ON THE RANK OF CONGRUENT ELLIPTIC CURVES FARZALI IZADI AND HAMID REZA ABDOLMALEKI Abstract. In this paper, p and q are two different odd primes. First, 7 We construct the congruent elliptic curves corresponding to p, 2p, pq, 1 and2pq,then,inthecasesofcongruentnumbers,wedeterminetherank 0 of the corresponding congruent elliptic curves. 2 n a J 8 1. introduction ] T N The rank of an elliptic curve is a measure of the size of the set of rational . points. However, the question is to ask how one can compute the exact size h t ofthisset ofrationalpoints. Ontheotherhand, itiseasytofindtherational a m points of a projective line or a plane curve defined by a quadratic equation. [ Having said that, there is no known guaranteed algorithm to determine the 1 rank and it is not known which numbers can occur as the rank of an elliptic v 6 curve. (see [5]). 8 6 The rational number n is a congruent number if there are positive ratio- 2 nal numbers a,b,c such that a2 +b2 = c2 and 1ab = n , equivalently, there 0 2 . 1 is a Pythagorean triangle with rational sides and the area equals to n. 0 In a modern language, n is a congruent number if and only if the elliptic 7 1 curve E : y2 = x3−n2x contains a rational point with y 6= 0, equivalently, a : v rational point of infinite order, i.e., the rank of E which is denoted by r(E) i X is nonzero, (see [9]). r a First we quote from Monsky [4]. Let p ,p ,p and p denote primes ≡ 1 3 5 7 1,3,5 and 7 (mod 8). Heegner [3], and Brich [1], proved that 2p and 2p 3 7 are congruent numbers. Heegner asserted without proof that p and p are 5 7 congruent numbers, and this claim is repeated in [7]. Monsky [4] has given a unified proof that the following are all congruent numbers : (1) : p , p , 2p , and 2p , 5 7 7 3 (2) : p p , p p , 2p p and 2p p , 3 5 3 7 3 5 5 7 (3) : p p provided (p1) = −1, and 2p p provided (p1) = −1 and p p , 1 5 p5 1 3 p3 1 7 2p p , provided (p1) = −1, 1 7 p7 2010 Mathematics Subject Classification. Primary 11G05; Secondary 14H52, 14G05, Key words and phrases. Elliptic curves, Rank, congruent numbers. 1 2 F. IZADIAND H.R. ABDOLMALEKI In other words, any N = 5,6 or 7 (mod 8) having at most 2 odd prime factors is a congruent number, with the following possible exception N = pq or 2pq with p ≡ 1 (mod 8) and (p1) = +1. p3 Remark 1.1. For N in the form of (1),(2) and (3), our proof shows that the rank of EQ : Y2 = X3 − N2X is 1. However, for N = pq or 2pq with N p ≡ 1 (mod 8) and (p) = +1, this result is not true. For example, for q N = (521).(5) , the rank of EQ is 3. N Insectiontwo,werecallthe2-descentmethodwhichisaclassicalmethod for finding the rank of elliptic curves. Section three includes our results and contains five parts. In part (1), Let E : y2 = x3 −p2x, be the corresponding congruent number elliptic curve for p. It is known that if p ≡ 5,7 (mod 8) then p is congruent number (see[4]). Using the 2-descent method, we show in this two cases, r(E) = 1. Moreover if p ≡ 3 (mod 8) then r(E) = 0, i.e., p is not congruent. In part (2), for p ≡ 1 (mod 8), we investigate that r(E) = 2. We show this is happen whenever p satisfy in the two following conditions: (1) ∃a,b ∈ N: p = a+b , a−b = (cid:3) and