Question

Let be an integer other than 0, ±1 with this property: Whenever$b$ and $c$are integers such that $p|bc$, then $p|b$or $p|c$. Prove that $p$is prime. [Hint: If $d$is a divisor of $p$, say $p=dt$, then $p|d$or $p|t$. Show that this implies $d=\pm p$or $p=\pm 1$.]

Short answer

It is proved that $p$is a prime.

Step-by-step solution

Use the given part

It is given that $p$is an integer other than$0,\pm 1$.

Then, assume that $p\ne 0,\pm 1$such that $p|bc\Rightarrow p|borp|c$.

Then, $p$is a prime.

Proof

$p=dt$Now assume that $d$is a divisor of $p$; then for some integer $t$, .

As$p|dt$, which implies $p|d$and $p|t$.

Since $d|p$and $p|d$, we get $d=\pm p$and $t=\pm 1$.

In the same way, if $p|t$, then $t=\pm p$and $d=\pm 1$.

This implies $p$is a prime.