tate pairing. The matrix H k is positive denite and hence ^h(p ) k M k (30) where k is the smallest eigenvalue of H k and M k = max 1irk jm i j.

19 On sums of consecutive squares 19 Suppose now P is an integral point of (7) with X(P ) > 0, that is, x(p ) > 6(k 1). Then P is a point on the component of E k (R) containing the identity 0 of E k (Q). The known boundedness of x(p ) can be expressed by saying that P cannot be too close to 0. In order to measure the distance between P and 0, we use the group isomorphism : E 0 k(r)! R=Z(circle group) where E 0 = k E k for k and E1(R) 0 is the noncompact component of E 1 (R), with given by (P ) 8 >< >: 0 mod 1 if P = 0; 1! Z 1 x(p ) dt p t3 36t + b k mod 1 if y(p ) 0; ( P ) mod 1 if y(p ) < 0: (see Zagier [14, p. 49]). Here b k = 864k(k 1)(k 1), and! = R 1 6(k 1) p dt t 3 36t+b k is the fundamental real period of the Weierstra }-function associated with (4). There is no loss of generality in assuming that (P ) [0; 1), so that (P ) [0; 1 ] when y(p ) 0, which henceforth we shall assume. The quantities u i =!(P i ) are known as the elliptic logarithms associated with the basis fp 1 ; : : : ; P rk g. Applying to (9) yields (P ) m 1 (P 1 ) + + m rk (P rk ) + 1 " mod 1; where " = 0; 1, according to whether P 0 = 0, P 0 6= 0, respectively. Hence there exists m 0 Zsuch that (P ) = m " + m 1(P 1 ) + + m rk (P rk ): (31) Clearly, jm 0 j 1 + jm 1 j + + jm rk j 1 + r k M k. Multiplying by u 0 =! and setting L(P ) =!(P ) yields L(P ) = m " u 0 + m 1 u m rk u rk : (3) It is now straightforward to obtain an upper bound for jl(p )j in terms of k and M k. Lemma 3. Let P = (x(p ); y(p )) E k (Q) be an integral point of (4) with x(p ) > 1k, y(p ) > 0. Suppose that P satises (9), and let L(P ) be as in (3). Then where jl(p )j d k p k exp( k M k ); d 1 = 53: and d k = 4:08 for k :

20 0 A. Bremner, R.J. Stroeker and N. Tzanakis Proof. From the denition of and by (3), it follows that jl(p )j = Z 1 X(P ) Z 1 X(P ) dt p t3 + 18(k 1)t + 7(6k 6k + 1)t t 3= dt X(P ) 1= : Using the estimates of Lemma applied to (30), together with x(p ) > 1k, we deduce for k, 6(k 1) log X(P ) = log(x(p ) 6(k 1)) = log x(p ) + log 1 x(p ) = h(p ) + log 1 k M k 3 log 3 6(k 1) x(p ) 1 + log (k 1) k M k log k 1 log 7; 3 and the result follows. For k = 1 the reasoning is similar. 1 k The upper bound for jl(p )j of Lemma 3, combined with Sinnou David's lower bound (Lemma 4) produces an upper bound for M k. We shall state this lower bound for jl(p )j as it applies to the curves (4), referring the reader to Stroeker & Tzanakis [14] for further details. Lemma 4. (David) Let P E k (Q) be as in Lemma 3. Put ( log 178; if k = 1; h k = log ((1k 1k 1)(6k 6k + 1) ) ; if k ; and for j = 0; : : : ; r k, let A j be a positive number satisfying (where P 0 = 0 by denition). If A j maxf^h(p j ); h k g (33) B k maxfexp(a 0 ); : : : ; exp(a rk ); jm 0 j + 1; jm 1 j; : : : ; jm rk j; 16g; (34) then a lower bound for jl(p )j is given by where jl(p )j exp c k (log B k + 1)(log log B k h k ) r k+ ; c k = 10 7r k+15 (rk +1) (r k + ) 4r k +18r k+14 e r k Y j=0 A j :

