If 2a+b is even, so is b.
If a+2b is even, so is a.
This contradicts gcd(a, b)=1.
ok - so after a long proof that I know is correct, I have come to conclude that:
given: gcd(2a+b, a+2b) = 1 or 2 or 3
I have to prove that this does not equal 2.
I also know (and have proven) that gcd(a,b)=1.
I know I need to show that for d to not equal 2, I would have to pretend and show that when d equals 2, d could not equal 1 - which is false since d has to equal 1.
So to start: Assume gcd(2a+b, a+2b) = 2, then it divides both a and b
He did. Reiterating what wisterville said, if the gcd of both numbers is 2, then that must mean they are both even. Since 2a + b is even, then so must b (even + even = even) and a + 2b being even also implies a is even.
So, 2 is a common factor of both a and b which contradicts that their gcd was 1 which you were given. So, our initial assumption that (2a+b, a+2b) = 2 was wrong.