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# completely multiplicative functions whose convolution inverses are completely multiplicative

###### Corollary 1.

The only completely multiplicative function whose convolution inverse is also completely multiplicative is $\varepsilon$, the convolution identity function.

###### Proof.

Let $f$ be a completely multiplicative function whose convolution inverse is completely multiplicative. By this entry, $f\mu$ is the convolution inverse of $f$, where $\mu$ denotes the Möbius function. Thus, $f\mu$ is completely multiplicative.

Let $p$ be any prime. Then

$\begin{array}[]{rl}(f(p))^{2}&=(f(p))^{2}(-1)^{2}\\ \\ &=(f(p))^{2}(\mu(p))^{2}\\ \\ &=(f(p)\mu(p))^{2}\\ \\ &=f(p^{2})\mu(p^{2})\\ \\ &=f(p^{2})\cdot 0\\ \\ &=0.\end{array}$

Thus, $f(p)=0$ for every prime $p$. Since $f$ is completely multiplicative,

$f(n)=\begin{cases}1&\text{if }n=1\\ 0&\text{if }n\neq 1.\end{cases}$ |

Hence, $f=\varepsilon$. ∎

## Mathematics Subject Classification

11A25*no label found*

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