Solving a Diaphontine Equation

Question

Find all integral solutions to the equation $(x^2 + 1)(y^2 + 1) + 2(x − y)(1 − xy) = 4(1 + xy)$.

I've gotten this far: $x^2y^2+x^2+y^2+1+2x-2y-2x^2y-2xy^2=4+4xy$

$(x-y)^2+x^2y^2-2x^2y-2xy^2+2x-2y-2xy=3$

What do I do from here?

Answer 1

Let $A=(x^2+1)(y^2+1)$, $B=2(y-x)(1-xy)$, $C=4(1+xy)$. If $|x|,|y| \geq 4$, then we have $A > 4C$ and if $|x|,|y| \geq 6$, then we have $3A>4B$. Thus for $|x|,|y| \geq 6$, there's no solution. Now the remaining cases can be solved by calculation.

Now I prove just one of the claims above, i.e. $A>4C$ for $|x|,|y| \geq 4$: Let $a=|x|,b=|y|$, then $4C \leq 16|C| \leq 16(1+ab) < (a^2+1)(b^2+1)$ because $a^2b^2 \geq 16ab$ and $a^2 \geq 16$.