Does it follow from $\gcd(a,c) \mid \gcd(b,c)$ that $a \mid b$? If not, under what conditions on $a,b,c$ does this implication hold? 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
$$\gcd(a,c) \mid \gcd(b,c),\qquad a \not\mid b\qquad\text{and}\qquad \gcd(a,b)>1$$
