Prove that for all $a,b\in\mathbb Z$ of opposite parity there exists a number $c\in\mathbb Z$ such that $c+ab$, $c+a$ and $c+b$ are perfect squares.

Question

Prove that for all $a,b\in\mathbb Z$ of opposite parity there exists a number $c\in\mathbb Z$ such that $c+ab$, $c+a$ and $c+b$ are perfect squares.

So we could prove that $c+ab=k^2$, $c+a=l^2$ and $c+b=m^2$, where $k,l,m\in\mathbb Z$.

This seems like an interesting problem and I can't see how I could solve it. I can't think of an idea for the solution, so some help would be great. Thanks.

Answer 1

Hint: If $a,b$ have opposite parity, then $a-b=2k+1$ for some $k\in\mathbb Z$. Now choose $c=k^2-b$.