TITLE: How do I show that $\frac{n}{\varphi(n)}=\sum\limits_{d \mid n} \frac{\mu^2(d)}{\varphi(d)}$?
QUESTION [5 upvotes]: As the title says. How do I get from  $\frac{n}{\varphi(n)}$ to $\sum\limits_{d \mid n} \frac{\mu^2(d)}{\varphi(d)}$?
I know that
$$\frac{n}{\varphi(n)}=\frac{\sum\limits_{d \mid n} \varphi(d)}{\sum\limits_{d \mid n} \mu(d)\frac{n}{d}},$$
and I suspect it is down this road I should go. But I get totally confused by the sums and I have no clue on how to divide them and "combine" them into one again. Any hints or advice would be much appreciated.

REPLY [6 votes]: Alternatively, you can use the fact that 
$$\frac{n}{\phi(n)} = \prod_{p \mid n} \left(\frac{p}{p-1}\right)$$
and expand the product by writing it as
$$\prod_{p \mid n} \left(1+\frac{1}{p-1}\right).$$

REPLY [3 votes]: Consider the prime powers.  For $n=p^k$ $$\sum_{d|p^k} \frac{\mu(d)^2}{\phi(d)}=1+\frac{1}{\phi(p)}=\frac{p}{p-1}=\frac{p^k}{p^{k-1}(p-1)}=\frac{p^k}{\phi(p^k)}.$$  Now use the fact that these functions are multiplicative.