Show: $\gcd(9k+4,2k+1)=1 ~~ \forall k\in \mathbb Z$
Indirect proof.
If $1\neq d=\gcd(9k+4,2k+1)~\exists k\in \mathbb Z$,
then $d$ has to be of the form $2m+1$ for an integer $m$.
That somehow throws me back at the beginning.
Answer
Please excuse me for this messed up question,
but I've come up with an answer as compensation:
Just apply the Euclidean algorithm, and in 3 lines you get 1 as GCD.
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