I was trying to prove that $n/(n,k)$ is relatively prime to $k/(n,k)$, if $n$ is a non-zero integer and $k$ is an integer. My approach was to set $(n,k) = p$, and let $(n/p, k/p) = m$, where $m$ is either $1$ or greater to $1$. If $m$ is $1$, the proof is done. Otherwise if $m$ is greater to $1$, I was trying to derive an contradiction, but I was stuck. Could someone help me out?
Here the notation $(n,k)$ refers to the $\gcd(n,k)$.
There are integers $u,v$ such that $un+vk=(n,k)$. Then $$ u\cdot\frac n{(n,k)}+v\cdot \frac k{(n,k)}=\frac{un+vk}{(n,k)}=1.$$
If $p = (n,k)$ and $m = (n/p, k/p)$, then $m$ divides $n/p$ and $k/p$, which implies $pm$ divides $n$ and $k$. Thus $pm$ divides $p$, which implies $m = 1$, as desired.
Let $\,g=(n,k).\ $ Then $\ d\,\mid n/g,\,k/g \iff dg\mid n,k \iff dg\mid (n,k)=g \iff d\mid 1$
Remark $\ $ This proof is more general than proofs uisng the Bezout identity or prime factors, since it is valid in any gcd domain.
Alternatively by the gcd distributive law $\ g(n/g,k/g) = (n,k)=g,\,$ so $\,(n/g,k/g)=1,\, $ which also is valid in any gcd domain.
A solution without using the representation of gcd:
Assume $n/(n,k)$ and $k/(n,k)$ are not coprime. It means that there exists a prime number $q$ such that $q$ divides both of them. Then obviously $q$ divides $n$ and $k$.
Let $q^a$ and $q^b$ be maximal powers of $q$ dividing $n$ and $k$ respectively; without loss of generality $a\ge b$. Then the maximal power of $q$ dividing $(n,k)$ is $\min(a,b)=b$, what means that $k/(n,k)$ is not divisible by $q$ - contradiction.
