Theorem   Let $f$ be a multiplicative function with convolution inverse $g$ Then $f$ is completely multiplicative if and only if $g(p^k)=0$ for all primes $p$ and for all $k \in \mathbb{N}$ with $k>1$

Proof. Note first that, since $f(1)=1$ and $f*g=\varepsilon$ where $\varepsilon$ denotes the convolution identity function, then $g(1)=1$ Let $p$ be any prime. Then

$$0=\varepsilon(p)=(f*g)(p)=f(1)g(p)+f(p)g(1)=g(p)+f(p).$$

Thus, $g(p)=-f(p)$

Assume that $f$ is completely multiplicative. The statement about $g$ will be proven by induction on $k$ Note that:

Let $m \in \mathbb{N}$ with $m>2$ such that, for all $k \in \mathbb{N}$ with $1<k<m$ $g(p^k)=0$ Then:

Conversely, assume that $g(p^k)=0$ for all $k \in \mathbb{N}$ with $k>1$ The statement $f(p^k)=(f(p))^k$ will be proven by induction on $k$ The statement is obvious for $k=1$ Let $m \in \mathbb{N}$ such that $f(p^{m-1})=(f(p))^{m-1}$ Then:

Thus, $f(p^m)=(f(p))^m$ It follows that $f$ is completely multiplicative.