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[数论] 一道五元二次丢番图方程

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青青子衿 Posted at 2023-1-16 10:20:50 |Read mode
Last edited by 青青子衿 at 2023-1-16 10:26:00对于丢番图方程
$\color{black}{a^2+b^2+c^2+d^2=e^2}$
存在如下恒等式
\begin{align*}
\left(t_1^2+t_2^2+t_3^2+t_4^2\right)^2&=\left(t_1^2-t_2 t_3-t_2 t_4-t_3 t_4\right)^2
+\left(t_2^2-t_1 t_3-t_1 t_4-t_3 t_4\right)^2\\
&\qquad+\left(t_3^2-t_1 t_2-t_1 t_4-t_2 t_4\right)^2
+\left(t_4^2-t_1 t_2-t_1 t_3-t_2 t_3\right)^2\\
\end{align*}

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