TITLE: Suppose $(a,b)=1$, then $(2a+b,a+2b)=1\text{ or }3$.
QUESTION [5 upvotes]: Suppose $(a,b)=1$.  Let $d=(2a+b,a+2b)$.  
Then $d=(2a+b)u+(a+2b)v=a(2u+v)+b(2v+u)$ where $u,v \in \mathbb{Z}$.  
Since $(a,b)=1$, then $a(2u+v)+b(2v+u)=1$.
I'm not sure if I'm going in the right direction to solve this problem.

REPLY [2 votes]: Direct manipulation of gcd
$\gcd(a+2b,2a+b) = \gcd(a+2b,(2a+b)-2(a+2b)) = \gcd(a+2b,-3b)$
$ | \gcd(a+2b,3) \gcd(a+2b,b) | 3 \gcd((a+2b)-2(b),b) = 3 \gcd(a,b) = 3$