30th Junior Turkish Mathematical Olympiad

Answer: $5$. The quadruple $(m,n,a,k)=(1,3,3,5)$ satisfies the equation. We will show the cases $k\le 4$ are not possible.

For $k=2$, we will look at the equation in modulo $7$. $5^m\equiv a^2(\mathrm{mod}7)$ hence $m$ should be even. Therefore $$5^m+63n+49\equiv 2(\mathrm{mod}3)$$ and LHS cannot be a perfect square. Similarly the case $k=4$ is also eliminated.

For $k=3$, we again consider modulo $7$ and $5^m\equiv a^3(\mathrm{mod}7)$ hence $m$ should be a multiple of $3$, say $m=3l$. Then $5^{3l}+63n+49\equiv 125^l+4\equiv 3,5(\mathrm{mod}9)$ however $a^3\equiv 0,1,8(\mathrm{mod}9)$ so there are no solutions in this case as well.

Since $5^m\in \{1,5,25,-1,-5,-25\}(\mathrm{mod}63)$ we get $$a^k\in \{11,24,44,48,50,54\}(\mathrm{mod}63)$$ Then $a^k\equiv 2(\mathrm{mod}3)$ or $a^k\in \{3,5,6\}(\mathrm{mod}7)$. Therefore, $a^k$ is not a perfect square: $k$ is odd. Moreover $a^k\in \{2,3,4,5\}(\mathrm{mod}7)$ or $a^k\in \{3,5\}(\mathrm{mod}9)$. Therefore, $a^k$ is not a perfect cube: $3\nmid k$. Thus, $k\ge 5$.