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If $\nu(n)=$ Number of positive divisors of $n,$ $\mu$ is the Möbius function and $\sigma(n)$ is the sum of positive divisors.

show that;

  1. $\sum\limits_{d|n} \mu(n/d)\nu(d)=1$ for all $n.$

  2. $\sum\limits_{d|n} \mu(n/d)\sigma(d)=n$ for all $n.$

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    These are both Dirichlet convolutions. See [this](http://math.stackexchange.com/a/83812) for instance.2012-04-28
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    The first case corresponds to getting the coefficients of $\zeta(s)$, and the second case corresponds to the coefficients of $\zeta(s-1)$.2012-04-28
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    @J.M. Given the (elementary-number-theory) tag, I think an approach through analytic number theory might be overkill.2012-04-28
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    @anon: I was about to write an answer until I saw the tag. That's why I restricted myself to comments instead. :)2012-04-28

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