On the Rank of the Elliptic Curve y 2 = x 3 nx

International Journal of Algebra, Vol. 6, 2012, no. 18, 885-901 On the Rank of the Elliptic Curve y 2 = x 3 nx Yasutsugu Fujita College of Industrial Technology, Nihon University 2-11-1 Shin-ei, Narashino, Chiba 275 8576, Japan fujita.yasutsugu@nihon-u.ac.jp Nobuhiro Terai Division of General Education, Ashikaga Institute of Technology 268-1 Omae, Ashikaga, Tochigi 326 8558, Japan terai@ashitech.ac.jp Abstract Let n be a positive integer and E an elliptic curve over Q defined by y 2 = x 3 nx. In this paper, we give several parameterizations of n such that E has rank greater than or equal to three and four. Some of them are constructed by using the method of Tate, and the others are obtained by showing the independence of certain points on E. Mathematics Subject Classification: 11D25, 11G05 Keywords: Elliptic curves, rank, independent points 1 Introduction Let E be the elliptic curve given by E : y 2 = x 3 nx, where n is a positive integer. In 2007, Spearman [11] used the method of Tate (cf. [10, Chapter 3]) to show that if n = u 4 + v 4 is an odd prime for positive integers u, v, then rank E(Q) = 2. Spearman [12] also showed that if n = (u 2 +2v 2 ) 4 +(u 2 2v 2 ) 4 is twice a prime, then rank E(Q) = 3. Extending this Spearman s result, Walsh [13] has recently given analogous sufficient conditions for the elliptic curve with n twice a prime to have rank 3, where the conditions are given in terms of rational points on the curve. In our previous paper [4], we showed that if n is an odd prime of the form n = a 2 + b 4 with some conditions, then rank E(Q) = 2.

886 Y. Fujita and N. Terai In the above cases, we obtain rank E(Q) 2 (or 3) if we do not assume that n is a prime or twice a prime. Moreover, by using the method of Tate, it is not difficult to find that if n can be expressed as n = a 4 4b 4, then rank E(Q) 2. The aim of this paper is to obtain several parameterizations of n such that rank E(Q) 3 and 4, and our strategy is to consider those integers n that can be expressed in two ways as follows: n = a 4 + b 4 = c 4 4d 4, n = a 2 + b 4 = c 2 + d 4, n = a 4 + b 4 = c 2 + d 4, n = a 4 + b 4 = c 4 + d 4. Some of the parameterizations of the above n can be derived from the wellknown identities by people such as Fauquembergue and Euler. The organization of this paper is as follows. In Section 2, we briefly explain the method of Tate, and using it we construct some families of elliptic curves of rank E(Q) 3 and 4. In Section 3, we construct some families of elliptic curves of rank E(Q) 3 and 4 by showing that certain points are independent modulo E(Q) tors. This is done by checking that all the linear combinations of the points and the torsion point (0, 0) are not in 2E(Q) (for the reason this implies the independence of the points, see the proof of [2, Proposition 2] or of [4, Main Theorem (ii)]). At the end of Section 3, examples of high rank elliptic curves obtained by specializing our families are listed. Notation. Let E be an elliptic curve defined by y 2 = x 3 nx with a positive integer n, and E the isogenous curve to E defined by y 2 = x 3 +4nx. For these models, x(p ) (resp. x(p )) denotes the x-coordinate of a point P (resp. P )on E (resp. E). Denote by T (resp. T ) the torsion point (0, 0) on E (resp. E). Let ϕ : E E and ψ : E E be the isogenies defined by ( ) y 2 ϕ(p )= x, y(x2 + n) if P =(x, y) {O, T}, 2 x 2 O if P {O, T} and ( ) ȳ2 ψ(p )= 4 x, ȳ( x2 + n) if P =( x, ȳ) {O, T }, 2 8 x 2 O if P {O, T }. E 0 (R) denotes the identity component of E(R). Finally, denotes the square of a rational number.

On the rank of the elliptic curve y 2 = x 3 nx 887 2 The method of Tate We describe the method for computing rank E(Q) of the above elliptic curve E, following [10, Chapter 3]. Denote by Γ the groups E(Q) of Q-rational points of E. Then, there exists a homomorphism α :Γ Q /(Q ) 2 defined by x (mod (Q ) 2 ) if P =(x, y) with x 0; α(p )= n (mod (Q ) 2 ) if P =(0, 0); 1 (mod (Q ) 2 ) if P = O. Let E be the elliptic curve given by E : y 2 = x 3 +4nx. Denoting E(Q) byγ, we can define a homomorphism α : Γ Q /(Q ) 2 in the same way as α. Then, examining the orders α(γ) and α(γ) reveals the rank r of Γ. In fact, we have α(γ) α(γ) 4 =2 r. (1) As seen in [10, Chapter 3], one may choose a square-free divisor of n as a representative of an element in α(γ). Moreover, a square-free divisor n of n which equals neither 1 nor the square-free part of n belongs to α(γ) if and only if the equation n S 4 n n T 4 = U 2 has an integer solution (s, t, u) with st 0 and (s, t) = (t, u) = (u, s) = (n,t)=(n, s) = 1. Then, the point (n s 2 /t 2, n su/t 3 ) is in Γ. The same is true for α(γ). Lemma 2.1. (cf. Spearman [11]) Let n be a positive integer of the form n = a 4 + b 4, where a and b are distinct positive integers. Then, 1 α(γ), 2 α(γ) and rank E(Q) 2.