a2 +b2 = (cid:3). (2) ∃a,b ∈ N: p = a2 +b2 , (a,2b) = 1 and (a±2b)2 +b2 = (cid:3). In part (3), Let E : y2 = x3 − 4p2x, be the corresponding congruent number elliptic curve for 2p. It is known that if p ≡ 3,7 (mod 8) then 2p is congruent number (see[4]). Using the 2-descent method, we show in this two case, r(E) = 1. Moreover if p ≡ 5 (mod 8) then r(E) = 0, i.e., 2p is not congruent. In part (4), Let E : y2 = x3 − p2q2x, be the corresponding congruent number elliptic curve for pq. It is known that if [p ≡ 3 and q ≡ 5,7 (mod 8)] or [p ≡ 1 and q ≡ 5,7 (mod 8) such that (p) = −1] then pq is a congruent q number (see[4]). Using 2-descent method, we show inthis four cases, r(E) = 1. Moreover if p,q ≡ 3 (mod 8) then r(E) = 0, i.e., pq is not congruent. In part (5), Let E : y2 = x3 −4p2q2x. We know If [p ≡ 5 and q ≡ 3,7 (mod 8)] or [p ≡ 1 and q ≡ 3,7 (mod 8) such that (p) = −1] then 2pq is q congruent number (see[4]). Using the 2- descent method, we show in this four cases, r(E) = 1. Moreover if [p,q ≡ 5 (mod 8)] or [p ≡ 1 and q ≡ 5 (mod 8) such that (p) = −1], then r(E) = 0, i.e., 2pq is not congruent. q 2. 2-descent method In this section we describe the 2- descent method for determining the rankofanellipticcurve,(see[2],Chapter8formoredetails).LetE(Q)isthe ON THE RANK OF CONGRUENT ELLIPTIC CURVES 3 group of rational points on the elliptic curve E : y2 = x3 +ax2 +bx, where a,b ∈ Q. Let Q∗ is the multiplicative group of nonzero rational numbers and Q∗2 is the subgroup of squares of elements of Q∗. Define the 2-descent homomorphism α from E(Q) to Q∗ as follows : Q∗2 1 (mod Q∗2) if P = O = ∞, α(P) =  b (mod Q∗2) if P = (0,0),  x (mod Q∗2) if P = (x,y),x 6= 0.  Similarly, take the isogenous curve E : y2 = x3−2ax2+(a2−4b)x with the group of rational points E(Q). The 2-descent homomorphism α from E(Q) to Q∗ as follows : Q∗2 1 (mod Q∗2) if P = O = ∞, α(P) =  a2 −4b (mod Q∗2) if P = (0,0),  x (mod Q∗2) if P = (x,y),x 6= 0.  The rank of E(Q) which is denoted by r is determined by | Imα | · | Imα | (2.1) 2r = . 4 The group α(E(Q)) equals to the classes modulo squares of 1, b and the positive and negative divisors of b such that b 2 4 2 2 4 N = b m +am e + e , 1 b 1 b gcd(m,e) = gcd(m,N) = gcd(e,N) = gcd(b ,m) = gcd( ,e) = 1,me 6= 0. 1 b 1 b m2 b mN 1 1 If (m,N,e) is a solution, then P = ( , ) belongs to E(Q) and we e2 e3 have the same for α as well. 3. our results Part (1). According the 2-descent method, for the elliptic curves 2 3 2 (3.1) E : y = x −p x and 2 3 2 2 (3.2) E : y = x +4p q x. Wehaverespectively{±1} ⊆ Imα ⊆ {±1,±p}and{1} ⊆ Imα ⊆ {1,2,p,2p}. Therefore, according to (2.1), the maximum rank of (3.1) is 2. Moreover, the homogeneous equation of E is (a ) : N2 = ±(pm4−pe4), gcd(m,pe) = gcd(e,pm) = gcd(N,me) = 1, 1 4 F. IZADIAND H.R. ABDOLMALEKI and the homogeneous equations of E are (b ) : N2 = pm4 +4pe4, gcd(m,2pe) = gcd(e,pm) = gcd(N,me) = 1, 1 (b ) : N2 = 2m4 +2p2e4, gcd(m,2pe) = gcd(e,2m) = gcd(N,me) = 1, 2 (b ) : N2 = 2pm4 +2pe4, gcd(m,2pe) = gcd(e,2pm) = gcd(N,me) = 1. 3 First we study the solvability of the above homogeneous equations. Proposition 3.1. If (b ) has integer solution then p ≡ 1 (mod 4). 