Prove or disprove: $a\mid(bc)$ if and only if $a\mid b$ and $a\mid c$ 
Prove $a\mid(bc)$ if and only if $a\mid b$ and $a\mid c$.

My attempt is proving the converse first so if $a|b$ and $a|c$ then $a|bc$
So since $a\mid b$ and $a\mid c$, 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. 

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 $a|bc$.
