A FAMILY OF INTEGER SOMOS SEQUENCES

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1 A FAMILY OF INTEGER SOMOS SEQUENCES BETÜL GEZER, BUSE ÇAPA OSMAN BİZİM Communicated by Alexru Zahărescu Somos sequences are sequences of rational numbers defined by a bilinear recurrence relation. Remarkably, although the recurrences describing the Somos sequences are rational, some Somos sequences turn out to have only integer terms. In this paper, a family of Somos 4 sequences is given it is proved that all Somos 4 sequences associated to Tate normal forms with h 1 = ± 1 consist entirely of integers for n 0. It is also shown that there are infinitely many squares infinitely many cubes in Somos 4 sequences associated to Tate normal forms. AMS 2010 Subject Classification: 11B37, 11B83, 11G07. Key words: Somos sequences, elliptic curves, torsion points. 1. INTRODUCTION For an integer k 4, a Somos k sequence is a sequence (h n ) which satisfies the recurrence relation (1) h n h n k = k/2 i=1 τ i h n i h n k+i, for n Z where the coefficients τ i the initial values h 0,..., h k 1 are given integers. Especially, a Somos(k) sequence is a sequence whose coefficients initial values are all equal to 1. These sequences were named after Micheal Somos. He first introduced the sequence Somos(6) which begins 1, 1, 1, 1, 1, 1, 3, 5, 9, 23, 75,... observed that the terms of the sequence consist entirely of integers even though the terms are obtained from a rational recursion, see [19] for more details. Note that the recurrences describing the Somos sequences involve divisions by another term. It is clear that these sequences turn out to have rational terms. The surprising fact is that there are sequences which contain only integer terms. Indeed, the Somos(k) sequences have only integer terms for k 7 MATH. REPORTS 18(68), 3 (2016),

2 418 Betül Gezer, Buse Çapa Osman Bizim 2 but not for k = 8 [6, 26]. The question of when a Somos sequence has only integer terms has received much attention in the literature [5 7, 13, 16]. Some history about the integrality properties of these sequences can be found in [6,7]. Gale also mentioned that there are many families of Somos sequences that appear to have only integer terms. In this work, we are interested in Somos 4 sequences which satisfy a recurrence relation of the form (2) h n+2 h n 2 = τ 1 h n+1 h n 1 + τ 2 h 2 n, for n Z where τ 1, τ 2 are given integers. These sequences are generalisations of elliptic divisibility sequences (EDSs) which were first introduced by M. Ward [30]. He also showed that the terms of EDSs consist entirely of integers if the sequence begins 1, h 2, h 3, h 2 c,...(h 2, h 3, c Z). For more details see [4, 30]. In [26], Robinson was interested in the properties of the Somos(4) sequence reduced modulo a prime power p r where r 1. Swart [29] extended his results to Somos 4 sequences proved some of his conjectures. In particular, Somos 4 sequences are quite interesting because of the close relation with elliptic curves. In fact, Somos 4 sequences with the first coefficient square can be expressed in terms of the x-coordinates of the points (x n, y n ) = Q + np where P = (0, 0) Q = (x, y) is a suitable point on an elliptic curve. Somos 4 sequences are also closely related to cluster algebras [5], to integrable systems [12] to continued fractions [20,21]. Fomin Zelevinsky [5] used the theory of cluster algebras to prove that all elements of the Somos 4 sequences are Laurent polynomials, i.e., h n Z[τ 1, τ 2, h ±1 1, h±1 2, h±1 3, h±1 4 ] for all n Z. Hone Swart [13] extended the known results on integrality of Somos 4 sequences. They used the relation between Somos 4 sequences sequences of points on elliptic curves. This gave a stronger Laurent phenomenon for Somos 4 sequences. Hence they obtained integrality results for Somos 4 sequences. In this paper, using the Tate normal form having one parameter α Z of elliptic curves with torsion points, a family of Somos 4 sequences is given by means of Mazur s theorem. It is shown that all elements of the Somos 4 sequences associated to Tate normal forms with h 1 = ± 1 are elements of the ring of polynomials in x, y, α h 0 with integral coefficients, i.e., h n Z[x, y, α, h 0 ] for all n 0. Our first main theorem shows that all Somos 4 sequences associated to Tate normal forms with h 1 = ± 1 consist entirely of integers for n 0.

