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Let $a,b,c \in\mathbb Z$ where $a \neq0$ or $b \neq 0$. Suppose that $c \neq 0$ and is a common divisor of $a$ and $b$.

Prove that:

${gcd(a,b)\over |c|}= gcd ({a \over c}, {b \over c})$

What I have so far is

Let $a,b \in \mathbb Z$ , both not zero. If $d = gcd(a,b)$, then $gcd({a \over d},{b \over d})=1$.

But I am not sure how I can use this in the equation we have to prove. Can anyone help please?

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  • $\begingroup$ It's also true if $a=0$ or $b=0$. See this question. $\endgroup$ – user236182 Feb 14 '16 at 11:36
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It is of course not restrictive to assume $a$, $b$ and $c$ non negative.

Suppose $d$ is a common divisor of $a/c$ and $b/c$; then $cd$ is a common divisor of $a$ and $b$; therefore $cd$ is a divisor of $\gcd(a,b)$, so $d$ is a divisor of $\gcd(a,b)/c$.

Suppose $d$ is a divisor of $\gcd(a,b)/c$. Then $cd$ is a divisor of $\gcd(a,b)$, hence a common divisor of $a$ and $b$. Therefore $d$ is a common divisor of $a/c$ and $b/c$.

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