
N4. Find all integers $x, y$ such that

$$
x^{3}(y+1)+y^{3}(x+1)=19
$$


Solution. Substituting $s=x+y$ and $p=x y$ 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 $D0$.

For $s \geqslant 11$ and $s \leqslant-8$ 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 $s \geqslant 11$ or $s \leqslant-8$, 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 $-7 \leqslant s \leqslant 10$, then $D$ is directly checked to be 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 the four cases of the form $s \equiv 0 \bmod 5$.
