Let $a, b,$ and $c$ be positive (odd) integers.
I know that the implication $$a \mid b \implies \gcd(a,c) \mid \gcd(b,c)$$ holds.
Here is my:
INITIAL QUESTION: Does the converse $$\gcd(a,c) \mid \gcd(b,c) \implies a \mid b$$ also hold? If not, under what conditions does the converse hold?
MY ATTEMPT
Consider $a = 1$. Then for any positive (odd) integer $b$, then $$1 = \gcd(a, c) \mid \gcd(b, c) \implies 1 = a \mid b,$$ whence the implication holds.
Consider $a = 3$, $b = 5$, and $c = 7$. (Note that $\gcd(a,b)=1$.) Then we have $$1 = \gcd(a, c) \mid \gcd(b, c) = 1 \implies 3 = a \nmid b = 5,$$ whence the implication does not hold.
Here is my:
FINAL QUESTION: As it is easy to cook counterexamples for the implication $$\gcd(a,c) \mid \gcd(b,c) \implies a \mid b$$ when $\gcd(a,b)=1$, can you think of a counterexample for which $\gcd(a,b)>1$?
There are infinitely many counterexamples with $\gcd(a,b)>1$.
For $(a,b,c)=(PQ,RQ,Q)$ where $P,Q,R$ are positive odd integers satisfying $Q\gt 1$ and $P\not\mid R$, we have $$\gcd(a,c) \mid \gcd(b,c),\qquad a \not\mid b\qquad\text{and}\qquad \gcd(a,b)>1$$