21 On sums of consecutive squares 1 This is a special case of David [3, Theoreme.1]). Remark. Following closely Stroeker & Tzanakis [14, Appendix], we have, for k, taken the following fundamental periods for E k :! 1 = 1 + ;! = 1 and =! 1 =! ; where 1 = = p6 q M 4p p 3(k 1) 1; 1 18(k 1) + 1 3(k 1) 1 i p6 q M 4p p 3(k 1) 1; 1 18(k 1) + 1 3(k 1) 1 ; : Here M(u; v) denotes the AGM (Arithmetic-Geometric Mean) of u and v. Also note that 1 = 1! is the real period of the Weierstra } function associated with (4). Then satises the requirements < = 1, = > 0, jj 1, and for j = 0; : : : ; r k we have For k = 1, we have chosen Then! =! 1 and! 1 = 3u j j! 1 j = < 3 < min(h k): k M( p 1; p 6) ;! =! 1 i; and =! =! 1 = i: 3u j j! 1 j = < 3 4 < min (h k ): k Moreover, the number E of [14, Appendix] has been chosen equal to e. Corollary 5. If B k satises the inequality (34) then k M k < c k (log B k + 1)(log log B k h k ) r k+ + log(d k p k): (35) Proof. Combine Lemmas 3 and 4. Remark. If we take B k = r k M k + 3 then B k maxfjm 0 j + 1; jm 1 j; : : : ; jm rk jg. Furthermore, if M k is taken suciently large to meet the remaining conditions of (34), then (35) says that M k cannot be too large. It is clear from (35) that r k is the dominant factor in the calculation of the upper bound for M k. For this reason we shall put the curves E k into classes, depending on their

22 A. Bremner, R.J. Stroeker and N. Tzanakis rank. Only the rst curve (k = 1) will be treated separately. Second to the rank, k is the major contributor to the size of the upper bound for M k ; the values of h k, d k and the A j have only a minor inuence. Therefore it is not necessary to distinguish between the curves where these quantities are concerned. Since from (33) max h k k100 max log k100 we replace h k in (35) by Furthermore, max k (k 1)6 < 33:67; max ^h(pj ) = 8:3739 : : : ; 1jr k and hence A j may be chosen as 8.4 for all k 100, 1 j r k. Finally, p max log(d k k) < log 40:8 = 3:708 : : : ; 1k100 p which makes it possible to replace log(d k k) by 3.71 in (35) for all k 100. For k = 1 we have ^h(p 1 ) = 0:4443 : : : and h 1 = 7:454 : : :, so that we may choose A 0 = A 1 = 7:46. The following table gives particulars about the calculations of the upper bound for M k, broken into cases for k = 1 and for each of the rank-classes. Upper bound K for M k by (35) in the range 1 k 100 r k k k K 1 1 0:444 5:80 10 r k # k min k K :800 8: :607 5: :740 : :705 : LLL-reduction Clearly the resulting upper bounds K are far too large to be of practical use, and it is necessary now to apply the LLL-reduction process described in detail in Stroeker & Tzanakis [14, Sec. 5] in order to reduce the magnitude of the bounds. See also de Weger [16, Chap. 3]. A brief description of the procedure should suce here.

23 On sums of consecutive squares 3 From Lemma 3, j(p )j < K 1 exp( K M k ) and M k < K; (36) p where K 1 = d k k=!, and K = min k. But we can bound j(p )j from below, as follows. Let L be the (r k +1) dimensional lattice, generated by the columns of the integral matrix A L = 0 1 : : : : : : : : : 1 0 [K 0 (P 1 )] : : : [K 0 (P r )] K 0 (here [] means rounding towards 0, that is, [] = de if 0, and [] = bc if > 0) where K 0 will be a large integer that will be conveniently chosen later. If the vector (m 1 ; : : : ; m rk ; m 0 ) Z r k+1 satises jm j j < K for j = 1; : : : ; r k, put `k = A L 0 m 1. m rk m 0 + " 1 C A = 0 m 1. m rk t k 1 C A ; 1 C A ; with t k = r k X j=1 m i [K 0 (P j )] + (m 0 + ")K 0 : Then using (31), t k = r k X j=1 m i [K 0 (P j )] K 0 (P j ) + K 0 (P ) so that jt k j r k K + K 0 (P ) and k`kk = 4 r k X i=1 m i + t k 4r k K + 4(r k K + K 0 j(p )j) : (37) On the other hand, if fb 1 ; : : : ; b rk +1g is an LLL-reduced basis of L, then kb 1 k r k k`kk : (38)

