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Math Help - Sum of Mobius and Divisor functions

  1. #1
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    Sum of Mobius and Divisor functions

    Prove that for every positive integer n, we have
    Sum[for all m|n](μ(m)*σ(n/m)) = n.
    Here, μ(x) is the mobius function and σ(x) is the sum of all positive divisors of x.


    I'm not sure where to begin with this. Since μ and σ are multiplicative, I think I could get
    Sum[for all m|n](μ(m)*σ(n/m)) = Sum[for all m|n](μ(n/m)*σ(m))
    but I don't know how useful that is.
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  2. #2
    MHF Contributor chisigma's Avatar
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    The 'Moebious inversion formula' extablishes that if ...

    f(n)= Sum[for all d|n] g(d) (1)

    ... then...

    g(n)= Sum[for all d|n] f(d) μ (n/d) (2)

    Now if You set in (1) f(n)= σ (n) is...

    σ (n)= Sum[for all d|n] d (3)

    ... so that g(d)=d and (2) becomes...

    n= Sum[for all d|n] σ (d) μ (n/d) = Sum[for all d|n] μ (d) σ (n/d) (4)

    Kind regards

    chi sigma
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