Finding a "Closed Form" Expression for $\sum_{d|n}\mu(n/d)\tau(d)$

Question

I recently got this question on a homework, and after some initial experiments got the hypothesis that $\sum_{d|n}\mu(n/d)\tau(d)=1$. I did prove this, however along the way I also encountered a proof which I could not finish. Below I have written out the partial proof, up until where I got stuck and would love any input on if it is even possible. I have also included my finished proof. Any other proofs of this hypothesis, or methods to find the closed form of the expression are also appreciated. (Additional info: $\tau(k)$ is the number of divisors of k, $\mu$ is the mobius function)

Partial Proof

A. We will first consider the case $n=1$:

$\sum_{d|1}\mu(1/d)\tau(d)=\mu(1/1)\tau(1)=1\cdot1=1$

B. We will now consider $n>1$:

This proof will be inductive.

Claim 1: if n is a prime, then $\sum_{d|n}\mu(n/d)\tau(d)=1$.

The divisors of a prime $p$ are $1,p$, which means that $\sum_{d|p}\mu(n/d)\tau(d)=\mu(p)\tau(1)+\mu(1)\tau(p)$. Since $\mu(p)=-1,\mu(1)=1$, and $\tau(1)=1,\tau(p)=2$, we have that $\mu(p)\tau(1)+\mu(1)\tau(p)=-1\cdot1+1\cdot2=1$. $\square$

Claim 2: If $n>1$, $p$ is a prime, and $p\nmid n$, then $\sum_{d|np}\mu(np/d)\tau(d)=\sum_{d|n}\mu(n/d)\tau(d)$.

The divisors of $np$ are the divisors $d$ of $n$ and the divisors $d$ of $n$ times $p$. This means that we can write: $\sum_{d|np}\mu(np/d)\tau(d)=\sum_{d|n}\mu(np/dp)\tau(dp)+\sum_{d|n}\mu(np/d)\tau(d)$.

We will $\mu(np/d)=-\mu(n/d)$:

1. If $n/d=1$, then $\mu(n/d)=1$ and $\mu(np/d)=\mu(p)=-1$
2. If $n/d$ is k distinct primes, then $\mu(n/d)=(-1)^k$ and $pn/d$ is also k+1 distinct primes so $\mu(np/d)=(-1)^{k+1}$
3. $n/d$ has a prime with exponent $a\geq 2$, then $\mu(n/d)=0$ and np/d also has that prime with exponent $a\geq 2$, so $\mu(np/d)=0$ This proves that $\mu(np/d)=-\mu(n/d)$ and then gives us:

$\sum_{d|n}\mu(n/d)\tau(dp)-\sum_{d|n}+\mu(n/d)\tau(d)=\sum_{d|n}\mu(n/d)(\tau(dp)-\tau(d))$

Finally, $\tau(dp)=2\tau(d)$ because the divisors of $dp$ as already discussed above are all the divisors of $d$ and all the divisors of $d$ times $p$. This then gives us:

$\sum_{d|np}\mu(np/d)\tau(d)=\sum_{d|n}\mu(n/d)\tau(d)$, proving the claim. $\square$

Claim 3: If $n>1$, $p$ is a prime, and $p\mid n$, then $\sum_{d|np}\mu(np/d)\tau(d)=\sum_{d|n}\mu(n/d)\tau(d)$ or $=1$.

This is the part where I get stuck. I do have the following:

Since $p$ divides $n$ we know that for $\mu(np/d)$ to not be $0$, $d$ must be a multiple of $p$ and so we have $\sum_{d|np}\mu(np/d)\tau(d)=\sum_{d|n}\mu(n/d)\tau(dp)$. I have verified that this works but have not been able to finish the proof.

Similar Full Proof

A. Identical to A. in the previous part

B. We will now consider $n>1$:

Claim 1: if p is a prime, and $k\geq 1$ then $\sum_{d|p^k}\mu(p^k/d)\tau(d)=1$.

The divisors of $p^k$ are $d\in\{1,p^1,...,p^k\}$, and $\mu(p^k)\neq0$ only if $k<2$, which means that we only need to consider the divisors $p^k,p^{k-1}$. $\sum_{d|p^k}\mu(p^k/d)\tau(d)=\mu(1)\tau(p^k)+\mu(p)\tau(p^{k-1})$. Since $\mu(p)=-1,\mu(1)=1$, and $\tau(p^k)=k+1,\tau(p^{k-1})=k$, we have that $\mu(1)\tau(p^k)+\mu(p)\tau(p^{k-1})=1\cdot(k+1)-1\cdot k=1$. $\square$

Claim 2: If $n>1$, $p$ is a prime, $k\geq1$ and $p\nmid n$, then $\sum_{d|np^k}\mu(np^k/d)\tau(d)=\sum_{d|n}\mu(n/d)\tau(d)$.

