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Prove $a|(bc)$ if and only if $a|b$ and $a|c$

My attempt is proving the converse first so if $a|b$ and $a|c$ then $a|bc$

So since $a|b$ and $a|c$ then $b=ax$ and $c=ay$ for some integers $x$ and $y$.

So $bc=a(xy)$ therefore $a|bc$. Now the forward direction if $a|bc$ then $bc=az$ for some integer $z$. Letting $z=xy$ implies that $bc=(ax)(ay)$ so $b=ax$ and $c=ay$ thus $a|b$ and $a|c$. I'm not confident with the forward direction.

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    This is not true, $3 | 5 \cdot 3 $ but clearly $3 \not | 5$. – 2017-01-23
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    What about $6|2\times3$? – 2017-01-23
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    the theorem isn't true in general. for example, $6 | (3 \cdot 2)$ yet neither is it that $6 | 3$ nor $6 | 2$ – 2017-01-23
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    Oh wow I made a mistake reading the problem that I could disprove and didn't consider counter examnples – 2017-01-23
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    Hint: Consider **composite** $\, a = bc,\ \ b,c > 1.\ $ Then $\,a\mid bc\, $ but .... Or consider $\ b = a \,$ and $\,c=1.\ \ $ – 2017-01-23

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Counter-Example to prove the $\Rightarrow$ statement is not true for every value $a, b, c \in \mathbb Z$.

Put $a = 6,\; b=3,\; c = 4$

$a \mid (bc),\;\;$ but $\;a\not \mid b\;$ and $\;a \not \mid c$.

On the other hand, if $\;(a\mid b$ and $a\mid c),\rightarrow\,a\mid (bc)$.

Like you've shown: "So since $a|b$ and $a|c$ then $b=ax$ and $c=ay$ for some integers $x$ and $y$."

From there we have $bc= (ax)(ay).\,$ So $bc=a^2(xy)$ therefore $a|bc$.