lssl wrote: Find all ($x,y$) ($x,y \in \mathbb{Z}$) satisfies : $x^2-2xy=126y^2=2009$. There is no $y\in\mathbb Z$ such that $126y^2=2009$ since LHS is even while RHS is odd.

Ver sorry it should be $ x^{2}-2xy+126y^{2}=2009 $ , I have edited .

lssl wrote: Ver sorry it should be $ x^{2}-2xy+126y^{2}=2009 $ , I have edited . So $(x-y)^2+125y^2=2009$ and $y^2\le \frac {2009}{125}\sim 16.07$ and so $y^2\in\{0,1,4,9,16\}$ $y^2=0$ $\implies$ $(x-y)^2=2009$, impossible since $2009$ is not a perfect square $y^2=1$ $\implies$ $(x-y)^2=1884$, impossible since $1884$ is not a perfect square $y^2=4$ $\implies$ $(x-y)^2=1509$, impossible since $1509$ is not a perfect square $y^2=9$ $\implies$ $(x-y)^2=884$, impossible since $884$ is not a perfect square $y^2=16$ $\implies$ $(x-y)^2=9$, and so the solutions $\boxed{(x,y)\in\{(-7,-4),(-1,-4),(1,4),(7,4)\}}$