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Prove or disprove (by providing a counter-example) that
if $a \mid bc$ then $a \mid b$ or $a \mid c$ for $a, b, c$ positive integers where $a$ is not zero.

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    $\begingroup$ It's always a good idea to put more than a bald problem statement into your Question's body text. What motivates your question? What have you been able to relate this property to? Such information helps Readers to gauge where you may be stuck in finding your own solution and thus supply a skillful note that lets a light bulb go on. $\endgroup$ – hardmath Mar 6 '16 at 1:15
  • $\begingroup$ Because of Euclid's lemma the property you wrote is often used as a definition of "prime". $\endgroup$ – user228113 Mar 6 '16 at 2:14
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$4$ certainly divides $36$ as $4(9)=36$
However $4$ does not divide $6$.

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your statement holds only in the case that $a$ is a prime. In this case, $a\mid bc$ implies that $a$ is "contained" multiplicatively in $bc$. But since $a$ is a prime thus contains no other factors than $1$ and $a$, then it must either be contained (the whole of $a$) in $b$ or in $c$. Consequently either $a\mid b$ or $a\mid c$.

However, it does not hold if $a$ is a composite. For example consider $a=9$, $b=3$, $c=15$.

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  • $\begingroup$ I like this answer because of the intuition it provides. $\endgroup$ – Fryie Mar 7 '16 at 18:36
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$8\mid(4\times6)$ but $8\nmid 4$ and $8\nmid 6$.

But if $a$ is prime, then it's true. That is called Euclid's lemma. Google that term.

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    $\begingroup$ $8$ certainly does not divide $18$. $\endgroup$ – KonKan Mar 6 '16 at 0:50
  • $\begingroup$ @KonstantinosKanak0glou : Correct. I intended $4$ rather than $3$. I've fixed it. $\endgroup$ – Michael Hardy Mar 6 '16 at 3:10
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False; note that:

$$6 \mid 12 \Longrightarrow 6 \mid (3 \cdot 4)$$

but

$6 \mid 3$ and $6 \mid 4$ are both false.

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