Divisor Count of Square-Free Integer is Power of 2

Theorem

Let $n$ be a square-free integer.

Let $\sigma_0: \Z \to \Z$ be the divisor count Function.

Then $\map {\sigma_0} n = 2^r$ for some $r \ge 1$.

The converse is not true in general.

That is, if $\map {\sigma_0} n = 2^r$ for some $r \ge 1$, it is not necessarily the case that $n$ is square-free.

Proof

Let $n$ be an integer such that $n \ge 2$.

Let the prime decomposition of $n$ be:

$n = p_1^{k_1} p_2^{k_2} \dotsm p_r^{k_r}$

Then from Divisor Count Function from Prime Decomposition we have that:

$\ds \map {\sigma_0} n = \prod_{i \mathop = 1}^r \paren {k_i + 1}$

Let $n$ be square-free.

Then by definition:

$\forall i: 1 \le i \le r: k_i = 1$

So:

$\ds \map {\sigma_0} n = \prod_{i \mathop = 1}^r \paren {1 + 1} = 2^r$

The statement about the converse is proved by counterexample:

Let $n = p^3$ where $p$ is prime.

Then $n$ is not square-free as $p^2 \divides n$.

However:

$\map {\sigma_0} n = 3 + 1 = 2^2$

$\blacksquare$