Lucas sequences whose 12th or 9th term is a square

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1 Journal of Number Theory 107 (2004) Lucas sequences whose 12th or 9th term is a square A. Bremner a, and N. Tzanakis b a Department of Mathematics, Arizona State University, Box , Tempe AZ , USA b Department of Mathematics, University of Crete, Iraklion, Greece Received 20 May 2002; revised 18 February 2004 Communicated by D.J. Lewis Abstract Let P and Q be non-zero relatively prime integers. The Lucas sequence fu n ðp; QÞg is defined by U 0 ¼ 0; U 1 ¼ 1; U n ¼ PU n 1 QU n 2 ðnx2þ: We show that the only sequence with U 12 ðp; QÞ a perfect square is the Fibonacci sequence fu n ð1; 1Þg; and we show that there are no non-degenerate sequences fu n ðp; QÞg with U 9 ðp; QÞ a perfect square. The argument involves finding all rational points on several curves of genus 2. r 2004 Elsevier Inc. All rights reserved. Keywords: Lucas sequence; Elliptic curve; Formal group 1. Introduction Let P and Q be non-zero relatively prime integers. The Lucas sequence fu n ðp; QÞg is defined by U 0 ¼ 0; U 1 ¼ 1; U n ¼ PU n 1 QU n 2 ðnx2þ: ð1þ Corresponding author. Fax: addresses: bremner@asu.edu (A. Bremner), tzanakis@math.uoc.gr (N. Tzanakis). URLs: X/$ - see front matter r 2004 Elsevier Inc. All rights reserved. doi: /j.jnt

2 216 A. Bremner, N. Tzanakis / Journal of Number Theory 107 (2004) The sequence fu n ð1; 1Þg is the familiar Fibonacci sequence, and it was proved by Cohn [12] in 1964 that the only perfect square greater than 1 in this sequence is U 12 ¼ 144: The question arises, for which parameters P; Q; can U n ðp; QÞ be a perfect square? This has been studied by several authors: see for example [13 15,22,25]. Using Baker s method on linear forms in logarithms, work of Shorey and Tijdeman [26] implies that there can only be finitely many squares in the sequence fu n ðp; QÞg: Ribenboim and McDaniel [23] with only elementary methods show that when P and Q are odd, and P 2 4Q40; then U n can be square only for n ¼ 0; 1; 2; 3; 6 or 12; and that there are at most two indices greater than 1 for which U n can be square. They characterize fully the instances when U n ¼ &; for n ¼ 2; 3; 6; and observe that U 12 ¼ & if and only if there is a solution to the Diophantine system P ¼ &; P 2 Q ¼ 2&; P 2 2Q ¼ 3&; P 2 3Q ¼ &; ðp 2 2QÞ 2 3Q 2 ¼ 6&: ð2þ When P is even, a later paper of Ribenboim and McDaniel [24] proves that if Q 1 ðmod 4Þ; then U n ðp; QÞ ¼& for n40 only if n is a square or twice a square, and all prime factors of n divide P 2 4Q: Further, if p 2t jn for a prime p; then U p 2u is square for u ¼ 1; y; t: In addition, if n is even, then U n ¼ & only if P is a square or twice a square. A remark is made that no example is known of an integer pair P; Q; and an odd prime p; such that U p 2 ¼ & (note none can exist for P; Q odd, P 2 4Q40). In this paper, we complete the results of Ribenboim and MacDaniel [23] by determining all Lucas sequences fu n ðp; QÞg with U 12 ¼ & (in fact, the result is extended, because we do not need the restrictions that P; Q be odd, and P 2 4Q40): it turns out that the Fibonacci sequence provides the only example. Moreover, we also determine all Lucas sequences fu n ðp; QÞg with U 9 ¼ &; subject only to the