China Southeastern Mathematical Olympiad

Suppose that the integers $x, y$ satisfy
$$x^2 - 2xy + 126y^2 - 2009 = 0.$$
Looking at this as a quadratic function of $x$,
$$\Delta = 4y^2 - 4 \times (126y^2 - 2009) = 500(4^2 - y^2) + 36$$
should be a square number.
If $y^2 > 4^2$, then $\Delta < 0$. So $y^2 < 4^2$, when $y^2 \in \{0, 1^2, 2^2, 3^2\}$, $\Delta \in \{8036, 7536, 6036, 3536\}$ is not a square number.
For $y^2 = 4^2$, $\Delta = 500(4^2 - y^2) + 36 = 6^2$. According to $y = 4$, one can get $x = 1$ or $7$; according to $y = -4$, one can get $x = -1$ or $-7$.
All the integer solutions are $(x, y) = (1, 4), (7, 4), (-1, -4), (-7, -4)$.