1 Proof. If (b ) has integer solution then N2 = p(m4 + 4e4), gcd(m,p) = 1. 1 There is an integer u such that pu2 = m4+4e4. Hence m4 ≡ −4e4 (mod p). Therefore (2e2m∗2)2 ≡ −1 (mod p). So (−1) = +1, i.e., p ≡ 1 (mod 4). (cid:3) p Proposition 3.2. If (b ) has integer solution then p ≡ ±1 (mod 8). 2 Proof. If (b ) has integer solution then N2 = 2(m4 +p2e4), gcd(m,p) = 1. 2 There is an integer u such that 2u2 = m4 + p2e4. So 2u2 ≡ m4 (mod p). Hence (2um∗2)2 ≡ 2 (mod p). So (2) = +1, i.e., p ≡ ±1 (mod 8). (cid:3) p Remark 3.3. If u is an integer number then u2 ≡ 0,1,4 (mod 8). Hence, gcd(u,8) = 1 if and only if u2 ≡ 1 (mod 8). Proposition 3.4. If (b ) has integer solution then p ≡ 1 (mod 8). 3 Proof. If (b ) has integer solution then N2 = 2p(m4 +e4), gcd(em,2) = 1. 3 There is an integer u such that 2pu2 = m4 +e4. Hence 2pu2 ≡ 2 (mod 16). So pu2 ≡ 1 (mod 8). We have (u,8) = 1. Therefore u2 ≡ 1 (mod 8). Con- sequently p ≡ 1 (mod 8). (cid:3) Corollary 3.5. The above propositions are summarized in the followings: (1) ±p ∈ Imα if and only (a ) has integer solution. 1 (2) p ∈ Imα if and only if (b ) has integer solution. In this case p ≡ 1,5 1 (mod 8). (3) 2 ∈ Imα if and only if (b ) has integer solution. In this case p ≡ 1,7 2 (mod 8). (4) 2p ∈ Imα if and only if (b ) has integer solution. In this case p ≡ 1 3 (mod 8). Corollary 3.6. By using previous corollary, we have: (1) Let p ≡ 3 (mod 8). Then (b ),(b ),(b ) do not have any integer solutions, so Imα = 1 2 3 {1}. As (2.1), implies | Imα |≥ 4. ON THE RANK OF CONGRUENT ELLIPTIC CURVES 5 So Imα = {±1,±p}. Hence r = 0, i.e., p is not a congruent. (2) Let p ≡ 5 (mod 8). It is possible for (b ) to have integer solution. So Imα ⊆ {1,p}. 1 p is a congruent number, (see [4]), we have r ≥ 1. As (2.1), implies Imα = {±1,±p} and Imα = {1,p}. Hence r = 1. (3) Let p ≡ 7 (mod 8). It is possible for (b ) to have integer solution. So Imα ⊆ {1,2}. 2 p is a congruent number, (see [4]), we have r ≥ 1. As (2.1), Imα = {±1,±p} and Imα = {1,2}. Hence r = 1. Part (2). In this section, for p ≡ 1 (mod 8) we investigate that r(E) = 2. • First we study the solvability of the homogeneous equation (b ). We 1 need some definitions. Definition 3.7. p is a α−Pythagorean whenever, there are coprime integer numbers a,b and c such that pc2 = a2 +b2. − Definition 3.8. p is a α Pythagorean whenever there are coprime integers − a,b and c such that pc2 = a2 +b2 and (a−2b)2 +b2 = (cid:3). + Definition 3.9. p is a α Pythagorean whenever there are coprime integers − a,b and c such that pc2 = a2 +b2 and (a+2b)2 +b2 = (cid:3). ± − Definition 3.10. p is a α Pythagorean whenever p is a α Pythagorean or − − + α Pythagorean. − Remark 3.11. Considering the results of part 2, we know p ≡ 5 (mod 8) ± is α Pythagorean. − − Example 3.12. 37 with (a,b,c) = (22,21,5) is a α Pythagorean. − − Example 3.13. 41 with (a,b,c) = (5,4,1) is a α Pythagorean. − + Example 3.14. 149 with (a,b,c) = (10,7,1) is a α Pythagorean. − ± Proposition3.15. (b ) has integersolution if andonlyifp is a α Pythagorean. 1 − Proof. If (b ) has integer solution then N2 = p(m4 + 4e4),gcd(m,2e) = 1. 1 There is an integer number u such that pu2 = m4 +4e4. 