Let $n$ be a natural number greater than $1$.
Prove the following:
$$\sum_{d|n} \mu(d) \sigma(d)=(-1)^rp_1...p_r$$where $p_1,...,p_r$ are the distict prime factors of $n$ and $\mu$ is the Mobius function.
I am really stuck on this problem. The only thing I can think is since $\mu(n)=0$ when $p^2|n$ for some prime $p$ we have a lot of terms in this sum equal to $0$.
In order to complete the question with an answer, see the comments, let me mention that in this duplicate it is shown that $$ \sum_{ d \mid n} \mu(d)f(d)=\prod_{p\mid n} (1-f(p)) $$ for a multiplicative arithmetic function $f$. For $f(n)=\sigma(n)$, the sum of the positive divisors of $n$, we obtain $$ \sum_{ d \mid n} \mu(d)\sigma(d)=\prod_{p\mid n} (1-\sigma(p))=\prod_{p\mid n}(1-(1+p))=\prod_{p\mid n}(-p). $$
