How to show $\sum_{d\mid n} \frac{\mu^2(d)}{d} =\prod_{p|n} \left(1+\frac{1}{p}\right)$?

Question

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} $

Answer 1

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 $n$.

Answer 2

Let $ n=p_1^{r_1}\ldots p_k^{r_k}$, now on R.H.S. one needs to look at only those $d/n$ such that $d= p_1^{a_1}\ldots p_k^{a_k}$ where $a_i =0$ or $1$. So R.H.S. is sum of reciprocals of all combinations of the primes $p_1 \ldots p_k$ . Now if one expands L.H.S its the same.