Prove that if $a \mid bc$ then $a \mid b$ or $a \mid c$ for $a, b, c$ positive integers where a is not zero

Question

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.

Answer 1

$4$ certainly divides $36$ as $4(9)=36$
However $4$ does not divide $6$.

Answer 2

False; note that:

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

but

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

Answer 3

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 $a=9$, $b=3$, $c=15$.

Answer 4

$8\mid(4\times6)$ 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.