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.
Best Answer
Hint: Line up all the prime divisors of $n$, say $p_1$, $p_2$, and so on up to $p_k$, in a row. To make a square-free divisor of $n$, stop in front of each prime and say YES or NO (if you prefer, say $1$ or $0$). All NO gives you the square-free divisor $1$. All YES gives you $p_1p_2\cdots p_k$. Some YES some NO gives you stuff in between. How many ways can you make your decisions?
Another way: Note that $\sum_{d|n}|\mu(n)|$ and $2^{\omega(n)}$ are both multiplicative functions of $n$. So to verify they are the same, we need only verify equality at $1$ and at prime powers. For any prime $p$, and any $e>0$, if $n=p^e$, each of our functions is $2$ at $n$. This proof is probably the intended one.