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

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

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- Dec 28th 2010, 02:35 PMchiph588@Sum of Möbius Function
Let $\displaystyle \mu $ be the Möbius function and $\displaystyle k,n\in\mathbb{N} $, evaluate

$\displaystyle \displaystyle \sum_{d^k\mid n} \mu(d) $ - Dec 28th 2010, 02:51 PMBruno J.
We can factor this as $\displaystyle \prod_{p^{km}||n}(\mu(1)+\mu(p)+\dots+\mu(p^m))$ where $\displaystyle p^{km} || n$ means that $\displaystyle p^{km}|n$ but $\displaystyle p^{k(m+1)}\nmid n$. Clearly the product is $\displaystyle 0$ unless $\displaystyle p^{0}||n$ for each $\displaystyle p$. Hence the sum is $\displaystyle 1$ if $\displaystyle n$ is not divisible by a $\displaystyle k$-power (other than $\displaystyle 1$), and $\displaystyle 0$ otherwise.

- Dec 29th 2010, 12:26 PMBruno J.
So, is my solution accepted? :)

- Dec 29th 2010, 12:29 PMchiph588@
- Dec 29th 2010, 12:37 PMBruno J.
Well it's more that I'm assuming the reader will have no trouble convincing him/herself of it! (Happy)

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