888 Y. Fujita and N. Terai Proof. We first need Proth s identity (cf. [1, p. 657]) Since n = a 4 + b 4, the equation a 4 + b 4 +(a b) 4 =2(a 2 ab + b 2 ) 2. S 4 + nt 4 = U 2 has a solution (S, T, U) =(a, 1,b 2 ). By definition of α(γ), we see that {1, n, 1,n} α(γ). By Proth s identity, the equation 2S 4 +2nT 4 = U 2 has a solution (S, T, U) =(a b, 1,a 2 ab + b 2 ). It is easy to see that n can be neither a square nor twice a square, since the Diophantine equations X 4 + Y 4 = Z 2 and X 4 + Y 4 =2Z 2 have no positive integer solutions X, Y, Z with X Y, respectively (cf. [8, p. 16, p. 18]). By definition of α(γ), we see that {1,n,2, 2n} α(γ). Therefore we conclude that rank E(Q) 2. Theorem 2.2. Let n>17 be a positive integer of the form (i) n =(u 2 +2v 2 ) 4 +(u 2 2v 2 ) 4 or (ii) n =(u 2 +2uv v 2 ) 4 +(u 2 2uv v 2 ) 4, where u and v are positive coprime integers, then rank E(Q) 3. Proof. In the case of (i), from the identity we have (u 2 +2v 2 ) 4 +(u 2 2v 2 ) 4 4(2uv) 4 =2(u 4 4v 4 ) 2. 2(2uv) 4 n 2 = 2(2uv)4 (u2 +2v 2 ) 4 +(u 2 2 v 2 ) 4 which shows that the equation 2 = (u 4 4v 4 ) 2, 2S 4 + n 2 T 4 = U 2 has a solution (S, T, U) =(2uv, 1,u 4 4v 4 ). Hence, 2 α(γ). Similarly, in the case of (ii), using the identity (u 2 +2uv v 2 ) 4 +(u 2 2uv v 2 ) 4 64u 2 v 2 (u 2 v 2 ) 2 =2(u 4 6u 2 v 2 + v 4 ) 2, we see that 2 α(γ). Lemma 2.1. Therefore, in any case rank E(Q) 3 follows from

On the rank of the elliptic curve y 2 = x 3 nx 889 Remark 2.3. (1) If n = 17, then E(Q) has rank 2 with generators ( 1, 4), ( 4, 2) and the torsion T =(0, 0). (2) The proof of the case (i) of Theorem 2.2 is essentially due to Spearman. More precisely, in [12] using the above argument he showed that if p is an odd prime of the form 2p =(u 2 +2v 2 ) 4 +(u 2 2v 2 ) 4 for some integers u, v, then rank E(Q) = 3. (3) Using the method of Tate, Johnstone-Spearman [7] constructed an infinite family of congruent number elliptic curves each with rank at least three. In order to construct elliptic curves of rank E(Q) 4, we need the following lemmas. Lemma 2.4. Let n be a positive integer of the form n = AB, A B = M 2, A+ B =2N 2, where A, B, M, N are positive integers. Then, A α(γ) and 2A α(γ). Moreover, if A, 2A, B, 2B and n are non-square, then rank E(Q) 2. Proof. Since A B = M 2, the equation AS 4 BT 4 = U 2 has a solution (S, T, U) = (1, 1, M). By definition of α(γ), we see that {1,A, B, n} α(γ). Since A + B =2N 2, the equation AS 4 + BT 4 =2U 2 has a solution (S, T, U) =(1, 1,N). By definition of α(γ), we see that {1, 2A, 2B, n} α(γ). Therefore, if A, 2A, B, 2B and n are non-square, then rank E(Q) 2. Lemma 2.5. The Diophantine equation X 4 +4X 2 Y 2 ± 8XY 3 +4Y 4 = Z 2 has no solutions in positive integers X, Y, Z with X odd, gcd(x, Y )=1 and X Y.

