\( x^2 + 5y^2 = p \) 有整数解

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令 $p$ 为一素数且 $p \nmid 20$。当且仅当 $p \equiv 1,9 \pmod{20}$ 时，方程 $x^2 + 5y^2 = p$ 有整数解。

青青子衿

Corollary 11. For an odd prime \( p \neq 5 \), the following are equivalent:  
(i) \(\exists x, y \in \mathbb{Z} \) with \( x^2 + 5y^2 = p \).  
(ii) \(\left(\frac{-5}{p}\right) = 1\) and \(\left(\frac{-1}{p}\right) = 1\).  
(iii) \(p \equiv 1, 9 \pmod{20}\).

citeseerx.ist.psu.edu/document?repid=rep1& … cf39230e6fb0ff80796c

hagedorn.pages.tcnj.edu/files/2022/08/Geometr … nvenient-Numbers.pdf