3 3 A family of integer Somos sequences 419 Theorem 1. Let E N denote a Tate normal form of an elliptic curve having integral points P = (0, 0) Q = (x, y) with P torsion point of maximal order N Q + np O for all n Z. Let (h n ) denote a Somos 4 sequence associated to a Tate normal form with h 1 = ± 1. Then the Somos 4 sequence (h n ) consists entirely of integers for n 0. The question of when a term of a Lucas sequence can be square has generated interest in the literature [1, 2, 24, 25]. Similar results concerning cubes were also obtained for specific sequences such as Fibonacci, Lucas Pell numbers [18, 23]. It has been proved that the only perfect powers in the Fibonnaci sequence are 1, [3]. Authors wrote many papers when a term of an integer sequence generated by linear recurrence can be a perfect power. However, not so much is known about nonlinear recurrence sequences. In [9, 10], we describe when a term of an elliptic divisibility sequence can be a square or a cube, if one of the first six terms is zero. We [11] extended these results to elliptic divisibility sequences associated to Tate normal forms. Reynolds [22] showed that there are finitely many perfect powers in an elliptic divisibility sequence whose first term is divisible by 2 or 3. The following question arises: Are there finitely or infinitely many squares (perfect powers) in a Somos 4 sequence? In the second main theorem we gave a partial answer to this question: There are infinitely many squares infinitely many cubes in Somos 4 sequences associated to Tate normal forms. Theorem 2. Let E N denote a Tate normal form of an elliptic curve let P, Q be points as defined in Theorem 1.1. Let (h n ) denote a Somos 4 sequence associated to a Tate normal form with h 1 = ± 1. There are infinitely many squares infinitely many cubes in (h n ). In Lemma 3.1, we give explicit formulas for the terms of h n in terms of x, y, α, h 1, h 0. Theorem 1.1 Theorem 1.2 are proven by using these formulas. 2. SOME PRELIMINARIES Let E denote an elliptic curve given by a Weierstrass equation (3) E : y 2 + a 1 xy + a 3 y = x 3 + a 2 x 2 + a 4 x + a 6 with coefficients a 1,..., a 6 in Q. Let E(Q) denote the set of rational points on E together with a point O, called the point at infinity. The set E(Q) forms an abelian group, with the point at infinity as the identity. For more details on elliptic curves in general, see [27, 28]. One of the most important theorems

4 420 Betül Gezer, Buse Çapa Osman Bizim 4 in the theory of elliptic curves is the Mordell-Weil theorem, which implies that, if K is a number field containing Q, then E(K) is a finitely generated abelian group. Also, the Mordell-Weil theorem shows that E tors (K), the torsion subgroup of E(K), is finitely generated abelian, hence it is finite, since its generators are of finite order. It is always interesting to characterize the torsion subgroup of a given elliptic curve. The question of a uniform bound on E tors (Q) was studied from the point of view of modular curves by Shimura, Ogg, others. In 1976, B. Mazur proved the following strongest result which had been conjectured by Ogg: Theorem 3 (Mazur [17]). Let E be an elliptic curve defined over Q. Then the torsion subgroup E tors (Q) is either isomorphic to Z/NZ for N = 1, 2,..., 10, 12 or to Z/2Z Z/2NZ for N = 1, 2, 3, 4. Further, each of these groups does occur as an E tors (Q). It is well known that all elliptic curves with a torsion point of order N lie in one parameter family where N {4,..., 10, 12}. The Tate normal form of an elliptic curve E with a torsion point P = (0, 0) is defined by E N : y 2 + (1 c)xy by = x 3 bx 2. If an elliptic curve in Weierstrass form has a point of order N > 3, then there is an admissible change of variables that transforms the curve to the Tate normal form. In this case the point P = (0, 0) is a torsion point of maximal order. Especially, if we want a classification with respect to the order of the torsion points, the use of Tate normal form of elliptic curves is unavoidable. In [15], Kubert listed one parameter family of elliptic curves E defined over Q with a torsion point of order N where N = 4,..., 10, 12. Most cases can be found in [14]. Also some algorithms are given by using the existence of such a family [8]. To decide when an elliptic curve defined over Q has a point of given order N, we need a result on parametrization of torsion structures: Theorem 4 ([8]). Every elliptic curve with a point P of order N = 4,..., 10, 12 can be written in the following Tate normal form E N : y 2 + (1 c)xy by = x 3 bx 2, with the following relations: 1. If N = 4, b = α, c = If N = 5, b = α, c = α. 3. If N = 6, b = α + α 2, c = α. 4. If N = 7, b = α 3 α 2, c = α 2 α. 5. If N = 8, b = (2α 1)(α 1), c = b/α. 6. If N = 9, c = α 2 (α 1), b = c(α(α 1) + 1).