890 Y. Fujita and N. Terai Proof. The equation can be written as Hence we have X 4 +(2Y (X ± Y )) 2 = Z 2. X 2 = r 2 s 2, 2Y (X ± Y )=±2rs, where r, s are positive integers such that gcd(r, s) =1,r s (mod 2) and r>s. Since X is odd and gcd(r, s) = 1, we obtain with X = mn, and so r + s = m 2, r s = n 2 r = m2 + n 2, s = m2 n 2 2 2 Substituting these into Y 2 ± XY = ±rs gives. 4Y 2 ± 4mnY = m 4 n 4. (Interchange m and n, if necessary.) This implies that W 2 = m 4 + m 2 n 2 n 4 with W =2Y ± mn. Put t = m/n and W 1 = W/n 2.Now(x, y) =(t 2, tw 1 ) is a rational point on the elliptic curve y 2 = x(x 2 + x 1), which has rank 0 and torsion subgroup {O, (0, 0), (1, 1), (1, 1), ( 1, 1), ( 1, 1)}. This is impossible, since X, Y > 0, gcd(x, Y ) = 1 and X Y. Now we use Fauquembergue s identity (cf. [1, p. 658]) (u 4 2v 4 ) 4 +(2u 3 v) 4 =(u 4 +2v 4 ) 4 4(2uv 3 ) 4 (2) to construct elliptic curves of rank E(Q) 4. Theorem 2.6. Let n>17 be a positive integer of the form (i) n =(u 4 2v 4 ) 4 +(2u 3 v) 4 or (ii) n =(u 4 8v 4 ) 4 +(8uv 3 ) 4, where u and v are positive coprime integers with u odd. Then rank E(Q) 4.

On the rank of the elliptic curve y 2 = x 3 nx 891 Proof. First consider rank E(Q) in the case of (i). Put a = u 4 2v 4, b =2u 3 v, c = u 4 +2v 4, d =2uv 3 ; A = c 2 +2d 2, B = c 2 2d 2. Then, A B =(2d) 2, A+B =2c 2, and (2) shows that n = a 4 +b 4 = c 4 4d 4 = AB. Hence, A α(γ) and 2A α(γ) by Lemma 2.4. Since n = a 4 + b 4,we know by Lemma 2.1 that 1 α(γ) and 2 α(γ). Hence, {1, 1, A, A, B, B, n, n} α(γ) and {1, 2, A, 2A, B, 2B, n, 2n} α(γ). (3) Since n and 2n are non-square, it suffices to show that A, 2A, B, 2B are non-square. Putting X = u 2 and Y = v 2, we have A =(u 4 +2v 4 ) 2 + 2(2uv 3 ) 2 = X 4 +4X 2 Y 2 +8XY 3 +4Y 4, B =(u 4 +2v 4 ) 2 2(2uv 3 ) 2 = X 4 +4X 2 Y 2 8XY 3 +4Y 4. By Lemma 2.5, we see that A and B are non-square. If 2A (resp. 2B) isa square, then A (resp. B) is even, which is impossible, since u is odd. Therefore we conclude that rank E(Q) 4. Next consider rank E(Q) in the case of (ii). Since 4A 4B =(4d) 2, 4A +4B = 2(2c) 2, n = 16(a 4 + b 4 ) = 16(c 4 4d 4 )=16AB, we have (3) in the same way as above. non-square, we obtain rank E(Q) 4. Since A, 2A, B, 2B, n, 2n are 3 The independence of points Lemma 3.1. Let P O be a point in E(Q). (1) P ψ(e(q)) if and only if x(p ) is a square. In this case, putting P =(x, y) with x = x 2 0, one can express P E(Q) with ψ(p )=P as P = ( 2 (x 20 ± yx0 ), ±2x 0 x(p ) where the signs are taken simultaneously. (2) P 2E(Q) if and only if both x(p ) and x(p ) are squares for some P E(Q) with ψ(p )=P. ),

