Determine all integer solutions $(x, y)$ of the Diophantine equation

$$
x^{3}-y^{3}=2 x y+8
$$
Let $(x, y)$ be a solution.

If $x=0$ then $-y^{3}=8$ and thus $y=-2$.

If $y=0$ then $x^{3}=8$ and thus $x=2$.

Let's distinguish 5 cases:

- If $x=0$ then $-y^{3}=8$ and thus $y=-2$.
- If $y=0$ then $x^{3}=8$ and thus $x=2$.
- If $x>0$ and $y<0$ then we have $y^{3}-x^{3}=-2 x y-8$ $y^{2}+(-x)^{2} \geq-2 x y$ which is impossible.
- If $x y>0$, then $x^{3}-y^{3}=2 x y+8>0$, so $x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)>0$ and this implies that $x-y>0$. We then distinguish two more cases:
- If $x-y=1$ then

$$
2 x y+8=(x-y)\left(x^{2}+x y+y^{2}\right)=x^{2}+x y+y^{2}=(x-y)^{2}+3 x y=1+3 x y
$$

and since $x=y+1$, then:

$$
2(y+1) y+8=1+3(y+1) y
$$

and thus $y^{2}+y-7=0$ but this equation has no integer solution.

- If $x-y \geq 2$ then

$2 x y+8=x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)=(x-y)\left((x-y)^{2}+3 x y\right) \geq 2(4+3 x y)=8+6 x y$

We then have that $2 x y+8 \geq 8+6 x y$ and thus $x y \leq 0$ which contradicts the hypothesis.

The only solutions are therefore $(0,-2)$ and $(2,0)$.