5 5 A family of integer Somos sequences If N = 10, c = (2α 3 3α 2 + α)/(α (α 1) 2 ), b = cα 2 /(α (α 1) 2 ). 8. If N = 12, c = (3α 2 3α+1)(α 2α 2 )/(α 1) 3, b = c( 2α 2 +2α 1)/(α 1). Theorem 2.2 states that, if any elliptic curve has a point of finite order then this curve is birationally equivalent to one of the Tate normal forms given in the theorem above. Therefore, in this work, we are only interested in the elliptic curves in Tate normal forms with one parameter α Z. The relation between an elliptic curve a Somos 4 sequence is established independently by N. Elkies (for more details see [19]) N. Stephens. In [29], some unpublished works of N. Stephens are given. See also [12] for a different approach. Let P = (0, 0) Q = (x, y) be integral points on E as in Theorem 1.1. Then the terms of Somos 4 sequence (h n ) can be defined as follows: Let h 1 h 0 arbitrary non-zero integers (4) h n+1 = x nh 2 n h n 1 for all n 0 (see [29] for more details). The following result due to Nelson Stephens. Theorem 5 ([29]). Let E denote an elliptic curve given by a Weierstrass equation (5) E : y 2 + a 1 xy + a 3 y = x 3 + a 2 x 2 + a 4 x with integral coefficients a 1,..., a 4. Let P = (0, 0) Q = (x, y) be integral points on E such that Q + np O for all n Z write Q + np = (x n, y n ). Then the coefficients τ 1, τ 2 the initial values of the Somos 4 sequence associated to the elliptic curve E given by τ 1 = a 2 3, τ 2 = a 4 (a 4 + a 1 a 3 ) a 2 3a 2 h 1 h 0 arbitrary non-zero integers, h 1 = xh 2 0/h 1, h 2 = (a 4 x a 3 y)h 3 0/h 2 1. Let E N denote a Tate normal form of an elliptic curve E with a point P of order N. We can use E N to give general terms of associated Somos 4 sequences. In Theorem 2.3 we assume that the coefficients of E N are chosen to lie in Z. However for N = 8, 10 or 12, E N has rational coefficients. In these cases, we transform E N into a birationally equivalent curve E N having an equation with integral coefficients. The equations of the birationally equivalent curves for N = 8, 10 or 12 are given as follows. (6) E 8 : y 2 +(α β)xy α 3 βy = x 3 α 2 βx 2,

6 422 Betül Gezer, Buse Çapa Osman Bizim 6 (7) E 10 : y 2 +(ζ 2 αβζ)xy α 3 βζ 4 y = x 3 α 3 βζ 2 x 2, (8) E 12 : y 2 +(α 1)((α 1) 3 λ)xy (α 1) 8 λθy = x 3 (α 1) 4 λθx 2, where (9) β = (2α 1)(α 1), ζ = α (α 1) 2, λ = (3α 2 3α + 1)(α 2α 2 ), θ = 2α 2α 2 1. From now on, for simplicity of notation, we continue to write E 8, E 10, E 12 for E 8, E 10, E 12, respectively. 3. PROOF OF THEOREM 1.1 Let P Q be points as in Theorem 1.1, let h 1, h 0 be arbitrary non-zero integers. Let β, ζ, λ, θ be as in (2.7) let γ = α 2 (α 1)(α 2 α + 1). We present below all the initial values h 1, h 2 the coefficients τ 1, τ 2 of the Somos 4 sequences associated to E N by using Theorem 2.3. Table 1 The values τ 1, τ 2, h 1 h 2 N τ 1 τ 2 h 1 h 2 4 α 2 α 3 xh 1 1 h2 0 5 α 2 α 3 xh 1 1 h2 0 6 α 2 (α + 1) 2 α 3 (α + 1) 3 xh 1 1 h2 0 7 α 4 (α 1) 2 α 6 (α 1) 3 xh 1 1 h2 0 8 α 6 β 2 α 8 β 3 xh 1 1 h2 0 9 γ 2 γ 3 xh 1 1 h α 6 β 2 ζ 8 α 9 β 3 ζ 10 xh 1 1 h (α 1) 16 λ 2 θ 2 (α 1) 20 λ 3 θ 3 xh 1 1 h2 0 αyh 2 1 h3 0 αyh 2 1 h3 0 α(α + 1)yh 2 1 h3 0 α 2 (α 1)yh 2 1 h3 0 α 3 βyh 2 1 h3 0 γyh 2 1 h3 0 α 3 βζ 4 yh 2 1 h3 0 (α 1) 8 λθyh 2 1 h3 0 We consider only the case N = 8. The remaining cases can be dealt with similarly. If N = 8 then the coefficients τ 1, τ 2 the initial values h 1, h 2 of the Somos 4 sequences associated to E 8 are τ 1 = α 6 β 2, τ 2 = α 8 β 3 h 1 = xh 2 0/h 1, h 2 = α 3 βyh 3 0/h 2 1 as shown in Table 1. Using the relation (2) we obtain h 3 = α 8 β 3 (x 2 αy)h 4 0/h 3 1,