6 F. IZADIAND H.R. ABDOLMALEKI pu2 = (m2 −2me+2e2)(m2 +2me+2e2). As gcd(m2−2me+2e2,m2+2me+2e2) = 1, there are coprime integers c and w such that (1) m2−2me+2e2 = pc2 andm2+2me+2e2 = w2.So(m−e)2+e2 = pc2 and [(m−e)+2e]2 +e2 = w2. Hence p is a α+Pythagorean. − (2) m2−2me+2e2 = w2 andm2+2me+2e2 = pc2.So(m+e)2+e2 = pc2 and [(m+e)−2e]2 +e2 = w2. Hence p is a α−Pythagorean. (cid:3) − Remark 3.16. If p ≡ 1 (mod 4) ≡ 1,5 (mod 8) then there are unique positive integers a and b such that p = a2 +b2. p ≡ 1 (mod 8) if and only if there are unique integers k and t such that p = 16k2 +t2,(t,2k) = 1. p ≡ 5 (mod 8) if and only if there are unique integers k and t such that p = 4k2 +t2,(t,2k) = (k,2) = 1. ± Lemma 3.17. If p with (a,b,c) is a α Pythagorean, then − a2 ≡ 1 , b2 ≡ 0 (mod 8) or a2 ≡ 4 , b2 ≡ 1 (mod 8). Proof. We have pc2 = a2 +b2,(a,b) = 1 and (a±2b)2 +b2 = (cid:3). Since pc2 is sum of two primitive squares, then p and all factors of c are form 4k + 1, where k is integer, (see[10]). We know squares in mod 8 are 0,1 or 4. Suppose otherwise if a2 ≡ 0,b2 ≡ 1 or a2 ≡ 1,b2 ≡ 4 (mod 8), then (cid:3) = a2 +5b2 ±4ab ≡ 5 (mod 8), which is a contradiction. (cid:3) ± Corollary 3.18. Suppose p is a α Pythagorean: − (1) If p = 4k2+t2 ≡ 5 (mod 8), then a = 2k and b = t, where kt is odd. (2) If p = 16k2+t2 ≡ 1 (mod 8), then a = t and b = 4k, where t is odd. • Next theorem tell us that the prime number p ≡ 1 (mod 8) is a ± α Pythagorean if and only if c = 1, (see the above Definition). − Theorem 3.19. (1) If p = a2 +b2 and 2 k b, then p is congruent number. (2) If 4 | b and (a±2b)2 +b2 6= (cid:3), then p is not a α±Pythagorean. − Proof. (1) If 2 k b, then there is odd integer number b such that b = 2b . 0 0 So p = a2 +4b2 ≡ 5 (mod 8), therefore p is congruent number, (see[4]). 0 (2)If4 | b,(a±2b)2+b2 6= (cid:3)andSupposeotherwisepisaα±Pythagorean − then there are coprime integers a , b and c 6= 1 such that 0 0 0 pc2 = a2 +b2 and 4a2 +5b2 ±4a b = (cid:3). 0 0 0 0 0 0 0 ON THE RANK OF CONGRUENT ELLIPTIC CURVES 7 Therefore pc2 is sum of two primitive numbers, then p and all factors of c2 0 0 are form 4k + 1. Consequently there are integers m and n such that m is odd number and n is nonzero even number such that c2 = m2 +n2, then 0 pc2 = (b2 +a2)(m2 +n2) = (bm+an)2 +(bn−am)2. 0 We have [a = bm+an , b = bn−am] or [a = bm−an , b = bn+am]. 0 0 0 0 If a = bm+an and b = bn−am, then a2n2 ±5a2m2 ≡ (cid:3) (mod 8). 0 0 If n2 ≡ 4 and m2 ≡ 1 (mod 8), then c2 ≡ 5 (mod 8), which is a contra- 0 diction. If n2 ≡ 0,m2 ≡ 1 and a2 ≡ 1 (mod 8), then (cid:3) ≡ ±5 (mod 8), which is a contradiction. proof for a = bm−an and b = bn+am is similar. (cid:3) 0 0 Example 3.20. 17 = 16×12+12 ≡ 0+1 (mod 8),then a2 = 1andb2 = 16. As (a±2b)2 +b2 6= (cid:3), hence 17 is not a α±Pythagorean. − Example 3.21. 41 = 16×12+52,with(a,b,c) = (5,4,1),isaα−Pythagorean. − • Now, we study the solvability of the homogeneous equation (b ), how- 2 ever we first need a definitions. Definition 3.22. p is a β−Pythagorean whenever there are integers a,e,m and u such that pe2m2 = 2a2 −u2 , pe2 = 2a−m2 and (e,m) = 1. Remark 3.23. Considering the results of part 2, we know p ≡ 7 (mod 8) is β−Pythagorean. Proposition3.24. (b2) has integersolution if andonlyifp is a β−Pythagorean. Proof. If (b ) has integer solution then N2 = 2(m4 +p2e4),gcd(m,e) = 1. 