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?

It's possible to assume $a>0$.
Saying that $a\nmid b$ means $b=ax+r$, with $0<r<a$, with no condition on $x$. Similarly, you can certainly write $c=ay+s$, with $0<s<a$. Then, upon multiplying, you get
$$
bc=(ax+r)(ay+s)=a(axy+r+s)+rs
$$
but in order to conclude along your line of attack you should have $0<rs<a$, which is impossible to prove. The alternative attack by proving $a\nmid rs$ also fails. The simplest counterexample for both lines of attack is $a=4$, $b=c=2$.
