How to show that

Question

If $f$ is a multiplicative function proof that:

i) $f^{-1}(p^{2})= [f(p)]^{2}-f(p^{2})$

ii) $f$ is completely multiplicative $\Longleftrightarrow f^{-1}(p^{\alpha}) = 0; \forall p $ prime $\alpha \geq 2$

Answer 1

The Dirichlet inverse is what is meant here. It can be computed directly from the definition $\sum_{d\mid n} f^{-1}(d)f(n/d)=\delta_{1,n}$, by setting $n=1,p,p^2$.

We can also calculate its generating function, which is $\prod_p (1+\frac{f^{-1}(p)}{p^s}+\frac{f^{-1}(p^2)}{p^{2s}}+\cdots) = \prod_p (1+\frac{f(p)}{p^s}+\frac{f(p^2)}{p^{2s}}+\cdots)^{-1} = \prod_p (1-\frac{f(p)}{p^s}+\frac{f(p)^2 - f(p^2)}{p^{2s}} + \cdots)$