一道四元二次丢番图方程

青青子衿

$\color{black}{a^2+b^2=2(c^2+d^2)}$
math.stackexchange.com/questions/2847898/

\begin{align*}
&\color{black}{\left\{
\begin{split}
a&=2mp+2np+nq\\
b&=2mp+2mq+nq\\
c&=2mp+\phantom{1}mq+np\\
d&=\phantom{1}mq+\,\,\phantom{1}np+nq
\end{split}
\right.}\\
\\
&\color{black}{\left\{
\begin{split}
a&=mp+mq-np+nq\\
b&=mp-mq+np+nq\\
c&=mp-nq\\
d&=mq+np
\end{split}
\right.}
\end{align*}

...

1. a^2 + b^2 - 2 c^2 - 2 d^2 /. {a -> 2 m*p + 2 n*p + n*q,
2.    b -> 2 m*p + 2 m*q + n*q, c -> 2 m*p + m*q + n*p,
3.    d -> m*q + n*p + n*q} // FullSimplify
4. a^2 + b^2 - 2 c^2 - 2 d^2 /. {a -> m*p + m*q - n*p + n*q,
5.    b -> m*p - m*q + n*p + n*q, c -> m*p - n*q,
6.    d -> m*q + n*p} // FullSimplify

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青青子衿

\begin{align*}
&\color{black}{\left\{
\begin{split}
a&=m^2+2 m n-n^2+2 p^2-4 p q-2 q^2\\
b&=m^2-2 m n-n^2-2 p^2-4 p q+2 q^2\\
c&=2(m p-m q+n p+n q)\\
d&=m^2+n^2-2 p^2-2 q^2\\
\end{split}
\right.}
\end{align*}

1. a^2 + b^2 - 2 c^2 - 2 d^2 /. {
2.    a -> m^2 + 2 m n - n^2 + 2 p^2 - 4 p q - 2 q^2,
3.    b -> m^2 - 2 m n - n^2 - 2 p^2 - 4 p q + 2 q^2,
4.    c -> 2 (m p - m q + n p + n q),
5.    d -> m^2 + n^2 - 2 p^2 - 2 q^2} // FullSimplify

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青青子衿

\begin{align*}
&\qquad\qquad\qquad\,\color{black}{a X_1^2+b X_2^2+c X_3^2+d X_4^2}\\
\\
&\color{black}{=\left(a t_1^2+b t_2^2+c t_3^2+d t_4^2\right){}^2 \left(a x_1^2+b x_2^2+c x_3^2+d x_4^2\right)}\\
\\
&\qquad\qquad\color{black}{\left\{\begin{split}
X_1&=x_1 \left(b t_2^2+c t_3^2+d t_4^2-a t_1^2\right)\\
&\qquad-2 t_1 \left(b t_2 x_2+c t_3 x_3+d t_4 x_4\right)\\

X_2&=x_2 \left(a t_1^2+c t_3^2+d t_4^2-b t_2^2\right)\\
&\qquad-2 t_2 \left(a t_1 x_1+c t_3 x_3+d t_4 x_4\right)\\

X_3&=x_3 \left(a t_1^2+b t_2^2+d t_4^2-c t_3^2\right)\\
&\qquad-2 t_3 \left(a t_1 x_1+b t_2 x_2+d t_4 x_4\right)\\

X_4&=x_4 \left(a t_1^2+b t_2^2+c t_3^2-d t_4^2\right)\\
&\qquad-2 t_4 \left(a t_1 x_1+b t_2 x_2+c t_3 x_3\right)\\
\end{split}\right.}\\
\\
&\qquad\qquad\quad\color{black}{
\left\{\begin{split}
X_1&=u_1x_1-2t_1u_2\\
X_2&=u_1x_2-2t_2u_2\\
X_3&=u_1x_3-2t_3u_2\\
X_4&=u_1x_4-2t_4u_2\\
u_1&=at_1^2+bt_2^2+ct_3^2+dt_4^2\\
u_2&=at_1x_1+bt_2x_2+ct_3x_3+dt_4x_4
\end{split}\right.}
\end{align*}