Showing $\sum_{d\mid n} \mu(d)\tau(n/d)=1$ and $\sum_{d\mid n} \mu(d)\tau(d)=(-1)^r$

Question

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.

Trying to search for the first one, I found these questions: An application of Mobius Inversion $\sum_{d \mid n} \mu(\frac{n}{d})\nu(d) = 1$ and Sum of Positive Divisors: $\sum_{d|n} \mu(n/d)\nu(d)=1$ and $\sum_{d|n} \mu(n/d)\sigma(d)=n$. Perhaps with a bit of searching you can find more copies of that question. — Martin Sleziak, Jan 15 '17 at 11:17
Similarly, with a bit of searching, you will probably be able to find several posts about the other one. Like here or here. 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. — Martin Sleziak, Jan 15 '17 at 11:22
The sad thing is, the fact that it's two questions in one makes it difficult to mark it as a duplicate. — Bart Michels, Jan 15 '17 at 20:12
Forgive me this time . Will try to search from the next time , newbie mistake . Also thnx alot for citing the answers . :) — da6932, Jan 16 '17 at 19:19

Chain Markov · Answer 1 · 2017-01-15T20:05:58.587

1

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) => $\sum_{d|n} μ(d)τ(d) = \sum_{d|rad(n)} μ(d)τ(d) = (-1)^r \sum_{d|rad(n)} μ(rad(n)/d)τ(d)= (-1)^r \sum_{d|rad(n)} μ(d)τ(rad(n)/d)=(-1)^r.$

edited Jan 15 '17 at 20:05

answered Jan 15 '17 at 17:24

Chain Markov

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Really nice observation for tau(n) , and indeed nice proof for first . Thnx . – da6932 Jan 16 '17 at 19:23

Marko Riedel · Accepted Answer · 2017-01-16T21:08:08.690

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 $\tau(n/d) = \tau(n) \prod_{p_q\in S} \frac{e_q}{e_q+1}.$

Showing $\sum_{d\mid n} \mu(d)\tau(n/d)=1$ and $\sum_{d\mid n} \mu(d)\tau(d)=(-1)^r$

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