Proof that $\sum_{d|m} |\mu(d)|=2^n$, where $n$ is the number of distinct prime divisors of $m$?

Question

Given an integer $m$ such that $n$ is denoting the distinct prime divisors of $m$, is there a proof that the sum over the divisors of m of the absolute value of the Möbius function $\mu(d)$ is equal to

\begin{equation} \sum_{d|m} \left|\mu(d)\right|=2^n, \end{equation}

where the Möbius function is defined as
\begin{equation} %\small \mu(d) = \left\{ \begin{array}{ll} 1 & \text{if $d = 1$} \\ (-1)^k & \text{if $d$ is the product of $k$ distinct primes} \\ 0 & \text{if $d$ has one or more repeated prime factors}\,. \end{array} \right. \end{equation}

I checked the relation in Maple empirically and it seems to be correct, however I could not come up with a formal proof for this result. Any help is greatly appreciated.

Thanks.

Answer 1

Note that $|\mu(d)|$ is non-zero and equal to one when $d$ is a product of primes, i.e. there is a bijection between those $d$ and the subsets of the set of prime divisors of $m$. There are precisely $2^n$ of these, done.

Answer 2

Outline: The function given by the sum on the left is multiplicative. So basically you only need to verify that the equality holds for prime powers.

Answer 3

Denoting the distinct prime factors of $m$ as $p_1, \cdots, p_n$: $ \sum_{d|m} \lvert \mu(d)\rvert = \lvert \mu(1)\rvert + \sum_i \lvert \mu(p_i)\rvert + \sum_{1\leq i \leq j \leq n} \lvert \mu(p_i p_j) \rvert + \cdots + \lvert \mu(p_1 p_2\cdots p_n)\rvert $ $= 1 + \binom{n}{1} 1 + \binom{n}{2} 1^2 + \cdots + \binom{n}{n}1^n $