## Task A-2.1.

Determine all pairs $(m, n)$ of natural numbers such that $n$ divides $2m-1$ and $m$ divides $2n-1$.
## Solution.

From the conditions of the problem, there exist natural numbers $k, l$ such that

$$
2 m-1=k n, \quad 2 n-1=l m
$$

We have $2 m=k n+1, 2 n=l m+1$, so it follows that

$$
4 n-2=2(2 n-1)=2 \cdot m l=2 m \cdot l=(n k+1) l
$$

From $4 n-2=(n k+1) l$ it follows that

$$
(4-k l) n=l+2
$$

Since the right-hand side is positive, it must be that $4-k l>0$, or $k l<4$. Since $k$ and $l$ are natural numbers, we have the following possibilities:

1. $k l=1$, i.e., $k=l=1$.

It must hold that $2 m-1=n, 2 n-1=m$, from which it follows that $m=n=1$.
2. $k l=2$

This is not possible, because from $2 m-1=k n, 2 n-1=l m$ it follows that $k$ and $l$ are odd numbers.
3. $k l=3$, i.e., $k=3, l=1$ or $k=1, l=3$.

In the first case, from $2 m-1=3 n, 2 n-1=m$ we get $m=5, n=3$.

Similarly, in the second case, $m=3, n=5$.

Therefore, the solutions are $(m, n) \in\{(1,1),(3,5),(5,3)\}$.