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This question already has an answer here:

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?

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marked as duplicate by Bill Dubuque elementary-number-theory Apr 2 '17 at 15:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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

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

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    $\begingroup$ I understand how you conclude that $\gcd(a,m) \cdot \gcd(a,n) \leq \gcd(a,mn)$, but how do you jump from that to $\gcd(a,m) \cdot \gcd(a,n) = \gcd(a,mn)$? $\endgroup$ – SBS Oct 19 '16 at 13:40
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You can also try looking at gcd's in terms of the prime factorizations of the integers, from which the desired result follows readily.

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If you know some ring theory: The greatest common divisor of $r,s$ is the smallest positive integer in the ideal $(r,s) := \{ kr+ls \ | \ k,l\in \mathbb{Z} \}.$

We have $$(a,m)(a,n) = (a^2,am,an,mn) = (a,mn)$$

where the last equal followed from $(m,n)=1.$ Reading off the least positive integer from both sides gives the result.

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