Prove that if $\gcd(a,b)=1$, then $\gcd(a\cdot b,c) = \gcd(a,c)\cdot \gcd(b,c)$.

Question

Let $a,b,c \in \mathbb{Z}$, prove that if $\gcd(a,b)=1$, then $\gcd(a\cdot b,c) = \gcd(a,c)\cdot \gcd(b,c)$.

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score 3 · Answer 1 · answered Oct 22 '13 at 18:24

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 $c$

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