The Möbius function is the inverse of the constant function 1

Statement

Let $\mathbf{1}$ be the arithmetic function with $\mathbf{1}(n) = 1$ for all $n$, and let $\varepsilon$ be the function with $\varepsilon(1) = 1$ and $\varepsilon(n) = 0$ for $n > 1$. Then $\mu * \mathbf{1} = \varepsilon$, where $\mu$ is the Möbius function and $*$ is the Dirichlet convolution. That is, for each positive integer $n$, $$\sum_{d \mid n} \mu(d) = \begin{cases} 1 & \text{if } n = 1, \\ 0 & \text{if } n > 1. \end{cases}$$

Remarks

The proof for $n > 1$: let $k = \omega(n) \ge 1$. Only squarefree divisors count, so the sum is $\sum_{j=0}^{k} \binom{k}{j} (-1)^j = (1 - 1)^k = 0$. For example, the divisors of $12$ give $\mu(1) + \mu(2) + \mu(3) + \mu(6) = 1 - 1 - 1 + 1 = 0$. This identity is the key step in the Möbius inversion formula.

Explanations

L1 human

The signs $\mu(d)$ over the divisors of $n$ cancel out, except for $n = 1$. This is the same cancellation as in the principle of inclusion and exclusion.