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?



2 Answers 2


There is a counter-example. Take $x=1,y=49$. Then, your numbers are $0^2,7^2,10^2$ are all squares. This is not the only example, as any root of the Pell-like equation $x^2-2y^2=-1$ gives a solution. I'd guess there is no solution for $x,y>2$, but, modular arithmetic or qudratic residue will not give a solution; as 0,1,7 is a solution in any modulo.


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 $x=y$.


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