Integer solutions to $x^3 + x^2y + xy^2 + y^3 = 8(x^2 + xy + y^2 + 1)$ 
Find all pairs of integer solutions $(x,y)$:
$$x^3 + x^2y + xy^2 + y^3 = 8(x^2 + xy + y^2 + 1).$$

Let $s = x+y$ and $p = xy$. We then reduce the given equation to $2p(s-4) = s^3-8s^2-8$. Thus, $s-4 \mid s^3-4s^2-4(s^2+2) \implies s-4 \mid 4$ or $s-4 \mid s^2+2$.
What do I do from here?

HINT:
After your hard work, we notice that $s$ must be even $=2t$(say)
$$p=\dfrac{2t^3-8t^2-2}{t-2}=2t^2-4t-8-\dfrac{18}{t-2}$$ 
So, $t-2$ must divide $18$ 
