Lecture 13: Counting Divisors

Summary

In today's class we started by giving an alternate proof of the fact that $\sum_{d \mid n} \phi(d) = n$. We then discussed the function $\nu(n)$, which counts the number of positive divisors of n. We saw that the function is multiplicative (but not completely multiplicative), and we used this function to give an expression for the product of divisors of a number.

Using Multiplicativity to Compute an Arithmetic Function

Last class period we finished by showing that whenever f is a multiplicative function, so too is the function $F(n) = \sum_{d \mid n}f(d)$. We have already seen such a function when we were dealing with the Euler $\phi$ function; indeed, we proved that $\sum_{d\mid n} \phi(d)=n$. We're going to now give an alternate proof of that fact. Our strategy is to use the fact that $F(n) = \sum_{d\mid n}\phi(d)$ is multiplicative to evaluate F easily.

Since we know that F is multiplicative, we know that F is determined by its values on prime powers. So let's compute $F(p^a)$. We have

(1)

\begin{align} F(p^a) = \sum_{d \mid p^a} \phi(d). \end{align}

Notice, however, that the divisors of $p^a$ are just the numbers $1,p,p^2,\cdots,p^a$. Hence we have

(2)

\begin{align} \sum_{d \mid p^a} \phi(d) = \sum_{i=0}^a \phi(p^i). \end{align}

We already know that $\phi(p^i) = p^i - p^{i-1}$ and $\phi(1) = 1$, and hence we have

(3)

\begin{align} \sum_{i=0}^a \phi(p^i) = \sum_{i=0}^a p^i-p^{i-1} = (p^a-p^{a-1})+(p^{a-1}-p^{a-2})+\cdots+(p^2-p)+(p-1) + 1. \end{align}

This sum "telescopes" to just be $p^a$, and so we get $F(p^a) = p^a$.

Hence we have

(4)

\begin{align} F(n) = F(p_1^{a_1}\cdots p_k^{a_k}) = F(p_1^{a_1})\cdots F(p_k^{a_k}) = p_1^{a_1}\cdots p_k^{a_k}. \end{align}

The Divisor Counting Function

Having dealt with arithmetic functions in the abstract, it's now time for us to start introducing some specific arithmetic functions. Today our function will be the function that counts the number of divisors of a given integer.

Definition: The function $\nu(n)$ is defined as the number of positive divisors of n.

We'll begin by showing that $\nu$ is multiplicative. You might think that this would be tricky, but in fact it's super simple.

Lemma: The function $\nu(n)$ is multiplicative.

Proof: Notice that we have $\nu(n) = \sum_{d \mid n} 1$. Notice also that the constant function $f(n) = 1$ is a multiplicative function (you have this as homework!). By our lemma concerning functions of the form $F(n) = \sum_{d\mid n}f(n)$, we have that $\nu(n)$ is multiplicative. $\square$

Now that we know $\nu$ is multiplicative, we can use this fact to give an easy proof of its value on a given integer n.

Theorem: If $n = p_1^{a_1}\cdots p_k^{a_k}$, then $\nu(n) = \prod_{i=1}^k (a_i+1)$.

Proof: To see that this is true, notice that since $\nu$ is multiplicative we only need to know how to evaluate $\nu$ on prime powers. But notice that for a prime power $p^a$, the only divisors of $p^a$ are $1,p,p^2,\cdots,p^a$. There are $a+1$ many divisors, and so we have $\nu(p^a) = a+1$. Hence we can compute

(5)

\begin{align} \nu(n) = \nu(p_1^{a_1}\cdots p_k^{a_k}) = \nu(p_1^{a_1})\cdots \nu(p_k^{a_k}) = (a_1+1)\cdots(a_k+1) = \prod_{i=1}^k(a_i+1). \end{align}

$\square$

The Product of Divisors of a Function

At the end of class you all split up into groups and tried to come up with a simple formula for

(6)

\begin{align} \prod_{d \mid n} d. \end{align}