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Does $a\mid(bc)$ imply that $a\mid b$ or $a\mid c$?

Can someone elaborate on this a bit?

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It follows if $a$ is prime. If it is not look at tetori's answer. If $a$ is prime, then at least one of $b,c$ has to be a multiple of $a$. –  Apostolos Jul 16 '13 at 1:21
The smallest counterexample is $a=4$, $b=c=2$. –  André Nicolas Jul 16 '13 at 1:25
How on Earth is this off-topic? –  Pedro Tamaroff Jul 16 '13 at 2:43

3 Answers 3

The statement

$$\tag 1 \forall a,b\in \Bbb Z\;\;,\;\;p\mid ab\implies p\mid a\text{ or }p\mid b$$

can be considered a fundamental property of prime numbers.

If $p$ is allowed to be any number, the result is false. Take $p=4,a=b=2$. Even if $(a,b)=1$, the statement needn't hold: let $p=15,a=3,b=5$.

There is a slight generalization of $(1)$ which is

$$\tag 2 a\mid bc \text{ and } (a,c)=1\implies a\mid b$$

Since when $p$ is prime $p\not\mid a\iff (p,a)=1$

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It is not true. Consider in the case of $a=bc$, $a,b,c>1$ (e.g. $a=15$, $b=5$, $c=3$).

And $a=15$, $b=6$, $c=10$ is another counterexample. Even if you assume $b$ and $c$ are relatively prime this statement falls. (consider $a=6$, $b=4$, $c=9$.)

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False. Consider

$$a = 2 \cdot 3, \ \ \ \ \ \ b = 2^2, \ \ \ \ \ \ c = 3^2$$

We have that $a|(bc)$ but it is not true that $a|b$ or that $a|c$. Therefore, $a|(bc) \kern.6em\not\kern -.6em \implies a|b \ \text{ or } \ a|c$.

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Even simpler: $a = 2\cdot 3$, $b = 2$, $c = 3$. –  Cameron Williams Jul 16 '13 at 2:43
@CameronWilliams Of course, but I wanted to avoid $a=bc$. –  Gamma Function Jul 16 '13 at 2:46

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