Let p be an integer other than 0, $\pm$1. Prove that p is prime if whenever r and s are integers such that p=rs, then r=$\pm$1 or s=$\pm$1 Let $p$ be an integer other than $0, \pm 1$. Prove that $p$ is a prime if whenever $r$ and $s$ are integers such that $p=rs$, then $r=\pm1$ or $s=\pm1$
Since this is an if and only if proof, I started off trying to prove this statement "forward" and then "backwards". However I am not sure of my work is correct thus far.
This is what I have: 
Proof:
(Forward)
Let $p$ be an integer and not equal to $0, \pm 1$ and $p$ is prime.
Suppose $r,s$ are integers such that $p=rs$
(NTS: $r= \pm 1$ or $s=\pm 1$)
Case $1$:
$$r \neq \pm1$$
$$p=rs $$
$p$ is prime and therefore only divisible by $\pm 1$ or $\pm p$ 
In this case $r$ divides $p$, but $r$ cannot equal $\pm1$. 
This forces $r=\pm p$ 
Plugging this back into the equation: $p=\pm s$ then $s=\pm 1$
Case $2$:
$$s\neq \pm 1$$
$$p=rs$$ 
$p$ is prime and therefore only divisible by $\pm 1$ or $\pm p$ 
In this case $s$ divides $p$, but $s$ cannot equal $\pm 1$. 
This forces $s=\pm p$ 
Plugging this back into the equation: $p=(\pm p)r$ then $r=\pm 1$
(Backward)
Let $r,s$ be integers such that $p=rs$, $r=\pm 1$ or $s=\pm 1$ 
(NTS: $p$ is prime)
$$p=rs$$
Therefore $r$ divides $p$. We also know that every divisor ($r$) of a non zero integer ($p$) must be less than or equal to $|p|$, however since $r$ is an integer, $r$ must equal $1$ or $p$. 
In the case that $r=1$
$$p=(1)s$$
$$s=p$$
In the case that $r=p$
$$p=(p)s$$
$$s=1$$
This is where I get really stuck because although I showed that $r$ and $s$ are essentially $1$ or $p$, I don't necessarily prove that p is prime because of the condition that that there could be more divisors than just $1$ and $p$ with what I have shown a number like $p=4$ would still work, but $4$ is not prime. What am I missing? Any help appreciated!!
 A: Wrong statement:


*

*We also know that every divisor $t$ of a non zero interger $p$ must be less than or equal to |p|, however since $r$ is an integer, $r$ must be equal to $1$ or $p$.


This is not true as we have not shown that $p$ is a prime.
Another wrong statement is the statement that you are trying to prove, the statement is true for negative of prime numbers as well.
Let any factorization of $|p|$ be $|p|=|r||s|$. Using the property that is stated, if $|r|=1$ then $|s|=|p|$. If $|s|=1$, then $|r|=|p|$. Hence $|p|$ must be a prime number.
A: The second implication  is a false result. If it were true we would conclude that $1$ and $-2$ are prime.  What is true is that if $p>1$, and if $|r|=1$ or $|s|=1$ whenever $rs=p,$ then $p$ is prime. Your attempt suffers from too much abbreviation.
State the initial assumptions: Let $p>1$ have the property that $|r|=1$ or $|s|=1$ whenever $rs=p.$
State the conclusion to be proved: Then $p$ is prime.
Then give the proof of the conclusion: Proof:
(I). If $r>1$ and $s>1$ then $rs\ne p,$ because otherwise $rs=|r|\cdot |s|=p$ with $|r|\ne 1\ne |s|.$ And also $p>1.$ 
(II). The definition of "$p$ is prime" is that $p>1$ and $p\ne rs$  whenever $r>1$ and $s>1.$ So by (I),  $p$ is prime. 
