Exercise 4.3.5 ($\sum_{d \mid n} | \mu(d) | = 2^{\omega(n)}$)

Question

Prove that for every positive integer $n$, $\sum_{d \mid n} | \mu(d) | = 2^{\omega(n)}$.

Richard Ganaye · Accepted Answer

{\bf First Proof.}
\begin{proof} 
As in the alternative formulation of the proof of Theorem 4.7, consider those square-free divisors $d$ of $n$ with exactly $k$ prime factors. There are $\binom{\omega(n)}{k}$ such divisors, each contributing $|\mu(d)| = 1$. Thus by the binomial theorem, the sum in question is
\begin{align*}
\sum_{d \mid n} | \mu(d) |& = \sum_{k=0}^{\omega(n)} \binom{\omega(n)}{k} \
&= (1 + 1)^{\omega(n)}\
&= 2^{\omega(n)}.
\end{align*}
\end{proof}

{\bf Second Proof.}
\begin{proof}
We define for all positive integers $n$
$$F(n) = \sum_{d \mid n} | \mu(d) | ,\qquad G(n) = 2^{\omega(n)}.$$

Since $\mu$ is a multiplicative function, so is $|\mu|$, and by Theorem 4.4, $F$ is a multiplicative function. If $n \wedge m = 1$, then $\omega(nm) = \omega(n) + \omega(m)$, thus $G$ is a multiplicative function.

Moreover, if $p$ is a prime number, $G(p^0) = G(1) = F(1) = F(p^0)$, and if $\alpha >0$,
$$F(p^\alpha) = \sum_{i=0}^\alpha |\mu(p^i)| = 1 + |\mu(p)| = 2,$$
and $\omega(p^\alpha) = 1$, so
$$G(p^\alpha) = 2.$$
Since $F$ and $G$ are multiplicative functions, this shows that $F = G$, so for all poisitive integers $n$,
$$\sum_{d \mid n} | \mu(d) | = 2^\omega(n).$$

\end{proof}

Exercise 4.3.5 ($\sum_{d \mid n} | \mu(d) | = 2^{\omega(n)}$)

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Exercise 4.3.5 ($\sum_{d \mid n} | \mu(d) | = 2^{\omega(n)}$)

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