Non-congruent numbers, odd graphs and the Birch Swinnerton-Dyer conjecture

Transcription

1 ACTA ARITHMETICA LXXV.1(1996 Non-congruent numbers, odd graphs and the Birch Swinnerton-Dyer conjecture by Keqin Feng (Hefei 1. Introduction. The aim of this paper is twofold. One is to present more non-congruent numbers n with arbitrarily many prime factors. Another is to verify the (even part of the Birch Swinnerton-Dyer conjecture on the elliptic curve E n : y = x 3 n x for several series of integer n. Congruent numbers. A positive integer n is called a congruent number (CN if n is the area of a rational right triangle. Otherwise n is called a non-congruent number (non-cn. It is well known that n is non-cn iff the rank of the rational point group E n (Q is zero (see Koblitz [4], for instance. From now on we assume without loss of generality that n is square-free. The congruent number problem is very old and was discussed by Arab scholars in the tenth century. By the author s (incomplete knowledge, the following CN and non-cn have been determined (p, q and r denote distinct prime numbers, p i means a prime number congruent to i modulo 8. For CN: n = p 3 (Heegner (195, Birch (1968, n = p 5,p 7 (Stephens, 1975, n = p u q v 5,6,7 (mod 8, 0 u,v 1 (B. Gross, 1985, n = p 3 p 5,p 5 p 7 n = p 1 p 7, ( p 1 p 7 = 1 (Monsky, 1990, n = p 1 p 3, ( p 1 p 3 = 1 where ( p q is the Legendre symbol. Supported by the National Natural Science Foundation of China and the National Education Committee of China. [71]

2 7 K. Feng For non-cn: n = p 3,p 3 q 3,p 5,p 5 q 5 (Genocchi, 1855, n = p 1, p 1 = a + 4b, ( a+b } p 1 = 1 (Bastien, n = p, p 9 (mod 16 Lagrange [5] (1974 presented many non-cn n with at most three odd prime factors by using the -descent method to prove ranke n (Q = 0. Some of them are: ( p r n = p 1 p 3, ( p 1 p 3 = 1, n = p 1 p 5, ( p 1 p 5 = 1, n = p 1 p 3 q 1, with the condition ( (see below, n = p 1 p 5 q 1, with the condition (. Condition (. n can be written as n = pqr or pqr such that ( p q = = 1. A well-known conjecture made by Alter, Curtz and Kubota [1] says that n is CN if n 5,6,7 (mod 8. Several particular cases of this conjecture has been verified (see the above-mentioned n. Moreover, the whole ACK conjecture can be derived from the BSD conjecture on the elliptic curve E n. Birch and Swinnerton-Dyer conjecture. Let L En (s be the L-function of the elliptic curve E n. The BSD conjecture says that: (BSD1 ranke n (Q = ord s=1 L En (s. (BSD If L En (1 0, then (1.1 L En (1/A = X(E n, where X(E n is the Tate Shafarevich group of E n, and A is a certain nonzero number which we do not want to describe exactly. K. Rubin ([8], [9] proved that if L En (1 0, then the group X(E n is finite and the odd parts of both sides of (1.1 are equal. It is well known that L En (1 = 0 for n 5,6,7 (mod 8. Therefore the Alter Curtz Kubota conjecture can be derived from the BSD conjecture (BSD1. A remarkable step was made by Tunnell [1] in 1983 who presented an elementary formula for L En (1/A by using modular forms with weight 3/. For n odd, let (1. a(n = 1 ζ(x + iy (i = 1, x +y +z =n y

