# Dirichlet product of distinct prime factors and the Möbius function

In text book Analytic number theory by Apostol on page $47$, Exercise $5$ we have the following:

Define $v(1)=0$, and for $n>1$ let $v(n)$ be the number of distinct prime factors of $n$. Let $f=\mu*v$ and prove that $f(n)$ is either $0$ or $1$.

The symbol $*$ is taken as the Dirichlet product.

I tried to divide cases into whether $n$ is square free or not. But I can't find anything. Please help me to solve it.

Let us call $\omega\left(n\right)$ the function that count the number of distinct prime factors of $n$. We have that $$\sum_{d\mid n}\delta_{p}\left(d\right)=\omega\left(n\right)$$ where $$\delta_{p}\left(d\right)=\begin{cases} 1, & d=p,\, p\textrm{ is a prime number}\\ 0, & \textrm{otherwise} \end{cases}$$ and we have done since, by Möbius inversion formula, we get $$f\left(n\right)=\sum_{d\mid n}\mu\left(d\right)\omega\left(\frac{n}{d}\right)=\delta_{p}\left(n\right).$$

$$L_1(s) = \sum_{n\ge 1}\frac{\mu(n)}{n^s} = \prod_p\left(1-\frac{1}{p^s}\right) = \frac{1}{\zeta(s)}.$$

Furthermore introduce

$$L_2(s) = \sum_{n\ge 1}\frac{\omega(n)}{n^s} = \left. \left(\prod_p\left(1+\frac{a}{p^s}+\frac{a}{p^{2s}}+\cdots\right) \right)'\right|_{a=1}.$$

Here the derivative is with respect to $a.$ Continuing,

$$\left. \left(\prod_p\left(1+a\frac{1/p^s}{1-1/p^s}\right) \right)'\right|_{a=1} \\ = \left. \left(\prod_p\left(1+a\frac{1/p^s}{1-1/p^s}\right) \sum_p \frac{\frac{1/p^s}{1-1/p^s}} {1+a\frac{1/p^s}{1-1/p^s}} \right)\right|_{a=1} \\ = \zeta(s) \sum_p \frac{1/p^s}{1-1/p^s+1/p^s} = \zeta(s) \sum_p \frac{1}{p^s}.$$

In fact this last equation could have been obtained by inspection. It follows that

$$L(s) = \sum_{n\ge 1} \frac{(\mu\star\omega)(n)}{n^s} \\ = L_1(s) L_2(s) = \sum_p \frac{1}{p^s} = \sum_{n\ge 1} \frac{1}{n^s} [[n\;\text{is prime}]]$$

as claimed.

• Thank you very much. (+1.) I was somewhat concerned that the OP might not have seen Dirichlet series in their coursework yet. Commented Aug 12, 2016 at 19:46

One can argue by cases if the nice idea of $$\delta_p$$ elludes us. Here is how it can be done:

Let $$n$$ be made up of the primes $$p_1,...,p_k$$.

Define $$A=\{j \: : \:p_j\| n \}$$ and $$B=\{j\::\: p_j^2|n \}$$. Upon reindexing the primes, we may suppose that $$j_o,j_o+1,...,i_k$$ are all the primes in $$A$$. It is clear that:

$$f(n)=\sum_{(i_1,...,i_k)\in\{0,1\}^k}\mu(p_1^{i_1}...p_k^{i_k})\nu(n/p_1^{i_1}...p_k^{i_k})=$$ $$\sum_{t=0}^{k-j_o+1}\left(\sum_{i_{j_o}+...+i_k=t\\i_1+...+i_k\equiv0 \mod 2\\i_j\in\{0,1\}\quad \forall j}\nu(n/p_1^{i_1}...p_k^{i_k})-\sum_{i_{j_o}+...+i_k=t\\i_1+...+i_k\equiv 1 \mod 2\\i_j\in\{0,1\}\quad \forall j}\nu(n/p_1^{i_1}...p_k^{i_k})\right)=: \sum_{t=0}^{k-j_o+1}S_t$$

I now separate in two cases:

CASE I: $$j_o\not=1$$. In this case, I claim that $$S_t=0$$ for all $$t$$ and so $$f(n)=0$$. Because $$i_{j_o}+...+i_k=t$$ in $$S_t$$, then $$\nu$$ is always $$k-t$$ there and we have:

$$S_t=(k-t)\left(\sum_{i_{j_o}+...+i_k=t\\i_1+...+i_k\equiv0 \mod 2\\i_j\in\{0,1\}\quad \forall j}1-\sum_{i_{j_o}+...+i_k=t\\i_1+...+i_k\equiv 1 \mod 2\\i_j\in\{0,1\} \quad \forall j}1\right)$$

However we are summing over sets of the same size! Indeed:

$$\Psi(i_1,...,i_k)=\begin{cases}(0,i_2,...,i_k)\quad \text{if i_1=1}\\ (1,i_2,...,i_k) \quad \text{if i_1=0}\end{cases}$$

Is easily seen to be a bijection between $$C_1=\{i_{j_o}+...+i_k=t\:,\:i_1+...+i_k\equiv0 \mod 2\:,\:i_j\in\{0,1\}\: \forall j\}$$ and $$C_2=\left\{i_{j_o}+...+i_k=t\:,\:i_1+...+i_k\equiv0 \mod 2\:,\:i_j\in\{0,1\}\: \forall j\right\}$$.

Hence, $$f(n)=0$$ in CASE I. We notice that for $$\Psi$$ to be well defined it is indeed necessary for $$j_o>1$$.

CASE II: $$j_o=1$$. In this case, all primes appear only once in $$n$$. We also remember that: $$f(n)=\sum_{(i_1,...,i_k)\in\{0,1\}^k}\mu(p_1^{i_1}...p_k^{i_k})\nu(n/p_1^{i_1}...p_k^{i_k})$$ If $$i_1+...+i_k=0$$, we have that these contribute with a $$\nu(n)=k$$ in the sum above. If $$i_1+...+i_k=1$$, we have that these contribute with a $$\sum_i\nu(n/p_i)={k \choose 1} (k-1)$$. Continuing like this it is easy to see we have that:

$$f(n)=\sum_{j=0}^k {k \choose j}(-1)^j (k-j)$$

If $$k=1$$, then $$f(n)=1$$.

Otherwise we can argue with our nice "moment generating functions" which are quite natural if one knows some basic probability:

$$f(n)=\sum_{j=0}^k {k \choose j}(-1)^j (k-j)=\left.\frac{d}{dz}\sum_{j=0}^k {k \choose j}(-1)^j e^{(k-j)z}\right|_{z=0}=$$ $$\left.\frac{d}{dz}(e^z-1)^k\right|_{z=0}=\left.k(e^z-1)^{k-1}e^z\right|_{z=0}=0$$

Hence, case I yields only $$0$$ and case II may yield $$1$$ or $$0$$.