(Find all prime numbers $p$ such that there exist $x, y \in \mathbb{N}^{*}, x\left(y^{2}-p\right)+y\left(x^{2}-p\right)=5 p$.)
We write $(x+y+5) p=x y(x+y) . p \mid x y(x+y)$, so $p$ divides one of the factors. If, for example, $p \mid x$ and $p \neq x$, we have $x \geqslant 2 p$ and $x y(x+y) \geqslant 2 p(x+y)>p(x+y+5)$ unless $x+y=5$, which provides $p=2$ and $(x, y)=(4,1)$. If $x=p$, it follows that $p+y+5=y(p+y)$ and by the same argument, $p+y=5, y=2$ and $p=3$. Otherwise, $p \mid x+y=k p$. It follows that $k p+5=k x y$, so $5=k(x y-p)$. If $k=1, x y=p+5=x+y+5$, otherwise $x y=x+y+1$. In the first case, $(x-1)(y-1)=5$ so $x+y=6$ and $p=2$ or $p=3$, in the second, $(x-1)(y-1)=2$ so $x+y=3$ and $p=3$. Therefore, the solutions are 2 and 3.