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Re:本原勾股数的构造法，（a,b,c）令 a = st, b=(s*s-t*t)/2, c = (s*s+t*t)/2, 其中s > t >= 1, 是没有公因数的奇数In Reply To:本原勾股数的构造法，（a,b,c）令 a = st, b=(s*s-t*t)/2, c = (s*s+t*t)/2, 其中s > t >= 1, 是没有公因数的奇数 Posted by:DarkArmed at 2009-07-23 17:09:45 这方法怎么得不到881个。。汗 Followed by: Post your reply here: |
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