I have to proof the following statement:
Let $S(n)$ the number of (positive) squarefree divisors of $n$, thus $S(n)=\sum \limits_{d \mid n} \left|{\mu(d)}\right|$, and let $\omega$ be the number of different prime integers of $n$. Then follows $S(n)=2^{\omega(n)}$.
Any help is appreciated.
[Edit]
Ok, I'll try the proof:
Let $n=p_{1}^{e_1} * p_{2}^{e_2}* \ldots *p_{k}^{e_k}$
Both functions are multiplicative, so we get the following
$\sum \limits_{d \mid p_{1}^{e_1}} \left| {\mu(d)} \right|=\left| 1+\mu(p_{1}^{e_1}) \right|$
$\sum \limits_{d \mid p_{2}^{e_2}} \left| {\mu(d)} \right|=\left| 1+\mu(p_{2}^{e_2}) \right|$
$\vdots$
$\sum \limits_{d \mid p_{k}^{e_k}} \left| {\mu(d)} \right|=\left| 1+\mu(p_{k}^{e_k}) \right|$
When $p_i$ is squarefree, we get for each addend as result a "2". So we have $2^k$ possibilities. If $p_i$ is not squarefree, the addend is "1". That finishs the proof.