7 7 A family of integer Somos sequences 423 h 4 = α 14 β 5 (x 3 αxy y 2 )h 5 0/h 4 1, (10) h 5 = α 23 β 8 (αβx 4 + x 3 y 2α 2 βx 2 y αxy 2 + α 3 βy 2 y 3 )h 6 0/xh 5 1. Note that both h 3 h 4 Z[α, x, y, h ±1 1, h 0], but h 5 / Z[α, x, y, h ±1 1, h 0] since x appears in the denominator of h 5. Now using the equation (6) we see that x divides the numerator of h 5. The equation (10) may therefore be rewritten as h 5 = α 23 β 9 (αx 3 α 2 xy y 2 )h 6 0/h 5 1. Similarly using equation (2) we find that (11) h 6 = α 33 β 12 ( x 6 + α 2 βx 5 + 2αx 4 y + 2x 3 y 2 2α 3 βx 3 y αβx 2 y 2 α 2 x 2 y 2 + α 4 βxy 2 2αxy 3 + α 2 βy 3 y 4 )h 7 0/yh 6 1. In this case, the value y appears in the denominator of h 6. Now applying (6) to (11), we obtain h 6 = α 33 β 13 (α 2 βx 3 + (α 2 α β)x 2 y α 3 βxy (α 3 α 2 )y 2 )h 7 0/h 6 1. In the same manner we see that h 7 = α 45 β 18 ((α 4 β α 5 β)x 3 + (α 4 α 3 + α 3 β 2α 2 β)x 2 y (α 3 α 2 β)xy 2 + (α 6 β α 5 β)xy (α 5 α 4 α 3 β)y 2 )h 8 0/h 7 1. Now since P is a point of order 8 every subsequent term of the sequence can be expressed as products of the previous 9 terms. Thus, the general term of the Somos 4 sequence associated to E 8 can be given as (12) where h n = εα {(15n2 p)/16} (α 1) {(7n2 6n q)/16} (2α 1) {(3n2 r)/8} P 8 (α, x, y)[q 8 (α, x, y)] {(n m)/8} h n 1 hn+1 0 ; { +1 if n 0, 3, 6, 8, 9, 13, 14, 15 (16) 1 if n 1, 2, 4, 5, 7, 10, 11, 12 (16), 0 if n 0 (8) 7 if n 3, 5 (8) p = 12 if n 2, 6 (8) 15 if n 1, 7 (8) 16 if n 4 (8), 0 if n 0, 2 (8) 3 if n 3 (8) q = 8 if n 4, 6 (8) 1 if n 1, 5 (8) 13 if n 7 (8), 0 if n 0 (8) 3 if n 1, 3, 5, 7 (8) r = 4 if n 2, 6 (8) 8 if n 4 (8),

8 424 Betül Gezer, Buse Çapa Osman Bizim 8 1 if n 0, 7 (8) x if n 1 (8) y if n 2 (8) P 8 (α, x, y) = x 2 αy if n 3 (8) x 3 αxy y 2 if n 4 (8) αx 3 α 2 xy y 2 if n 5 (8) α 2 βx 3 (α 1) 2 x 2 y α 3 βxy α 2 (α 1)y 2, if n 6 (8), Q 8 = α 4 β(α 1)x 3 + 2α 2 (α 1) 3 x 2 y + α 5 β(α 1)xy (α 1) 3 xy 2 + α 3 (α 1) 2 y 2, { k if n k (8) k 7 m = 1 if n k (8) k = 7. As an example, the 1222 th,..., 1231 th terms of the Somos 4 sequences associated to E 8 can be obtained by using the general term formula (12). The values in (12) can be found in Table 2. Table 2 The values in (12) for the 1222 th,..., 1231 th terms n (16) (8) ε p q r P 8(α, x, y) m α 2 βx 3 (α 1) 2 x 2 y α 3 βxy α 2 (α 1)y x y x 2 αy x 3 αxy y αx 3 α 2 xy y α 2 βx 3 (α 1) 2 x 2 y α 3 βxy α 2 (α 1)y Thus by using the equation (12) we find the 1222th term in the sequence as h 1222 = εα {( p)/16} (α 1) {( q)/16} (2α 1) {( r)/8} P 8 (α, x, y)[q 8 (α, x, y)] {(1222 m)/8} h n 1 hn+1 0. Consider the elliptic curve E 8 : y 2 199xy y = x x 2,

