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I have to prove for $n \in \mathbb{N}>1$ with $n=\prod \limits_{i=1}^r p_i^{e_i}$. $f$ is a multiplicative function with $f(1)=1$:

$$\sum_{ d \mid n} \mu(d)f(d)=\prod_{i=1}^r (1-f(p_i))$$

How I have to start? Are there different cases or can I prove it in general?

Any help would be fine :)

  • 2
    Hint 1: If you can show that both sides define a [multiplicative function](http://en.wikipedia.org/wiki/Multiplicative_function) of $n$, then it suffices to verify this for prime powers $n=p^a$. Hint 2 (different approach): Notice that only squarefree $d$'s have non-zero contribution to the LHS. Can you get something similar from the RHS by expanding it?2011-11-15
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    It is recommended not to use titles consisting only of formula (with no text). http://meta.math.stackexchange.com/questions/3135/why-no-use-displaystyle-in-titles/3136#31362011-11-15
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    In case the above hints are not sufficient, it might help you further if you have a look at Theorem 1 in [Chapter 2](http://rutherglen.science.mq.edu.au/wchen/lnentfolder/ent02-af.pdf) of W.W.L.Chen's [Elementary number theory notes](http://rutherglen.science.mq.edu.au/wchen/lnentfolder/lnent.html).2011-11-15
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    There is no theorem 1 in chapter two.2011-11-15
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    I think he means the 1st Theorem.2011-11-15
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    You're right, I meant Theorem 2A (the first one in that chapter). Thanks for noticing my mistake @Ali.2011-11-15
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    The solution should be easy now!2011-11-15

1 Answers 1

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Please see Theorem 2.18 on page $37$ in Tom Apostol's Introduction to analytic number theory book.

The proof goes as follows:

Define $$ g(n) = \sum\limits_{d \mid n} \mu(d) \cdot f(n)$$

  • Then $g$ is multiplicative, so to determine $g(n)$ it suffices to compute $g(p^a)$. But note that $$g(p^a) = \sum\limits_{d \mid p^{a}} \mu(d) \cdot f(d) = \mu(1)\cdot f(1) + \mu(p)\cdot f(p) = 1-f(p)$$
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    @Martin: I can't view the page in google books.2011-11-15