This is my attempt to prove the following (excerpt from the exercises in my textbook):
Let $p$ be an integer other than $0, \pm 1$ with this property: Whenever $b$ and $c$ are integers such that $p \vert bc$, then $p\vert b$ or $p\vert c$. Prove that $p$ is prime.
Here is my attempt:
If $d\vert p$ then we have $p=dt$ for some $t\in \mathbb Z$. This implies that $p\vert d$ or $p\vert t$. If $p\vert d$ then $\vert p \vert \leq \vert d \vert$ and since we have assumed $d \vert p$ we have $\vert d \vert \leq \vert p \vert$. This implies that $\vert d \vert = \vert p \vert$ so $d= \pm p$.
If $p$ does not divide $d$ then $p \vert t$ which implies $\vert p \vert \leq \vert t \vert$. Together with $p=dt$ this implies that $d=\pm 1$. This shows that the only possible divisors of $p$ are $\pm 1$ and $\pm p$, meaning that $p$ is prime.
I would greatly appreciate any constructive feedback/criticism on my attempted proof.
I have trouble understanding your logic, but here is my version. Assume on the contrary that $p$ is not prime, then $p=rs$, now p|rs implies $p|r$ or $p|s$, but both are false.