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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.

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    @Soarer: Thanks, I have some reading to do it seems!2011-07-27

3 Answers 3

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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.

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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

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

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    Ah, I guess I should have seen that. Thanks!2011-07-27
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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...

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    I'll dive into multiplicative functions right away! Thanks!2011-07-27