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 $(x,y)=(1,4),(7,4),(-1,-4),(-7,-4)$.