892 Y. Fujita and N. Terai Proof. (1) The assertion follows from the same argument as the proof of (iii) in [10, p. 83]. (2) (iii) in [10, p. 83] itself says that for P E(Q), P ϕ(e(q)) if and only if x(p ) is a square. The assertion now follows from the identity [2] = ψ ϕ as isogenies on E. Lemma 3.2. Assume that a non-square integer n can be expressed as n = a 2 + b 4 = c 2 + d 4 for nonzero integers a, b, c, d with { a,b 2 } { c,d 2 }. Let P 1 =( b 2,ab) and P 2 =( d 2,cd). (1) If neither 2(n b 2 d 2 + ac) nor 2(n b 2 d 2 ac) is a square, then P i,p i + T,P 1 + P 2,P 1 + P 2 + T 2E(Q) for i {1, 2}. Hence, P 1 and P 2 are independent modulo E(Q) tors. (2) If n is square-free, then the same assertion as (1) holds. Proof. (1) Since P 1,P 2,P 1 + P 2 + T E 0 (R), they cannot have 2-division points by Lemma 3.1 (2). Since x(p + T )= n/x(p ) for P {O, T}, the non-squareness of n implies that P 1 + T,P 2 + T 2E(Q). It suffices to check that P 1 + P 2 2E(Q). By the addition formula, we have ( (ad bc) 2 P 1 + P 2 = (b 2 d 2 ), (ad bc) {2nbd ac(b2 + d 2 ) 2b 3 d 3 } 2 (b 2 d 2 ) 3 From Lemma 3.1 (1), we see that there exists P E(Q) with ψ(p )=P 1 + P 2 such that x(p )= 2(n b2 d 2 ± ac). (b ± d) 2 It follows from Lemma 3.1 (2) that if 2(n b 2 d 2 ± ac) are non-square, then P 1 + P 2 2E(Q). (2) We only have to check that P 1 + P 2 2E(Q). Suppose that P 1 + P 2 2E(Q). Put A =2(n b 2 d 2 + ac) and B =2(n b 2 d 2 ac). If A and B, then P 1 + P 2 2E(Q) by (1), a contradiction. Otherwise, since AB =4 { (n b 2 d 2 ) 2 a 2 c 2} = 4 { (n b 2 d 2 ) 2 (n b 4 )(n d 4 ) } = 4(b 2 d 2 ) 2 n and n is square-free, either A = and n B or B = and n A holds. Hence, ). either n b 2 d 2 + ac or n b 2 d 2 ac holds. (4) If b 2 d 2 = ac, then a 2 + b 4 = c 2 + d 4 = b 4 d 4 /a 2 + d 4 = d 4 (a 2 + b 4 )/a 2, and thus a = d 2 and c = b 2, which contradicts { a,b 2 } { c,d 2 }. Hence, b 2 d 2 ±ac 0. Moreover, since 2{n (b 2 d 2 ± ac)} =(a c) 2 +(b 2 d 2 ) 2 > 0, we must have n>b 2 d 2 ± ac 0, which contradicts (4). Therefore, P 1 + P 2 2E(Q).

On the rank of the elliptic curve y 2 = x 3 nx 893 3.1 n = a 2 + b 4 = c 2 + d 4 In Theorem 2.6, using the method of Tate, we constructed a family of elliptic curves of rank 4 for n of the the form n = a 4 + b 4 = AB with A B = and A + B =2. In this subsection, we construct a family of elliptic curves of rank 3 for n of the the form n = a 2 + b 4 = c 2 + d 4 = AB with A B =. For example, one parametrization of the above n can be derived from the following identities: n = u 8 +2αu 4 v 4 + γ 2 v 8 = (u 4 + αv 4 ) 2 +(2mv 2 ) 4 = (u 4 γv 4 ) 2 +(2muv) 4 = (u 4 +2u 2 v 2 + γv 4 )(u 4 2u 2 v 2 + γv 4 ), where α, β, γ are Pythagorean numbers satisfying α =4m 4 1, β =4m 2, γ =4m 4 +1 with m positive integer. Then the points P 1 = ( 4m 2 v 4, 2mv 2 (u 4 + αv 4 )), P 2 = ( 4m 2 u 2 v 2, 2muv(u 4 γv 4 )), P 3 = (u 4 +2u 2 v 2 + γv 4, 2uv(u 4 +2u 2 v 2 + γv 4 )) are in E(Q). When m = 1, we show the following: Theorem 3.3. Let n = u 8 +6u 4 v 4 +25v 8 with distinct positive integers u and v. Then rank E(Q) 3. Proof. Let P 1 = ( 4v 4, 2v 2 (u 4 +3v 4 )), P 2 = ( 4u 2 v 2, 2uv(u 4 5v 4 )) and P 3 =(u 4 +2u 2 v 2 +5v 4, 2uv(u 4 +2u 2 v 2 +5v 4 )). Then P 1,P 2,P 3 E(Q). It suffices to show that P 1,P 2,P 3 are independent modulo E(Q) tors. We will do this by checking that P i,p i + T,P i + P j,p i + P j + T,P 1 + P 2 + P 3,P 1 + P 2 + P 3 + T 2E(Q) for i, j {1, 2, 3} with i j. Now we show that the Diophantine equations u 4 +6u 2 v 2 +25v 4 = w 2,