24 4 A. Bremner, R.J. Stroeker and N. Tzanakis Combining (37) with (38) yields K 0 j(p )j p r k kb 1 k r k K r k K: (39) From (36), (39), we have a new upper bound for M k, given implicitly by the inequality M k K 1 p log(k 0 K 1 ) log( r k kb 1 k r k K r k K) ; (40) provided that the right-hand side of (39) is positive, that is, provided q kb 1 k > 1+r k = K r + k r k: (41) It is reasonable to expect that kb 1 k (det A L ) 1=(r k+1) = K 1=(r k+1) 0. Therefore, if we choose K 0 such that K 1=(r k+1) 0 is slightly larger than the right-hand side of (41), then it is likely (41) will be satised. In that case, (40) produces a new upper bound which is of the size of p log K, a considerable improvement. There is nothing to prohibit using this reduction process as many times as possible; we found in practice that at most three reduction steps were needed to bring the upper bound for M k down to a manageable size (between 3 and 6). In order to execute the reduction process, it is necessary to know the values of (P j ) to a great number of decimal places (600 in the case of rank 4). Zagier [17] describes a very ecient algorithm to compute these values to the precision needed, and this was programmed in the very fast Ubasic language. The LLL-reduction step was carried out using the integral LLL algorithm of Pari/gp. Below we list the outcome of each reduction process. Reduction process for 1 k 100 (d stands for the number of digits precision) step 1 step step 3 k K 0 d K K 0 d K K 0 d K step 1 step step 3 r k K 0 d K K 0 d K K 0 d K ; ; : The in the nal column indicates one exception at k = 79, in which case the process stopped at step with K value of 6.

25 On sums of consecutive squares All integral points for 1 k 100 When completing the nal search for integral points with the K values from the above table, the worst instances occur for rank 4 curves. Corresponding to m 1 > 0; m 1 = 0, m > 0; m 1 = m = 0, m 3 > 0; m 1 = m = m 3 = 0, m 4 > 0 in (9), there are respectively , 6 13, 6 13, 6 cases to consider, a total of = 8560 points to check for integrability. For each k-value, this search took about three hours using Apecs on a 486 desktop with 16 Mbytes of extended memory. For the other ranks, this nal search was signicantly shorter. A list of all the integral points found is given in Table. In fact the coecients m j corresponding to integral points rarely exceed unity; there are only ve instances where this is not true (for k = 1; ; 7; 9 and 0), and then the largest (absolute) coecient is. Table : Integer points on (7), k = 1; : : : ; 100 All integer points (X; Y ) with Y 0 for the curves (7), k = 1; : : : ; 100, omitting in each case the point (0; 0) k (X; Y ) 1 (-1,0), (-6,0), (-9, 9), (-8, 8), (6, 36), (1, 7), (88, 5040) (1, 144), (78, 936) 3 (1, 16), (18, 88), (4, 360), (111, 1665), (148, 368), (, 3996), (178, 10375), (6936, ), (1153, 1434) 4 (4, 15), (1, 88), (147, 583), (438, 1051), (1314, 4993), (883, ) 5 (1, 360), (76, 1780) 6 (1, 43), (1086, 39096) 7 (1, 504), (46, 1196), (66, 1584), (88, 04), (07, 4761), (76, 664), (396, 1096), (600, 17640), (88, 9736), (1518, 63756), (08, 10996), (18975, 69935), (6508, ) 8 (1, 576), (196, 4760), (363, 9603), (0, 97056) 9 (1, 648), (88, 763), (94, 781), (598, 1409), (664588, ) 10 (1, 70), (75, 95), (196, 530), (346, ) 11 (1, 79), (3966, 61756), (74776, ) 1 (1, 864), (76, 5344), (4758, 34576) 13 (1, 936), (150, 4860), (115, 46944), (56, ) 14 (4, 568), (1, 1008), (53, ), (6558, 55087), (19674, ), (38307, ) 15 (1, 1080), (78, 376), (1164, 48888), (3744, 453), (7566, ) continued on next page