3 Non-congruent numbers 73 where ζ is the character of (Z[i]/(4(1 + i such that { 1 for α = 1,7,3 + i,1 + i, ζ(α = 1 for α = 3,5,7 + i,5 + i. For n, let (1.3 b(n/ = 1 x +y +z =n/ z ζ (x + y, where ζ is the character of (Z[ ]/(4 such that ζ (α = ζ ( α, ζ (1 = 1, ζ (1 + = ζ (3 + = 1. Let w(n be the number of distinct prime factors of n. For the left-hand side of (1.1, Tunnell [1] proved that { (a(n/ (1.4 L En (1/A = w(n if n, (b(n// w(n/ if n. We are now ready to explain the title and philosophy of this paper. The sums (1. and (1.3 extend over the solutions of x +y +z = n (or n/ with y. We have a one-to-one correspondence between the solutions of x + y + z = n and X + Y + Z = n (with Z as follows: (1.5 (X,Y,Z = (x + y,x y,z, (x,y,z = ( X+Y, X Y, Z. A well-known Gauss result says that the number of solutions of X + Y + Z = n (with Z is 4h( n, where h( n is the class number of Q( n. Since Rubin proved that the odd parts of both sides of (1.1 are equal provided a(n 0 or b(n/ 0, we need to determine the Sylow -subgroup C ( K of the class group C K for K = Q( n. Gauss genus theory says that -rankc K = w(n 1 = w(n. For each n we can define a graph G(n. Rédei and Reichardt ([6], [7] essentially proved that w(n h( n iff G(n is an odd graph (for the definition of G(n and odd graph see Section. It turns out that for a series of n we can show by the -descent method that ranke n (Q = 0 and the order of X(E n is odd provided the graph G(n is odd (see Section 3. Therefore we present a series of non-congruent numbers n with arbitrarily many prime factors. By using the above-mentioned Rédei Reichardt result we can show that a(n/ w(n or b(n// w(n/ is an odd integer so that the BSD conjectures (BSD1 and (BSD are true for such n (see Section 4. This is the relation between non-congruent numbers, odd graphs, the -parts of the class numbers of imaginary quadratic fields and the BSD conjecture on E n.

4 74 K. Feng. Odd graphs and the -class number of Q( n. We use standard terminology of graph theory. Let G = (V,E be a (simple directed graph, V = {v 1,...,v m } the vertices of G, and E ( V V the arcs of G. The adjacency matrix of G is defined by A(G = (a ij 1 i,j m, where { a ij = 1 if i j and vi v j E, 0 otherwise. Let d i = m j=1 a ij be the outdegree of the vertex v i (1 i m, and M(G = diag(d 1,...,d m A(G. Then the sum of each row of M(G is zero, so that det M(G = 0. Let M ij = M ij (G be the (i,j co-factor of M(G; we have M ij = ( 1 j+k M ik. If the matrix A(G, and so M(G, is symmetric, we view G as a non-directed graph and the two-direction arc v i v j as an edge. For a non-directed graph G, we have M 11 = ( 1 i+k M ik (1 i,k m and it is well known that the absolute value of M 11 is the number of spanning trees of G. Definition.1. Let G = (V, E be a directed graph. A partition V = V 1 V is called odd if either there exists v 1 V 1 such that #{v 1 V } (the number of arcs from v 1 to vertices of V is odd, or there exists v V such that #{v V 1 } is odd. Otherwise the partition is called even. G is called odd if each non-trivial partition of V is odd. Let r = rank F M(G be the rank of the matrix M(G over F. Then r rank Q M(G m 1. We have the following nice criterion for oddness of G. Lemma.. Let G = G(V,E be a directed graph with m vertices, r = rank F M(G. Then the number of even partitions of V is m r 1. In particular, G is an odd graph iff r = m 1. For G non-directed, G is odd iff the number t(g of spanning trees of G is odd. P r o o f. Consider the following homogeneous linear equations over F : x 1 0 (.1 M(G.. =... 0 x m Each vector (c 1,...,c m over F corresponds to a partition of V = {v 1,...,v m } by V 1 = {v i : c i = 0}, V = {v i : c i = 1}. The vectors (c 1,...,c m and (c 1 + 1,...,c m + 1 correspond to the same partition of V up to interchanging V 1 and V. A vector (c 1,...,c m is a

