58th Ukrainian National Mathematical Olympiad

Rewrite first equation in the form $25x+5y=5p^k$ and subtract the second equation: $24x=p^k(5-p^n)$. Thus, it is obvious, that $5-p^n>0$, so there are only the following cases.

Case 1. $p=1$, $n\in \mathbb{N}$. It is obvious, that $5x+y\ge 6$, so there are no solutions.

Case 2. $p=2$, $n=1$. Then we have an equality $24x=2^k\cdot 3$ or $x=2^{k-3}$. Thus, $k\ge 3$. Then from the first equality $y=2^k-5\cdot 2^{k-3}=3\cdot 2^{k-3}$. It is not hard to see, that these values satisfy also the second equality. Thus, for any $k\ge 3$ the answer is a set of numbers: $x=2^{k-3}$, $y=3\cdot 2^{k-3}$, $p=2$ and $n=1$.

Case 3. $p=2$, $n=2$ or $p=4$, $n=1$. Then we have equality $24x=2^k$ and so there are no answers.

Case 4. $p=3$, $n=1$. Then we have an equality $24x=3^k\cdot 2$ and then there are no answers.