3
$\begingroup$

Given an integer $m$ such that $n$ is denoting the distinct prime divisors of $m$, is there a proof that the sum over the divisors of m of the absolute value of the Möbius function $\mu(d)$ is equal to

\begin{equation} \sum_{d|m} \left|\mu(d)\right|=2^n, \end{equation}

where the Möbius function is defined as
\begin{equation} %\small \mu(d) = \left\{ \begin{array}{ll} 1 & \text{if $d = 1$} \\ (-1)^k & \text{if $d$ is the product of $k$ distinct primes} \\ 0 & \text{if $d$ has one or more repeated prime factors}\,. \end{array} \right. \end{equation}

I checked the relation in Maple empirically and it seems to be correct, however I could not come up with a formal proof for this result. Any help is greatly appreciated.

Thanks.

$\endgroup$
2
$\begingroup$

Note that $|\mu(d)|$ is non-zero and equal to one when $d$ is a product of primes, i.e. there is a bijection between those $d$ and the subsets of the set of prime divisors of $m$. There are precisely $2^n$ of these, done.

$\endgroup$
2
$\begingroup$

Denoting the distinct prime factors of $m$ as $p_1, \cdots, p_n$: $ \sum_{d|m} \lvert \mu(d)\rvert = \lvert \mu(1)\rvert + \sum_i \lvert \mu(p_i)\rvert + \sum_{1\leq i \leq j \leq n} \lvert \mu(p_i p_j) \rvert + \cdots + \lvert \mu(p_1 p_2\cdots p_n)\rvert $ $= 1 + \binom{n}{1} 1 + \binom{n}{2} 1^2 + \cdots + \binom{n}{n}1^n $

$\endgroup$
  • $\begingroup$ you're welcome :) $\endgroup$ – Damodar Apr 18 '16 at 6:49
1
$\begingroup$

Outline: The function given by the sum on the left is multiplicative. So basically you only need to verify that the equality holds for prime powers.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.