The question is to prove:
$\sum_{\phi(n)=k}\mu(n) = 0$
where $\phi(n)$ is the Euler totient function and $\mu(n)$ is the Mobius function.
I have tried various approaches but nothing seems to be working. I am not sure but someone recommended using the fact that $\sum_{d|n}\mu(d)*n/d = \phi(n)$