5 Non-congruent numbers 75 solution of (.1 iff m j=1 a ijc j + d i c i = 0 (1 i m. But in F we have m m a ij c j + d i c i = a ij (c j + c i j=1 j=1 { m j=1 = a ijc j = m j=1,c j =1 a ij if c i = 0, m j=1 a ij(c j + 1 = m j=1,c j =0 a ij if c i = 1, { #{vi V = } if v i V 1, #{v i V 1 } if v i V. Therefore (x 1,...,x m = (c 1,...,c m is a solution of (.1 over F iff the partition V = V 1 V is even. So the number of even partitions of V is half of the number of solutions of (.1 over F, which is 1 m r = m r 1. For G non-directed, we know that r = m 1 iff t(g = M 11 = 1 F. This completes the proof. Many odd non-directed graphs can be found easily from Lemma.. Corollary.3. (1 The following non-directed graphs are odd: a tree T; a cycle C n with an odd number n of vertices; a perfect graph K n with an odd number n of vertices (for each pair of distinct vertices v i and v j there exists an edge v i v j in K n. ( Suppose that G 1 and G are non-directed graphs. Let G be a glue of G 1 and G as shown in Fig. 1. Fig.1 Then G is odd iff both G 1 and G are odd. (3 Every odd non-directed graph is connected. P r o o f. (1 follows from t(t = 1, t(c n = n and t(k n = n n by Cayley. ( comes from t(g = t(g 1 t(g. (3 For a disconnected nondirected graph G, t(g = 0. The concept of odd graph has been used to determine the solvability of the Pell equation x dy = 1 (see [], for instance. For our purpose, we now describe a relation between an odd graph and the Sylow -subgroup C K of the class group C K of an imaginary quadratic field K. Let K = Q( D (D be an imaginary quadratic field, D = disc(k the discriminant of K, h K = C K the class number of K, r = dim F C K /C K the -rank of C K. Gauss genus theory says that r = t 1 so

6 76 K. Feng that t 1 h K, where t = w(d is the number of distinct prime factors of D. Now we define the directed graph G(D in the following way. The vertices of G(D are all prime factors of D. For distinct vertices p i and p j, there is an arc p i p j in G(D iff ( p j ( p i = 1, where p q is the Legendre symbol but we assume that ( n = 1 for each odd integer n. Theorem.4. Let K = Q( D (D be an imaginary quadratic field, D = disc(k, and t the number of distinct prime factors of D. Then (1 t 1 h K the directed graph G(D is odd. ( If D = 8p...p t (t, p ±3 (mod 8 and p i 1 (mod 8 for i 3, then t 1 h K iff G(D/8 is odd. P r o o f. (1 Let p 1,...,p t be the distinct prime factors of D. For each subset S of {1,...,t} with 1 S t 1, let Q S = i S p i and Q S be the square-free part of D/Q S. We denote by α S the ambiguous ideal in O K with N(α S = Q S. Then genus theory says that the set {[α S ] = [α S ] : S {1,...,t}, 1 S t 1} consists of r 1 = 1 (t ideal classes in C K with order, where [α] denotes the class of the ideal α, and S = {1,...,t} S. Rédei and Reichardt ([6], [7] proved that [α S ] CK iff the equation u Q S + v Q S w = 0 has a non-trivial Q-solution (u,v,w (0,0,0. By Legendre, the last statement is equivalent to the existence of X,Y Z such that Therefore X Q S (mod Q S and Y Q S (mod Q S. t 1 h K [α S ] CK for each S {1,...,t} with 1 S t 1, for each S {1,...,t} with 1 S t 1, either there exists a prime number p Q S such that ( Q S p = 1, or there exists a prime number q Q S such that ( Q S q = 1, G(D is an odd graph. ( In this case G(D/8 is a non-directed graph by the quadratic reciprocity law and G(D is as in Fig. since ( ( p = 1 and p i = 1 for i 3. Fig.

