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Section 6.4 The Möbius Function

Definition 6.4.1.

For \(n \in\mathbb{N},\) the Möbius function, \(\mu\) is defined by

\begin{equation*} \begin{cases} 1 \amp \hspace{5mm} if \hspace{2mm} n=1 \\ 0 \amp \hspace{5mm} if \hspace{2mm} p^2\vert n \hspace{2mm} for \hspace{2mm} some \hspace{2mm} prime \hspace{2mm} p \\ (-1)^r \amp \hspace{5mm} if \hspace{2mm} n=p_1p_2\cdots p_r, \hspace{2mm} where \hspace{2mm} p_i \hspace{2mm} are \hspace{2mm} distinct \hspace{2mm} primes. \end{cases} \end{equation*}

Stated in words, we have that \(\mu\) is a function with the property that \(\mu(1)=1,\) if \(n\) is not square-free then \(\mu(n)=0,\) and if \(n\) is square-free then \(\mu(n)= (-1)^r\) where \(n\) has \(r\) prime factors.

We have \(\mu(60) = 0\) since \(2^2\vert 60,\) but \(\mu(30) =-1,\) since \(30 = 2\cdot 3\cdot 5.\) For the first few values we have

\begin{equation*} \mu(1) = 1, \mu(2) = -1, \mu(3) = -1, \mu(4) = 0, \mu(5) = -1, \mu(6) = 1,... \end{equation*}

Remark 6.4.3.

It is clear from the definition that for \(p\) a prime

\begin{equation*} \mu(p)=-1 \hspace{5mm} \mathrm{and} \hspace{5mm} \mu(p^k)=0 \hspace{2mm} (k\geq 2). \end{equation*}

Let \(m, n \in\mathbb{N}\) with \(\gcd(m, n) = 1.\) If \(p^2\vert mn\) for some \(p,\) then either \(p^2\vert m\) or \(p^2\vert n.\) In either case we have \(\mu(mn) = 0,\) and either \(\mu(n) = 0\) or \(\mu(m) = 0,\) so that

\begin{equation*} \mu(mn) = 0 = \mu(m)\mu(n). \end{equation*}

So assume that \(m\) and \(n\) are square free, say \(m = p_1p_2\cdots p_r\) and \(n =q_1q_2\cdots q_s,\) where \(p_i\) and \(q_i\) are all distinct. Then

\begin{equation*} \mu(mn) = \mu(p_1p_2\cdots p_rq_1q_2\cdots q_s) = (-1)^{r+s} = (-1)^r(-1)^s = \mu(m)\mu(n), \end{equation*}

so that \(\mu\) is multiplicative.

If we consider \(F(n) = \sum_{d\vert n}\mu(d),\) we see something very interesting.

First if \(n = 1,\) then

\begin{equation*} F(1) =\sum_{d\vert 1}\mu(d)=\mu(1)=1. \end{equation*}

If \(n = p^k\) for some \(k \geq 1,\) then

\begin{equation*} F(p^k)=\sum_{d\vert p^k} \mu(d)=\mu(1)+\mu(p)+\mu(p^2)+\cdots +\mu(p^k)=1+(-1)+0+\cdots +0=0. \end{equation*}

Recall that \(F\) is a multiplicative function, since \(\mu\) is multiplicative. Thus if \(n\gt 1,\) then \(n = p_1^{k_1}p_2^{k_2}\cdots p_r^{k_r},\) and so

\begin{equation*} F(n)=F(p_1^{k_1})F(p_2^{k_2})\cdots F(p_r^{k_r})=0 \end{equation*}

since at least one of the \(k_i\)'s is at least \(1.\) Thus we have the following theorem.

Let \(n = 15,\) then

\begin{equation*} M(15) = \sum_{d\vert 15} \mu(d) = \mu(1) + \mu(3) + \mu(5) + \mu(15) = 1 + (-1) + (-1) + 1 = 0. \end{equation*}

This following conjecture is worth \($1,000,000 US\) if you can prove it (or disprove it). But don't get your hopes up; it's been an open question for almost \(150\) years.

This conjecture seems out of reach at this point in time. In fact, even proving that

\begin{equation*} \lim_{n\rightarrow\infty}\frac{\mu(1)+\mu(2)+\cdots +\mu(n)}{n^{\vartheta+\varepsilon}}=0, \end{equation*}

for some \(\vartheta\lt 1,\) has yet to be tackled. This would be a very powerful new result and would validate a large number of conjectures.