2

Let $\mu$ be the Möbius function, and let $\nu(n)$ be the number of distinct prime factors of $n$. Then we can define $p = \mu * \nu$, i.e. $$ p(n) = \sum_{d \mid m} \mu(d) \nu(n/d). $$ An exercise in an introductory textbook I already put back on the library shelf and can't find anymore asks:

Prove that $p$ only takes the values 0 and 1.

The obvious solution seems to be through Möbius inversion: if we define $q(n)$ to be 1 when $n$ is prime and $0$ otherwise, then $\nu = 1 * q$ so that $q = \mu * \nu = p$. However, the formulation of the question suggests that there is an easy way to prove that $p$ only takes values $0, 1$ without identifying it entirely.

How might one do that?

Mees de Vries
  • 27,550

1 Answers1

0

There is the generalization of your first solution, showing $\mu \ast f$ for additive functions is always easy to find :

Let $f(n)$ be additive that is $f(nm) =f(n)+f(m)$ for $gcd(n,m)=1$.

Let $g(p^k) = f(p^k)-f(p^{k-1})$ and $g(n) = 0$ if $n$ isn't a prime power

Then $f(n) = \sum_{p^k \| n} f(p^k) = \sum_{p^k | n} g(p^k) = \sum_{d | n} g(d)$

Whence $\sum_{d | n} \mu(d) f(n/d)= g(n)$

reuns
  • 79,880
  • This is a nice observation, but it doesn't really answer my question. – Mees de Vries Feb 01 '19 at 21:07
  • @MeesdeVries Can you rephrase your question to make clear how it doesn't answer to it ? Your function is additive thus we can easily find the values taken by $\mu\ast f$ – reuns Feb 01 '19 at 21:08
  • But I already know what $p$ is -- it's in the question body. I'm specifically asking for an argument that does not show what $p$ is precisely, but only demonstrates the fact that $p$ only takes the values $0, 1$ because I'm curious what the intention was of the person who set the question. – Mees de Vries Feb 01 '19 at 21:10
  • @MeesdeVries I think you asked what special property of $\nu$ implies $\sum_{d | n} \mu(d) \nu(n/d) \in {0,1}$ and the answer is that $\nu$ is additive with $\nu(p^k) - \nu(p^{k-1}) \in {0,1}$. – reuns Feb 01 '19 at 21:16
  • Yes, but from that statement it is also immediate that $p$ is the indicator function of the primes. The question in the textbook did not ask what $\mu * \nu$ was, only to prove that one property. That suggests to me that there should be some easy, by-inspection method to see that $p$ should only take the values $0, 1$, without identifying the function entirely. (I could be wrong of course, but that would just mean my question has no answer.) – Mees de Vries Feb 01 '19 at 21:19