9 9 A family of integer Somos sequences 425 for α = 9 the points P = (0, 0), Q = ( 210, 3900) on this curve. Now, the 1222th term of the Somos 4 sequence associated to E 8 is given by h 1222 = (+1)( 9) {( )/16} ( 10) {( )/16} ( 19) {( )/8} P 8 ( 9, 210, 3900) [Q 8 ( 9, 210, 3900)] {(1222 6)/8} = for h 1 = h 0 = 1. In the same manner, the other nine terms are obtained as follows: h 1223 = , h 1224 = , h 1225 = , h 1226 = , h 1227 = , h 1228 = , h 1229 = , h 1230 = , h 1231 = The general terms of Somos 4 sequences associated to Tate normal forms are given in the following lemma. For the convenience of the reader, we have given the polynomials P N (α, x, y), Q N (α, x, y) the values ε, p, q, r, s, t, m in Appendix A. Lemma 1. Let E N be a Tate normal form of an elliptic curve. Let P Q be points on a Tate normal form as in Theorem 1.1 let ζ, λ, θ, γ be as defined above. Let (h n ) be a Somos 4 sequence associated to a Tate normal form with the initial values the coefficients as above. Then the general terms of (h n ) are given by the following formulas: 1. If N = 4, 2. if N = 5, 3. if N = 6, h n = εα {(3n2 p)/8} P 4 (α, x, y)[q 4 (α, x, y)] {(n m)/4} h n 1 hn+1 0, h n = εα {(2n2 p)/5} P 5 (α, x, y)[q 5 (α, x, y)] {(n m)/5} h n 1 hn+1 0, h n = εα {(5n2 2n p)/12} (α + 1) {(n2 q)/3} P 6 (α, x, y)[q 6 (α, x, y)] {(n m)/6} h n 1 hn+1 0,

10 426 Betül Gezer, Buse Çapa Osman Bizim if N = 7, h n =εα {(5n2 p)/7} (α 1) {(3n2 n q)/7} P 7 (α, x, y)[q 7 (α, x, y)] {(n m)/7} h n 1 hn+1 0, 5. if N = 8, 6. if N = 9, h n = εα {(15n2 p)/16} (α 1) {(7n2 6n q)/16} (2α 1) {(3n2 r)/8} P 8 (α, x, y)[q 8 (α, x, y)] {(n m)/8} h n 1 hn+1 0, h n = εα {(7n2 2n p)/9} (α 1) {(4n2 2n q)/9} (α 2 α + 1) {(n2 r)/3} 7. if N = 10, P 9 (α, x, y)[q 9 (α, x, y)] {(n m)/9} h n 1 hn+1 0 h n = εα {(21n2 p)/20} (α 1) {(9n2 2n q)/20} (2α 1) {(2n2 r)/5} ζ {(5n2 s)/4} 8. if N = 12, P 10 (α, x, y)[q 10 (α, x, y)] {(n m)/10} h n 1 hn+1 0, h n = εα {(n2 2n p)/12} (α 1) {(59n2 q)/24} (2α 1) {(n2 r)/24} λ {(3n2 s)/8} θ {(n2 t)/3} P 12 (α, x, y)[q 12 (α, x, y)] {(n m)/12} h n 1 hn+1 0, where P N (α, x, y), Q N (α, x, y) are polynomials in Z[α, x, y]. Proof. We give the proof only for the case N = 8 by induction on n as follows. It is clear that the result is true for n = 7. Hence we assume that n > 7. First suppose that n 0 (8), n 0 (16) (12) is true for n + 1. Then we have h n+2 = α 60k2 +30k+3 (α 1) 28k2 +11k+1 (2α 1) 24k2 +12k+1 y Q k 8 h (8k+2) 1 h 8k+3 0 (k N) by (12). Indeed, we see that h n 2 = α 60k2 30k+3 (α 1) 28k2 17k+2 (2α 1) 24k2 12k+1 (α 2 βx 3 (α 1) 2 x 2 y α 3 βxy α 2 (α 1)y 2 ) Q k 1 8 h (8k 2) 1 h 8k 1 0 h n = α 60k2 (α 1) 28k2 3k (2α 1) 24k2 Q k 8 h 8k+1 0 h 1 8k h n 1 = α 60k2 15k (α 1) 28k2 10k (2α 1) 24k2 6k Q k 8 h (8k 1) 1 h 8k 0 h n+1 = α 60k2 +15k (α 1) 28k2 +4k (2α 1) 24k2 +6k x Q k 8 h (8k+1) 1 h0 8k+2 Substituting these expressions τ 1 = α 6 β 2, τ 2 = α 8 β 3 into (2) then using equation (6) we obtain h n+2 = α 60k2 +30k+3 (α 1) 28k2 +11k+1 (2α 1) 24k2 +12k+1 y Q k 8 h (8k+2) 1 h 8k+3 0.

