<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] HINT:
$$x^2-2x(1+y)+4y^2-4y-8=0$$
$$x=\dfrac{2(1+y)\pm\sqrt{\{2(1+y)\}^2-4(4y^2-4y-8)}}2=1+y\pm\sqrt{9+5y-3y^2}$$
We need
$$0\le9+5y-3y^2\iff3y^2-5y-9\le0$$
Now roots of $3y^2-5y-9=0$ are $\dfrac{5\pm\sqrt{25+108}}2$
Now $(x-a)(x-b)\le0, a\le b\implies a\le x\le b$
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