If $\gcd(a,b) = 1$, show that $\gcd(2a+b, a+2b)=1 \mbox{ or } 3$
Solution 1
Since $\gcd(a,b) = 1$, there exist integers $x,y$ such that $ax + by = 1.$
Let $d = \gcd(2a+b, a+2b).$
Then $d\,{\large{}}(2a+b)$ and $d\,{\large{}}(a + 2b)$, hence
$$d{\large{{\;\,}}}\bigl(2(2a+b)  (a + 2b)\bigr) \implies d{\,\,}3a$$
$$d{\large{{\;\,}}}\bigl(2(a+2b)  (2a + b)\bigr) \implies d{\,\,}3b$$
Then
$$d{\,\,}3a \implies d{\,\,}3ax$$ $$d{\,\,}3b \implies d{\,\,}3by$$
Hence
\begin{align*} &d\,{\large{}}(3ax + 3by)\\[4pt] \implies\; &d\,{\large{}}\bigl(3(ax + by)\bigr)\\[4pt] \implies\; &d{\,\,}3\;\;\;\;\text{[since$\;ax + by = 1$]}\\[4pt] \implies\; & d = 1\;\,\text{or}\;\,d = 3\\[4pt] \end{align*}
Solution 2
Hint: $$\require{cancel} \gcd(2a+b,a+2b) \mid \gcd\big(2 \cdot(2a+\bcancel{b})  (a+\bcancel{2b}), 2\cdot (\cancel{a}+2b)(\cancel{2a}+b)\big)=\gcd(3a, 3b)$$
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Guerlando OCs
Updated on January 22, 2021Comments

Guerlando OCs almost 3 years
If $\gcd(a,b) = 1$, show that $\gcd(2a+b, a+2b)=1 \mbox{ or } 3$.
What I tried:
Suppose:
$d \  \ 2a+b$ and $d \  \ a+2b$, then:
$$2a+b = d\cdot k_1$$ $a+2b = d\cdot k_2$$
therefore:
$$3(a+b) = d(k_1+k_2)$$
I also noted that if $\gcd(a,b)=1$ then $a+b$ has some interesting property for division but I cannot figure out exactly

Asim over 2 yearshelpful answer, thanks