Prove that for some $x, y \in \mathbb{Z}^+$, if $(x-1)(y-1), xy, (x+1)(y+1)$ are all squares then $x = y$.
I tried taking all possible combinations $\bmod 3$ and $\bmod 4$ and it has a solution only when $x \equiv y$ for both of them (all the above 3 are priper quadratic residues only then). Doesn't this imply $x \equiv y \pmod{12}$?
Now, I am stuck with this problem. How do I proceed?
Thanks.