Question

How many integer pairs $(x,y)$ satisfy $x^2+4y^2-2xy-2x-4y-8=0$?

Answer 1

Rewrite the equation as

$\frac{1}{4}\left(\right(x+2y-4{)}^2+3(x-2y{)}^2-48)=0.$

Setting $a:=x+2y-4$ and $b:=x-2y$, we must have that $a$ and $b$ are integers such that

$a^2+3b^2=48$.....$\left(1\right)$

The only integer solutions to $\left(1\right)$ satisfy $(|a|,|b|)=(0,4)$ and $(|a|,|b|)=(6,2)$.

This gives six possibilities for $a$ and $b\to (a,b)=\left\{\right(0,-4),(0,4),(-6,-2),(-6,2),(6,-2),(6,2\left)\right\}$,

By solving for this values of $a$ and $b$ and putting in equation $a:=x+2y-4$ and $b:=x-2y$ will yield six solutions$:(x,y)=(4,0),(0,2),(6,2),(4,3),(0,-1),$ and $(-2,0).$

Hence $6$ integer pairs $(x,y)$ satisfy $x^2+4y^2-2xy-2x-4y-8=0$