2 N2 = 2(m4 +p2e4) if and only if there is a integer number u such that 2u2 = m4+p2e4 ifandonlyif (m2+pe2)2 = 2(u2+pm2e2) ifandonlyif there is an integer number a such that u2+pm2e2 = 2a2 and m2+pe2 = 2a. (cid:3) • Next proposition tell us that there is a relationship between solutions (b ) and Pythagorean triples such that difference of the two smaller sides is 2 square. Proposition 3.25. If pe2 = 2a − m2, then (m,e,u), is a solution (b ) if 2 and only if (a−m2,a,u), is a Pythagorean triple. Proof. (a − m2)2 + a2 = u2 if and only if m2(2a − m2) = 2a2 − u2 if and only if [pe2m2 = 2a2 −u2 and pe2 = 2a−m2]. (cid:3) 8 F. IZADIAND H.R. ABDOLMALEKI Remark 3.26. (a,b,u),a < b is primitive Pythagorean triple if and only if there are coprime positive integers s and t such that a = s2 −t2 , b = 2st or a = 2st , b = s2 −t2. Lemma 3.27. If a = s2 −t2,b = 2st,(s,t) = 1 and a−b = m2, then there are integers x and y such that s = 2x2 +y2+2xy,t = 2xy and (2x,y) = 1. Proof. We have (s−t)2−2t2 = m2. Let k = s−t, therefore 2t2 = k2−m2. From(s,t) = 1 we have (k,m) = 1. So m , k are oddand (k−m,k+m) = 2. Consequently, there are coprime integers x and y such that k +m = 4x2, k −m = 2y2 or k −m = 4x2, k +m = 2y2. Then k = y2 + 2x2 and m = ±(y2 − 2x2). Hence t = ±2xy and s = 2x2 +y2±2xy. (cid:3) Corollary 3.28. a = 4x4 +y4 +4x2y2+8x3y +4xy3 and b = 8x2y2 +8x3y +4xy3. Lemma 3.29. If a = 2st,b = s2 −t2,(s,t) = 1 and a−b = m2 then there are integers x and y such that s = 2xy,t = 2x2 +y2+2xy and (2x,y) = 1. Proof. The proof is similar to the previous lemma by letting k = s+t. (cid:3) Corollary 3.30. b = −4x4 −y4−4x2y2+8x3y +4xy3 and a = −8x2y2 +8x3y +4xy3. Proposition 3.31. (b ) has integer solution if and only if p(cid:3) ∈ Imf ∪ 2 1 Imf , where 2 f (x,y) = 4x4 +y4 +12x2y2 +16x3y +8xy3, (2x,y) = 1, 1 f (x,y) = −4x4 −y4−12x2y2 +16x3y +8xy3,(2x,y) = 1. 2 Proof. (b ) has integer solution if and only if pe2 = a+b, a−b = m2 and 2 a2 +b2 = u2, then the above above corollaries yield the result. (cid:3) Example 3.32. f (1,1) = 41,f (−1,7) = 137,f (1,1) = 7, hence for p = 1 1 2 41,137,7 equation (b ) has an integer solution. 2 Remark 3.33. As y is odd we have f (x,y) ≡ (2x2+y)2 ≡ 1 (mod 8) and 1 f (x,y) ≡ −(2x2 +y)2 ≡ 7 (mod 8). 2 Proposition 3.34. If pe2 is β−Pythagorean then p is β−Pythagorean. ON THE RANK OF CONGRUENT ELLIPTIC CURVES 9 Proof. If pe2 is β−Pythagorean if and only if there are integers m,E and u such that 2u2 = m4 +(pe2)2E4,(m,E) = 1. Consequently m and e are odd. If (m,e) = d, then d is odd and also d2 | u. There are integers m ,u and e such that e = e d, m = m d and 0 0 0 0 0 u = u d2. Weknow (m ,e ) = 1. Hence m4+p2(e E)4 = 2u2,(m ,e E) = 1. 0 0 0 0 0 0 0 0 Consequently p is β−Pythagorean. (cid:3) Corollary 3.35. (b ) has an integer solution if and only if p ∈ Imf ∪Imf . 2 1 2 i.e., for solving the equation (b ), we can choose e = 1. 2 Example 3.36. 17 is not β−Pythagorean because, there are not positive integers a and b such that 17 = a+b, a−b = m2 and a2 +b2 = u2. Example 3.37. [41 = 21 + 20, 21 − 20 = (cid:3) and 212 + 202 = (cid:3)], i.e., f1(1,1) = 41, therefore 41 is β−Pythagorean. • Now, we study the solvability of the homogeneous equation (b ). 