Find the primes $p$ such that there exist $x, y \in \mathbb{N}^{*}$ such that

$$
x\left(y^{2}-p\right)+y\left(x^{2}-p\right)=5 p
$$
Let's rewrite the equation $(x+y)(xy-p)=5p$. $x+y$ divides $5p$ and thus equals $1, 5, p,$ or $5p$.

- If $x+y=1$: impossible because $x, y>0$
- If $x+y=5$: We have $xy-p=p$, so $xy=2p$, and thus $x$ equals $1, 2, p,$ or $2p$. If $x=1$, then $y=2p=5-x=4$ so $p=2$. Conversely, $(x, y, p)=(1,4,2)$ is a solution. If $x=2$, then $y=p=3$, and conversely $(x, y, p)=(2,3,3)$ is a solution. Finally, by symmetry, the only solutions with $x=p, x=2p$ are respectively $(x, y, p)=(4,1,2)$ and $(x, y, p)=(3,2,3)$.

- If $x+y=p$: We have $xy-p=5$, so $xy=p+5$. And $(x-1)(y-1)=xy-x-y+1=p+5-p+1=6$. Thus $(x, y)=(2,7),(3,4),(4,3)$ or $(7,2)$. If it's $(2,7)$ or $(7,2)$, then $p=x+y=9$ is not prime, which is absurd. If it's $(3,4)$ or $(4,3)$, then $p=x+y=7$ is prime, and we have $(x, y, p)=(3,4,7),(4,3,7)$ as solutions.

- If $x+y=5p$: We have $xy-p=1$, so $xy=p+1$. And $(x-1)(y-1)=xy-x-y+1=p+1-5p+1=2-4p<0$, which is absurd.