Cesko-Slovacko-Poljsko 2013

Let us examine the diophantine equation $x^2+ax+b=y^2$ with unknown integers $x$ and $y$. It can be transformed into the form $(2x+2y+a)(2x-2y+a)=a^2-4b$. Since we assume $b$ is not a perfect square, $a^2-4b\ne 0$. There are only finitely many ways to write $a^2-4b$ as a product of two integers. Each such factorization gives two linear equations for $x,y$ which have at most one integer solution. Thus there are only finitely many such $x$.