Diophantine equation: $x^2 + 4y^2- 2xy -2x -4y -8 = 0$ How many integer pairs $ (x, y) $ satisfy $$x^2 + 4y^2- 2xy -2x -4y -8 = 0 ?$$
As it seems at first glance, and hence, my obvious attempt was to make perfect squares but the $2xy$ term is causing much problem.

As a comment has said, $$(x-y-1)^2 + 3(y-1)^2 = 12$$
We have $a^2\ge 0$ for all $a\in\mathbb R$, therefore $(x-y-1)^2\ge 0$ and $(y-1)^2\ge 0$.
Notice that if $(y-1)^2\ge 9$, then $$(x-y-1)^2 + 3(y-1)^2 \ge 0+3\cdot 9>12$$
You'll get $(y-1)^2\in\{0,1,4\}$, i.e. $y-1\in\{0,\pm 1,\pm 2\}$,
i.e. $y\in\{-1,0,1,2,3\}$. Checking these $5$ cases leaves $5$ quadratic equations.