restriction that ðp; QÞ ¼1: Throughout this paper the symbol & means square of a non-zero rational number. Theorem 1. Let ðp; QÞ be any pair of relatively prime non-zero integers. Then, * U 12 ðp; QÞ ¼& iff ðp; QÞ ¼ð1; 1Þ (corresponding to the Fibonacci sequence). * U 9 ðp; QÞ ¼& iff ðp; QÞ ¼ð72; 1Þ (corresponding to the sequences U n ¼ n and U n ¼ð 1Þ nþ1 n). The remainder of the paper is devoted mainly to the proof of this theorem. Theorems 3 and 6 of [23] combined with the first statement of Theorem 1 imply the following. Theorem 2. Let P; Q be relatively prime odd integers, such that P 2 4Q40: Then the nth term, n41; of the Lucas sequence U n ¼ U n ðp; QÞ can be a square only if n ¼ 2; 3; 6 or 12. More precisely: 1 1 Below it is understood that parameters a; b are in every case chosen so that P; Q are odd, relatively prime and P 2 4Q40:

3 A. Bremner, N. Tzanakis / Journal of Number Theory 107 (2004) * U 2 ¼ & iff P ¼ a 2 : * U 3 ¼ & iff P ¼ a; Q ¼ a 2 b 2 : * U 6 ¼ & iff P ¼ 3a 2 b 2 ; Q ¼ a8 þ12a 4 b 4 9b 8 2 : * U 12 ¼ & iff ðp; QÞ ¼ð1; 1Þ: Moreover, this result is also valid even if we remove all restrictions on P; Q except for gcdðp; QÞ ¼1: The proof of Theorem 1 hinges, in both cases, upon finding all rational points on a curve of genus 2. When the rank of the Jacobian of such a curve is less than 2, then methods of Chabauty [11], as expounded subsequently by Coleman [16], Cassels and Flynn [10] and Flynn [17] may be used to determine the (finitely many) rational points on the curve. When the rank of the Jacobian is at least 2, as is the case here, a direct application of these methods fails. In order to deal with such situations, very interesting methods have been developed recently by a number of authors; see Chapter 1 of Wetherell s Ph.D. thesis [31], Bruin [2 4], Bruin and Flynn [7,8], Flynn [18], and Flynn and Wetherell [19,20]. For the purpose of this paper, the method of [18,19] is sufficient. 2. The Diophantine equations 2.1. The case U 12 For U 12 ðp; QÞ to be square, we have from (1) U 12 ðp; QÞ ¼PðP 2 3QÞðP 2 2QÞðP 2 QÞðP 4 4P 2 Q þ Q 2 Þ¼&: ð3þ Now ðpðp 2 3QÞðP 2 QÞ; ðp 2 QÞðP 4 4P 2 Q þ Q 2 ÞÞ divides 2, so that U 12 ¼ & implies PðP 2 3QÞðP 2 QÞ ¼d&; ðp 2 2QÞðP 4 4P 2 Q þ Q 2 Þ¼d&; where d ¼ 71; 72: With x ¼ Q=P 2 ; we deduce ð1 2xÞð1 4x þ x 2 Þ¼d& and of these four elliptic curves, only the curve with d ¼ 2 has positive rational rank. Torsion points on the three other curves do not provide any solutions for P; Q: We are thus reduced to considering the equations PðP 2 3QÞðP 2 QÞ ¼2&; ðp 2 2QÞðP 4 4P 2 Q þ Q 2 Þ¼2&: From the first equation, PðP 2 3QÞ ¼72&; P 2 Q ¼ 7&; or PðP 2 3QÞ ¼7&; P 2 Q ¼ 72&:

4 218 A. Bremner, N. Tzanakis / Journal of Number Theory 107 (2004) The former case implies one of P ¼ d& P 2 3Q ¼ 2d& P 2 Q ¼ &; P ¼ d& P 2 3Q ¼ 2d& P 2 Q ¼ &; P ¼ 2d& P 2 3Q ¼ d& P 2 Q ¼ &; where d ¼ 71; 73: The latter case implies one of P ¼ 2d& P 2 3Q ¼ d& P 2 Q ¼ &; ð4þ P ¼ d& P 2 3Q ¼ d& P 2 Q ¼ 2&; P ¼ d& P 2 3Q ¼ d& P 2 Q ¼ 2&; ð5þ where d ¼ 71; 73: Solvability in R or elementary congruences shows impossibility of the above equations (4), (5), except in the following instances: P ¼ &; P 2 3Q ¼ 2&; P 2 Q ¼ &; P ¼ 3&; P 2 3Q ¼ 6&; P 2 Q ¼ &; P ¼ 6&; P 2 3Q ¼ 3&; P 2 Q ¼ &; P ¼ &; P 2 3Q ¼ 2&; P 2 Q ¼ &; P ¼ 6&; P 2 3Q ¼ 3&; P 2 Q ¼ &; P ¼ &; P 2 3Q ¼ &; P 2 Q ¼ 2&; Recall now that P ¼ 3&; P 2 3Q ¼ 3&; P 2 Q ¼ 2&: ð6þ ðp 2 2QÞðP 4 4P 2 Q þ Q 2 Þ¼2&;

