# Given that $\gcd(a,b) =1$, show that $\gcd(a+2b,b)=1$ without using prime factorization theorem

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.

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 $\gcd(a,b)=1$.
• @Nizar of course, $dm=b$ for some $m$ so $2b=2dm=d(2m)$. It's immediate from the definition of divides. – Adam Hughes Oct 22 '15 at 7:10