Problem 984

If $a^2 - 4b = c^2 - 4d$, then $a^2 \equiv c^2 \pmod{4}$, so $a$ and $c$ have the same parity. So if we take any integer $x$ and then $y$ to be the integer $x - \frac{a-c}{2}$ we have $x - \frac{a}{2} = y - \frac{c}{2}$ and hence $(x - \frac{a}{2})^2 - \frac{a^2-4b}{4} = (y - \frac{c}{2})^2 - \frac{c^2-4d}{4} \iff x^2 + a x + b = y^2 + c y + d$. Thus the equation has infinitely many integer solutions.

Conversely, suppose $k = (a^2 - 4b) - (c^2 - 4d) \neq 0$, then we have $(2x - a)^2 - (2y - c)^2 = k$, so $(2x + 2y - a - c)(2x - 2y - a + c) = k$. But $k$ has only finitely many factorizations, so there are only finitely many possible values for the pair $(2x + 2y - a - c, 2x - 2y - a + c)$ and hence for the pair $(x, y)$.