<s>[INST] 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.
 [/INST] Too long for a comment.
$$x^2 + 4y^2- 2xy -2x -4y -8 = (x-y)^2 + 3(y-1)^2 -2x + 2y - 11\\ = (x-y-1)^2 + 3(y-1)^2 - 12 = 0,$$
$$(x-y-1)^2+3(y-1)^2=12,$$
$$(x-y-1,y-1)\in\{(0,2), (0,-2), (3,1), (3,-1), (-3,1), (-3,-1)\},$$
$$(x,y)\in\{(4,3), (0,-1), (6,2), (4,0), (0,2), (-2,0)\}.$$
</s>