Prime Number has 4 Integral Divisors

Theorem
Let $p$ be an integer.

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

Necessary Condition
Let $p$ be a prime number from the definition that $p$ has exactly $2$ divisors which are positive integers.

From One Divides all Integers and Integer Divides Itself those positive integers are $1$ and $p$.

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

Aiming for a contradiction suppose:
 * $\exists x < 0: x \mathop \backslash p$

where $x \ne -1$ and $x \ne -p$.

Then:
 * $\left|{x}\right| \mathop \backslash x \mathop \backslash p$

and so $\left|{x}\right|$ is therefore a positive integer other than $1$ and $p$ that divides $p$.

This is a contradiction of the condition for $p$ to be 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 $p$ has exactly two divisors which are positive integers.

Also see

 * Definition:Negative Prime