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.

  • $\begingroup$ thanks, I'll edit $\endgroup$ Mar 8 '19 at 1:08

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.

  • $\begingroup$ I believe if $p$ is not prime, then $p=rs$ does not imply $p\vert r$ or $p\vert s$ consider $p=12$, $r=3$ and $s=4$. Then we have $12=3 \cdot 4$, but $12$ does not divide $3$ and $12$ does not divide $4$. $\endgroup$ Mar 8 '19 at 11:41
  • $\begingroup$ I said "if p=rs, then ..." by your assumption. $\endgroup$ Mar 8 '19 at 12:38
  • $\begingroup$ I am using proof by contradiction . $\endgroup$ Mar 8 '19 at 12:39

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