Diophantine equation: $x^2 + 4y^2- 2xy -2x -4y -8 = 0$

Question

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.

Answer 1

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

Answer 2

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.