Find all integers $x, y$ such that

$$
x^{3}(y+1)+y^{3}(x+1)=19
$$
Substituting $s=x+y$ and $p=xy$ we get

$$
2 p^{2}-\left(s^{2}-3 s\right) p+19-s^{3}=0
$$

This is a quadratic equation in $p$ with discriminant $D=s^{4}+2 s^{3}+9 s^{2}-152$.

For each $s$ we have $D \geq 0$.

For $s \geq 11$ and $s \leq -8$ we have $D > \left(s^{2}+s+3\right)^{2}$ as this is equivalent to $2 s^{2}-6 s-161 > 0$, and thus also to $2(s+8)(s-11) > -15$.

We have the following cases:

- If $s \geq 11$ or $s \leq -8$, then $D$ is a perfect square only when $D = \left(s^{2}+s+4\right)^{2}$, or equivalently, when $s = -21$. From (1) we get $p = 232$ (which yields no solution) or $p = 20$, giving the solutions $(-1, -20)$ and $(-20, -1)$.
- If $-7 \leq s \leq 10$, then $D$ is directly checked to be a perfect square only for $s = 3$. Then $p = \pm 2$ and only $p = 2$ gives solutions, namely $(2, 1)$ and $(1, 2)$.

Remark by PSC. In the second bullet point, one actually needs to check 18 possible values of $s$ which is actually quite time consuming. We did not see many possible shortcuts. For example, $D$ is always a perfect square modulo 2 and modulo 3, while modulo 5 we can only get rid of the four cases of the form $s \equiv 0 \bmod 5$.