62nd Ukrainian National Mathematical Olympiad

Let $2^n-13^m=k^3$. Then $2^n\equiv k^3(\mathrm{mod}13)$. If $n$ is not divisible by $3$, then $2^n$ can be written as $4l^3$ or $2l^3$. Then either $4$ or $2$ is a cube of some integer modulo $13$. But by simple checking, we can see that only remainders $0$, $\pm 1$, $\pm 5$ can be cubes of integers modulo $13$. Therefore, $n$ is divisible by $3$, so let $n=3t$. The condition can be rewritten as $$13^m=(2^t-k)(4^t+k\cdot 2^t+k^2).$$ Then both factors are powers of $13$, and the left one is smaller than the right one, so $2^t-k$ divides $4^t+k\cdot 2^t+k^2$. Since $2^t\equiv k(\mathrm{mod}(2^t-k))$, we have $4^t+k\cdot 2^t+k^2\equiv 3\cdot 4^t(\mathrm{mod}(2^t-k))$. Therefore, $3\cdot 4^t$ is divisible by $2^t-k$, which is a power of $13$, so $2^t-k=13^0=1$ and $13^m=3\cdot 4^t-3\cdot 2^t+1$. Suppose that $t\ge 4$. Then $13^m\equiv 1(\mathrm{mod}16)$, so $m$ is divisible by Therefore, $m=4s$ and $$(13^{2s}+1)(13^{2s}-1)=3\cdot 2^t\cdot (2^t-1).$$ But $13^{2s}+1$ is not divisible by $4$ and $3$, so $13^{2s}+1$ divides $2\cdot (2^t-1)$. If these numbers are not equal, then $$13^{2s}+1\le 2^t-1<\sqrt{3\cdot 2^t\cdot (2^t-1)}<13^{2s}.$$ Otherwise, $13^{2s}+1=2\cdot (2^t-1)$ and $13^{2s}-1=3\cdot 2^{t-1}$. We will subtract the second equation from the first equation and get $2^{t-1}-2=2$, so $t=3$, which contradicts the assumption that $t>3$. By checking for $t=1$, $t=2$, and $t=3$, we verify that only $t=3$ satisfies the condition. Therefore, $n=9$ and $m=2$ is the unique solution.