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? 
 A: 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 $s^2+t^2+(s-t)^2=6$.
A: 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 $(-2,0)$.
A: 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 $(x-a)(x-b)\le0, a\le b\implies a\le x\le b$
A: As the equation given is that of a conic curve (an ellipse), one can enlarge the issue, and look for all points with rational (instead of integer) coordinates. It is maybe worth to recall a classical method for doing so.
(a) Choose a point with rational coordinates; I take here $P(0,2)$.
(b) Consider all lines through $P$, with general equation $y-2=t(x-0)$. The second point of intersection with the conic curve has coordinates:
$$x=\dfrac{6\,\left( 1 - 2\,t \right) }{1 - 2\,t + 4\,t^2} \ , y = \dfrac{2\,\left( 1 + t - 2\,t^2 \right) }{1 - 2\,t + 4\,t^2}$$
(c) Impose a rational slope $t=p/q$, $p$ and $q$ integers. 
One obtains in this way the general form of all rational points on the cubic:
$$x=\dfrac{6 q \,\left( q - 2\,p \right)}{4\,p^2 - 2\,p\,q + q^2} \ , \ y=\frac{2\,\left( -2\,p^2 + p\,q + q^2 \right) }{4\,p^2 - 2\,p\,q + q^2}$$
