Determine all prime numbers $p$ and all positive integers $x$ and $y$ satisfying $x^{3}+y^{3}=$ $p(x y+p)$.
. Up to a swap of the first two entries, the only solutions are $(x, y, p)=(1,8,19)$, $(x, y, p)=(2,7,13)$ and $(x, y, p)=(4,5,7)$. The verification is routine.

Set $s=x+y$. Rewrite the equation in the form $s\left(s^{2}-3 x y\right)=p(p+x y)$, and express $x y$ :

$$
x y=\frac{s^{3}-p^{2}}{3 s+p}
$$

In particular,

$$
s^{2} \geq 4 x y=\frac{4\left(s^{3}-p^{2}\right)}{3 s+p}
$$

or

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

so $s<2 p$.
If $p \mid s$, then $s=p$ and $x y=p(p-1) / 4$ which is impossible for $x+y=p$ (the equation $t^{2}-p t+p(p-1) / 4=0$ has no integer solutions).

If $p \nmid s$, rewrite $(*)$ in the form

$$
27 x y=\left(9 s^{2}-3 s p+p^{2}\right)-\frac{p^{2}(p+27)}{3 s+p}
$$

Since $p \nmid s$, this could be integer only if $3 s+p \mid$ $p+27$, and hence $3 s+p \mid 27-s$.

If $s \neq 9$, then $|3 s-27| \geq 3 s+p$, so $27-3 s \geq$ $3 s+p$, or $27-p \geq 6 s$, whence $s \leq 4$. These cases are ruled out by hand.

If $s=x+y=9$, then $(*)$ yields $x y=27-p$. Up to a swap of $x$ and $y$, all such triples $(x, y, p)$ are $(1,8,19),(2,7,13)$, and $(4,5,7)$.