TITLE: How to show $\sum_{d\mid n} \frac{\mu^2(d)}{d} =\prod_{p|n} \left(1+\frac{1}{p}\right)$?
QUESTION [5 upvotes]: how to prove:
$$\sum_{d\mid n} \frac{\mu^2(d)}{d} =\prod_{p|n} \left(1+\frac{1}{p}\right)$$
$\mu : \Bbb N\rightarrow \Bbb R$
$\mu(1)=1$
$ \mu(n)=
\begin{cases}
0  &,\;\;\; \text{if $\,n\,$ is divisible by a square prime number} \\{}\\
(-1)^r &,\;\;\; \text{if $\,n=p_1^1\cdots p_r^1\,$}  \\
\end{cases}
 $

REPLY [3 votes]: The function
$$
\sum_{d\mid n} \frac{\mu(d)^2}{d}
$$
is multiplicative (since both $\mu$ and $d$ are multiplicative and multiplication is distributive over addition). That is, if $(n_1,n_2)=1$, then
$$
\sum_{d\mid n_1n_2} \frac{\mu(d)^2}{d}=\sum_{d\mid n_1} \frac{\mu(d)^2}{d}\sum_{d\mid n_2} \frac{\mu(d)^2}{d}
$$
Obviously
$$
\prod_{p\mid n}\left(1+\frac1p\right)
$$
is multiplicative.
The equation is true for powers of primes; that is, both sides send $p^n\mapsto1+\frac1p$. Therefore, it is true for all $n$.