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Thread: Number theory Question 1

  1. #1
    Member
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    Sorry It should be:

    How can i show that

    $\displaystyle

    \frac{n}{\phi(n)} = \sum_{d|n} \frac{\mu(d)^2}{\phi(d)}
    $

    here $\displaystyle \mu , \phi$ are mobius function and Euler's phi function.
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  2. #2
    Super Member PaulRS's Avatar
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    Quote Originally Posted by peteryellow View Post
    $\displaystyle

    \frac{n}{\phi(n)} = \sum_{d|n} \frac{\mu^2(d)}{\phi(d)}
    $

    Remember that both, $\displaystyle \mu$ and $\displaystyle \phi$ are multiplicative, hence $\displaystyle \frac{\mu^2(n)}{\phi(n)}$ is multiplicative.

    Now, recall the fact that if $\displaystyle f(n)$ is a multiplicative arithmetic function then so is $\displaystyle \sum_{d|n} f(d)$

    From now on I will denote: $\displaystyle F(n)=\sum_{d|n} \frac{\mu^2(d)}{\phi(d)}$

    Now let's compute $\displaystyle F(p^s)=\sum_{d|p^s} \frac{\mu^2(d)}{\phi(d)}$ where $\displaystyle p$ is a prime number ( and s is greater than 0): $\displaystyle \sum_{d|p^s} \frac{\mu^2(d)}{\phi(d)}=\frac{\mu^2(1)}{\phi(1)}+ \frac{\mu^2(p)}{\phi(p)}=1+\frac{1}{p-1}$

    Now suppose $\displaystyle n=\prod_{p|n}p^s$ is the prime descomposition of n.

    Then $\displaystyle F(n)=\prod_{p|n}F(p^s)=\prod_{p|n}\left(1+\frac{1} {p-1}\right)=\prod_{p|n}\left(\frac{1}{1-\tfrac{1}{p}}\right)$

    Now recall the fact that: $\displaystyle \frac{\phi(n)}{n}=\prod_{p|n}\left(1-\tfrac{1}{p}\right)$ and we are done
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  3. #3
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    Thanks for veryyyyyyyy detalied reply. :-)
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