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[数论] $x^2+y^2\equiv 1\pmod p$的解的数量

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hbghlyj posted 2024-11-14 18:36 |Read mode
A Classical Introduction to Modern Number Theory by Ireland and Rosen
对任意奇素数 $p$,集合 $$A_p=\{(x,y):0\le x\le(p-1)/2,\, 0\le y\le(p-1)/2,\, x^2+y^2\equiv 1\pmod p\}$$ 中的元素数量为
$$\begin{cases}(p+3)/4&\text{ if } p\equiv 1\pmod 4\\(p+5)/4&\text{ if
}p\equiv 3\pmod 4\end{cases}$$

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original poster hbghlyj posted 2025-1-13 17:10
顶一下

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