Proof: divisibility

Question:
For all $a, b, c \in \mathbb{Z}$, if $a\mid bc$, then $a\mid b$ or $a\mid c$. Is this true?

My answer: True. (Proof by contrapositive) Proof that if $a \nmid b$ and $a \nmid c$, then $a \nmid bc$.

Suppose $b = ax+r$ and $c = ay+s$, where $x,y,r,s \in \mathbb{Z}$ and $x,y,r,s \neq 0$

Then multiply $b$ by $c$:

$bc = (ax+r)(ay+s)=a(ayz+sy+rz) + rs$

$bc = az + t$, where $z = ayz+sy+rz$ and $t = rs$

Because $a,r,s,y,z$ are all integers $\mathbb{Z}$that are non-zero, both $z$ and $t$ are integers that is not zero.

Therefore, $bc$ is not divisible by $a$ because of the reminder $t$. This completes the proof.

Actual answer: False (Disproof by example) Consider $a=6, b=3, c=4$, $a\mid bc$, but $a \nmid b$ or $a \nmid c$.

My question: What is wrong with my proof?