Divisor Count Function of Prime Number

Theorem

Let $p \in \Z_{> 0}$.

Then $p$ is a prime number if and only if:

$\map {\sigma_0} p = 2$

where $\map {\sigma_0} p$ denotes the divisor count function of $p$.

Proof

Necessary Condition

Let $p$ be a prime number.

Then, by definition, the only positive divisors of $p$ are $1$ and $p$.

Hence by definition of the divisor count function:

$\map {\sigma_0} p = 2$

$\Box$

Sufficient Condition

Suppose $\map {\sigma_0} p = 2$.

Then by One Divides all Integers we have:

$1 \divides p$

Also, by Integer Divides Itself we have:

$p \divides p$

So if $p > 1$ it follows that $\map {\sigma_0} p \ge 2$.

Now for $\map {\sigma_0} p = 2$ it must follow that the only divisors of $p$ are $1$ and $p$.

That is, that $p$ is a prime number.

$\blacksquare$