Integer solutions to $x^3 + x^2y + xy^2 + y^3 = 8(x^2 + xy + y^2 + 1)$

Question

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?

Answer 1

Hint: If $f(x)\in\mathbb{Z}[x]$ and $n\in\mathbb{N}$, then $(n-k)\mid f(n)$ implies $(n-k)\mid f(k)$. This shows that $s-4\mid 72$. Since $s$ must be even, $s-4$ can be one of the $18$ possible values.

Furthermore, we need $s^2-4p=(x-y)^2\geq 0$. Therefore, $$0\leq s^2-4p=s^2-4\left(\frac{s^3-8s^2-8}{2(s-4)}\right)=-\left(\frac{s^3-12s^2-16}{s-4}\right)\,.$$ This shows that $4< s<13$. There are now only $4$ cases. All solutions are $(x,y)=(2,8)$ and $(x,y)=(8,2)$.

Answer 2

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 $18$