If $ \gcd(a,b) = 1$ prove that $ \gcd(2a+b, a+2b) = 1$ or $3$?

Question

I have seen this question, some other related questions and answers for solving this problem. However, I tried to solve it using a different approach.

Let, $ \gcd(2a+b, a+2b) = d$
Assume $2a+b = qd\tag{1}$
and so $b = qd - 2a$
If we replace b with this in $a+2b$, we get $a+2b = 2qd - 3a$

We know, $\gcd(a,b) = 1 $. Let, $\gcd(a,qd)=m$. So $a=mn$ and $qd=lm$. In equation (1), $2a+b=qd$ or $b=qd−2a=lm−2mn=m(l−2n)$. That means $m|b$ and so $\gcd(a,b)=m$ which is not true. So, $ \gcd(a, qd) = 1$.

Thus $\gcd(2a+b,a+2b)$ = $\gcd(qd,2qd-3a)$ = $\gcd(qd,2qd-3)$. Since $\gcd(2,3) = 1, \gcd(qd,3) = 1$ or $3$ will be the answer.
Is this correct?

Answer 1

The OP's analysis seems valid to me. However, I don't think it is the intended answer. If $d$ is a common factor of both $2a + b$ and $a + 2b$, then it is a factor of their sum, which equals $3a + 3b = 3(a+b).$

Thus any common factor of $2a + b$ and $a + 2b$ will either be a factor of $3$ or a factor of $(a+b).$

Answer 2

Choose integers $r,s$ such that $ar+bs=1$. Then

$$ (2a+b)(2r-s) + (a+2b)(2s-r) = 3(ar+bs) = 3. $$

So if $g=\gcd(2a+b,a+2b)$, then $g \mid 3$. Hence, $g=1$ or $3$.

Moreover, from $3 \mid \big((2a+b)+(a+2b)\big)$, we have $3 \mid (2a+b)$ if and only if $3 \mid (a+2b)$. Further, note that $3 \mid (2a+b)$ if and only if $3 \mid \big(3a-(2a+b)\big)=a-b$. Therefore,

$$ \gcd(2a+b,a+2b) = \begin{cases} 1 & \:\mbox{if}\: 3 \nmid (a-b); \\ 3 & \:\mbox{if}\: 3 \mid (a-b). \end{cases} $$