5 A. Bremner, N. Tzanakis / Journal of Number Theory 107 (2004) from which P 2 2Q ¼ Z&; P 4 4P 2 Q þ Q 2 ¼ 2Z& or P 2 2Q ¼ 2Z&; P 4 4P 2 Q þ Q 2 ¼ Z&; where Z ¼ 71; 73: The only locally solvable equations are P 2 2Q ¼ &; P 2 2Q ¼ 3&; P 4 4P 2 Q þ Q 2 ¼ 2&; P 4 4P 2 Q þ Q 2 ¼ 6&; P 2 2Q ¼ 2&; P 4 4P 2 Q þ Q 2 ¼ &: ð7þ It is straightforward by elementary congruences to deduce from (6), (7), that we must have one of the following: P P 2 3Q P 2 Q P 2 2Q P 4 4P 2 Q þ Q 2 & 2& & & 2& 6& 3& & 2& & & 2& & & 2& & & 2& 3& 6& Now the rational ranks of the following elliptic curves are 0: ð x þ 1Þðx 2 4x þ 1Þ ¼ 2&; ð 3x þ 1Þðx 2 4x þ 1Þ ¼3&; ð x þ 1Þðx 2 4x þ 1Þ ¼2&; and consequently the rational points on the curves corresponding to the first three rows of the above table are straightforward to determine: they are ðp; QÞ ¼ð 1; 1Þ; ð0; 1Þ; and ð1; 1Þ; respectively. These lead to degenerate Lucas sequences with U 12 ¼ 0: It remains only to find all rational points on the following curve: P ¼ &; P 2 3Q ¼ &; P 2 Q ¼ 2&; P 2 2Q ¼ 3&; P 4 4P 2 Q þ Q 2 ¼ 6&; satisfying ðp; QÞ ¼1: Note that this is the curve (2), though we have removed the restriction that P and Q be odd, and P 2 4Q40: Put Q=P 2 ¼ 1 2u 2 ; so that 3u 2 1 ¼ 2&; 4u 2 1 ¼ 3&; 2u 4 þ 2u 2 1 ¼ 3&: ð8þ Eq. (8) defines a curve of genus 9, with certainly only finitely many points. We restrict attention to the curve of genus 2 defined by 4u 2 1 ¼ 3&; 2u 4 þ 2u 2 1 ¼ 3&:

6 220 A. Bremner, N. Tzanakis / Journal of Number Theory 107 (2004) pffiffiffi pffiffiffi p ffiffi Define K ¼ Qð 3 Þ; with ring of integers OK ¼ Z½ 3 Š; and fundamental unit 2 þ 3 : Observe that ðu 2 1Þ 2 3u 4 ¼ 3& implies pffiffiffi pffiffiffi u 2 1 þ u 2 3 ¼ e 3 g 2 ; p for e a unit of O K of norm þ1; and gao K : If e ¼ 2 þ ffiffi 3 ; the resulting equation is locally unsolvable above 3, and so without loss of generality, e ¼ 1: Consider now p u 2 ð4u 2 1Þðu 2 ð1 þ ffiffi pffiffi 3 Þ 1Þ ¼3 3 V 2 ; VAK: p In consequence, ðx; yþ ¼ðð12 þ 4 ffiffi p 3 Þu 2 ; ð36 þ 12 ffiffi 3 ÞVÞ is a point defined over K on the elliptic curve p E 1 : y 2 ¼ xðx ð3þ ffiffiffi pffiffi 3 ÞÞðx 4 3 Þ ð9þ ffiffi satisfying ð3 p 3 Þ 24 xaq 2 : We shall see that the K-rank of E 1 is equal to 1, with p ffiffiffi pffiffiffi generator of infinite order P ¼ð 3 ; 3 3 Þ: 2.2. The case U 9 For U 9 ðp; QÞ to be square, we have from (1) U 9 ðp; QÞ ¼ðP 2 QÞðP 6 6P 4 Q þ 9P 2 Q 2 Q 3 Þ¼&: ð10þ So P 2 Q ¼ d&; P 6 6P 4 Q þ 9P 2 Q 2 Q 3 ¼ d&; where 7d ¼ 1; 3: Put Q ¼ P 2 dr 2 : Then 3 d P6 9P 4 R 2 þ 6dP 2 R 4 þ d 2 R 6 ¼ &; ð11þ with covering elliptic curves 3 d 9x þ 6dx2 þ d 2 x 3 ¼ &; 3 d x3 9x 2 þ 6dx þ d 2 ¼ &: ð12þ For d ¼ 71; the first curve has rational rank 0, and torsion points do not lead to non-zero solutions for P; Q: For d ¼ 73; both curves at (12) have rational rank 1, so that the rank of the Jacobian of (11) equals 2. To solve the equation U 9 ðp; QÞ ¼&; it is necessary to determine all integer points on the two curves P 6 9P 4 R 2 þ 18P 2 R 4 þ 9R 6 ¼ & ð13þ

