elementary number theory - Let $a,m,n in mathbf{N}$. Show that if $gcd(m,n)=1$, then $gcd(a,mn)=gcd(a,m)cdotgcd(a,n)$.

Let $a,m,n\in\mathbf{N}$. Show that if $\gcd(m,n)=1$, then $\gcd(a,mn)=\gcd(a,m)\cdot\gcd(a,n)$.

Proof:

Let $u,v\in\mathbf{Z}$ such that $\gcd(m,n)=um+vn=1$. Let $b,c\in\mathbf{Z}$ such that $\gcd(m,a)=ab+cm$. Let $d,e\in\mathbf{Z}$ such that $\gcd(a,n)=ad+en$.

So $\gcd(a,m)\cdot\gcd(a,n)=a^2bd+cmen+aben+emad$.

Where do I go from here?

Answer

$\gcd(a,m)\cdot \gcd(a,n) = a(abd+ben+emd)+(mn)(ce) \ge \gcd(a,mn)$

Say $\gcd(\gcd(a,m),\gcd(a,n)) = p$ where $p>1$.
Then $p|gcd(a,m)$ and $p|gcd(a,n)$. Which means $p|m$ and $p|n$. So $p$ is a common divisor of $m$ and $n$. $\gcd(m,n) \ge p$. But this is impossible since $\gcd(m,n)=1$ and $p>1$. Thus, $p=1$

If $\gcd(a,m) = x$, this means $x|a$ and $x|m$. If $x|m$, then $x|mn$.
Thus $x$ is a common divisor of $a$ and $mn$. $x|\gcd(a,mn)$
If $\gcd(a,n) = y$, this means $y|a$ and $y|n$. If $y|n$, then $y|mn$.
Thus $y$ is a common divisor of $a$ and $mn$. $y|\gcd(a,mn)$
Because $\gcd(x,y) = 1, xy|\gcd(a,mn)$

So $\gcd(a,m) \cdot \gcd(a,n) \le \gcd(a,mn)$

Therefore, $$\gcd(a,m) \cdot \gcd(a,n) = \gcd(a,mn)$$