Lecture 13 - Counting Divisors


We started today by finishing a proof of a proposition mentioned last class period which lets us generate new multiplicative functions from old multiplicative functions. The benefit of this proposition is larger than one might expect; for example, it gives us a quick proof that the function $\nu$ is multiplicative. After we discovered $\nu$ is multiplicative, we used this fact to prove a handful of nice properties about the function.


The exam is next Wednesday, so we'll spend Monday's class reviewing for the test. You should come to class on Monday prepared to ask lots of questions, because I won't be preparing anything for the review. Also, if you'd like to show up 5 minutes before noon on Wednesday, I'll give you the chance to read over the test for a while before you actually start the midterm. This will give you a few minutes to plan a strategy of attack. If for some reason you can't show up to class by 11:55 but would still like to have 5 minutes to read the test without writing, I can arrange for this. If this is the case, though, you need to let me know by Monday so I can make appropriate arrangements.

Finishing a proof

Last class period we stated the following

Proposition: If f is a multiplicative function, then

$F(n) = \sum_{d \mid n} f(n)$

is also a multiplicative function.

Today we'll start class by proving this result. In order to prove this result, we'll need the following

Lemma: If $(m,n) = 1$ and d is a divisor of mn, then there exists divisors $d_1$ of n and $d_2$ of m which are relatively prime to each other and whose product is d. Hence divisors of mn corresponds to pairs of divisors $d_1,d_2$ of n and m (respectively).

We didn't prove this result in class, but you have it assigned for this week's homework.

Proof of Theorem: If we want to show that F is arithmetic, then we need to show that for any relatively prime integers n and m, we have

\begin{equation} F(nm) = F(n)F(m). \end{equation}

With this as our goal, we'll go ahead and try to compute $F(nm)$.

First, notice that by definition we have

\begin{align} F(nm) = \sum_{d\mid n}f(d). \end{align}

Now we can use our lemma above to express each divisor d of nm as a product of (relatively prime) divisors of n and m:

\begin{align} \sum_{d\mid n}f(d) = {\sum_{d_1\mid n}}_{d_2 \mid m}f(d_1d_2). \end{align}

Now since f is arithmetic, we know that $f(d_1d_2) = f(d_1)f(d_2)$. Hence, after reexpressing as a double sum, we get

\begin{align} {\sum_{d_1\mid n}}_{d_2 \mid m}f(d_1d_2) = \sum_{d_1 \mid n}\sum_{d_2 \mid m} f(d_1)f(d_2). \end{align}

Finally, notice that the quantity $f(d_1)$ is not dependent on $d_2$, and hence we can pull this quantity out of the second sum. This leaves us with

\begin{align} \sum_{d_1 \mid n}\sum_{d_2 \mid m} f(d_1)f(d_2)= \sum_{d_1 \mid n}f(d_1)\sum_{d_2 \mid m} f(d_2). \end{align}

But this is just $F(n)F(m)$, and so we've proven

\begin{equation} F(nm) = F(n)F(m). \end{equation}


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 completely multiplicative function. This is true because for integers any integers a and b with $(a,b) = 1$, we have

\begin{align} f(ab) = 1 = 1\cdot 1 = f(a)f(b). \end{align}

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

\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}


Example: Playing with our $\nu$ function

At the beginning of the course, Nick posted the following question: Suppose that a number n has 27 divisors and that $9 \mid n$ and $16 \mid n$. What is n? Today we can give a good answer to this question.

First, note that our hypothesis on the number of divisors says $\nu(n) = 27$. Now our number is divisible by both 2 and 3, so the prime factorization of n is $n = 2^{e_1}3^{e_2}p_3^{e_3}\cdots p_k^{e_k}$, where we have $e_1 \geq 4$ and $e_2 \geq 2$. Our formula for evaluating $\nu$ then says

\begin{align} \nu(n) = 27 = \prod_{i=1}^k (e_i+1). \end{align}

Since both $2 \mid n$ and $3 \mid n$, the product on the right hand side has at least 2 factors. On the other hand, we know that the factor $(e_1 + 1) \geq 5$ (since $e_1 \geq 4$), and likewise $(e_2+1) \geq 3$. Since the product of these two terms is already at least 15, and since another prime divisor would contribute at least another factor of 2 to $\nu(n)$, we know that there can't be any more prime divisors of n aside from 2 and 3.

Hence we have $n = 2^{e_1}3^{e_2}$, where

\begin{equation} 27 = (e_1+1)(e_2+1). \end{equation}

Both of the factors on the right side are non-trivial (i.e., at least 1), and since 27 has only one nontrivial factorization (as $27 = 9\cdot 3$), we must have that one of $e_1+1$ or $e_2+1$ is 9, while the other is 3. Since we alread know that $e_1+1 \geq 5$, this forces $e_1+1 = 9$ and $e_2+1 = 3$. Hence we get $e_1 = 8$ and $e_2 = 2$, so that

\begin{equation} n = 2^{8}3^2. \end{equation}


Example: Multiplying all divisors

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

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

Many groups were able to conjecture that this number should be

\begin{align} n^{\frac{\nu}{2}}. \end{align}

To prove this result, notice that for each divisor d of n there is a "partner" divisor given by $\frac{n}{d}$, and that each such pair multiplies to give

\begin{align} d\left(\frac{n}{d}\right) = n. \end{align}

Provided that every such pair consists of distinct numbers, this means we can group our $\nu(n)$ divisors into $\frac{\nu(n)}{2}$ pairs, each with product n. Multiplying all these pairs together then gives

\begin{align} \prod_{d \mid n} d = n^{\frac{\nu}{2}}. \end{align}

The only time this might fail is if there is a divisor d so that $d = \frac{n}{d}$. But this occurs only when $d^2 = n$, which means that n is a perfect square and that d is the divisor $\root\of{n}$. In that case, we can group the remaining $\nu(n)-1$ divisors into pairs whose product is n, and so we get

\begin{align} \prod_{d \mid n} d = \left(\prod_{\mbox{\tiny{pairs }}\left\{d,\frac{n}{d}\right\}} n\frac{n}{d}\right)\root\of{n} = n^{\frac{\nu(n)-1}{2}}n^{\frac{1}{2}} = n^{\frac{\nu(n)}{2}}. \end{align}

Hence our formula works out in both cases.$\square$

Implicit in the previous problem is determining when $\nu(n)$ is even (in which case $\nu(n)/2$ is an integer) or odd (in which case $(\nu(n)-1)/2$ is an integer. We have the following

Lemma: A number n has $\nu(n)$ odd if and only if ever exponent in the prime factorization of n is even. Equivalently, $\nu(n)$ is odd if and only if n is a perfect square.

Proof: The fact that perfect squares are precisely those numbers for which every exponent in the prime factorization just comes from rules about exponenents; if $n = p_1^{2f_1}\cdots p_k^{2f_k}$, then $n = \left(p_1^{f_1}\cdots p_k^{f_k}\right)^2$. So we'll now focus on the first statement.

Suppose that $n = \prod_{p_1}^{e_1}\cdots p_k^{e_k}$, so that $\nu(n) = \prod_{i=1}^k(e_i+1)$. Now if even one of the $e_j$ is odd, then the corresponding factor $e_j+1$ is even, and therefore the whole product $\prod_{i=1}^k(e_i+1)$ is even. On the other hand, if all the exponents are even, then each of the $e_i+1$ is odd, and so the whole product $\prod_{i=1}^{k} (e_i+1)$ is also odd.$\square$

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