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If $a,b,c\in\mathbb{Z}$, $\gcd(a,b)=1$ and $c \mid (a+b)$ then prove $$\gcd(a,c)=\gcd(b,c)=1$$

I think this can be proven with linear combinations but I'm not sure how to go about starting the proof.

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    $\begingroup$ Are you allowed to use Bezout's identity? $\endgroup$ – user258700 Dec 15 '15 at 0:45
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The $gcd(c,a)$ will divide both $a$ and $c$. $gcd(a,c) | c$ and $c| a + b$ so $gcd(a,c)|a + b$. As $gcd(a,c)|a$, $gcd(a,c)|(a + b) - a = b$. So $gcd(a,c)|a$ and $gcd(a,c)|b$. So $gcd(a,)|g(a,b)=1$.

See where I'm going?

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If $g=\gcd(a,c)$ then $g\mid a$, also $g\mid c$ so $g\mid a+b$. Can you finish it from here?

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