Lecture 12 - An Intro to Arithmetic Functions

# Summary

In today's class we finished our discussion of Euler's $\phi$ function, particularly how one goes about computing $\phi(n)$ for an arbitrary integer n. This left us considering a more general class of functions which share some of the properties of the $\phi$ function. We started investigating these so-called arithmetic functions, and finished with a proposition which gave us one method for constructing a "nice" arithmetic function out of an old "nice" arithmetic function.

# Computing Euler's Totient Function

Last class period we finished to computing values of the $\phi$ function for particular classes of integers. We saw

If p is prime, then $\phi(p) = p-1$ and $\phi(p^e) = p^e-p^{e-1}$.

We also found that

If p and q are distinct primes, then $\phi(pq) = \phi(p)\phi(q)$.

On the other hand, we realized that

(1)
\begin{align} \phi(12) = \phi(4)\phi(3) \neq \phi(6)\phi(2). \end{align}

Today we started by noting that the perhaps $\phi(12) \neq \phi(6)\phi(2)$ because $(6,2) \neq 1$, whereas perhaps $\phi(12) = \phi(4)\phi(3)$ since $(4,3) = 1$. As further evidence for this conjecture, note that

(2)
\begin{align} \phi(p^2) = p^2-p^1 \neq \phi(p)\phi(p), \end{align}

presumably because $(p,p) \neq 1$.

This let us conjecture the following

Theorem: If n and m are relatively prime integers, then $\phi(nm) = \phi(n)\phi(m)$.

Before proving this result, notice that it gives the following

Corollary: If $n = p_1^{e_1}\cdots p_k^{e_k}$, then

$\displaystyle \phi(n) = \phi(p_1^{e_1}\cdots p_k^{e_k}) = \phi\left(p_1^{e_1}\right)\cdots \phi\left(p_k^{e_k}\right).$

In particular, this tells us that

(3)
\begin{align} \begin{split} \phi(n) &= \phi\left(\prod_{i=1}^k p_1^{e_1}\cdots p_k^{e_k}\right)\\ &= \phi\left(p_1^{e_1}\right)\cdots\phi\left(p_k^{e_k}\right)\\ &= (p_1^{e_1}-p_1^{e_1-1})\cdots(p_k^{e_k}-p^{e_k-1})\\ &= p_1^{e_1}\cdots p_k^{e_k} \left(1-\frac{1}{p_1}\right)\cdots \left(1-\frac{1}{p_k}\right)\\ &= n \prod_{p \mid n} \left(1-\frac{1}{p}\right). \end{split} \end{align}

#### Example: Computing $\phi(360)$

Since $360 = 2^33^25$, we have

(4)
\begin{align} \phi(360) = \phi(2^3)\phi(3^2)\phi(5) = (2^3-2^2)(3^2-3^1)(5-1) = 96. \end{align}

Alternatively, we could compute $\phi(360)$

(5)
\begin{align} \phi(360) = 360 (1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{5}). \end{align}

Either way works, so feel free to compute $\phi(n)$ using whichever technique you prefer. $\square$

Proof of Theorem: Let $(n,m) = 1$. Using the Chinese Remainder Theorem, we know that an integer a with $1 \leq a \leq nm$ amounts to a pair of integers $a_1,a_2$ so that

(6)
\begin{split} a &\equiv a_1 \mod{n}\\ a &\equiv a_2 \mod{m}. \end{split}

We claim that, in fact, one has $(a,nm) = 1$ if and only if $(a_1,n) =1$ and $(a_2,m) = 1$.

To prove this claim, suppose first that $(a,nm) = 1$. Hence there exist integers x and y so that

(7)
$$xa + ynm = 1.$$

Since $a \equiv a_1 \mod{n}$, there exists some integer k with $a = a_1 + kn$, and plugging this into the above equation yields

(8)
$$x(a_1 + kn) + y(nm) = xa_1 + (xk + ym)n = 1.$$

Since we've written 1 as an integral linear combination of $a_1$ and n, this tells us that $(a_1,n) = 1$. In a similar way, we can also prove that $(a_2,m) = 1$. Hence we know that

(9)
\begin{align} (a,nm) = 1 \LongRightarrow (a_1,n) = (a_2,m) = 1. \end{align}

For the other direction, suppose we know that $(a_1,n) = (a_2,m) = 1$. This means we integers x,y, w and z so that

(10)
\begin{split} xa_1+yn &= 1\\ wa_2 + zm &=1. \end{split}

Since $a \equiv a_1 \mod{n}$ we also know that there exists an integer k so that $a_1 = a+kn$, and likewise $a \equiv a_2 \mod{m}$ means there is some integer s so that $a_2 = a+sm$. Plugging these into the previous equations then gives

(11)
\begin{split} xa_1+yn &= 1\\ wa_2 + zm &=1. \end{split}

Now we can multiply left and right hand sides, combining terms as appropriate, to get

(12)
$$a(xam+xaw+xknw+axmw+wyn) + nm(xkz+xkws+wsy+yz) = 1$$

Since this is an integral linear combination of a and nm which equals 1, we have that $(a,nm) = 1$ as desired.