7 Non-congruent numbers 77 It is easy to see that G(D is odd iff G(D/8 is odd. This completes the proof of Theorem descent method. The aim of this section is to show more integers n with arbitrarily many prime factors to be non-congruent numbers and #(X(E n for these n by the -descent method. First, we describe the -descent method briefly. (For details see the last chapter of Silverman s book [11]. Let a,b Z and E : y = x 3 + ax + bx be an elliptic curve over Q. The -dual curve of E is E : Y = X 3 ax + (a 4bX. We have the -isogeny φ : E E, φ(x,y = (y /x,y(b x /x. The kernel of φ is E[φ] = {0,(0,0}, where 0 denotes the infinite point of E as the identity of the Q-point group E(Q. Let φ : E E be the dual of φ so that φ φ = [] and φφ = []. We have the following exact sequences: (3.1 0 E (Q[ φ] φ(e(q[] E (Q φ(e(q ˆφ E(Q E(Q E(Q φ(e (Q 0, (3. 0 E (Q φ(e(q S(φ f (E X(E[φ] 0, (3.3 0 E(Q φ(e (Q S(ˆφ (E ˆf X(E [ φ] 0, (3.4 0 X(E[φ] X(E[] φ X(E [ φ] 0, where S (φ (E is the φ-selmer group of E/Q which is a finite subgroup of Q /Q and can be calculated in the following way. Let S = { } {prime factors of b(a 4b}. Let M be the subgroup of Q /Q generated by 1 and all prime factors of b(a 4b. For each d M, we have the curves (homogeneous spaces of E/Q and E /Q c d : dw = d t 4 adt z + (a 4bz 4, c d : dw = d t 4 + adt z + bz 4. Then we have the following isomorphisms of groups S (φ (E = {d M : c d (Q v for each v S}, S (ˆφ (E = {d M : c d (Q v for each v S}, where c d (Q v means that the curve c d has a non-trivial solution (w,t,y (0,0,0 in Q v. With these isomorphisms, the kernels of f and f in the

8 78 K. Feng exact sequences (3. and (3.3 are (3.5 ker f = {d M : c d (Q }, ker f = {d M : c d (Q }. and For our case, the elliptic curve E n and its dual E n have the equations E n : y = x 3 n x, E n : Y = X 3 + 4n X S = { } {prime factors of n}. Moreover, M is the subgroup of Q /Q generated by 1 and the prime factors of n. For each d M, the homogeneous spaces are c d : dw = d t 4 + 4n z 4, c d : dw = d t 4 n z 4. From (3.5 we know that (3.6 1 ker f, ±1, ±n ker f. Since E n (Q[] = {0,(y,x = (0,0,(0, ±n} and E n(q[ φ] = φ(e n (Q[] = {0,(0,0}, the exact sequences (3.1 (3.3 imply that (3.7 + ranke n (Q = dim F ker f + dim F ker f = dim F S (φ (E n dim F X(E n [φ] which together with (3.6 implies that + dim F S (ˆφ (E n dim F X(E n[ φ], ranke n (Q = 0 ker f = {1} and ker f = {±1, ±n}. In particular, if S (φ (E n = {1} and S (ˆφ (E n = {±1, ±n}, then we have ranke n (Q = 0 and X(E n [φ] = X(E n [ φ] = {1}. Then (3.4 implies X(E n [] = {1}, which means that the order of the group X(E n is odd. Theorem 3.1. We have S (φ (E n = {1} and S (ˆφ (E n = {±1, ±n} in the following two cases (p 1,...,p t are distinct odd prime numbers. (I n = p 1 p...p t (t 1, p 1 3 (mod 8, p i 1 (mod 8 for i, and G(n is an odd graph. (II n = p 1 p...p t (t 1, p 1 5 (mod 8, p i 1 (mod 8 for i, and G(n/ is an odd graph. Therefore ranke n (Q = 0 so that n is a non-congruent number, and the order of the Tate Shafarevich group X(E n is odd. P r o o f. (I Note that the graph G(n is non-directed by the quadratic reciprocity law, and G(n odd implies that G(n is odd. Moreover, M = 1,,p 1,...,p t Q /Q, S = {,,p 1,...,p t }, c d : dw = d t 4 + 4n z 4, c d : dw = d t 4 n z 4.

