How do I show that $\frac{n}{\varphi(n)}=\sum\limits_{d \mid n} \frac{\mu^2(d)}{\varphi(d)}$?

Question

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.

Answer 1

HINT:

These functions are multiplicative. So we consider the case only for $n = p^n$ for a prime p. Here, $\mu (p^a) $ is 1 only once in the sum $\sum_{d|p^a} \frac{\mu ^2 (d)}{\varphi(d)}$. Now we use that $\varphi$ is multiplicative too, and...

Answer 2

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.

Answer 3

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

$$\prod_{p \mid n} \left(1+\frac{1}{p-1}\right).$$