Only prime numbers $a$ (and $1$) have the property that $a\mid bc$ implies $a\mid b$ or $a\mid c$

Question

I'm trying to prove this statement but I don't know where to start.
The problem says, Let $a$,$b$, and $c$ be positive integers. Suppose that, for any $b$ and $c$, whenever $a\mid bc$, either $a\mid b$ or $a\mid c$. Show that $a=1$ or $a$ is prime.

Answer 1

Assume $a \neq 1$, we show its a prime. Suppose to the contrary that its not a prime, then $a = m\cdot n$, where we can assume that $m$ is a prime. Thus $a \mid m$, or $a\mid n$. If $a \mid m \Rightarrow a = m$ and is a prime. If not, $a \mid n$, and since $n \mid a \Rightarrow a = n \Rightarrow m = 1$, contradiction it being a prime (factor) of $a$.

Answer 2

Show that if $a$ is composite then it is possible to find a $b$ and $c$ such that $a|bc$ but $a$ divides neither $b$ nor $c$.

In fact. Let $c$ and $b$ be factors of $a$...

If $a = m*n$ (neither $m$ nor $n$ equal to 1 or to $a$) then $a|mn$ but neither $a|m$ nor $a|n$. So the condition is not true.

So if the condition is true, it has to be that $a$ is either prime or $a = 1$.