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[数论] 二次不定方程

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力工 Posted 2024-8-16 16:02 |Read mode
求$x^2-4x=3y^2+1$的正整数解。

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abababa Posted 2024-8-16 17:43
这个无解吧。$y^2=\frac{1}{3}(x^2-4x-1)$,左边是整数,右边也要是整数,所以右边的分子只能是$3$的倍数。把$x$用模$3$分类,当$x=3k,3k+1,3k+2$时分子都不是$3$的倍数。

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