How many integer pairs (x, y) satisfy $x^2 + 4y^2 − 2xy − 2x − 4y − 8 = 0$? How many integer pairs (x, y) satisfy $x^2 + 4y^2 − 2xy − 2x − 4y − 8 = 0$?  
My Attempt  
Let $f(x,y)=x^2 + 4y^2 − 2xy − 2x − 4y − 8$ . So $f(x,0)=x^2 − 2x − 8$ . $f(x,0)$ has two roots $x=4 , -2$ . So (4,0),  (-2,0) are solution of the given equation. Same way solving for $f(0,y)=0$ we get $ y=2 , -1$ are roots and hence (0,2),  (0,-1) are solutions. I have tried to factorize $f(x,y)$ or writing it as sum of squares but could not succeed. Is there any other solutions? How do I find them? 

Any $x$ that satisfies the equation must be even. Let $x=2z$. Substituting and dividing by $2$ we get
$$2z^2+2y^2-2zy-2z-2y-4=0.$$
We can rewrite this as 
$$(z-1)^2+(y-1)^2+(z-y)^2=6.$$
The only way the sum of three squares is $6$ is if the squares are, in some order, $1$, $1$, and $2$. Now it is a matter of examining a small number of cases. For example $z=0$, $y=2$ gives a solution, as does $z=2$, $y=3$, as does $z=3$, $y=2$. 
We could cut down on the arithmetic by letting $s=z-1$ and $t=y-1$. Then we are looking at $s^2+t^2+(s-t)^2=6$.
