How to show that $ \sum_{d/n} \mu^{2}(d)/\phi(d) = n/\phi(n)$?

Question

$\forall n, n\in\mathbb{N}$ $\frac{n}{\phi{(n)}} = \sum_{d/n} \frac{\mu^{2}(d)}{\phi(d)}$

Where $\mu$ is the Möbius function.

Answer 1

Write $n = p_1^{r_1} \cdots p_k^{r_k}$ with the $p_i$ distinct primes and $r_i > 0$. Then $\mu(d)^2$ is $1$ if $d$ is square-free and $0$ otherwise, so \begin{align*} \sum_{d | n} \frac{\mu(d)^2}{\phi(d)} &= \sum_{n_i = 0, 1} \phi\left(p_1^{n_1} \cdots p_k^{n_k}\right)^{-1} \\ &= \sum_{n_i = 0, 1} \phi\left(p_1^{n_1}\right)^{-1} \cdots \phi\left(p_k^{n_k}\right)^{-1} \\ &= \left(1 + \frac{1}{p_1 - 1}\right) \cdots \left(1 + \frac{1}{p_k - 1}\right) \\ &= \frac{p_1}{p_1 - 1} \cdots \frac{p_k}{p_k - 1} \\ &= \frac{n}{\phi(n)}. \end{align*}

Answer 2

A start: The functions $\mu^2$, $\varphi$, and the identity function $f(n)=n$ are multiplicative.

Also, if $g$ is multiplicative, then $h(n)=\sum_{d|n}g(d)$ is multiplicative.

So we only need to establish the result for $n$ of the form $p^k$, where $p$ is prime. That will be straightforward.