2013/14 AHSMC Part 1 Problem 14

The equation can be written as $x^2 + (y-1)x + y^2 + y - 2 = 0$. By the Quadratic Formula, the solutions are $$x = \frac{-(y-1) \pm \sqrt{(y-1)^2 - 4(y^2 + y - 2)}}{2} = \frac{-(y-1) \pm \sqrt{-3(y-1)(y+3)}}{2}.$$ These are real if and only if $(y+3)(y-1) \le 0$. Since $y$ is an integer, it must be one of $-3$, $-2$, $-1$, $0$ and $1$. If $y = -3$, then $x = 2$. If $y = -2$, then $x = 0$ or $3$. If $y = -1$, $x$ is not an integer. If $y = 0$, then $x = -1$ or $2$. Finally, if $y = 1$, then $x = 0$. The answer is (D).