If $d=(m, n)$, with $m, n>0$, do $a, b$ integers exist such that $1=(\frac{m}{a}, \frac{n}{b})$ and $d=ab$?
If $n,m \in \mathbb{N}$ then there are $c,d$ such that $cd = (m,n)$, $(c,d) = 1$ and $(m/c,n/d) = 1$.