7 A. Bremner, N. Tzanakis / Journal of Number Theory 107 (2004) and P 6 9P 4 R 2 18P 2 R 4 þ 9R 6 ¼ &: ð14þ To this end, let L ¼ QðaÞ be the number field defined by a 3 3a 1 ¼ 0: Gal(L=Q) is cyclic of order 3, generated by s; say, where a s ¼ 1=ð1 þ aþ ¼ 2 a þ a 2 : The ring of integers O L has basis f1; a; a 2 g; and class number 1. Generators for the group of units in O L are e 1 ¼ a; e 2 ¼ 1 þ a; with norms Normðe 1 Þ¼1; Normðe 2 Þ¼ 1: The discriminant of L=Q is 81, and the ideal (3) factors in O L as ð 1 þ aþ 3 : Eq. (13) may be written in the form and it follows that Norm L=Q ðp 2 þð 5 þ a þ a 2 ÞR 2 Þ¼S 2 ; say; P 2 þð 5 þ a þ a 2 ÞR 2 ¼ lu 2 ; ð15þ with lao L squarefree and of norm þ1 modulo L 2 : Applying s; P 2 þð 5 2a þ a 2 ÞR 2 ¼ l s V 2 : ð16þ Suppose P is a first degree prime ideal of O L dividing ðlþ: Then for the norm of l to be a square, l must also be divisible by one of the conjugate prime ideals of P: It follows that P; or one of its conjugates, divides both l and l s : Then this prime will divide ðð 5 þ a þ a 2 Þ ð 5 2aþa 2 ÞÞ ¼ ð3aþ ¼ð1 aþ 3 : So P has to be ð1 aþ; with ð1 aþ 2 dividing l; contradicting l squarefree. If the residual degree of P is3, then the norm of l cannot be square. Finally, the residual degree of P cannot be 2, otherwise y ¼ 5 a a 2 m ðmod PÞ; for some rational integer m; so that a ¼ 1 2y þ 1 3 y2 is congruent to a rational integer modulo P; impossible. In consequence, l is forced to be a unit, of norm þ1: Without loss of generality, the only possibilities are l ¼ 1; e 1 ; e 2 ; e 1 e 2 : However, specializing the left-hand side of (15) at the root a 0 ¼ 1: ::: of x 3 3x 1 ¼ 0 shows that P 2 þ 0:4114::R 2 ¼ lða 0 ÞUða 0 Þ 2 ; so that lða 0 Þ40; giving unsolvability of (15) for l ¼ e 2 ; e 1 e 2 : There remain the two cases l ¼ 1; with solution ðp; R; UÞ ¼ð1; 0; 1Þ; and l ¼ e 1 ; with solution ðp; R; UÞ ¼ð0; 1; 4 a 2 Þ: From (15) and (16) we now have P 2 ðp 2 þð 5 þ a þ a 2 ÞR 2 ÞðP 2 þð 5 2a þ a 2 ÞR 2 Þ¼mW 2 ; with m ¼ ll s ¼ 1or1þ a a 2 : Accordingly, X ¼ P 2 =R 2 gives a point on the elliptic curve: XðX þð 5 þ a þ a 2 ÞÞðX þð 5 2a þ a 2 ÞÞ ¼ my 2 : ð17þ

8 222 A. Bremner, N. Tzanakis / Journal of Number Theory 107 (2004) Now when m ¼ 1; a relatively straightforward 2-descent argument shows that the QðaÞ-rank of (17) is equal to 0 (we also checked this result using the Pari-GP software of Denis Simon [30]). The torsion group is of order 4, and no non-zero P; Q arise. When m ¼ 1 þ a a 2 ; then ðx; yþ ¼ curve E 2 over QðaÞ; where m P 2 ; ð1 aþ 2 R 2 m 2 W ð1 aþ 3 R 3 is a point on the elliptic E 2 : y 2 ¼ xðx þð 2 a þ a 2 ÞÞðx þð 1 þ a þ a 2 ÞÞ; ð18þ satisfying ð1 aþ2 m x ¼ð4 þ a 2a 2 ÞxAQ 2 : We shall see that the QðaÞ-rank of E 2 is 1, with generator of infinite order equal to ð1; aþ: Eq. (14) may be written in the form so that Norm L=Q ð P 2 þð 5 þ a þ a 2 ÞR 2 Þ¼S 2 ðsayþ; P 2 þð 5 þ a þ a 2 ÞR 2 ¼ lu 2 ; ð19þ with lao L squarefree and of norm þ1 modulo L 2 : Arguing as in the previous case, l must be a squarefree unit of norm þ1; so without loss of generality equal to 1, e 1 ; e 2 ; e 1 e 2 : Only when l ¼ e 1 is (19) solvable at all the infinite places. Thus P 2 þð 5 þ a þ a 2 ÞR 2 ¼ au 2 ; P 2 þð 5 2a þ a 2 ÞR 2 ¼ð 2 a þ a 2 ÞV 2 and x ¼ ð1þa a2 Þ P 2 is the x-coordinate of a point on the elliptic curve ð1 aþ 2 R 2 y 2 ¼ xðx þð2 þ a a 2 ÞÞðx þð1 a a 2 ÞÞ ð20þ satisfying ð1 aþ2 ð1þa a 2 