prove that $\sum_{d|n}{|\mu(d)|} = 2^{k}$

Question

Let k be the number of prime factors other than a positive integer n. Prove that $$\sum_{d|n}{|\mu(d)|} = 2^{k}$$ I'm not sure how to approach this problem. Can anyone give me a hint about how to start/approach this proof?

Answer 1

$\mu(m)=1$ if $m$ is square-free and has an even number of prime factors, and $-1$ if $m$ is square-free and has an odd number of prime factors. Otherwise it is 0.

So suppose $n$ has exactly $k$ prime factors. The only divisors $d$ of $n$ which have $\mu(d)\ne0$ are those which are products of distinct prime factors of $n$. There are exactly $2^k$ such factors, because each prime factor can either divide $d$ or not. For example, if $n=2\cdot3\cdot5$, then there are 8 divisors with $\mu(d)\ne0$, namely 1, 2, 3, 5, 6, 10, 15, 30. If $\mu(d)\ne0$ then $|\mu(d)|=1$, so $$\sum_{d|n}|\mu(d)|=2^k$$