894 Y. Fujita and N. Terai u 4 +2u 2 v 2 +5v 4 = w 2, u 4 +2u 2 v 2 5v 4 = w 2, u 4 2u 2 v 2 +5v 4 = w 2, has no positive integer solutions u, v, w with u v, respectively. Indeed, the above quartic curves can be reduced to the following elliptic curves of rank 0, respectively: E 1 : Y 2 = X(X 2 +6X + 25), E 2 : Y 2 = X(X 2 +2X +5), E 3 : Y 2 = X(X 2 +2X 5), E 4 : Y 2 = X(X 2 2X +5) with X = u 2 /v 2 and Y = uw/v 3 for t = u/v. Also, E 1 (Q) tors = {O, (0, 0), (5, 20), (5, 20)}, E 2 (Q) tors = {O, (0, 0)}, E 3 (Q) tors = {O, (0, 0)}, E 4 (Q) tors = {O, (0, 0), (1, 2), (1, 2), (5, 10), (5, 10)}. Since P 1,P 2,P 3 +T,P 1 +P 3,P 2 +P 3,P 1 +P 2 +T,P 1 +P 2 +P 3 +T E 0 (R), they cannot have any 2-division points in E(Q). Since x(p 3 )=u 4 +2u 2 v 2 +5v 4 is non-square, we have P 3 2E(Q). Since x(p i )= and x(p i +T )= n/x(p i ) for i {1, 2}, the non-squareness of n implies that P 1 + T,P 2 + T 2E(Q). Moreover, since 2(n b 2 d 2 ± ac) = if and only if u 4 +2u 2 v 2 ± 5v 4 =, Lemma 3.2 (2) shows that P 1 + P 2 2E(Q). Hence, we only have to check that P 1 + P 3 + T,P 2 + P 3 + T,P 1 + P 2 + P 3 2E(Q). We see from the addition formula that x(p 1 + P 2 + T ) = (u4 2u 2 v 2 +5v 4 )(u 2 +2uv +3v 2 ) 2, (u + v) 4 x(p 2 + P 3 + T ) = (u4 2u 2 v 2 +5v 4 )(u 2 + v 2 ) 2. (u 2 v 2 ) 2 Since u 4 2u 2 v 2 +5v 4 is non-square, we have P 1 +P 3 +T,P 2 +P 3 +T 2E(Q) by Lemma 3.1 (2). We further have ( A 2 P 1 + P 2 = 4v 2 (u v), AB 2 8v 3 (u v) 3 ),

On the rank of the elliptic curve y 2 = x 3 nx 895 where A = u 4 2u 3 v +2u 2 v 2 2uv 3 +5v 4, B = u 8 4u 7 v +8u 6 v 2 12u 5 v 3 +14u 4 v 4 28u 3 v 5 +40u 2 v 6 20uv 7 15v 8, and x(p 1 + P 2 + P 3 )= (u4 +2u 2 v 2 +5v 4 )(u 4 +3v 4 ) 2 (u 4 4uv 3 v 4 ) 2. Since u 4 +2u 2 v 2 +5v 4 is non-square, we have P 1 + P 2 + P 3 2E(Q). Remark 3.4. When m 2, we can not reduce the quartic curves to the elliptic curves of rank 0 in the same way as above. For example, E 1,m : Y 2 = X(X 2 +2αX + γ 2 ) ; rank E 1,m (Q) 1 for m =7, 8, 9, E 2,m : Y 2 = X(X 2 +2X + γ) ; rank E 2,m (Q) 1 for m =4, 6, 8, 9, E 3,m : Y 2 = X(X 2 +2X γ) ; rank E 2,m (Q) 1 for m =4, 5, 7, E 4,m : Y 2 = X(X 2 2X + γ) ; rank E 4,m (Q) 1 for 2 m 20. 3.2 n = a 4 + b 4 = c 2 + d 4 In this subsection, we construct a family of elliptic curves of rank 3 for n of the the form n = a 4 + b 4 = c 2 + d 4. One of the parameterizations of such n is given by Fauquembergue: a =17u 2 12uv 13v 2, b =17u 2 +12uv 13v 2 c = 289u 4 +14u 2 v 2 239v 4, d =17u 2 v 2 (5) (cf. [9, Part 7]). The points P 1 =( b 2,a 2 b), P 2 =( d 2,cd), P 3 =( a 2,ab 2 ) are in E(Q). Then we show the following: Theorem 3.5. Assume that an integer n can be expressed as n = a 4 + b 4 = c 2 + d 4 for nonzero integers a, b, c, d with a 2 {b 2, c,d 2 }. (1) If n is square-free, then rank E(Q) 3. (2) Let a, b, c, d be as in (5) for nonzero coprime integers u, v with u v. If u is even with u 0 (mod 13) and v is odd, then rank E(Q) 3. Note that in the assertion (2), the assumption a 2 {b 2, c,d 2 } implies uv 0and u v.

