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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 $\gcd(a,b)>1$?

Bill Dubuque
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Jose Arnaldo Bebita Dris
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    Consider pairwise coprime numbers $a,b,c$, then $2a, 2b, 2c$ is a counterexample with gcd$(a,b)=2$. – Severin Schraven Dec 26 '22 at 06:11
  • Thank you for your time and attention, @SeverinSchraven! How is that a counterexample? Note that I am only considering odd integers $a, b, c$ such that $\gcd(a, c) \mid \gcd(b, c)$. – Jose Arnaldo Bebita Dris Dec 26 '22 at 06:20
  • If $a, b, c$ are pairwise coprime, then surely $\gcd(a,b)=1$, @SeverinSchraven? – Jose Arnaldo Bebita Dris Dec 26 '22 at 06:22
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    My point was not so much that you multiply by $2$, you can pick anything you like. Clearly you have for $a,b,c$ pairwise coprime and $0<n\in \mathbb{N}$ that gcd$(na,nc)=n =\text{gcd}(nb,nc)$. However, $na$ still does not divide $nb$. – Severin Schraven Dec 26 '22 at 20:04
  • By the GCD Universal Property in the linked dupe $(a,c)\mid (b,c)\iff (a,c)\mid b,c\iff (a,c)\mid b,,$ which does not imply $,a\mid b,,$ as very obvious counterexamples show. – Bill Dubuque Jan 10 '23 at 17:05

1 Answers1

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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 $$\gcd(a,c) \mid \gcd(b,c),\qquad a \not\mid b\qquad\text{and}\qquad \gcd(a,b)>1$$

mathlove
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