TITLE: Prove that for some $x, y \in \mathbb{Z}^+$, if $(x-1)(y-1), xy, (x+1)(y+1)$ are all squares then $x = y$.
QUESTION [8 upvotes]: 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.

REPLY [3 votes]: If $x \geq y > 1$ then one has $x=y$. To see this write
$y-1=D_1 A^2$, $x-1= D_1 B^2$, $y= D_2 C^2$, $x = D_2 D^2$, $y+1 = D_3 E^2$ and $x+1= D_3 F^2$. 
Then we have two solutions to the equation
$$
D_2^2 X^4 - D_1D_3 Y^2 =1,
$$
given by $(X,Y)=(D,BF)$ and $(C,AE)$. 
If $D_1D_3>1$ (which follows from $x>1$) there is at most one solution to this equation in positive integers by old work of Walsh et al and hence we have $C=D$ and  so $x=y$.