26 6 A. Bremner, R.J. Stroeker and N. Tzanakis continued from previous page (Table ) All integer points (X; Y ) with Y 0 for the curves (7), k = 1; : : : ; 100, omitting in each case the point (0; 0) k (X; Y ) 16 (1, 115), (1156, 49096), (8646, ) 17 (1, 14), (07, 7659), (76, 9936), (46, 15336), (568, 1016), (9798, ), (63948, ) 18 (1, 196), (13, 5544), (100, 4084), (475, ), (110, ) 19 (1, 1368), (1318, 14045) 0 (1, 1440), (4, 088), (16, 700), (88, 11376), (94, 1159), (1587, 7765), (178, ), (3468, 516), (6843, ), (13686, 16430), (1873, ), (0164, 9137), (85698, ), ( , ) 1 (1, 151), (115, 518), (1516, ) (1, 1584), (8, 11844), (376, 15416), (531, 1771), (708, 9736), (13818, ), (16638, 19616), (8668, ), ( , ) 3 (1, 1656), (18, ) 4 (1, 178), (19878, 8643) 5 (6, 160), (1, 1800), (39, 3393), (88, 13104), (31, 14040), (588, 5704), (831, 37395), (1300, 63440), (1734, 91188), (548, 15144), (6648, ), (700, 64870), (18954, ), (1606, ), (431, ), (5900, ), (177679, ), (496999, ) 6 (1, 187), (3406, ) 7 (1, 1944), (88, 5896), (16, 8856), (396, 1816), (766, 3536), (1650, 8710), (3447, 30949), (4056, 894), (578, ) 8 (1, 016), (150, 8640), (600, 770), (115, 56736), (0667, ), (7, ) 9 (1, 088), (938, ) 30 (1, 160), (08, ), (3136, ) 31 (1, 3), (33486, 68396) 3 (1, 304), (18, 844), (6936, 65464), (381, 37696), (30603, ), (35718, ), (4867, ), (735000, ), ( , ) 33 (1, 376), (380, ) 34 (1, 448), (40398, 84119) 35 (1, 50), (4846, ), (138750, ) continued on next page

27 On sums of consecutive squares 7 continued from previous page (Table ) All integer points (X; Y ) with Y 0 for the curves (7), k = 1; : : : ; 100, omitting in each case the point (0; 0) k (X; Y ) 36 (1, 59), (1875, 11005), (45366, ) 37 (1, 664), (47958, ) 38 (1, 736), (78, 7488), (13, 1096), (531, 905), (708, 383), (858, 4633), (1144, 690), (475, ), (3744, 71440), (460, ), (7788, ), (506, ), (1457, ) 39 (1, 808), (53358, ), (5848, ) 40 (1, 880), (56166, ) 41 (1, 95), (59046, ) 4 (1, 304), (61998, ) 43 (1, 3096), (650, ) 44 (1, 3168), (147, 1537), (88, 1978), (600, 35640), (1734, ), (178, ), (4483, ), (68118, ) 45 (1, 340), (460374, ), (7186, 19470) 46 (1, 331), (363, 455), (7456, ) 47 (1, 3384), (77838, ) 48 (1, 3456), (53, ), (0667, ), (81, ), (581597, ) 49 (1, 358), (84678, ) 50 (1, 3600), (73, 45549), (964, 59768), (1098, 68076), (1464, 93), (8806, ), (187500, ) 51 (1, 367), (91806, ) 5 (1, 3744), (16, 15444), (196, 17416), (115, 7864), (4056, ), (11163, 17861), (95478, ) 53 (1, 3816), (99, ) 54 (1, 3888), (103038, ) 55 (1, 3960), (41748, ), (10696, ) 56 (1, 403), (110886, ) 57 (1, 4104), (114918, ) 58 (1, 4176), (4900, ), (1190, ) 59 (1, 448), (13198, ) 60 (1, 430), (99, 13167), (396, 31680), (386, ), (15448, ), (0164, ), (38148, ), (17446, ) 61 (1, 439), (131766, ) 6 (1, 4464), (136158, ) 63 (1, 4536), (07, 131), (815, ), (1406, ) continued on next page

28 8 A. Bremner, R.J. Stroeker and N. Tzanakis continued from previous page (Table ) All integer points (X; Y ) with Y 0 for the curves (7), k = 1; : : : ; 100, omitting in each case the point (0; 0) k (X; Y ) 64 (1, 4608), (88, 635), (904, 0968), (145158, ) 65 (1, 4680), (94, 7468), (687, 5155), (616, 19600), (10368, ), (1603, ), (37098, ), (149766, ) 66 (1, 475), (154446, ) 67 (1, 484), (600, 47160), (178, ), (3077, ), (101400, ), (159198, ) 68 (1, 4896), (1640, ) 69 (1, 4968), (168918, ) 70 (1, 5040), (173886, ) 71 (1, 511), (17896, 76476), (144160, ) 7 (1, 5184), (76, 57816), (184038, ) 73 (1, 556), (376, 3570), (1098, 83448), (1464, ), (1551, 11635), (068, ), (6039, ), (794, 86804), (189, ), (7361, ) 74 (1, 538), (1587, ), (194478, ), (43777, ) 75 (1, 5400), (199806, ) 76 (1, 547), (88, 30960), (14406, ), (0506, ) 77 (1, 5544), (10678, ) 78 (1, 5616), (16, ) 79 (1, 5688), (1838, ) 80 (1, 5760), (756, ) 81 (1, 583), (3386, ) 8 (1, 5904), (39118, ) 83 (1, 5976), (163, 10303), (1684, 13470), (1746, ), (38, ), (450, ), (31105, ) 84 (1, 6048), (50998, ) 85 (1, 610), (57046, ) 86 (1, 619), (63166, ) 87 (1, 664), (600, 5740), (7938, ), (69358, ) 88 (1, 6336), (756, ) 89 (1, 6408), (81958, ) 90 (1, 6480), (88366, ) 91 (1, 655), (88, 7963), (904, 49480), (94846, ) 9 (1, 664), (301398, ) continued on next page

