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 theseproperties. $\mu$ clearly satisfies (1), so take some $n>1$. Let $p$ besome prime factor of $n$, and let $m$ be the product of all the primefactors 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$