Dirichlet, Möbius and Primes

Question

Is it correct that for the Möbius function, for $p$ prime and $m>1$, $μ(p^m)= 0$ because there are repeated prime factors? So how would I use this if I have $f = μ ∗ μ$ to find $f(p^m)$ using Dirichlet convolution? Could that then help me to find $(1 ∗ 1)(p^m)$ where $1$ is the constant function: $1(n)=1$?

Answer 1

We have $\mu(p)=-1$ and only for $m\ge 2$ (not for $m>0$) we have $\mu(p^m)=0$.

Recall that $(f*g)(n)=\sum_{d\mid n}f(d)g(n/d)=\sum_{ab=n}f(a)g(b)$, so $$f(n)=(\mu*\mu)(n)=\sum_{ab=n}\mu(a)\mu(b).$$ Apparently only those summands where both $a$ and $b$ are square-free contribute. Thus, if $n$ is divisible $p^m$ with $m>2$ for some prime $p$, certainly $f(n)=0$. We can directly verify that $f(p)=-1$, $f(p^2)=1$ and $f(p^m)=0$ for $m>2$, and then obtain $f$ by multipicativity: $f(n)=(-1)^k$ where $k$ is the number of primes occuring to exactly the first power in $n$, except that $f(n)=0$ if $n$ is divisible by a nontrivial cube.

For the other part, $$(1*1)(p^m)=\sum_{ab=p^m}1 = m+1 $$ and moregenerally, $(1*1)(n)$ is the number of divisors of $n$.