Prove that if $\gcd(a,b)=1$, then $\gcd(a\cdot b,c) = \gcd(a,c)\cdot \gcd(b,c)$. [duplicate]
Solution 1:
Let the highest power of prime $p$ in $a,b,c$ be $A,B,C$ respectively
Clearly, at least one of $A,B$ is zero,
Let $A=0,B\ge0$
The highest power of prime $p$ in $(ab,c)=$min$(A+B,C)=$min$(B,C)$
The highest power of prime $p$ in $(a,c)=$min$(A,C)=0$
The highest power of prime $p$ in $(b,c)=$min$(B,C)$
The highest power of prime $p$ in $(b,c)\cdot(a,c)=$min$(B,C)+0$
This holds true for all prime that divides $c$