The divisors of $np^k$ are the divisors $d$ of $n$ and the divisors $d$ of $n$ times $p^i$, for each $i\in\{0,1,...,k\}$. As shown earlier the divisors must be multiples of either $p^k$ or $p^{k-1}$(since $p$ and $n$ are coprime), which means that we can write: $\sum_{d|np^k}\mu(np^k/d)\tau(d)=\sum_{d|n}\mu(np^k/dp^k)\tau(dp^k)+\sum_{d|n}\mu(np^k/dp^{k-1})\tau(dp^{k-1})=\sum_{d|n}\mu(n/d)\tau(dp^k)+\sum_{d|n}\mu(np/dp)\tau(dp^{k-1})$.

Finally, $\mu(np/d)=-\mu(n/d)$ as shown in the previous proof, and $\tau(dp^r)=(r+1)\tau(d)$ since the divisors as stated above are the divisors of $d$ times the prime with any exponent $i<r$. This then gives us:

$\sum_{d|n}\mu(n/d)\tau(dp^k)+\sum_{d|n}\mu(np/dp)\tau(dp^{k-1})=\sum_{d|n}\mu(n/d)\tau(dp^k)-\sum_{d|n}\mu(n/dp)\tau(dp^{k-1}))=\sum_{d|n}\mu(n/d)(\tau(dp^k)-\tau(dp^{k-1}))=\sum_{d|n}\mu(n/d)((k+1)\tau(d)-k\tau(d))=\sum_{d|n}\mu(n/d)\tau(d)$

proving the claim. $\square$

Finally, by the fundamental theorem of arithmetic, we have that any $n>1$ can be created as the product of primes, and utilising part A, and the two claims above inductively, we get that $\sum_{d|n}\mu(n/d)\tau(d)=1$.

Answer 1

We consider some basics of arithmetical functions, i.e. functions from $f:\mathbb{N}\to\mathbb{C}$ and show this way the claim. We want to show \begin{align*} \sum_{d|n}\mu(n/d)\tau(d)=1\qquad\qquad n\geq 1\tag{1} \end{align*}

It is convenient to use the operator $\star$ to denote the Dirichlet Convolution \begin{align*} (f\star g)(n)=\sum_{d|n}f(d)g(n/d) \end{align*}

Some preliminaries:

Denoting the arithmetical function $\zeta$ which evaluates to $1$ for each input value $n$ \begin{align*} \zeta(n)=1\qquad\qquad n\geq 1 \end{align*} we can write the claim (1) as \begin{align*} \color{blue}{\mu\star\tau=\zeta}\tag{2} \end{align*} where we use the commutativity of the $\star$ operator since \begin{align*} (f\star g)(n)&=\sum_{d|n}f(d)g(n/d)=\sum_{a\cdot b=n}f(a)g(b)\\ &=\sum_{d|n}g(d)f(n/d)=(g\star f)(n) \end{align*} In a similar way we can show associativity. We also have a neutral element $\delta$, namely the arithmetical function \begin{align*} \delta(n)=\begin{cases} 1\qquad& n=1\\ 0\qquad& \mathrm{otherwise} \end{cases} \end{align*} so that \begin{align*} \color{blue}{f\star \delta = f=\delta\star f}\tag{3} \end{align*}

It is not difficult to show that arithmetic functions $f$ with $f(1)\ne 0$ have an inverse $f^{-1}$, so that these functions form an abelian group with respect to $\star$.

The Möbius-function $\mu$

From the definition of the Möbius function $\mu$ we find \begin{align*} \sum_{d|n}\mu(d)=\begin{cases} 1&\qquad n=1\\ 0&\qquad \mathrm{otherwise}\tag{4} \end{cases} \end{align*} This is obvious if $n=1$ and for $n=p_1p_2\cdots p_k>1$ we have \begin{align*} \sum_{d|n}\mu(d)&=\mu(1)+\mu(p_1)+\cdots +\mu(p_k)\\ &\qquad +\mu(p_1p_2)+\cdots +\mu(p_{k-1}p_k)\\ &\qquad+\cdots\\ &\qquad +\mu(p_1\cdots p_k)\\ &=1+\binom{k}{1}(-1)^1+\binom{k}{2}(-1)^2+\cdots+\binom{k}{k}(-1)^k\\ &=(1-1)^k\\ &=0 \end{align*} Using the $\star$ operator we can write (4) as $\color{blue}{\mu \star \zeta=\delta}$ and it follows from (3) that $\color{blue}{\mu^{-1}=\zeta}$. Since \begin{align*} \tau(n)=\sum_{d|n}1 \end{align*} we can write $\tau$ in the form \begin{align*} \color{blue}{\tau=\zeta\star\zeta}\tag{5} \end{align*}

Proof of the claim:

With all these preliminary derivations, we can now easily prove the claim. We obtain with the help of (3), (4) and (5) \begin{align*} {\color{blue}{\mu\star \tau}}&=\mu\star\left(\zeta\star\zeta\right)=\left(\mu\star\zeta\right)\star\zeta\\ &=\left(\zeta^{-1}\star\zeta\right)\star\zeta=\delta\star\zeta\\\ &\,\,\color{blue}{=\zeta} \end{align*} and the claim (1) follows.