Integer solutions of $x^2+5y^2=231^2$

Solution 1:

If $x^2+5y^2=z^2$ and $(x,y,z)=1$ then $x,y,z$ are pairwise coprime and $5y^2=(z+x)(z-x)$. Divide into four cases:

$z\equiv x \pmod 2$, $5|z+x$

$$\left(\frac y2\right)^2=\frac{z+x}{10}\frac{z-x}2$$

If the product of coprime squares is a square then both are squares. Let $z+x=10a^2$, $z-x=2b^2$. Then $z=5a^2+b^2$, $x=5a^2-b^2$ and $y=2ab$.

$z\equiv x \pmod 2$, $5|z-x$

This case is the same as the previous one.

$z\equiv x+1 \pmod 2$, $5|z+x$

$$y^2=\frac{z+x}5(z-x)$$

Applying the same substitution you find $z=\frac{a^2+5b^2}2$, $x=\frac{a^2-5b^2}2$ and $y=ab$.

$z\equiv x+1 \pmod 2$, $5|z-x$

This case is the same as the previous one.

Which means if $z^2$ is of the form $x^2+5y^2$ for coprime integers $x,y$, then $z$ or $2z$ are too.

Back to the problem, if $(x,y)=d$ and $x^2+5y^2=231^2=3^27^211^2$ then $\left(\frac xd\right)^2+5\left(\frac yd\right)^2=\left(\frac{231}d\right)^2$ then the solutions to $a^2+5b^2=\frac{231}d$ and $a^2+5b^2=\frac{462}d$ over all 8 possible values of $d$ will generate all possible pairs $x,y$.