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hzsh2009

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[求助] X^2+Y^2+Z^2=7W^2

如题,证明此等式没有整数解。       (提示,用模8来做)
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hzsh2009

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引用回帖:
2楼: Originally posted by acmuser at 2012-05-06 04:22:19:
Without loss of generality, suppose the common divisor of X, Y, Z, W is 1, then some of them have to be odd numbers. Notice that mod((2k+1)^2,8)=1, if W is odd, then mod(RHS, 8)=7, no such X,Y,Z th ...

就是不可能X, Y, Z, W都是even的吗?
4楼2012-05-06 04:51:15
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acmuser

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【答案】应助回帖

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感谢参与,应助指数 +1
hzsh2009: 金币+5, ★★★★★最佳答案 2012-05-06 06:32:19
Without loss of generality, suppose the common divisor of X, Y, Z, W is 1, then some of them have to be odd numbers. Notice that mod((2k+1)^2,8)=1, if W is odd, then mod(RHS, 8)=7, no such X,Y,Z that mod(X^2+Y^2+Z^2,8)=7, so W has to be even. Since the common divisor of X, Y, Z, W is 1, two of X, Y, Z have to be odd, therefore, mod(LHS, 8)=2 or 6, which is also impossible.

QED.
2楼2012-05-06 04:22:19
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hzsh2009

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引用回帖:
2楼: Originally posted by acmuser at 2012-05-06 04:22:19:
Without loss of generality, suppose the common divisor of X, Y, Z, W is 1, then some of them have to be odd numbers. Notice that mod((2k+1)^2,8)=1, if W is odd, then mod(RHS, 8)=7, no such X,Y,Z th ...

为什么能假设common divisor of X, Y, Z, W is 1呢?
3楼2012-05-06 04:44:47
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hzsh2009

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引用回帖:
2楼: Originally posted by acmuser at 2012-05-06 04:22:19:
Without loss of generality, suppose the common divisor of X, Y, Z, W is 1, then some of them have to be odd numbers. Notice that mod((2k+1)^2,8)=1, if W is odd, then mod(RHS, 8)=7, no such X,Y,Z th ...

actually, never mind, I see what you saying there. THX
5楼2012-05-06 06:32:10
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