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Section 6.2 The \(d\)- and \(\sigma\)-functions

Definition 6.2.1.

Given \(n\in \mathbb{N},\) let \(d(n)\) denote the number of positive divisors of \(n\) and \(\sigma(n)\) the sum of these divisors; that is,

\begin{equation*} d(n)=\sum_{d\vert n} 1 \hspace{5mm} \mathrm{and} \hspace{5mm} \sigma(n)=\sum_{d\vert n} d. \end{equation*}

For the first \(n \in\mathbb{N}\) we have

\begin{equation*} d(1) = 1, d(2) = 2, d(3) = 2, d(4) = 3, d(5) = 2, d(6) = 4, d(7) = 2,..., \end{equation*}

and

\begin{equation*} \sigma(1) = 1, \sigma(2) = 3, \sigma(3) = 4, \sigma(4) = 7, \sigma(5) = 6, \sigma(6) = 12, \sigma(7) = 8,.... \end{equation*}

This example lets us see a patterns that gives the following theorems.

According to the previous theorem all divisors of \(n\) are precisely those integers \(d\) of the form

\begin{equation*} d = p_1^{a_1}p_2^{a_2}\cdots p_r^{a_r} \end{equation*}

where \(0 \leq a_i\leq k_i.\) There are \(k_1 + 1\) choices for \(a_1, k_2 + 1\) choices for \(a_2,...,\) and \(k_r + 1\) choices for \(a_r.\) Hence, there are

\begin{equation*} (k_1 + 1)(k_2 + 1)\cdots (k_r + 1) \end{equation*}

possible divisors of \(n,\) thus

\begin{equation*} d(n) = (k_1 + 1)(k_2 + 1)\cdots (k_r + 1) = \prod_{i=1}^{r}(k_i+1). \end{equation*}

To find \(\sigma(n),\) consider the product

\begin{equation*} (1 + p_1 + p_1^2+\cdots p_1^{k_1})(1 + p_2 + p_2^2+\cdots p_2^{k_2})\cdots (1 + p_r + p_r^2+\cdots p_r^{k_r}). \end{equation*}

Each positive divisor of \(n\) occurs once and only once as a term of the expansion of the product. Thus

\begin{equation*} \sigma(n) = \prod_{i=1}^{r}(1+p_i+p_i^2+\cdots p_i^{k_i})=\prod_{i=1}^r \frac{p_i^{k_i+1}-1}{p_i-1}, \end{equation*}

since

\begin{equation*} 1 + p_i + p_i^2+\cdots p_i^{k_i}=\frac{p_i^{k_i+1}-1}{p_i-1} \end{equation*}

by a previous theorem.

For the number \(60 = 2^2\cdot 3\cdot 5,\) we have \(k_1 = 2, k_2 = k_3 = 1,\) so that

\begin{equation*} d(60) = (2 + 1)(1 + 1)(1 + 1) = 12. \end{equation*}

These are the divisors

\begin{equation*} 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60. \end{equation*}

The sum of these integers is

\begin{equation*} \sigma(60) = \frac{2^3-1}{2-1}\cdot\frac{3^2-1}{3-1}\cdot\frac{5^2-1}{5-1}=7\cdot 4\cdot 6=168. \end{equation*}

Let \(n\in\mathbb{N}.\) We know that there are \(d(n)\) divisors of \(n\) and for each divisor \(d_i\) there is a \(d_i'\) such that \(n = d_id_i'.\) So if we take the equations

\begin{align*} n\amp =d_1d_1'\\ n\amp =d_2d_2'\\ n\amp =d_3d_3'\\ \amp\vdots \\ n\amp =d_{d(n)}d_{d(n)}' \end{align*}

and multiply the left-hand sides together, and the right-hand sides together, we get

\begin{equation*} n^{d(n)}=\prod_{d\vert n} d\cdot\prod_{d'\vert n} d'. \end{equation*}

But \(\prod_{d\vert n} d=\prod_{d'\vert n} d',\) so that

\begin{equation*} n^{d(n)}=\left(\prod_{d\vert n} d\right)^2. \end{equation*}

which when we take the square root yields the desired result.