Prime Number has 4 Integral Divisors

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Theorem

Let $p$ be an integer.


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


Proof

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 $-1 \divides p$ and $-p \divides p$ from One Divides all Integers and Integer Divides its Negative.


Aiming for a contradiction, suppose:

$\exists x < 0: x \divides p$

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

Then:

$\size x \divides x \divides p$

and so $\size x$ 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.

$\Box$


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.

$\blacksquare$


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