3 Proposition 3.38. If (b ) has an integer solution, then p(cid:3) ∈ Imf , where 3 3 f (x,y) = 16x4 +y4 +24x2y2,(2x,y) = 1. 3 Proof. If (b ) has an integer solution, then 3 N2 = 2p(m4 +e4) , gcd(m,2pe) = gcd(e,2p) = 1. There is an integer number u such that 2pu2 = m4 +e4. So (m2−e2)2 = 2(pu2−e2m2). Again there is an integer number a such that pu2 −e2m2 = 2a2,m2 −e2 = 2a, from which one gets integers y and t such that m −e = 2y,m+ e = 2t and a = 2yt. Therefore m = t + y and e = t−y, where y or t is odd and the other is even. We have pu2 = t4 +y4+6y2t2. Suppose t is even and y is odd. There is an integers x such that t = 2x. This yields the result. (cid:3) Example 3.39. f (1,1) = 41, f (2,1) = 353, hence for p = 41, 353, the 3 3 equation (b ) has an integer solution. 3 Proposition 3.40. If (b ) or (b ) has an integer solution then equation (a) 2 3 has an integer solution. Proof. If (b ) has an integer solution, then N2 = 2(p2e4 +m4),(me,2) = 1. 2 There is an integer number u such that 2u2 = p2e4 + m4. Let 2c = pe2 + m2, 2d = pe2 − m2. We have c + d = pe2, c − d = m2. This implies that c2 −d2 = pe2m2, c2 +d2 = u2. Hence c4 −d4 = p(emu)2. If (b ) has an integer solution, then N2 = 2p(m4 + e4),(me,2) = 1. 3 Therefore, there is an integer number u such that 2pu2 = m4 + e4. Let 10 F. IZADIAND H.R. ABDOLMALEKI 2c = e2 +m2, 2d = e2 −m2. We have c+d = e2, c−d = m2. This implies that c2 −d2 = e2m2, c2 +d2 = pu2. Hence c4 −d4 = p(emu)2. (cid:3) Corollary 3.41. If (b ) or (b ) has an integer solution, then p is congruent. 2 3 Proof. If(b )or(b )hasanintegersolutionthen| Imα |≥ 2and| Imα |= 4. 2 3 As (2.1), we have r(E) ≥ 1. Consequently p is a congruent number. (cid:3) ± Main Theorem 3.42. r(E) = 2 if and only if p is α Pythagorean and − β−Pythagorean. ± Proof. p is α−Pythagorean and β−Pythagorean if and only if p ∈ Imα and 2 ∈ Imα if and only if (b ) and (b ) have integer solutions if and only if 1 2 Imα = {1,2,p,2p} and Imα = {±1,±p} if and only if r(E) = 2. (cid:3) Corollary 3.43. If p ≡ 1 (mod 8) then r(E) = 2, whenever p satisfies in the two following conditions: (1) ∃a,b ∈ N: p = a+b , a−b = (cid:3) and a2 +b2 = (cid:3). (2) ∃a,b ∈ N: p = a2 +b2 , (a,2b) = 1 and (a±2b)2 +b2 = (cid:3). Example 3.44. For p = 41 we have r(E) = 2. Part (3). According to the 2-descent method, for the elliptic curves 2 3 2 (3.3) E : y = x −4p x and 2 3 2 (3.4) E : y = x +16p x. We have respectively {±1} ⊆ Imα ⊆ {±1,±2,±p,±2p} and {1} ⊆ Imα ⊆ {1,2,p,2p}. Therefore, according to (2.1), the maximum rank of (3.3) is 3. In fact, we showed that Imα ⊆ {1,p}. Hence the maximum rank is 2. The homogeneous equations of E are (a ) : N2 = ±(m4 −4pe4), gcd(m,2pe) = gcd(e,p) = gcd(N,em) = 1, 1 (a ) : N2 = ±(2m4 −2p2e4), gcd(m,2pe) = gcd(e,2) = gcd(N,em) = 1, 2 (a ) : N2 = ±(2pm4 −2pe4), gcd(m,2pe) = gcd(e,2p) = gcd(N,em) = 1 3 and the homogeneous equations of E are (b ) : N2 = 2m4 +8p2e4, gcd(m,2pe) = gcd(e,2) = gcd(N,em) = 1, 1 (b ) : N2 = 2pm4 +8pe4, gcd(m,2pe) = gcd(e,2p) = gcd(N,em) = 1, 2 (b ) : N2 = pm4 +16pe4, gcd(m,2pe) = gcd(e,p) = gcd(N,em) = 1. 3