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Math Help - Number theory Question 1

  1. #1
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    Sorry It should be:

    How can i show that

    <br /> <br />
\frac{n}{\phi(n)} = \sum_{d|n} \frac{\mu(d)^2}{\phi(d)}<br />

    here  \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
    <br /> <br />
\frac{n}{\phi(n)} = \sum_{d|n} \frac{\mu^2(d)}{\phi(d)}<br />

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

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

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

    Now let's compute F(p^s)=\sum_{d|p^s} \frac{\mu^2(d)}{\phi(d)} where p is a prime number ( and s is greater than 0): \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 n=\prod_{p|n}p^s is the prime descomposition of n.

    Then 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: \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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