Question

Let p be an integer other than $0,\pm 1$. Prove that p is prime if and only if it has this property: Whenever r and s are integers such that $p=rs$, then $r=\pm 1$or $s=\pm 1$.

Short answer

It is proved that if p is prime if and only if it satisfies the property that if $p=rs$, then $r=\pm 1$or $s=\pm 1$.

Step-by-step solution

Prove that if p is prime, then it satisfies the property that if p=rs, then r=±1 or 

According to the given condition, we have $p=rs$,which is prime.

Then by definition, an integer p satisfies, $p\ne 0,\pm 1$and the only divisors of p are $\pm p$and then that integer p is said to be prime. This implies that s lie in $\left\{1,-1,p,-p\right\}$.

Therefore, $r=\pm 1$or $r=\pm p$,and we have

$\begin{array}{rcl}s& =& \frac{p}{r}\\ & =& \pm 1\end{array}$.

Hence, it is proved that if p is prime, then it satisfies the property that if $p=rs$, then $p=\pm 1$or $s=\pm 1$.

Prove that p is an integer, and it satisfies the property that if p=rs, then r=±1 or s=±1then p is the prime.

Assume that p is not a prime,then it has a divisor r that does not lie in $\left\{1,-1,p,-p\right\}$. This implies that $p=rs$for some integer s.

It is given that $s=\pm 1$or $s=\pm p$,then$r=\frac{p}{s}$ and role="math" localid="1646211083470" $r=\pm por\pm 1$. The obtained result contradicts the assumption.

Therefore, p is a prime.

Hence,it is proved that if p is prime if and only if satisfies the property that if $p=rs$, then $r=\pm 1$or $s=\pm 1$.