Þ x ¼ð4þa 2a2 ÞxAQ 2 : However, a straightforward calculation shows that the QðaÞ-rank of (20) is equal to 0; with torsion group the obvious group of order 4. There are no corresponding solutions for P; Q: 3. The Mordell Weil basis Here we justify our assertions about the elliptic curves E 1 at (9) and E 2 at (18). These curves are defined over fields F (where F=K or L; respectively) with unique factorization, and have F-rational two-torsion. So standard two-descents over F work analogously to the standard two-descent over Q for an elliptic curve with rational two-torsion; see for example Silverman [28, Chapter 10.4]. It is thus

9 A. Bremner, N. Tzanakis / Journal of Number Theory 107 (2004) straightforward to determine generators for E i ðfþ=2e i ðfþ; i ¼ 1; 2 (and in fact software packages such as that of Simon [30] written for Pari-GP, and ALGAE [5] of Bruin written for KASH, with m-algae [6] for MAGMA, also perform this p ffiffiffi pffiffiffi calculation effortlessly). Such generators are the classes of P 1 ¼ð 3 ; 3 3 Þ for the curve E 1 ; and P 2 ¼ð1; aþ for the curve E 2 : In fact P 1 and P 2 are generators for the pffiffi Mordell Weil groups E 1 ðqð 3 ÞÞ and E2 ðqðaþþ; respectively. To show this necessitates detailed height calculations over the appropriate number field, with careful estimates for the difference ĥðqþ 1 2 hðqþ where ĥðqþ is the canonical height of the point Q; and hðqþ the logarithmic height. The KASH/TECC package of Kida [21] was useful here in confirming calculations. The standard Silverman bounds [29] are numerically too crude for our purposes, so recourse was made to the refinements of Siksek [27]. Full details of the argument are given in Section 3 of [1]. Actually, determination of the full F-rational Mordell Weil groups of E 1 and E 2 may be redundant; it is likely that the subsequent local computations can be performed subject only to a simple condition on the index in EðFÞ of a set of generators for EðFÞ=2EðFÞ: The reader is referred to Bruin [4] or Flynn and Wetherell [20] for details and examples. This latter technique must be used of course when the height computations are simply too time consuming to be practical. 4. Finding all points on (9) and (18) under the rationality conditions 4.1. General description of the method The problems to which we were led in Section 2 are of the following shape. Problem. Let E : y 2 þ a 1 xy þ a 3 y ¼ x 3 þ a 2 x 2 þ a 4 x þ a 6 ; ð21þ be an elliptic curve defined over QðaÞ; where a is a root of a polynomial f ðxþaz½xš; irreducible over Q; of degree dx2 and baqðaþ an algebraic integer. Find all points ðx; yþaeðqðaþþ for which bx is the square of a rational number. For the solution of this type of problem we adopt the technique described and applied in [19]. 2 Several problems of this type have already been solved with a similar technique (besides [19], see also [4,8,18,20]); therefore we content ourselves with a rather rough description of the employed method and refer the interested reader to Section 4 of [1] for a detailed exposition. We assume the existence of a rational prime p with the following properties: (i) f ðxþ is irreducible in Q p ½XŠ: (ii) The coefficients of (21) are in Z p ½aŠ: 2 But see also the references at the end of Section 1.

