59th Ukrainian National Mathematical Olympiad

Answer: $(1,1)$, $(2,1)$, $(5,3)$.

By checking trivial cases $m=1$, $n=1$ and $n=2$, we find the first two of the solutions listed. From now on, we assume $m\ge 2$ and $n\ge 3$.

Step 1. We prove that $m>n$. Suppose $n\ge m$. Then, numbers $m!$, $n!$ and $m^n$ are divisible by $m$, hence, $m$ also divides $m!+n!-m^n=1$ – contradiction. In particular, it follows that $m\ge 4$.

Step 2. Now, we show that $n>\frac{1}{2}m$. From well-known inequality $m!>m^{\frac{1}{2}m}$, which follows from the Root-Mean Square-Arithmetic-Mean-Geometric-Mean-Harmonic-Mean Inequality, we obtain $m^n+1=m!+n!>m^{\frac{1}{2}n}+1$.

Step 3. We prove that $m$ is prime. Otherwise, $m$ has a prime factor $p$ not larger than $\frac{1}{2}m$ and $n$. In that case, $p$ is a factor of $m!+n!-m^n=1$ – contradiction.

Step 4. We prove that $m-1$ is either prime or a square of a prime. Suppose the contrary. Then, $m-1$ can be represented as a product of two factors $a\ne b$, both smaller than $\frac{1}{2}m$ and $n$. Therefore, $m!$ and $n!$ are divisible by $n-1=ab$, since $n!$ contains both these factors. $m^n+1=((m-1)+1{)}^n+1$ gives a remainder of 2 when divided by $m-1$. Hence, $m-1$ is a factor of 2, which is impossible, since $m\ge 4$.

From the results of Steps 3 and 4 and from parity, we get that the only possible case is $m=5$. Indeed, since $m\ge 4$ is prime, it is odd, hence, $m-1$ is even and can be a square of a prime, if that prime is 2. From the results of Steps 1 and 2, we get that possible solutions are only $(5,3)$ and $(5,4)$. By checking, we see that only the first pair satisfies the initial equation.