# uniqueness of Moebius function

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

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

 $\displaystyle\mu(1)$ $\displaystyle=$ $\displaystyle 1$ (1) $\displaystyle\sum_{d|n}\mu(d)$ $\displaystyle=$ $\displaystyle 0\textrm{ for all }n>1$ (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 $n$.

 $\displaystyle\sum_{d|n}\mu(d)$ $\displaystyle=$ $\displaystyle\sum_{d|m}\mu(d)$ $\displaystyle=$ $\displaystyle\sum_{\begin{subarray}{c}d|m\\ p\nmid d\end{subarray}}\mu(d)+\sum_{\begin{subarray}{c}d|m\\ p\mid d\end{subarray}}\mu(d)$ $\displaystyle=$ $\displaystyle\sum_{d|m/p}\mu(d)-\sum_{d|m/p}\mu(d)$ $\displaystyle=$ $\displaystyle 0$
Title uniqueness of Moebius function UniquenessOfMoebiusFunction 2013-03-22 14:17:25 2013-03-22 14:17:25 mathcam (2727) mathcam (2727) 6 mathcam (2727) Definition msc 11A25