Let k be a positive integer. I need to prove that the sum $$\sum_{n \in \mathbb{N} : \phi(n)=k} \mu(n)=0$$ where $\phi$ is the Euler Totient function and $\mu$ is the Mobius function.
Here is what I have: $n:\phi(n)=k$ has to be written as the product of primes, where these are selected from a finite set of primes such that $p-1$ divides $k$. Now, I wanted to prove that for every $n$ such that $\mu(n)=-1$ there is an $m$ such that $\mu(m)=1$. That is, for every n that can be written as a multiplication of an even number of these primes (each of multiplicity 1), there is an m that can be written as a multiplication of an odd number of these primes (each of multiplicity $1$).
Fix $k$ and consider the set $S_k$ of square free natural numbers $n$ such that $\varphi(n)=k$. Note: we only care about the square free $n$ as the rest only contribute $0$ to the sum. There is a pairing within $S_k$ which maps the even members to the odd ones: pick $n\in S_k$ if $n$ is odd, pair it with $2n$. If $n$ is even pair it with $\frac n2$ (as $n$ is square free, $n$ even implies that $\frac n2$ is odd). We note that in each such pair, one member of the pair has an even number of prime factors while the other has an odd number of prime factors. This suffices to prove your claim.