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 01:21
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7The smallest counterexample is $a=4$, $b=c=2$. – André Nicolas Jul 16 '13 at 01:25
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3How on Earth is this off-topic? – Pedro Jul 16 '13 at 02:43
3 Answers
11
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 $p\not\mid a\iff (p,a)=1$
Pedro
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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 $a=6$, $b=4$, $c=9$.)
Hanul Jeon
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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, $a|(bc) \kern.6em\not\kern -.6em \implies a|b \ \text{ or } \ a|c$.
Gamma Function
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