Now that we've justified our claim, let's use it to prove that

(13)
\begin{align} \phi(nm) = \phi(n)\phi(m). \end{align}

Now the left hand side of this equation counts the numbers a so that $1 \leq a \leq nm$ and so that $(a,nm) = 1$. On the other hand, the previous claim says that a number a with $1 \leq a \leq nm$ and $(a,nm) = 1$ corresponds to numbers $a_1,a_2$ with $1 \leq a_1 \leq n$ and $1 \leq a_2 \leq m$ and so that $(a_1,n) = (a_2,m) = 1$. How many choices of $a_1,a_2$ can we make? By definition, there are $\phi(n)$ many choices for $a_1$ and $\phi(m)$ many choices for $a_2$. Since the choice of $a_1$ can be made independently of the choice of $a_2$, this tells us that we have a total of

(14)
\begin{align} \phi(n)\phi(m) \end{align}

choices for a. This gives the desired equality. $\square$

# Arithmetic Functions

The $\phi$ function is a very powerful number-theoretic function, but it is only one example of a class of functions which number theorists are interested in.

# Arithmetic functions

Having played with Euler's $\phi$ function for a while, we're now ready to think about more functions on the integers. We start with a basic

Definition: A function whose domain is the positive integers is called an arithmetic function.

There are, of course, loads and loads of arithmetic functions. For instance, your old favorite $f(x) = x^2$ becomes an arithmetic function when you only evaluate it on integers. Typically, though, number theorists are more interested in the functions which are connected to number-theoretic properties. For instance, here are a smattering of arithmetic functions which number theorists get really excited about

Function Name What it does
$\phi$ Counts congruence classes relatively prime to n
$\nu$ Counts the number of divisors
$\sigma$ Computes the sum of divisors
$\sigma_k$ Computes the sum of kth powers of divisors

#### Example: Computing the value of some arithmetic functions

Just to get some practice, let's evaluate some of the functions above at $n = 6$

(15)
\begin{split} \phi(6) &= 2\\ \nu(6) &=4 \quad \mbox{since the divisors of 6 are 1,2,3 and 6}\\ \sigma(6) &= 1+2+3+6 = 12 \\ \sigma_2(6) &= 1^2+2^2+3^2+6^2 \end{split}

$\square$

We'll eventually talk about all these functions, but first we want to discuss arithmetic functions more generally.

# Properties of Arithmetic Functions

One of the important properties that nearly all the arithmetic functions we'll deal with share is that they somehow respect multiplication, by which I mean one can often evaluate an arithmetic function for a composite number by evaluating it on primes (or powers of primes).

Definition: An arithmetic function f is called multiplicative if $f(nm) = f(n)f(m)$ whenever $(n,m) = 1$. An arithmetic function g is called completely multiplicative if $g(nm) = g(n)g(m)$ for all pairs of integers n and m.

We have already seen that $\phi$ is a multiplicative function. The benefit of multiplicativity is the following

Lemma: If f is an arithmetic function and $n = p_1^{a_1}\cdots p_k^{a_k}$, then

$f(n) = f(p_1^{a_1})\cdots f(p_k^{a_k})$.

If g is completely multiplicative, then we have

$g(n) = g(p_1)^{a_1}\cdots g(p_k)^{a_k}$

We'll finish our discussion of arithmetic functions by stated the following useful

Proposition: If f is a multiplicative function, then

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

is also a multiplicative function.

It's worth noting that our notation in the above sum means that d ranges over only positive divisors of n. We'll prove this result in class next time, but we'll use it now to verify the following

Corollary: For any integer n,

$\displaystyle n = \sum_{d \mid n} \phi(n)$

#### Example $\sum_{d \mid 12} \phi(d)$

The positive divisors of 12 are 1,2,3,4,6 and 12. So let's compute the sum of $\phi$ on each of these divisors:

(16)
\begin{align} \phi(1) + \phi(2) + \phi(3) + \phi(4) + \phi(6) + \phi(12) = 1 + 1 + 2 + 2 + 2 + 4 = 12. \end{align}

Just as the lemma predicted!$\square$

Proof of Corollay:
We'll start by letting $F(n)$ be the function

(17)
\begin{align} F(n) := \sum_{d \mid n} \phi(d). \end{align}

By the previous proposition, we know that $F(n)$ is multiplicative, and therefore it follows that if $n = p_1^{e_1}\cdots p_k^{e_k}$, we have

(18)
\begin{align} F(n) = F(p_1^{e_1})\cdots F(p_k^{e_k}). \end{align}

Therefore we can compute $F(n)$ simply by evaluating F at prime powers.

So let's do it. Let p be prime, and let's compute $F(p^e)$. Since the sum in the expression for $F(p^e)$ runs over positive divisors of $p^e$, we need to start by listing these divisors. Since p is prime, the divisors are just the powers of p between 1 and $p^e$: $\{1,p,p^2,\cdots,p^e\}$. Hence

(19)
\begin{split} F(p^e) &= \sum_{d \mid p^e} \phi(d) \\ &= \phi(1)+\phi(p)+\phi(p^2)+\phi(p^3) + \cdots + \phi(p^{e-1} + \phi(p^e)\\ &= 1 + (p-1) + (p^2-p) + (p^3-p^2) + \cdots + (p^{e-1}-p^{e-2}) + (p^e-p^{e-1}). \end{split}

In this last sum, notice that we get lots of cancellation: the 1 from $\phi(1)$ cancels the -1 from $\phi(p)$, the p in $\phi(p)$ is cancelled by the -p in $\phi(p^2)$, and so on down the line. The only thing that survives the massacre is the $p^e$ term from $\phi(p^e)$, and so we have

(20)
$$F(p^e) = p^e.$$

With the value of F at prime powers determined, we can now go back and compute

(21)
\begin{split} F(n) &= F(p_1^{e_1})\cdots F(p_k^{e_k}) \\ &= p_1^{e_1}\cdots p_k^{e_k} \\ &=n. \end{split}

$\square$