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Need some help on this question from Victor Shoup

Let $\tau(n)$ be the number of positive divisors of $n$. Show that:

  • $\sum_{d\mid n} \mu(d)\tau(n/d)=1$;
  • $\sum_{d\mid n} \mu(d)\tau(d)=(-1)^r$, where $n=p_1^{e_1}\cdots p_r^{e_r}$ is the prime factorization of $n$.

I have tried both of them but cant find any solution!We have to use Mobius Function properties to prove this Question.

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    Trying to search [for the first one](https://approach0.xyz/search/?q=%24%5Csum_%7Bd%5Cmid%20n%7D%20%5Cmu(d)%5Ctau(n%2Fd)%3D1%24&p=1), I found these questions: [An application of Mobius Inversion $\sum_{d \mid n} \mu(\frac{n}{d})\nu(d) = 1$](http://math.stackexchange.com/q/1960491) and [Sum of Positive Divisors: $\sum_{d|n} \mu(n/d)\nu(d)=1$ and $\sum_{d|n} \mu(n/d)\sigma(d)=n$](http://math.stackexchange.com/q/137971). Perhaps with a bit of searching you can find more copies of that question. – 2017-01-15
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    Similarly, [with a bit of searching](https://approach0.xyz/search/?q=%24%5Csum_%7Bd%5Cmid%20n%7D%20%5Cmu(d)%5Ctau(d)%24&p=1), you will probably be able to find several posts about the other one. Like [here](http://math.stackexchange.com/questions/847770/evaluating-a-sum-with-mobius-function) or [here](http://math.stackexchange.com/questions/351030/number-theory-exercise-sum-d-mid-n-mud-dd-1-omegan-and-s). In fact, you do not need to search at all, just look at the list of related questions in the sidebar on the right to find [this one](http://math.stackexchange.com/questions/871933). – 2017-01-15
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    The sad thing is, the fact that it's two questions in one makes it difficult to mark it as a duplicate. – 2017-01-15
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    Forgive me this time . Will try to search from the next time , newbie mistake . Also thnx alot for citing the answers . :) – 2017-01-16

2 Answers 2

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Using the fact that both $\mu$ and $\tau$ are multiplicative for the first one we get from first principles the value

$$\tau(n) \prod_{q=1}^r \left(1+(-1)\times\frac{e_q}{e_q+1}\right)$$

which simplifies to

$$\tau(n) \prod_{q=1}^r \frac{1}{e_q+1} = 1.$$

For the second one we may write

$$\prod_{q=1}^r (1+(-1)\times 2) = (-1)^r.$$

Here we have used that for a subset $S$ of the set of prime factors $P$ corresponding to a squarefree divisor $d$ of $n$ we get $\mu(d) = (-1)^{|S|}$ and $\tau(n/d) = \tau(n) \prod_{p_q\in S} \frac{e_q}{e_q+1}.$

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1) $τ(n)= \sum_{d|n} 1$ => According to the Moebius Inversion Formula $\sum_{d|n} μ(d)τ(n/d) = 1$

2) If d has a squared prime factor, then μ(d)=0 => $\sum_{d|n} μ(d)τ(d) = \sum_{d|rad(n)} μ(d)τ(d)$. If the number of prime factors of n is even, then μ(rad(n)/d)= μ(d). If it is odd, then μ(rad(n)/d)= -μ(d) => $\sum_{d|n} μ(d)τ(d) = \sum_{d|rad(n)} μ(d)τ(d) = (-1)^r \sum_{d|rad(n)} μ(rad(n)/d)τ(d)= (-1)^r \sum_{d|rad(n)} μ(d)τ(rad(n)/d)=(-1)^r.$

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    Really nice observation for tau(n) , and indeed nice proof for first . Thnx . – 2017-01-16