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 $p\ge 2$ the equation $x^2-x+6p=0$ has no solutions.

2. Let $x+y=5$ and $xy=2p$. For prime $p\ge 2$ the equation $x-5x+2p=0$ has the determinant $\Delta =p^2-4p-20$. The inequality $25-8p\ge 0$ 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 $p\ge 2$ the equation $x^2-px+p+5=0$ has the diskriminant $\Delta =p^2-4p-20$. The inequality $p^2-4p-20\ge 0$ implies $p\ge 7$. Let $p^2-4p-20=q^2$ with $1\le q\le p$. 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 $p\in \{2,3,7\}$.