<s>[INST] 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? 
 [/INST] Rewrite the equation as
$$\frac14\left((x+2y-4)^2 + 3(x-2y)^2-48\right)=0.$$
Setting $a:=x+2y-4$ and $b:=x-2y$, we must have that $a$ and $b$ are integers such that
$$a^2+3b^2=48.\tag1$$
The only integer solutions to (1) satisfy $(|a|,|b|)=(0,4)$ and $(|a|,|b|)=(6,2)$. This gives six possibilities for $a$ and $b$, which yields six solutions: $(x,y)$ = $(4,0)$, $(0,2)$, $(6,2)$, $(4,3)$, $(0,-1)$, and $(-2,0)$.
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