10 224 A. Bremner, N. Tzanakis / Journal of Number Theory 107 (2004) (iii) Eq. (21) is a minimal Weierstrass equation for E=Q p ðaþ at the unique discrete valuation v defined on QðaÞ with vðpþ ¼ 1: (iv) baq p ðaþ isap-adic unit. We work with both (21) and the associated model w ¼ z 3 þ a 1 zw þ a 2 z 2 w þ a 3 w 2 þ a 4 zw 2 þ a 6 w 3 ; which are related by means of the birational transformation ðx; yþ/ðz; wþ ¼ð x=y; 1=yÞ; ðz; wþ/ðx; yþ ¼ðz=w; 1=wÞ: We also need the formal group law which is defined by means of two p-adically convergent power series Fðz 1 ; z 2 ÞAZ½aŠ½½z 1 ; z 2 ŠŠ ( sum ) and iðzþaz½aš½½zšš ( inverse ), satisfying certain properties (see Section 2, Chapter IV of [28]). These series can be explicitly calculated up to any precision, and the operations ðz 1 ; z 2 Þ/Fðz 1 ; z 2 Þ; z/iðzþ make pz p ½aŠ a group # E (or, more precisely, # E=Zp ½aŠ), which is the formal group associated to E=Q p ðaþ: There is a group isomorphism between # E and the subgroup of EðQ p ðaþþ consisting of those points Q whose reduction mod p is the zero point of the reduced mod p curve, defined by z/q; where Q ¼ðz=wðzÞ; 1=wðzÞÞ if za0 and Q ¼ O if z ¼ 0; with wðzþ a p-adically convergent power series, that can be explicitly calculated up to any p-adic precision. The inverse map is given by zðoþ ¼0 and for QAEðQ p ðaþþ different from O whose reduction mod p is zero, zðqþ ¼ xðqþ yðqþ : The remarkable property relating the functions z and F is that, for any points Q 1 ; Q 2 as Q above, zðq 1 þ Q 2 Þ¼FðzðQ 1 Þ; zðq 2 ÞÞ: ð22þ With respect to E; a logarithmic function log is defined on pz p ½aŠ and an exponential function exp is defined on p r Z p ½aŠ; where r ¼ 1ifp42 and r ¼ 2ifp ¼ 2: These functions are mutually inverse and can be explicitly calculated as p-adic power series up to any precision. Moreover, if r is as above and z 1 ; z 2 Ap r Z p ½aŠ; then log Fðz 1 ; z 2 Þ¼log z 1 þ log z 2 and expðz 1 þ z 2 Þ¼Fðexp z 1 ; exp z 2 Þ: ð23þ Suppose now we know a point such that zðqþap r Z p ½aŠ and assume further that, for a certain specifically known point PAQðaÞ-EðZ p ½aŠÞ; or for P ¼ O; we want to find all naz for which bxðp þ nqþ is a rational number (or, more particularly, a square of a rational). According to whether P is a finite point or P ¼ O; we express bxðp þ nqþ or 1=bxðnQÞ first as an element of Z p ½aŠ½½zðnQÞŠŠ and then, using properties (22) and (23), as a sum y 0 ðnþþy 1 ðnþa þ? þ y d 1 ðnþa d 1 ; where each series y i ðnþ isapadically convergent power series in n with coefficients in Z p ; which can be explicitly calculated up to any desired p-adic precision. In order that this sum be a rational number we must have y i ðnþ ¼0 for i ¼ 1; y; d 1: At this point we use Strassman s Theorem, 3 which restricts the number of p-adic integer solutions n: If the maximum number of solutions implied by this theorem is equal to the number of solutions that 3 Theorem 4.1, in [9].

