If $a \mid c$ and $b \mid c$, does $ab \mid c^2$ automatically hold?

Question

So here goes:

If $a \mid c$ and $b \mid c$, does $ab \mid c^2$ automatically hold?

MY ATTEMPT AT A PROOF

$a \mid c$ means that there exists an integer $k$ such that $c = ka$.

$b \mid c$ means that there exists an integer $l$ such that $c = lb$.

This means that there exists an integer $m = kl$ such that $$c^2 = c \times c = {ka} \times {lb} = \left(k \times l\right) \cdot {ab} = m\cdot{ab}.$$

It follows that $ab \mid c^2$.

INQUIRY

Is my proof correct? If not, how can it be mended so as to produce a valid argument?

Thank you very much for your time and attention, @Jean-ClaudeArbaut! (Yes, I am aware of that caveat.) Please write out an answer, to that effect, so that this question does not remain in the unanswered queue, and furthermore, so that I could upvote and then accept your answer. — Jose Arnaldo Bebita Dris, Dec 05 '22 at 08:18
For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be an open-ended proof checking machine. Please do not use it that way. — Bill Dubuque, Dec 05 '22 at 08:59
Please delete the question since it is a dupe of a FAQ. It adds nothing new to the site but clutters serach results. — Bill Dubuque, Dec 05 '22 at 09:04
@BillDubuque: "It adds nothing new to the site..." I beg to disagree. =) — Jose Arnaldo Bebita Dris, Dec 05 '22 at 13:05

score 0 · Accepted Answer · answered Dec 05 '22 at 08:20

0

Yes, your proof is correct. (it's not in your question, but note that in general, you can't conclude $ab\,|\,c$, unless $a$ and $b$ are coprime)

answered Dec 05 '22 at 08:20

1 Answers1