896 Y. Fujita and N. Terai Proof. In order to show that P 1, P 2 and P 3 are independent modulo E(Q) tors, it suffices to check that P i,p i + T,P i + P j,p i + P j + T,P 1 + P 2 + P 3,P 1 + P 2 + P 3 + T 2E(Q) for i, j {1, 2, 3} with i j. We know that P i,p i +P j +T,P 1 +P 2 +P 3 2E(Q), since they are not in E 0 (R). It is also clear from x(p + T )= n/x(p ) for P {O, T} and the non-squareness of n that P i +T 2E(Q). Moreover, since ( (a 2 + ab + b 2 ) 2 P 1 + P 3 =, ab(a2 + ab + b 2 )(2a 2 +3ab +2b 2 ) (a + b) 2 (a + b) 3 Lemma 3.1 (1) implies that there exists P E(Q) with ψ(p ) = P 1 + P 2 such that x(p )=2n/(a + b) 2 or 2(a + b) 2. Since the Diophantine equation X 4 + Y 4 =2Z 2 has no solution with X Y (cf. [8, p. 18]) and a 2 b 2, the square-free part of n = a 4 + b 4 is not equal to 2, and hence x(p ) cannot be a square in either case. It follows from Lemma 3.1 (2) that P 1 + P 3 2E(Q). We further have x(p 1 + P 2 + P 3 + T )= ), n {(a 2 + ab + b 2 ) 2 + d 2 (a + b) 2 } 2 {abd(2a 2 +3ab +2b 2 ) c(a + b)(a 2 + ab + b 2 )} 2. Since n = a 4 + b 4 cannot be a square, P 1 + P 2 + P 3 + T 2E(Q). Thus, the remaining points to check are P 1 + P 2 and P 2 + P 3. (1) By Lemma 3.2 (2) we have P 1 + P 2,P 2 + P 3 2E(Q). (2) In this case, we have n =2AB, where A = 289u 4 612u 3 v + 638u 2 v 2 468uv 3 + 169v 4, B = 289u 4 + 612u 3 v + 638u 2 v 2 + 468uv 3 + 169v 4. Since u is even and v is odd, 2B 2 (mod 4) and so 2B. Next we show that A. IfA =, saya = C 2, then this becomes C 2 + 2(8uv) 2 = (17u 2 18uv +13v 2 ) 2. Hence we have C = k(m 2 2N 2 ), 8uv =2kMN and 17u 2 18uv +13v 2 = k(m 2 +2N 2 ), (6) where M,N are positive integers with gcd(m, N) = 1 and M N (mod 2), and k is a positive integer. Since u is even and v is odd, we see that M is odd, N is even and k =1, 17 from u 0 (mod 13). The left hand side of (6) is congruent to 5 modulo 8. On the other hand, the right hand side of (6) is congruent to 1 modulo 8. Therefore A.

On the rank of the elliptic curve y 2 = x 3 nx 897 Since 2(n b 2 d 2 + a 2 c)=a{2(17u 2 +6uv 7v 2 )} 2 and 2(n b 2 d 2 a 2 c)= 2B {24(u v)v} 2, Lemma 3.2 (1), together with the assumptions A and 2B, implies that P 1 + P 2 2E(Q). By interchanging a and b (i.e., substituting v into v), we obtain P 2 + P 3 2E(Q). Remark 3.6. If A = in Theorem 3.5 (2), then the quartic curve can be reduced to the cubic curve E A : Y 2 = X 3 + 638X 2 + 91052X + 1953640 with rank E A (Q) = 2 (cf. [8, p. 77]). In fact, there are many pairs of (u, v) such that A =. Thus we assume our conditions in Theorem 3.5. 3.3 n = a 4 + b 4 = c 4 + d 4 In this subsection, we construct a family of elliptic curves of rank 4 for n of the the form n = a 4 + b 4 = c 4 + d 4. The simplest parametrization known is a = u(u 6 + u 4 v 2 2u 2 v 4 +3uv 5 + v 6 ), b = v(u 6 3u 5 v 2u 4 v 2 + u 2 v 4 + v 6 ), c = u(u 6 + u 4 v 2 2u 2 v 4 3uv 5 + v 6 ), d = v(u 6 +3u 5 v 2u 4 v 2 + u 2 v 4 + v 6 ), which was first obtained by Euler. But this is not in any sense complete (cf. [5, p. 260]). When u =1,v= 2, it leads to 635318657 = 133 4 + 134 4 = 158 4 +59 4. This number is the smallest integer expressible as a sum of two fourth powers in two different ways (cf. [1, p. 646] and [14]). The points P 1 =( b 2,a 2 b), P 2 =( d 2,c 2 d), P 3 =( a 2,ab 2 ), P 4 =( c 2,cd 2 ) are in E(Q). Then we show the following: Theorem 3.7. Let A = u 4 +6u 2 v 2 + v 4, B = u 8 4u 6 v 2 +8u 4 v 4 4u 2 v 6 + v 8, C = u 8 u 4 v 4 + v 8, D = u 8 +2u 6 v 2 +11u 4 v 4 +2u 2 v 6 + v 8 and n = ABCD for positive coprime integers u and v with u v (mod 2). Then rank E(Q) 4.

