Is there an expression for $\mu(n)^2$ where $\mu$ is the mobius function?

Question

Is there an expression for $\mu(n)^2$ where $\mu$ is the mobius function?

I know that \begin{align} \sum_{d|n} \mu(d)=\left\{ \begin{array}{cc} 1 & \text{if }n=1\\ 0 & \text{if }n>1 \end{array}\right. \end{align}

and that

\begin{align} \mu(n)=\left\{ \begin{array}{ll} 0 & \text{if } n \text{ has one or more repeated prime factors}\\ 1 & \text{if }n=1\\ (-1)^k & \text{if } n \text{ is a product of } k \text{ distinct primes} \end{array}\right. \end{align}

I am trying to find an expression for $\mu(n)^2$ because I am trying to show that $$\sum_{d|n}\mu(d)^2 = 2^k$$ for some $k \in \mathbb{Z}$ ie. show that it is a power of 2.

Any hints or suggestions would be sincerely appreciated.

Answer 1

Hint:

Since $\mu(d)^2=0$ if $d$ has repeated prime factors, and $\mu(d)^2=1$ otherwise, you need to count how many $d$ dividing $n$ have no repeated prime factors.

Solution:

These $d$ are exactly the product of subsets of the prime factors of $n$, of which there are $2^k$ in total, where $k$ is the number of distinct prime factors of $n$.