Task 11.2. Find all pairs of natural numbers $m$ and $n$ such that $m^{2019} + n$ is divisible by $m n$.
Answer: $m=1, n=1$ and $m=2, n=2^{2019}$.

Solution. From the condition, it follows that $m^{2019}+n$ is divisible by $m$, hence $n$ is divisible by $m$. Denoting $n=m n_{1}$, we conclude that $m^{2018}+n_{1}$ is divisible by $m n_{1}$. Proceeding similarly: we deduce that $n_{1}$ is a multiple of $m$, introduce the notation $m n_{2}=n_{1}$, and arrive at the conclusion that $m^{2017}+n_{2}$ is a multiple of $m n_{2}$. Continuing in this manner, we arrive at the conclusion that $n$ can be represented as $k m^{2019}$ for some natural number $k$. In this case, the condition of the problem can be rewritten as: $(k+1) m^{2019}$ is divisible by $k m^{2020}$, which is only possible when $k=1$ and $m=2$.

## Criteria

76 . Any correct solution.

6 6. A generally correct solution is provided, but only one pair of numbers is indicated.

2 6. Both answers are given without a solution, or the solution is incorrect.

1 6. Only one answer is given without a solution, or the solution is incorrect.