Two quadratic functions with simillar coefficients and a square of their 'b' coefficient

Question

I have quite an interesting problem to share today. I have noticed, that when I pick two quadratic equations, say:
$x^2-mx-n=0$ and $x^2-mx+n=0$
Where $m, n$ are natural numbers, and I assume both of these equations have an integer root, most of the time $m^2$ will be a sum of two other squares.
The question is, how can I find parameters $m, n$ so that both functions have two integer roots? (Finding those equations is really hard by chance)
Is there any way to prove that if they do have the roots, $m^2$ is a sum of two squares?

Answer 1

Yes, there is.

The case that $n=0$ is trivial, so we asume that $n\neq 0$.

Indeed, we know that $m^2-4n$ and $m^2+4n$ (the discriminants of the equations) are perfect squares, so $2m^2=(m^2+4n)+(m^2-4n)$ is a sum of different squares, say $2m^2=a^2+b^2$ with $b<a$. But since $2m^2$ is even, this squares have the same parity. Now we have $$m^2=\left(\frac{a+b}2\right)^2+\left(\frac{a-b}2\right)^2$$