Question

Find the general expressions for two positive integers which are such that if their product is taken from the sum of their squares the difference is a perfect square.

Answer 1

Denote the integers by $x$ and $y$; then
$x^2-xy+y^2=z^2$ suppose;
$\therefore x(x-y)=z^2-y^2$.
This equation is satisfied by the suppositions
$mx=n(z+y)$, $n(x-y)=m(z-y)$,
where $m$ and $n$ are positive integers.
Hence $mx-ny-nz=0$, $nx+(m-n)y-mz=0$.
From these equations we obtain by cross multiplication
$\frac{x}{2mn-n^2}=\frac{y}{m^2-n^2}=\frac{z}{m^2-mn+n^2}$;
and since the given equation is homogeneous we may take for the general solution
$x=2mn-n^2$, $y=m^2-n^2$, $z=m^2-mn+n^2$.
Here $m$ and $n$ are any two positive integers, $m$ being the greater; thus if $m=7$, $n=4$, we have
$x=40$, $y=33$, $z=37$.