Sum of Möbius Function

**chiph588@** · December 28th 2010, 02:35 PM

Let $\mu$ be the Möbius function and $k,n\in\mathbb{N}$ , evaluate

$\displaystyle \sum_{d^k\mid n} \mu(d)$

**Bruno J.** · December 28th 2010, 02:51 PM

We can factor this as $\prod_{p^{km}||n}(\mu(1)+\mu(p)+\dots+\mu(p^m))$ where $p^{km} || n$ means that $p^{km}|n$ but $p^{k(m+1)}\nmid n$ . Clearly the product is $0$ unless $p^{0}||n$ for each $p$ . Hence the sum is $1$ if $n$ is not divisible by a $k$ -power (other than $1$ ), and $0$ otherwise.

**Bruno J.** · December 29th 2010, 12:26 PM

So, is my solution accepted?

**chiph588@** · December 29th 2010, 12:29 PM

Originally Posted by Bruno J.

So, is my solution accepted?

Haha! I forgot about this thread!

It appears you're assuming the sum i posted is multiplicative. Why is this so?

**Bruno J.** · December 29th 2010, 12:37 PM

Well it's more that I'm assuming the reader will have no trouble convincing him/herself of it!

Each term in the sum is $\mu(d)$ , where $d=p_1^{a_1}\cdots p_j^{a_j}$ and $d^k = p_1^{ka_1}\cdots p_j^{ka_j}$ is such that $d^k | n$ . In the product I wrote, the term $\mu(d)$ will be obtained as $\mu(p_1^{a_1})\cdots \mu(p_j^{a_j}) = \mu(d)$ once the product is expanded. Conversely, each term in the expansion of the product can be found in the sum.

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