三维本原勾股数的求解问题

着重探讨了三维本原勾股数的求解问题,证明了奇数5不能作为三维本原勾股数的弦数.以单质数表示为二数平方和的定理及行列式的运算形式,用实例演示了用奇数作为弦数,求解它所对应的三维本原勾股数的计算方法,由此提出了任何一个大于5的奇数作为弦数都可以求得它所对应的三维本原勾股数解的猜想.还证明了三维本原勾股数中,存在以下结论:当两个偶勾股数都是4的倍数时,一定存在模4余1的偶勾股数与弦数的关系;当两个偶勾股数是2的倍数而不是4的倍数时,一定存在模4余3的偶勾股数与弦数的关系.

Problem of Solving 3D Primitive Pythagorean Triple

This paper focus on how to solve 3D primitive Pythagorean triple. It proves that 5 can＇t be used as the string of Pythagorean triple. Using the theory of single prime number can presents as the sum of two squared numbers, the paper demonstrates the solution of solving the 3D primitive Pythagorean triple with examples of odd as string. Therefore, a conjecture is proposed that a 3D primitive Pythagorean triple with any odd（greater than 5） as string can be solved. It also draws conclusions as followings of 3D Pythagorean triple： when the two even Pythagorean numbers are multiples of 4, string number must be 1 modulo 4; when the two even are not the multiples of 4 but 2, the remainder must be 3 modulo 4.