Prove $p$ is prime

Question

Let $p$ be an integer with this property: whenever $b, c \in \mathbb Z$ such that $p\mid bc$, then $p\mid b$ or $p \mid c.$ Prove $p$ is prime. Here is my attempt at a proof:

Suppose $d \mid p$. Then $p = dt$. So, $p \mid dt$, implying $p \mid d$ or $p \mid t$.

Suppose $p \mid d$. Since $d \mid p $, then $d = \pm p.$

Suppose $p \mid t$. Since $p \mid dt$, then $pj = dt$. But we know $p \mid t$, so $d$ must be $\pm 1.$

Are you convinced?

Answer 1

Your last argument does not prove anything. The equation $pj=dt$ is possible with $j=1$ and does not imply that $d=\pm 1$.

What you need is this : since $p \, | \, t$, $t = pj$ for some $j$, hence $p = dt = pdj$, so that $dj = 1$. The solutions to this equation in $\mathbb Z$ are $d=j=1$ or $d=j=-1$, so now you are done.

Hope that helps,

Answer 2

Assume, $p=ab$ with $1<a,b<p$. Then, $p$ divides neither $a$ nor $b$, but $ab$.

If $p=-ab$ with $1<a,b<-p$, then $p$ divides neither $-a$ nor $b$, but $-ab$.