Euler's concordant forms
Euler's concordant forms
复制标题
欧拉的协调形式
DOI：
10.4064/aa-78-2-101-123
复制
发表时间：
1996
期刊：
影响因子：
--
通讯作者：
K. Ono
中科院分区：
文献类型：
--
作者：
K. Ono
这就是所谓的欧拉协调形式问题，当 M = N 时，欧拉问题就是全等数问题。 Tunnell 使用椭圆曲线和模形式给出了同余数问题的条件解。利用这些想法，我们考虑欧拉问题，该问题简化为对 Q 上的椭圆曲线的研究：EQ(M,N) : y 2 = x 3 + (M +N)x 2 +MNx。如果 EQ(M,N) 具有正秩，则 (1) 有无穷多个本原整数解；但如果 EQ(M,N) 的秩为 0，则可能存在一个非平凡的解。当且仅当扭转群是 Z2◊Z8 或 Z2◊Z6 时，这样的解才存在。我们对所有这些情况进行分类，从而将欧拉问题简化为等级问题。在某些情况下，EQ(M,N)的二次扭曲的等级由三元二次形式的整数表示来描述。因此，我们获得了关于欧拉问题的结果，以及一对佩尔方程解的存在性。此外，我们使用缺位模形式理论给出了一种新的基本方法，该方法通过算术级数中的判别式确定了 EQ(M,N) 存在无限多个 0 阶二次扭曲。
This is known as Euler’s concordant forms problem, and when M = N Euler’s problem is the congruent number problem. Tunnell gave a conditional solution to the congruent number problem using elliptic curves and modular forms. Using these ideas, we consider Euler’s problem which reduces to a study of the elliptic curve over Q : EQ(M,N) : y 2 = x 3 + (M +N)x 2 +MNx. If EQ(M,N) has positive rank, then there are infinitely many primitive integer solutions to (1); but if EQ(M,N) has rank 0, then there may be a non-trivial solution. Such a solution exists if and only if the torsion group is Z2◊Z8 or Z2◊Z6. We classify all such cases, thereby reducing Euler’s problem to a question of ranks. In some cases, the ranks of quadratic twists of EQ(M,N) are described by the representations of integers by ternary quadratic forms. Consequently, we obtain results regarding Euler’s problem, and the existence of solutions to a pair of Pell’s equations. Moreover, we give a new and elementary method, using the theory of lacunary modular forms, which establishes that there are infinitely many rank 0 quadratic twists of EQ(M,N) by discriminants in arithmetic progressions.