11 11 A family of integer Somos sequences 427 A similar result can be obtained when n 0 (16). Thus we proved the conclusion (12) is true for n + 2 which completes the proof for n 0 (8). The remaining parts of the theorem can be proved in a similar manner. From Lemma 3.1 we deduce that any Somos 4 sequence associated to a Tate normal form is determined by the values α, x, y, h 1, h 0. Corollary 1. Let (h n ) be a Somos 4 sequence associated to a Tate normal form with the initial values the coefficients as above. Then each h n can be expressed as elements of the ring R = Z[α, x, y, h ±1 1, h 0]. In particular if h 1 = ±1 the Somos 4 sequence consists entirely of integers for n 0. Remark 1. It is also possible to use the recursion (2) to extend the Somos 4 sequences to negative indices. However, for n < 0 the terms of the Somos 4 sequences associated to Tate normal forms may not be integral. For instance, consider the elliptic curve E 12 : y xy y = x x 2, the points P = (0, 0), Q = ( 21945, ) on this curve. The Somos 4 sequence (h n ) associated to E 12 is the sequence with coefficients the initial values which begins τ 1 = , τ 2 = h 1 = h 0 = 1, h 1 = 21945, h 2 = , 1, 21945, , ,... the sequence extends backward as..., , , , 1, 1, , CASES N = 2, 3 There is no Tate normal form of an elliptic curve with the torsion point of order two or three, but Kubert in [15] listed that the elliptic curves with torsion point of order two or three are (13) E 2 : y 2 = x 3 + a 2 x 2 + a 4 x

12 428 Betül Gezer, Buse Çapa Osman Bizim 12 (14) E 3 : y 2 + a 1 xy + a 3 y = x 3, respectively. Let h 1, h 0 arbitrary nonzero integers. The initial values h 1, h 2 the coefficients τ i of the Somos 4 sequences associated to the elliptic curves E 2 E 3 are τ 1 = 0, τ 2 = a 2 4 h 1 = xh 1 1 h2 0, h 2 = a 4 xh 2 1 h3 0, τ 1 = a 2 3, τ 2 = 0 h 1 = xh 1 1 h2 0, h 2 = a 3 yh 2 1 h3 0, respectively. Under these considerations, an easy computation gives the general terms of these sequences. Theorem 6. Let E N be an elliptic curve as in (13) or (14) let P, Q be points on E N as in Theorem 1.1. Let (h n ) be a Somos 4 sequence associated to E N with the initial values the coefficients as above. Then the general terms of (h n ) are given by the following formulas: i. If N = 2, h n = εa {(n2 p)/4} 4 x {(n q)/2} h n 1 hn+1 0 where { +1 if n 0, 3 (4) ii. If N = 3, where p = { 0 if n 0 (3) p = 1 if n 1, 2 (3), { 0 if n 0 (2) 1 if n 1 (2), 1 if n 1, 2 (4), q = { 0 if n 0 (2) 1 if n 1 (2). h n = εa {(n2 p)/3} 3 x q y {(n r)/3} h n 1 hn+1 0 { +1 if n 0, 2 (3) q = 1 if n 1 (3), { 0 if n 0, 2 (3) 1 if n 1 (3), r = 0 if n 0 (3) 1 if n 1 (3) 1 if n 2 (3). 5. PROOF OF THEOREM 1.2 Lemma 3.1 gives us the general terms of Somos 4 sequences associated to Tate normal forms. So, we can say there are infinitely many squares infinitely many cubes in (h n ) by using these explicit formulas. The symbols C mean a square a cube of a non-zero rational number.

13 13 A family of integer Somos sequences 429 Lemma 2. Let (h n ) denote a Somos 4 sequence associated to a Tate normal form with h 1 = ± 1. i. If n 1 (2N), then h n = for every non-zero α, x, y, ii. if n 1 (3N), then h n = C for every non-zero α, x, y, where N {4,..., 10, 12}. Proof. Let (h n ) be a Somos 4 sequence associated to a Tate normal form with h 1 = ± 1. Lemma 3.1 shows that each term of (h n ) can be expressed as products of the polynomials p(α, x, y, h 0 ). It is easy to check that these polynomials are pairwise relatively prime. This implies that if each factor of the product is a square then h n is a square. Clearly, the same is true for the cube case. We give the proof only for the case N = 8 based on this fact. For (i), if n 1 (16) then n = 16k 1(k N). Thus we have h n = α 30k(8k 1) (α 1) 4k(28k 5) (2α 1) 12k(8k 1) Q 2k 8 h 16k 0, by (12). Hence h n =. For(ii), if n 1 (24) then n = 24k 1(k N). So we have h n = εα 45k(12k 1) (α 1) 6k(42k 5) (2α 1) 18k(12k 1) Q 3k 8 h 24k 0. Therefore h n = C. The remaining cases can be proved in the same way as above. Lemma 5.1 tells us that there are infinitely many squares infinitely many cubes in (h n ) which proves the Theorem 1.2. Remark 2. The same conclusion can be drawn for the cases N = 2 or 3, i.e., let (h n ) denote a Somos 4 sequence associated to E N with h 1 = ± 1, i. if n 1 (2N), then h n = for every non-zero α, x, y, ii. if n 1 (3N), then h n = C for every non-zero α, x, y. APPENDIX A The polynomials P N (α, x, y), Q N (α, x, y) the values ε, p, q, r, s, t, m in Lemma 3.1 are given as follows. 1. If N = 4, 1 if n 0, 3 (4) P 4 = x if n 1 (4) Q 4 = x 2 y, y if n 2 (4), { 1 if n 1, 2, 4, 5 (8) +1 if n 0, 3, 6, 7 (8),