29 On sums of consecutive squares 9 continued from previous page (Table ) All integer points (X; Y ) with Y 0 for the curves (7), k = 1; : : : ; 100, omitting in each case the point (0; 0) k (X; Y ) 93 (1, 6696), (3080, ) 94 (1, 6768), (314718, ) 95 (1, 6840), (31486, ), (5786, ) 96 (1, 691), (3836, ) 97 (1, 6984), (33538, ) 98 (1, 7056), (507, 55809), (1100, 15810), (34, ) 99 (1, 718), (34978, ) 100 (1, 700), (183618, ), (356406, ) From Table, it is immediate to deduce a full list of integer solutions (n; t) to equation () in the range 1 k 100. These are listed in Table 3, in the form of triples (k; k + n 1; t). Table 3: Integer solutions of (1), k = 1; : : : ; 100 All integer solutions (k + n 1; t) with t > k of (1) No entry for k indicates no solution exists k (k + n 1; t) 1 (4, 70) 3 (4, 5), (580, 8075), (963, 1767) 7 (9, 9), (39, 143), (56, 45), (190, 1518), (15, 6007) 9 (3, 106), (55057, ) 11 (908, ) 13 (108, 65) 15 (111, 679), (36, 3406) 17 (39, 138), (5345, 5643) 18 (8, 77) 0 (1, 9), (43, 158), (308, 318), (11044, ) 1 (116, 74) (80, 413), (6910, ) 5 (48, 18), (50, 195), (73, 357), (578, 8033), (64, 9010), (365, 16035), (164, ) 7 (59, 53), (364, 4017) continued on next page

30 30 A. Bremner, R.J. Stroeker and N. Tzanakis continued from previous page (Table 3) All integer solutions (k + n 1; t) with t > k of (1) No entry for k indicates no solution exists k (k + n 1; t) 8 (77, 385), (13, 788) 30 (198, 161) 3 (609, 8687), (4087, ), (6181, ), (148856, ) 38 (48, 143), (96, 531), (349, 3770), (686, 10384), (11918, 7518) 44 (67, 74), (93, 495) 50 (171, 181), (15674, ) 5 (147, 101), (389, 4433) 55 (3533, 1168) 58 (13, 56855) 60 (9, 440), (338, ) 64 (305, 3069) 65 (8, 75), (98, 1633) 67 (116, 655), (8516, ) 73 (194, 155), (873, ) 74 (36554, ) 76 (99, 430) 83 (76, 619), (6003, 40957) 87 (136, 795) 91 (33, 3465) 3.4 The rank 1 case Now we restrict attention to the case where the rank of E k (Q) equals 1. The point Q k = (1; k) at (), respectively, Q k = (1k +6; 7k) on (4), is a point of innite order. We show here that for any integer k, then neither mq k (m > 1) nor mq k +T k (m 1) can be an integer point of (). By virtue of the previous determination of E k (Z) for 1 k 100, we could assume k > 100, but in fact we shall assume only k, implying that the torsion point T k is (0; 0) on () or as at (5) on (4). A consequence of this result is that if Q k is indeed a generator for the group E k (Q)=E k (Q) tors, then the only integer solution of () is (n; t) = (1; k). The approach of this section is much in the spirit of Ayad [1] and it is from that paper that it has been inspired. The idea is in essence quite simple. When expressing the x-coordinate of mq k as the quotient of two polynomials in Z[k], it turns out that the resultant of these two polynomials is an integer divisible by only and 3; so any common