XV-th Junior Balkan Mathematical Olympiad

Given equation is equivalent to
$$(x + y)(xy - p) = 5p.$$
We consider the following cases:
1. Let $x + y = 1$ and $xy = 6p$. For prime $p \geq 2$ the equation $x^2 - x + 6p = 0$ has no solutions.

2. Let $x + y = 5$ and $xy = 2p$. For prime $p \geq 2$ the equation $x - 5x + 2p = 0$ has the determinant $\Delta = p^2 - 4p - 20$. The inequality $25 - 8p \geq 0$ implies $p \in \{2, 3\}$. For $p = 2$ we obtain the solutions $(1, 4)$ and $(4, 1)$.
For $p = 3$ we obtain the solutions $(2, 3)$ and $(3, 2)$.

3. Let $x + y = p$ and $xy = p + 5$. For prime $p \geq 2$ the equation $x^2 - px + p + 5 = 0$ has the diskriminant $\Delta = p^2 - 4p - 20$. The inequality $p^2 - 4p - 20 \geq 0$ implies $p \geq 7$.
Let $p^2 - 4p - 20 = q^2$ with $1 \leq q \leq p$. We obtain the equation $(p-2)^2 - q^2 = 24$ which is equivalent to
$$(p + q - 2)(p - q - 2) = 24.$$
It follows that both numbers $p+q-2$ and $p-q-2$ shall be even. We have two subcases:
a) $p+q-2=12$ and $p-q-2=2$. We have $p=9$, which is no prime.
b) $p+q-2=6$ and $p-q-2=4$. We obtain $p=7$ and $q=1$. The equation has the solutions $(3, 4)$ and $(4, 3)$.

4. Let $x + y = 5p$ and $xy = p + 1$. It follows that $p \notin \mathbb{N}$.
So, the equation has natural solutions only for $p \in \{2, 3, 7\}$.