14 430 Betül Gezer, Buse Çapa Osman Bizim 14 p = 2. If N = 5, p = 0 if n 0 (4) 3 if n 1, 3 (4) 4 if n 2 (4), m = 1 if n 0, 4 (5) x if n 1 (5) P 5 = y if n 2 (5) x 2 y if n 3 (5), 0 if n 0 (5) 2 if n 1, 4 (5) 3 if n 2, 3 (5), { k if n k (4) k 3 1 if n k (4) k = 3. { 1 if n 1, 2, 4, 5, 8 (10) +1 if n 0, 3, 6, 7, 9 (10), 3. If N = 6, 1 if n 0, 5 (6) x if n 1 (6) P 6 = y if n 2 (6) x 2 y if n 3 (6) x 3 xy y 2 if n 4 (6), 4. If N = 7, P 7 = p = Q 5 = x 2 xy y, { k if n k (5) k 4 m = 1 if n k (5) k = 4. Q 6 = (α + 1)x 3 (α + 1)xy y 2 { +1 if n 0, 3, 6, 7, 11 (12) 1 if n 1, 2, 4, 5, 8, 9, 10 (12), 0 if n 0 (6) 3 if n 1, 3 (6) 4 if n 2 (6) 7 if n 5 (6) 12 if n 4 (6), m = 1 if n 0, 6 (7) x if n 1 (7) y if n 2 (7) x 2 y if n 3 (7) x 3 xy y 2 if n 4 (7) αx 3 αxy y 2 if n 5 (7), { 0 if n 0, 3 (6) q = 1 if n 1, 2, 4, 5 (6), { k if n k (6) k 5 1 if n k (6) k = 5. Q 7 = α 2 x 3 (α 1)x 2 y α 2 xy y 2,

15 15 A family of integer Somos sequences 431 p = { +1 if n 0, 3, 6, 7, 10, 13 (14) 1 if n 1, 2, 4, 5, 8, 9, 11, 12 (14), 0 if n 0 (7) 3 if n 3 (7) 5 if n 1, 6 (7) 6 if n 2, 5 (7) 10 if n 4 (7), m = q = 0 if n 0 (7) 2 if n 1 (7) 3 if n 2, 3 (7) 4 if n 6 (7) 7 if n 5 (7) 9 if n 4 (7), { k if n k (7) k 6 1 if n k (7) k = If N = 9, 1 if n 0, 8 (9) x if n 1 (9) y if n 2 (9) x 2 y if n 3 (9) P 9 = x 3 xy y 2 if n 4 (9) (α 2 α + 1)x 3 (α 2 α + 1)xy y 2 if n 5 (9) (α 3 α 2 +α)x 3 (α 1) x 2 y (α 3 α 2 + α)xy y 2 if n 6 (9) (α 4 α 3 + α 2 )x 4 +x 3 y (α 4 α 3 + α 2 )x 2 y (α 2 + 1)xy 2 y 3 if n 7 (9) Q 9 = α 2 (α 2 α + 1) 2 x 4 (α 4 2α 3 +α 2 1)x 3 y α 2 (α 2 α + 1) 2 x 2 y p = { +1 if n 0, 3, 6, 7, 10, 11, 13, 14, 17 (18) 1 if n 1, 2, 4, 5, 8, 9, 12, 15, 16 (18), 0 if n 0 (9) 5 if n 1 (9) 6 if n 2 (9) 3 if n 3, 5 (9) 14 if n 4, 7 (9) 3 if n 6 (9) 9 if n 8 (9), q = 0 if n 0 (9) 2 if n 1 (9) 3 if n 2, 3 (9) 11 if n 4, 7 (9) 9 if n 5 (9) 6 if n 6, 8 (9), (α 3 +1)xy 2 y 3, r = { 1 if n 1, 2, 4, 5, 7, 8 (9) 0 if n 0, 3, 6 (9), m = { k if n k (9) k 8 1 if n k (9) k = 8.

