Divisor Count Function of Prime Number
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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.
$\blacksquare$