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

Solution 1:

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

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