Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Can someone elaborate on this a bit?

share|improve this question
    
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
7  
The smallest counterexample is $a=4$, $b=c=2$. –  André Nicolas Jul 16 '13 at 1:25
3  
How on Earth is this off-topic? –  Pedro Tamaroff Jul 16 '13 at 2:43
add comment

3 Answers

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 $p\not\mid a\iff (p,a)=1$

share|improve this answer
add comment

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 $a=6$, $b=4$, $c=9$.)

share|improve this answer
add comment

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, $a|(bc) \kern.6em\not\kern -.6em \implies a|b \ \text{ or } \ a|c$.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.