11 A. Bremner, N. Tzanakis / Journal of Number Theory 107 (2004) we actually know, then we have explicitly all solutions. Sometimes, as in some instances of the sections below, even Strassman s Theorem is not necessary. In the following two sections we apply the above method to Eqs. (9) and (18). At the suggestion of the referee, we give only a few computational details; most of our computational results, including the explicit form of the functions Fðz 1 ; z 2 Þ; wðzþ; log and exp with the required precision, can be found in Section 5.1 of [1] Eq. (9) p For this section, let a ¼ ffiffi 3 : We write (9) as E: y 2 ¼ x 3 ð3þ5aþx 2 þ 12ð1 þ aþx and according to the discussion in Section 2.1 we must find all pointsðx; yþ on this curve, such that bx ¼ u 2 AQ 2 ; where b ¼ð3 aþ=24: We work p-adically with p ¼ 7: According to Section 3, any point on EðQðaÞÞ isof the form n 1 P 1 þ T; where P 1 ¼ða; 3aÞ and TAfO; ð0; 0Þ; ð4a; 0Þ; ð3 þ a; 0Þg: For Q ¼ 11P 1 we have zðqþa7z 7 ½aŠ; and any point of EðQðaÞÞ can be written in the form n 1 P 1 þ T ¼ð11n þ rþp 1 þ T ¼ nq þ P; with P ¼ rp 1 þ T; 5prp5 and T a torsion point as above. There are 44 possibilities for P; one of which is P ¼ O: (i) Consider first the case when P is one out of the 43 possible finite points. As noted in Section 4.1, we are led to a relation bxðp þ nqþ ¼y 0 ðnþþy 1 ðnþa; where the 7-adically convergent series y i ðnþ; i ¼ 0; 1 depend on P: In 35 out of the 43 cases, it turns out that y 1 ð0þc0 ðmod 7Þ; which, in particular, implies that bxðp þ nqþ ¼ y 0 ðnþþy 1 ðnþa y 0 ð0þþy 1 ð0þa ðmod 7Þ cannot be rational. The only cases that are not excluded in this way occur when P is one of the following points: 74P 1 þ ð0; 0Þ; 73P 1 þð0; 0Þ; 7P 1 þð0; 0Þ; ð0; 0Þ; ð3 þ a; 0Þ: We deal with these cases as follows: If P ¼ 74P 1 þð0; 0Þ; then y 0 ð0þ ¼5; a quadratic non-residue of 7; therefore, whatever n may be, bxðp þ nqþ ¼y 0 ðnþþ y 1 ðnþa cannot be the square of a rational number. In a completely analogous manner we exclude P ¼ 73P 1 þð0; 0Þ; since, in this case, y 0 ð0þ ¼6: Next, consider P ¼ 7P 1 þð0; 0Þ ¼ð12 þ 4a; 7ð36 þ 12aÞÞ: With the plus sign we compute y 1 ðnþ ¼7 94n þ n 2 þ 7 3 6n 3 þ?; and with the minus sign, y 1 ðnþ ¼ 7 249n þ n 2 þ 7 3 n 3 þ?: In both cases, if na0; then dividing out by 7n we are led to an impossible relation mod 7; hence n ¼ 0 and x ¼ xðp þ nqþ ¼xðPÞ ¼ 12 þ 4a; which gives u 2 ¼ bð12 þ 4aÞ ¼1 and u ¼ 1: If P ¼ð0; 0Þ; then we compute y 1 ðnþ ¼ n 2 þ n 4 þ 7 7 2n 6 þ 7 7 5n 8 þ? and if na0 we divide out by 7 3 n 2 and we are led to an impossible relation mod 7: Thus, n ¼ 0; which leads to x ¼ 0andu¼0: Finally, if P ¼ð3þa; 0Þ; then y 1 ðnþ ¼ n 2 þ 7 4 n 4 þ?; forcing again n ¼ 0: Thus, x ¼ 3 þ a and u 2 ¼ bð3 þ aþ ¼1=4; hence u ¼ 1=2: (ii) Assume now that P ¼ O: Then we have 1=bxðnQÞ ¼y 0 ðnþþy 1 ðnþa; where y 1 ðnþ ¼ n 2 þ 7 4 2n 4 þ?: Since we are interested in finite points ðx; yþ ¼nQ; n must be non-zero. Dividing out y 1 ðnþ ¼0by7 2 n 2 we obtain an impossible equality. Conclusion: The only points on (9) satisfying bx ¼ u 2 AQ 2 are those with x ¼ 12 þ 4a; 3 þ a; 0; corresponding to u ¼ 1; 1 2 ; 0: Only the first leads to a solution of (8) and this leads to the solution ðp; QÞ ¼ð1; 1Þ of (3) with U 12 ð1; 1Þ ¼12 2 :

