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I named $\gcd(a,m) = d$ and $\gcd(ab,m) = d' $
So I know that $d\mid a$, $d\mid m $ and $d'\mid ab $ , $d' \mid m$
But I can't use the transitive property of divisibility here.
How can I prove that $d \mid d'$?
For each $x\mid \gcd(a,m)$, $x\mid a,m$ so $x\mid ab,m$ and $x\mid\gcd(ab,m)$.
Required, but never shown