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Thread: Number Theory Question (Mobius Theorem)

  1. #1
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    Number Theory Question (Mobius Theorem)



    Hi guys, I just have a few problems with part b) and c), I can do part a) the following is my working but I'm just not sure how to solve b) and c). Any help would be appreciated, cheers!

    If we let $\displaystyle \displaystyle{f(d) = d^k}$ and $\displaystyle \displaystyle{F(n) = \sigma_k (n)}$ then since $\displaystyle \displaystyle{\sigma_k(n) = \sum_{d|n} d^k}$ we have $\displaystyle \displaystyle{F(n) = \sum_{d|n} f(d)}$

    Using the Mobius Inversion Theorem yields:

    $\displaystyle \displaystyle{f(n) = \sum_{d|n} \mu(d) F\left(\frac{n}{d}\right)}$

    $\displaystyle \displaystyle{\Rightarrow n^k = \sum_{d|n} \mu (d) \sigma_k \left(\frac{n}{d}\right)}$ as required.
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  2. #2
    Super Member PaulRS's Avatar
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    For (b) remember Fermat's Little Theorem, we have $\displaystyle a^p\equiv a (\bmod.p) $ for all $\displaystyle a\in \mathbb{Z}$
    For (c) rewite it as $\displaystyle \sigma_k ( n) = n^k + 1$, what are the divisors?
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