
3. Determine all pairs $m>n$ of positive integers such that

$$
1=\operatorname{gcd}(n+1, m+1)=\operatorname{gcd}(n+2, m+2)=\cdots=\operatorname{gcd}(m, 2 m-n)
$$


Solution: Observe that $1=\operatorname{gcd}(n+r, m+r)=\operatorname{gcd}(n+r, m-n)$. Thus each of the $m-n$ consecutive positive integers $n+1, n+2, \ldots, m$ is coprime to $m-n$. Since one of these is necessarily a multiple of $m-n$, this is possible only when $m-n=1$. Hence each pair is of the form $(n, n+1)$, where $n \in \mathbb{N}$.
