Find all prime numbers $p$ such that there exist positive integers $x, y$ that satisfy the relation $x\left(y^{2}-p\right)+y\left(x^{2}-p\right)=5 p$.
The given equation is equivalent to $(x+y)(x y-p)=5 p$. Obviously $x+y \geq 2$.

We will consider the following three cases:

Case 1: $x+y=5$ and $x y=2 p$. The equation $x^{2}-5 x+2 p=0$ has at least a solution, so we must have $0 \leq \Delta=25-8 p$ which implies $p \in\{2,3\}$. For $p=2$ we obtain the solutions $(1,4)$ and $(4,1)$ and for $p=3$ we obtain the solutions $(2,3)$ and $(3,2)$.

Case 2: $x+y=p$ and $x y=p+5$. We have $x y-x-y=5$ or $(x-1)(y-1)=6$ which implies $(x, y) \in\{(2,7) ;(3,4) ;(4,3) ;(7,2)\}$. Since $p$ is prime, we get $p=7$.

Case 3: $x+y=5 p$ and $x y=p+1$. We have $(x-1)(y-1)=x y-x-y+1=2-4 p<0$. But this is impossible since $x, y \geq 1$.

Finally, the equation has solutions in positive integers only for $p \in\{2,3,7\}$.