If $\gcd(a, m) = 1 = \gcd(a, n)$ then prove that $\gcd(a, mn) = 1$

Question

I tried using the definition of a GCD and got two equations;

$1 = ra + sm$

$1 = ta + zn$

I then combined the two equations(under multiplication) and simplified them to get: $1 = (ra + sm)(ta + zn)$

$1 = a(rta + tsm + zrn) + szmn$

[leading to] $1 = a(k) + pmn$

where $k = (rta + tsm + zrn), p = sz$ and both are integers. (I am unsure whether this last step is legitimate).

But I was unsure as to whether this was a strong enough proof, furthermore it seems more like a brute force method, and I was wondering if there is a better way?

Answer 1

Assume $\gcd(a,mn) = d$. If $d\neq 1$ there exists a prime such that $p|d$. Since $p|d$ and $d|mn$ it follows that $p|mn$. By $p$ being prime, we have that $p|m$ or $p|n$, thus $p$ is a common divisor of $a$ and $m$ or $p$ is a common divisor of $a$ and $n$. Since $1 = \gcd(a,m) = \gcd(a,n)$ it follows that $p|1$. Contradiction.

Answer 2

If prime factorizations of $a$ and $m$ have no prime in common, and also prime factorizations of $a$ and $n$ have no prime in common then clearly prime factorizations of $a$ and $mn$ have no prime in common.