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If $\gcd(a,b) =1$, show $\gcd(a+2b,b)=1$. I need help figuring how to showing from just that $\gcd(a,b) =1$. Does it have to do with Euclidean formula and that $\gcd(a,b) = am + bn$ for some $m,n$? Thanks.

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If $d|a+2b$ and $d|b$ then $d|a+2b-2(b)=a$ so $\gcd(a+2b,b)|a$ and $\gcd(a+2b,b)|b$. So that gcd divides $\gcd(a,b)=1$.

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  • $\begingroup$ a-2b? do you mean a+2b $\endgroup$ – lodnots3 Oct 22 '15 at 7:02
  • $\begingroup$ Your welcome. What do you mean gcd divides gcd(a,b)? $\endgroup$ – lodnots3 Oct 22 '15 at 7:04
  • $\begingroup$ All divisors of a+2b and b divide both a and b, in particular so does the gcd of the former two. $\endgroup$ – Adam Hughes Oct 22 '15 at 7:05
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    $\begingroup$ @Nizar of course, $dm=b$ for some $m$ so $2b=2dm=d(2m)$. It's immediate from the definition of divides. $\endgroup$ – Adam Hughes Oct 22 '15 at 7:10
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    $\begingroup$ @Nizar no problem, glad to help clear up the confusion. :) $\endgroup$ – Adam Hughes Oct 22 '15 at 7:12

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