12 226 A. Bremner, N. Tzanakis / Journal of Number Theory 107 (2004) Eq. (18) We write (18) as y 2 ¼ x 3 þð 3þ2a 2 Þx 2 þð2 a 2 Þx: According to the discussion in Section 2.2.1, it suffices to find all points ðx; yþ on this curve such that bxaq 2 ; where b ¼ 4 þ a 2a 2 : We work p-adically with p ¼ 2: According to Section 3, any point on EðQðaÞÞ is of the form n 1 P 1 þ T; where P 1 ¼ð1; aþ and TAfO; ð1 a a 2 ; 0Þ; ð0; 0Þ; ð2 þ a a 2 ; 0Þg: For the point Q ¼ 4P 1 we have zðqþa4z 2 ½aŠ and we write any point on EðQðaÞÞ in the form n 1 P 1 þ T ¼ð4nþrÞP 1 þ T ¼ nq þ P; with P ¼ rp 1 þ T; raf 1; 0; 1; 2g and T a torsion point as above. Therefore, there are 16 possibilities for P; one of which is P ¼ O: Working as in Section 4.2 we check that the only solutions ðx; yþ such that bxaq 2 are ðx; yþ ¼ð0; 0Þ; 2P 1 þð0; 0Þ; 2P 1 þð0; 0Þ: To give an idea of how we apply Strassman s Theorem, let us consider the instance when P ¼ 2P 1 þ T with T ¼ ð0; 0Þ: We compute y 1 ðnþ ¼2 6 7n þ 2 6 3n 2 þ 0 n 3 þ 0 n 4 þ 0 n 5 þ 2 7 n 6 ð?þ: By Strassman s Theorem, y 1 ðnþ ¼0 can have at most two solutions in 2-adic integers n: On the other hand, a straightforward computation shows that bxðp þ 0 QÞ ¼4 ¼ bxðp QÞ; which implies, in particular, that y 1 ð0þ ¼0 ¼ y 1 ð 1Þ: Hence, n ¼ 0; 1 are the only solutions obtained for the above specific value of P; leading to the points ðx; yþ ¼P þ 0Q ¼ 2P 1 þð0; 0Þ and ðx; yþ ¼P þð 1ÞQ ¼ 2P 1 þð0; 0Þ; both having x ¼ 4 3 ð1 a2 Þ and bx ¼ 4: Conclusion: The only points on (18) satisfying bxaq 2 are those with x ¼ 0 (leading to P ¼ 0), and x ¼ 4 3 ð1 a2 Þ giving successively (in the notation of Section 2.2.1) P2 ¼ 4 and ðp; QÞ ¼ð72; 1Þ; corresponding to degenerate Lucas sequences. R 2 Acknowledgments We are grateful to the referee for suggestions which have greatly improved the presentation of this paper. References [1] A. Bremner, N. Tzanakis, Lucas sequences whose 12th or 9th term is a square, Extended version, pdf. [2] N. Bruin, Chabauty methods and covering techniques applied to generalized Fermat equations, CWI Tract, Vol. 133, Stichting Mathematisch Centrum voor Wiskunde en Informatica, Amsterdam, 2002; Dissertation, University of Leiden, Leiden, [3] N. Bruin, The diophantine equations x 2 7y 4 ¼ 7z 6 and x 2 þ y 8 ¼ z 3 ; Compositio Math. 118 (1999) [4] N. Bruin, Chabauty methods using elliptic curves, J. Reine Angew. Math. 562 (2003) [5] N. Bruin, [6] N. Bruin, [7] N. Bruin, E.V. Flynn, Towers of 2-covers of hyperelliptic curves, PIMS-01-12, Pacific Institute Math. Sci., 2001, preprint.

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