898 Y. Fujita and N. Terai Proof. We will show that P 1,P 2,P 3,P 4 are independent modulo E(Q) tors. The difficulty is that P 1 + P 2 + P 3 + P 4 has a 2-division point in E(Q). In fact, P 1 + P 2 + P 3 + P 4 = 2Q with where Q = ( (u 2 + uv + v 2 ) 2 AB, uv(u2 + uv + v 2 )FAB ). (u + v) 2 (u + v) 3 F =2u 8 +3u 7 v 2u 6 v 2 12u 5 v 3 14u 4 v 4 12u 3 v 5 2u 2 v 6 +3uv 7 +2v 8. We obtained this as follows. Put P = P 1 + P 2 + P 3 + P 4. Adding the points P 1 + P 3 and P 2 + P 4, we have x(p )=N 2 /M 2, where M = 2uv(u + v)(u 2 + uv + v 2 )F, N = 2u 20 +8u 19 v +20u 18 v 2 +36u 17 v 3 +47u 16 v 4 +36u 15 v 5 4u 13 v 7 Put +122u 12 v 8 + 312u 11 v 9 + 410u 10 v 10 + 312u 9 v 11 + 122u 8 v 12 4u 7 v 13 +36u 5 v 15 +47u 4 v 16 +36u 3 v 17 +20u 2 v 18 +8uv 19 +2v 20. s = u 2 v 2 F 2, t =2(u + v) 2 (u 2 + uv + v 2 ) 2. Then we find that x(p ) ± n =(s ± t n) 2 /M 2. It follows from Lemma 2.2 in [3] that there exists Q E(Q) with 2Q = P such that x(q) = 2sN + s2 t 2 n + N 2 or x(q) = 2sN s2 + t 2 n N 2. M 2 M 2 We adopted the former expression. In view of definitions of A, B, C, D, we easily see that AB, CD, AD, BC with gcd(u, v) = 1 and u v (mod 2). Indeed, the following relations hold for A, B, C, D: B = A(A 16u 2 v 2 )+66u 4 v 4, D = A(A 10u 2 v 2 )+33u 4 v 4, D C =2u 2 v 2 A, C B = u 2 v 2 (4u 4 9u 2 v 2 +4v 4 ), (4u 4 9u 2 v 2 +4v 4 )(4u 4 +9u 2 v 2 +4v 4 )=16C 33u 4 v 4. Then gcd(a, B) =1, gcd(c, D) =1, gcd(a, D) = 1 and gcd(b,c) = 1, since A and C are indivisible by 3, 11 and gcd(u, v) = 1. Hence if AB =, CD =, AD =, BC =, then A =, C =, which are impossible from Mordell [8, p. 18, p. 20].

On the rank of the elliptic curve y 2 = x 3 nx 899 We now show that P 1,P 2,P 3,Qare independent modulo E(Q) tors (which is equivalent to that P 1,P 2,P 3,P 4 are independent modulo E(Q) tors ). It is clear that P i,p i + T,Q + T,P i + Q, P i + P j + T,P 1 + P 2 + P 3,P 1 + P 2 + P 3 + Q 2E(Q) for i, j {1, 2, 3} with i j, since they are not in E 0 (R) and x(p i + T )= n/x(p i ). We also have Q 2E(Q) by Lemma 3.1 (1) and AB. From the proof of Theorem 3.5 we see that P 1 + P 3,P 1 + P 2 + P 3 + T 2E(Q). Moreover, since 2(n b 2 d 2 + a 2 c 2 ) = {2u 2 (u 6 2u 4 v 2 + u 2 v 4 + v 6 )} 2 AD, 2(n b 2 d 2 a 2 c 2 ) = (12u 3 v 3 ) 2 BC, Lemma 3.2 (1) together with AD and BC implies that P 1 + P 2 2E(Q). Similarly we see that P 2 + P 3 2E(Q). Thus, it suffices to check that P i + Q + T, P i + P j + Q, P 1 + P 2 + P 3 + Q + T 2E(Q) (i, j {1, 2, 3}, i j). (7) The addition formula shows the following. x(p 1 + Q + T )= CD u, 2 (u 6 2u 4 v 2 +4u 2 v 4 + v 6 ) 2 CD x(p 2 + Q + T )= u 2 (u 2 + v 2 ) 2 (u 2 uv v 2 ) 2 (u 2 + uv v 2 ), 2 (u 6 +4u 4 v 2 2u 2 v 4 + v 6 ) 2 CD x(p 3 + Q + T )= v 2 (u 2 + v 2 ) 2 (u 2 uv v 2 ) 2 (u 2 + uv v 2 ), 2 x(p 1 + P 2 + Q) = (u2 uv + v 2 ) 2 AB, (u v) 2 x(p 1 + P 3 + Q) = (u6 + u 4 v 2 3u 3 v 3 + u 2 v 4 + v 6 ) 2 AB, 9u 4 v 4 (u v) 2 x(p 2 + P 3 + Q) = (u6 + u 4 v 2 +3u 3 v 3 + u 2 v 4 + v 6 ) 2 AB, 9u 4 v 4 (u + v) 2 x(p 1 + P 2 + P 3 + Q + T )= CD v. 2 It follows from AB, CD and Lemma 3.1 (1) that (7) holds. This completes the proof of Theorem 3.7. Remark 3.8. On the assumptions in Theorem 3.7, one can find that rank E(Q) 3 only by using the method of Tate. Indeed, the Diophantine equation CDS 4 ABT 4 = U 2 has a solution (S, T, U) =(1,v,u 2 (u 6 +u 4 v 2 +4u 2 v 4 5v 6 )),

