Prime Number has 4 Integral Divisors

Theorem
Let $p$ be an integer.

The $p$ is a prime number iff $p$ has exactly four integral divisors: $1, -1, p, -p$.

Necessary Condition
Let $p$ be a prime number.

From the definition of a prime number, $1$ and $p$ divide $p$.

Also, we have $-1 \mathop \backslash p$ and $-p \mathop \backslash p$ from One Divides all Integers and Integer Divides its Negative.

Now suppose $x < 0: x \mathop \backslash p$ where $x \ne -1$ and $x \ne -p$.

Then $\left|{x}\right| \mathop \backslash x \mathop \backslash p$ and $\left|{x}\right|$ is therefore a positive integer other than $1$ and $p$ that divides $p$, which is a contradiction of the conditions of $p$ being a prime.

So $-1$ and $-p$ are the only negative integers that divide $p$.

It follows that $p$ has exactly those four divisors.

Sufficient Condition
Suppose $p$ has the divisors $1, -1, p, -p$.

It follows that $1$ and $p$ are the only positive integers that divide $p$

Thus by definition $p$ is a prime number.

Also see

 * Definition:Negative Prime