uniqueness of Moebius function

Here is a sample result for the function, essentially classifying its uniqueness :

Proposition 1: $\mu$ is the unique mapping ${\mathbb{N}}^{*}\to \mathbb{Z}$ such that

$\mu (1)$ $=$ $1$ (1)

${\displaystyle \sum _{d|n}}\mu (d)$ $=$ $0\text{ for all }n>1$ (2)

Proof: By induction, there can only be one function with these properties. $\mu$ clearly satisfies (1), so take some $n>1$. Let $p$ be some prime factor of $n$, and let $m$ be the product of all the prime factors of $n$.

${\displaystyle \sum _{d|n}}\mu (d)$ $=$ ${\displaystyle \sum _{d|m}}\mu (d)$

$=$ ${\displaystyle \sum _{\begin{array}{c}d|m\\ p\nmid d\end{array}}}\mu (d)+{\displaystyle \sum _{\begin{array}{c}d|m\\ p\mid d\end{array}}}\mu (d)$

$=$ ${\displaystyle \sum _{d|m/p}}\mu (d)-{\displaystyle \sum _{d|m/p}}\mu (d)$

$=$ $0$

Title	uniqueness of Moebius function
Canonical name	UniquenessOfMoebiusFunction
Date of creation	2013-03-22 14:17:25
Last modified on	2013-03-22 14:17:25
Owner	mathcam (2727)
Last modified by	mathcam (2727)
Numerical id	6
Author	mathcam (2727)
Entry type	Definition
Classification	msc 11A25