
3. Find all positive integers $m$, $n$, such that $n$ divides $2 m-1$ and $m$ divides $2 n-1$.


Solution. We seek the pairs of positive integers $m, n$, such that

$$
2 m-1=k n \quad \text { a } \quad 2 n-1=l m \text {. }
$$

for some positive integers $k, l$.

Considering $k, l$ as parameters, we can eliminate the variable $n$, or we can eliminate the variable $m$ to get:

$$
(4-k l) m=k+2, \quad \text { or } \quad(4-k l) n=l+2 \text {; }
$$

Since the RHS of both equation is positive, the LHS must be positive as well and we get $4-k l>0$, which is $k l<4$. We go through cases $k l=1, k l=2$ a $k l=3$ one by one.

If $k l=1$, then $k=l=1$ and the equations (2) read $3 m=3$ and $3 n=3$, which is $m=n=1$.

Since $m$ and $n$ have to be odd, it cannot be $k l=2$.

If $k l=3$ then $\{k, l\}=\{1,3\}$, and we get $m=5$ and $n=3$, or $m=3$ and $n=5$.

Conclusion. The solution are pairs $(1,1),(3,5)$, and $(5,3)$.
