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If $a$ and $b$ are relatively prime, prove that $a + 2b$ and $2a+b$ are also relatively prime or have a $\gcd$ of $3$.

I'm very new to number theory so please don't assume I am familiar with some of the terminology.

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Hint $\ $ By Cramer's rule (or elimination) we infer

$$ \begin{eqnarray} \ a\, +\, 2\ b &=& A\\ \\ 2\ a\, +\, \ b &\ =\ & B\end{eqnarray} \quad\Rightarrow\quad \begin{array}\ 3 \ a\ = \, -A\, +\, 2\ B \\\\ 3\ b\ =\ \ 2\ A\, -\, \ B \end{array} $$

Hence, by the RHS system: $\ n\mid A,B\,\Rightarrow\,n\mid 3a,3b\,\Rightarrow\,n\mid (3a,3b) = 3(a,b) = 3$

Remark $\ $ In the same way we can prove more generally

Theorem $\ $ If $\rm\,(x,y)\overset{A}\mapsto (X,Y)\,$ is linear then $\: \rm\gcd(x,y)\mid \gcd(X,Y)\mid \Delta \gcd(x,y),\ \ \ \Delta := {\rm det}\, A$

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Hint: $ \gcd(a + 2b, 2a + b) = \gcd(a + 2b, (2a + b) - 2(a + 2b)) = \gcd(a + 2b, -3b) $.

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