Mobius inversion problem

Question

Prove by Mobius inversion formula if $\frac{n}{\phi(n)}=\sum_{d\mid n} f(d)$ then $f(d)=\frac{\mu^2(d)}{\phi(d)}.$

Answer 1

We assume here that $f$ is multiplicative. By Möbius inversion, $$f(n)= \sum_{d \mid n} \frac{n\mu(d)}{d\phi(n/d)}.$$ Since both sides are multiplicative, it is enough to compute $f$ on prime powers. If $r \geq 2$, we find $$f(p^r) = \frac{p^r}{\phi(p^r)} - \frac{p^{r-1}}{\phi(p^{r-1})} =\frac{p^r-p^r}{\phi(p^r)}=0= \frac{\mu^2(p^r)}{\phi(p^r)},$$ while if $r=1$, we find $$f(p) = \frac{p}{p-1} - 1 = \frac{1}{p-1} = \frac{\mu^2(p)}{\phi(p)}.$$

Extending by multiplicativity, we get that $$f(d)= \frac{\mu^2(d)}{\phi(d)}.$$