900 Y. Fujita and N. Terai which implies CD α(γ). Since 1 α(γ) and 2 α(γ) by Lemmas 2.1, we obtain rank E(Q) 3. Note that CD α(γ) corresponds to the rational point ( ) CD P 1 + P 2 + P 3 + Q + T = v, u2 (u 6 + u 4 v 2 +4u 2 v 4 3v 6 )CD. 2 v 3 We conclude this paper by giving high rank curves. We searched elliptic curves of rank E(Q) 7 by Magma in the ranges of 1 u, v 50 in Theorems 2.2, 3.3, 3.5; 1 u, v 30 in Theorem 2.6; 1 u, v 20 in Theorem 3.7, and the results are listed in the following table (where * implies rank E(Q) 7). Theorems rank E(Q) (u, v) 2.2 (i) 7 (8,25), (10,23), (13,35)*, (23,5), (25,4) 2.2 (ii) 7 (1,32), (1,48), (7,20), (13,27), (31,33), (47,49) 2.6 (ii) 7 (27,10)* 8 (1,6), (9,2) 3.3 7 (8,31), (19,48), (21,43), (24,19), (33,19), (48,23), (49,45) 8 (11,36), (25,48) 3.5 (2) 7 (28,11)* 3.7 7 (1,5)*, (2,3)* References [1] L.E. Dickson, Theory of Numbers, Vol. II, CHELSEA, 1971. [2] A. Dujella and A. Pethő, Integer points on a family of elliptic curves, Publ. Math. Debrecen, 56 (2000), 321-335. [3] Y. Fujita, Torsion subgroups of elliptic curves with non-cyclic torsion in elementary abelian 2-extensions of Q, Acta Arith., 115 (2004), 29-45. [4] Y. Fujita and N. Terai, Integer points and independent points on the elliptic curve y 2 = x 3 p k x, Tokyo J. Math., 34 (2011), 367-381. [5] G.H. Hardy and E.M. Wright, An introduction to the theory of numbers, sixth edition, Oxford Univ. Press, 2008.

On the rank of the elliptic curve y 2 = x 3 nx 901 [6] A.J. Hollier, B.K. Spearman and Q. Yang, On the rank and integral points of elliptic curves y 2 = x 3 px, Int. J. of Algebra, 3 (2009), 401-406. [7] J.A. Johnstone and B.K. Spearman, Congruent number elliptic curves with rank at least three, Canad. Math. Bull., 53 (2010), 661-666. [8] L.J. Mordell, Diophantine Equations, Academic Press, 1969. [9] T. Piezas, A collection of algebraic identities, http://sites.google.com/site/tpiezas/home. [10] J.H. Silverman and J. Tate, Rational points on elliptic curves, Springer- Verlag, 1992. [11] B.K. Spearman, Elliptic curves y 2 = x 3 px of rank two, Math. J. Okayama Univ., 49 (2007), 183-184. [12] B.K. Spearman, On the group structure of elliptic curves y 2 = x 3 2px, Int. J. Algebra, 1 (2007), 247-250. [13] P.G. Walsh, Maximal ranks and integer points on a family of elliptic curves II, Rocky Mountain J. Math., 41 (2011), 311-317. [14] P.G. Zajta, Solutions of the Diophantine equation A 4 + B 4 = C 4 + D 4, Math. Comp., 41 (1983), 635-659. Received: March, 2012