Suppose we have $\gcd(x,z) = 1$, prove that $\gcd(xy, z) = \gcd(y,z)$. This appears very intuitive to me however is there a formal way of proving this?
I could write the obvious statement that since $z$ does not have a common divisor as $x$, then the only common divisor comes from $y$. So this means that $\gcd(xy, z) = \gcd(y,z)$ directly. However, this does not seem mathematical enough, is it?
On
Let $a,b\in \mathbb{N}_0$, $a=\prod_{i=1}^np_{i}^{t_i}, b = \prod_{i=1}^np_{i}^{s_i}$, where $p_i$ are primes, $t_i,s_i$ are non-negative integers. Then using the following fact $$ \operatorname{GCD}(a,b) = \prod_{i=1}^np_{i}^{\operatorname{min}\{t_i,s_i\}} $$ you can easily show your equality.
Hint. We have to show that $xy$, $z$ and $y$, $z$ have exactly the same common divisors. It is clear that if $d$ is a common divisor of $y$, $z$ then $d\mid y$ implies $d\mid xy$. It remains to show that if $d\mid xy$ and $d\mid z$ then $d\mid y$. Note that $\gcd(x,z)=1$ implies that there are $a,b\in\mathbb{Z}$ such that $ax+bz=1$, and therefore $axy+bzy=y$.