I'm trying to prove $\sum_{m|n} \mu(m)^2/\phi(m) = n/\phi(n)$
My first realization was that $\mu(m)^2 = 1$ iff $m=1$ or $m$ is a squareless factor of $n$ and otherwise is 0. Let $\{1,m_1,m_2, ... m_k\}$ be the set of squarefree factors of $n$.
So now we have a sum $\frac{1}{\phi(1)} +\frac{1}{\phi(m_1)} + \cdots +\frac{1}{\phi(m_k)}.$
How do I show the sum equals $n/\phi(n)$? Or am I going in the wrong direction?