suppose $\omega(n)$ denote the number of distinct prime factors of n

Question

Suppose $\omega(n)$ denote the number of distinct prime factors of n. Prove that$$|\mu(n)|=\sum_{d|n}\mu(d)*2^{\omega(n/d)}$$ Can any one give me some hints about this problem? Is $\mu(n)$ a multiplicative function?

Any help would be appreciated.

Answer 1

This problem can be done in three steps. Put $$g(n) = \sum_{d|n} \mu(d) \times 2^{\omega(n/d)}.$$

The first step is to note that since $\mu(n)$ and $2^{\omega(n)}$ are both multiplicative so is $g(n).$

The second step is to apply Moebius inversion to obtain $$ 2^{\omega(n)} = \sum_{d|n} g(n).$$

The third is to calculate $g(n)$ from this equation. We start with $n=1$ $$g(1) = 2^{\omega(1)} = 1$$ and continue with $n=p$ where $p$ is a prime $$g(1) + g(p) = 2^{\omega(p)} = 2\quad\text{thus}\quad g(p) = 1.$$

Next is $n=p^2$ which gives $$g(1) + g(p) + g(p^2) = 2$$ so $$g(p^2) = 0.$$

Continuing we have by induction that $g(p^v) = 0$ for $v\ge 2.$ But since $g$ is multiplicative this means that $g(n)$ is zero if $n$ is not squarefree and one otherwise, which is precisely the defintion of $|\mu(n)|$ so that $$g(n) = |\mu(n)|$$ which was to be shown.