Find all prime numbers $p$ such that there exist strictly positive integers $x, y$ such that

$$
x\left(y^{2}-p\right)+y\left(x^{2}-p\right)=5 p
$$
When we have prime numbers, we always look for divisibilities and thus start by factoring.

We have:

$$
(x+y)(x y-p)=5 p
$$

and $x+y>0$ so there are 4 cases:

Case $1: x+y=1$ and $x y-p=5 p$

$x, y \geqslant 1$ so their sum is at least 2: absurd.

Case $2: x+y=5$ and $x y-p=p$

$x y=2 p$ and $(x, y) \in\{(1,4) ;(2,3)\}$ so $(1,4,2) ;(4,1,2) ;(2,3,3),(3,2,3)$ are solutions.

Case $3: x+y=p$ and $x y-p=5$

We have $x y-x-y+1=6$ and thus $(x-1)(y-1)=6$ then $(x, y) \in\{(2,7) ;(3,4)\}$ so $(3,4,7) ;(4,3,7)$ are the only solutions.

$$
\text{Case } 4: x+y=5 p \text{ and } x y-p=1
$$
$$
x, y \leqslant x y=p+1 \text{ so } 5 p=x+y \leqslant 2 p+2
$$
which is absurd since $p \geqslant 2$.