9 We need to show that: Non-congruent numbers 79 (Ia For each d M, d 1, there exists v S such that c d (Q v =. (Ib For each d M, d ±1, ±n, there exists v S such that c d (Q v =. To prove (Ia, let V = {,p 1,...,p t }. It is easy to see that c d (Q = for d<0. So we just need to consider the cases d= p V 1 p for each V 1 V, V 1. Suppose that V 1 V. Then V 1 and V = V V 1 is a non-trivial partition of V. Since G(n is an odd graph, we know that either there exists q V 1 such that ( n/d q = 1, or there exists p V such that ( d p = 1. Now we prove c d (Q p = c d (Q q =. Suppose that (w,t,z (0,0,0 is a non-trivial solution of the curve c d in Q p. Let w = dw. Then c d has the form dw = t 4 + (n/d z 4. For each l Z, (w p l,tp l,zp l is also a solution of the curve c d in Q p. So we can assume w,t,z Z p and v p (w = v p (t = 0, where v p is the exponential valuation of Q p normalized by v p (p = 1. Since p d and p n d, we know that dw t 4 (mod p, which contradicts ( d p = 1. Therefore cd (Q p =. On the other hand, q since we assume ( m = 1 for each odd m. If q = p 1 3 (mod 4, the equation of c d implies that t 4 (n/d z 4 (mod q since q d and q n d. This contradicts ( 1 q = 1. If q = pi 1 (mod 8 (i, then ( ( ( 1 n/d (n/d = 1 and = = 1, p 4 q q 4 which contradicts the assumption ( n/d q = 1. Therefore cd (Q q =. Next we consider the case V 1 = V so that d = n. The curve c n is nw = t 4 + z 4. By reduction mod p 1 we know that c n (Q p1 =. This completes the proof of (Ia so we have S (φ (E n = {1}. To prove (Ib let V = {p 1,...,p t }. Since ±1, ±n S (ˆφ (E n, S (ˆφ (E n is a group, and c (Q =, we need to show that d S (ˆφ (E n for each 1 < d < n, d n. We have d = p V 1 p where V 1 is a subset of V such that 1 V 1 < t. Since G(n is an odd graph, we know that either there exists q d such that ( n/d ( q = 1 or there exists p n/d such that d p = 1. Since G(n is a non-directed odd graph, there exist at least prime factors of n having the above properties of p or q. Therefore we can assume p p 1. Suppose that c d : dw = d t 4 n z 4 has a solution (w,t,z (0,0,0 in Q p ; we can assume that min{v p (w,v p (t,v p (z} = 0. If v p (w 1, then v p (t 1. Let w = n d w and t = n d t. Then c d has the equation dw = n t 4 d z 4. Therefore ( ( ±d p = 1, which contradicts d p = 1, so we have c d (Q p =. In the same way we can show that c d (Q q =. This completes the proof of S (ˆφ (E n = {±1, ±n}.

10 80 K. Feng (II It is easy to see that c d (Q = c d (Q = for each d n, d,d > 0. Therefore d S (φ (E n and d S (ˆφ (E n for such d. Since G(n/ is a nondirected odd graph, we can show d S (φ (E n for each d n/, 1 < d n/, and d S (ˆφ (E n for each d n/, 1 < d < n/, by the same argument as in the proof of (I. Therefore S (φ (E n = {1} and S (ˆφ (E n = {±1, ±n}. This completes the proof of Theorem 3.1. R e m a r k 3.. From the quadratic reciprocity law and Dirichlet s theorem on prime numbers in arithmetic progressions it is easy to show that for each directed graph G there exist inifinitely many D such that G(D = G. Therefore Theorem.4 yields many non-congruent numbers with any given number of prime factors. For the case of t 3, Theorem 3.1 was proved by Genocchi and Lagrange (see the list in Section BSD conjecture on E n : y = x 3 n x. For natural numbers a 1, a,...,a n, we denote by N(n;a 1,a,...,a n the number of integral solutions of the equation n = a 1 x 1 + a x a nx n. Theorem 4.1. (1 Suppose that n satisfies the condition (I of Theorem 3.1. Then the conjectures (BSD1 and (BSD are true for E n iff N(n;1,64, 0 (mod t+1. ( Suppose that n satisfies the condition (II of Theorem 3.1. Then (BSD1 and (BSD are true for E n iff N(n/;1,3,4 0 (mod t+1. P r o o f. (1 By Tunnell s result stated in Section 1, we know that L En (1/A = (a(n/ t, where a(n is given by (1.. Since n 3 (mod 8, it is easy to see that a(n = 1 ζ(x + 4y = 1 ( 1 ((x+4d 1/8 x +16y +z =n = 1 = 1 ( = x +16y +z =n x +16y +z =n y x +64y +z =n ( 1 (x 1/8+y ( 1 (x 1/8 ( 1 (x 1/8 1 x +16y +z =n x +16y +z =n y x +16y +z =n ( 1 (x 1/8 ( 1 (x 1/8. For n 3 (mod 16, we have 3 n x + z x + (mod 16. Therefore x 1 (mod 16, and (x 1/8 0 (mod. For n 11 (mod 16, we have 11 x + (mod 16. Therefore x 9 (mod 16 and

