Finding formulas for sums

Question

I know that $\sum_{d \mid n} \mu(d) = 0$ whenever $n >1$, and I know that $\sum_{d \mid n} \phi(d) = n$. How can I use this in order to give a formula for $\sum_{d \mid n} \mu(d)\phi(d)$?

Answer 1

Since I don't immediately know a solid convolution for the thing you have written, we'll do this the old-fasioned way:

By multiplicativity we write:

$$\sum_{d|n}\mu(d)\phi(d)=\prod_{p|n}\left(\sum_{\alpha}\mu(p^\alpha)\phi(p^\alpha)\right) = \prod_{p|n}\left(1-(p-1)\right)=\prod_{p|n}(2-p)$$

so your answer is just

$$\prod_{p|n}(2-p).$$