16 432 Betül Gezer, Buse Çapa Osman Bizim If N = 10, P 10 = 1 if n 0, 9 (10) x if n 1 (10) y if n 2 (10) x 2 ζ 2 y if n 3 (10) x 3 ζ 2 xy y 2 if n 4 (10) α 2 x 3 α 2 ζ 2 xy ζy 2 if n 5 (10) α 3 ζx 3 (α 1)x 2 y α 3 ζ 3 xy ζ 2 y 2 if n 6 (10) α 4 (2α 1)ζ 2 x 3 α(α 1)(3α 1)ζx 2 y +(α 1)xy 2 α 4 (2α 1)ζ 4 xy αζ 4 y 2 if n 7 (10) α 7 (2α 1)ζ 3 x 3 (α 1)x 3 y α 3 (α 1)(2α 2 + 2α 1)ζ 2 x 2 y α 7 (2α 1)ζ 5 xy +(α 1)(2α 2 + 2α 1)ζxy 2 α 3 ζ 5 y 2 if n 8 (10) Q 10 = α 9 (2α 1)ζ 4 x 3 + (α 1)(2α 2 + 2α 1)ζx 3 y +α 3 (α 1)(2α 4 + 6α 2 5α + 1)ζ 3 x 2 y (α 1)x 2 y 2 +α 9 (2α 1)ζ 6 xy + (α 4 10α 3 + 4α 1)(α 1)ζ 2 xy 2 +α 3 ζ 7 y 2, { +1 if n 0, 3, 6, 7, 9, 12, 13, 14, 17, 18, 19 (20) 1 if n 1, 2, 4, 5, 8, 10, 11, 15, 16 (20), 0 if n 0 (10) 7 if n 1 (10) 0 if n 0 (10) 12 if n 2 (10) 21 if n 1, 9 (10) 0 if n 0 (10) 15 if n 3 (10) 24 if n 2, 8 (10) 2 if n 1, 6, 9 (10) 36 if n 4 (10) p = 9 if n 3, 7 (10) q = r = 3 if n 2, 3, 7, 8 (10) 35 if n 5 (10) 36 if n 4 (10) 7 if n 4 (10) 32 if n 6 (10) 25 if n 5 (10) 5 if n 5 (10), 27 if n 7 (10) 16 if n 6 (10), 20 if n 8 (10) 11 if n 9 (10) 0 if n 0 (10) { 5 if n 1, 3, 5, 7, 9 (10) k if n k (10) k 9 s = m = 4 if n 2, 6, 8 (10) 1 if n k (10) k = 9. 8 if n 4 (10),

17 17 A family of integer Somos sequences If N = 12, 1 if n 0, 11 (12) x if n 1 (12) y if n 2 (12) x 2 (α 1) 4 y if n 3 (12) x 3 (α 1) 4 xy y 2 if n 4 (12) θx 3 θ(α 1) 4 xy (α 1)y 2 if n 5 (12) θ(α 1) 3 (3α 2 3α + 1)x 3 (2α 2 α)x 2 y θ(α 1) 7 (3α 2 3α + 1) xy (α 1) 6 y 2 if n 6 (12) P 12 = θ(α 1)x 4 + θ(α 2 + α 1)(α 1) 3 x 2 y (3α 2 3α + 1)xy 2 θα(2α 1)(α 1) 7 y 2 if n 7 (12) θ 2 (α 1) 3 x 4 + α 2 x 3 y + θ(α 2 α + 1)(α 1) 7 x 2 y +(2α 1)(α 1) 5 xy 2 + θα(α 1) 12 y 2 if n 8 (12) θ 2 (3α 2 3α + 1)(α 1) 6 x 4 + (6α 2 7α + 3)α 2 (α 1) 3 x 3 y + α 2 x 2 y 2 + θ(2α 2 α + 1)(3α 2 3α + 1)(α 1) 10 x 2 y if n 9 (12) (α 1) 7 (8α 3 11α 2 + 6α 1)xy 2 θα(3α 2 3α + 1)(α 1) 14 y 2 2α 2 (3α 2 4α + 2)(α 1) 3 x 5 +θ(3α 2 3α + 1) (14α 5 36α 4 +40α 3 23α 2 +7α 1)(α 1) 7 x 4 + α 2 x 4 y α 2 (3α 2 4α + 2) (12α 3 24α 2 +18α 5)(α 1) 4 x 3 y θ(3α 2 3α + 1) (12α 5 30α 4 +33α 3 if n 10 (12) 19α 2 +6α 1)(α 1) 11 x 2 y +θ(9α 3 13α 2 +7α 1)(α 1) 10 xy 2 +θ 2 α(3α 2 3α + 1)(α 1) 17 y 2 Q 12 = α 2 (12α 4 42α 3 +58α 2 37α + 10)(α 1) 6 x 5 θ(3α 2 3α + 1)(30α 5 66α α 3 31α 2 + 8α 1)(α 1) 11 x 4 +2α 2 (3α 2 4α + 2)(α 1) 3 x 4 y + α 2 x 3 y 2 α 2 (12α 6 138α α 4 457α α 2 121α + 20)(α 1) 8 x 3 y +θ(3α 2 3α + 1)(28α 5 60α α 3 27α 2 + 7α 1)(α 1) 15 x 2 y θ(12α 3 16α 2 + 8α 1)(α 1) 14 xy 2 θ 2 α(3α 2 3α + 1)(α 1) 21 y 2, { +1 if n 0, 3, 8, 9, 13, 14, 16, 17, 18, 19, 22, 23 (24) 1 if n 1, 2, 4, 5, 6, 7, 10, 11, 12, 15, 20, 21 (24),

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