11 (x 1/8 1 (mod. Thus we know that Since n 3 (mod 8 we have Non-congruent numbers 81 a(n = ± ( N(n;1,64, 1 N(n;1,16,. N(n;1,16, = #{(x,y,z Z 3 : x + y + z = n, y} = h( n (see Section 1 t+1 (mod t+ (Theorem.4. Theorem 3.1 says that ranke n (Q = 0 and #(X(E n. Therefore (BSD1 and (BSD are true for E n a(n/ t 1 (mod ( In this case we have N(n;1,64, N(n;1,16, t+1 (mod t+ N(n;1,64, 0 (mod t+1. L En (1/A = (b(n// t, where b(n/ is given by (1.3. The congruence n/ 5 (mod 8 implies that b(n/ = 1 ζ (x + y = 1 x +8y +4z =n/ ( N(n/;1,3,4 But n/ 5 (mod 8 implies that Therefore x +8y +4z =n/ y = N(n/;1,3,4 1 N(n/;1,8,4. N(n/;1,8,4 = #{(x,y,z Z 3 : x + y + z = n, z} 1 = h( n t+1 (mod t+ (by Theorem.4. (BSD1 and (BSD are true for E n b(n// t 1 (mod N(n/;1,3,4 1 N(n/;1,8,4 t (mod t+1 N(n/;1,3,4 0 (mod t+1. This completes the proof of Theorem 4.1. R e m a r k 4.. If n satisfies the condition (I of Theorem 3.1, then N(n;1,64 = 0 since n has a prime factor p 1 3 (mod 8 and N(n;1, =

12 8 K. Feng t+1 by considering the decomposition of p 1,...,p t in Z[ ]. Therefore N(n;1,64, #{(x,y,z Z 3 : x +64y +z = n,xyz 0} (mod t+1. In particular, N(n;1,64, 0 (mod 8 and (BSD1 and (BSD are true for such E n provided t = 1 and. Similarly, if n satisfies the condition (II of Theorem 3.1, then N(n;1,3,4 #{(x,y,z Z 3 : x + 3y + 4z = n,xyz 0} (mod t+1 and (BSD1 and (BSD are true for such E n provided t = 1,. For t 3, we do not know in general how to prove the congruences N(n;1,64, 0 (mod t+1 for n satisfying the condition (I of Theorem 3.1, and N(n/;1,3,4 0 (mod t+1 for n satisfying the condition (II of Theorem 3.1. The following formula is found in [3]: N(n;1,1,16,3 = ( ( ( 1 d + 8 y x y d 1 d =n d 1 n=x +4y x,y 1 for n 5 (mod 8. For n satisfying the condition (II of Theorem 3.1 and t = 3, the above formula gives that N(n/; 1, 1, 16, 3 16 (mod 3. Therefore 0 #{(x,y,z,w Z 4 : x + y + 16z + 3w = n/} (mod 3 = N(n/;1,1,16,3 N(n/;1,1,16 N(n/;1,1,3 + N(n/;1, N(n/;1,1,1 N(n/;1,4, (mod 3 3 4h( n N(n/;1,4,3 (mod 3 N(n/;1,4,3 (mod 3. This shows that N(n/;1,4,3 0 (mod 4 and (BSD1 and (BSD are true for n satisfying the condition (II of Theorem 3.1 and t = 3. Acknowledgements. The author thanks the referee for several corrections and the reference [10] where P. Serf finds several classes of noncongruent numbers with 4 6 odd prime factors. References [1] R.Alter,T.B.CurtzandK.K.Kubota,Remarksandresultsoncongruent numbers, in: Proc. 3rd South Eastern Conf. Combin., Graph Theory and Comput., 197, Florida Atlantic Univ., Boca Raton, Fla., 197, [] J.E.CremonaandR.W.Odoni,SomedensityresultsfornegativePellequations; an application of graph theory, J. London Math. Soc. 39(1989, [3] G.P.Gogišvili,Thenumberofrepresentationsofnumbersbypositivequaternary diagonalquadraticforms,sakharthssrmecn.math.inst.šrom.40(1971, (MR 49#536(=E8-03(in Russian.

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