Lecture 12: An Introduction to Arithmetic Functions


Today we talked a little more about the Euler $\phi$ function, then moved on to talk about arithmetic functions more generally. We saw a few examples of arithmetic functions, but we spent the majority of our time discussing the properties these functions have — especially when they respect multiplication.

Summing Up Euler's Phi

As a final remark on the Euler $\phi$ function, we give the following

Lemma: For any integer n, we have

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

Proof: We start the proof by partitioning the integers between 1 and n into groups based on their gcds. Namely, we define

\begin{align} S_d = \{a: 1 \leq a \leq n \mbox{ and }(a,n) = d\}. \end{align}

Since every integer between 1 and n belongs to exactly one set $S_d$, we know that

\begin{align} \sum_{d \mid n} |S_d| = n. \end{align}

Now we notice that elements a of the set $S_d$ must have the property $(\frac{a}{d},\frac{n}{d}) = 1$ — after all, once we divide through by the gcd of two elements, these quotients are relatively prime to each other. Hence every element of $S_d$ corresponds to an element between 1 and $\frac{n}{d}$ which is relatively prime to $\frac{n}{d}$. How many such elements are there? Exactly $\phi\left(\frac{n}{d}\right)$; after all, that's exactly what Euler's $\phi$ function counts! Hence we have

\begin{align} n = \sum_{d \mid n} |S_d| = \sum_{d \mid n} \phi\left(\frac{n}{d}\right). \end{align}

To finish off the proof, we notice that as d runs through divisors of n, so too does the quantity $\frac{n}{d}$. Hence the final sum in our expression is actually the same as $\sum_{d \mid n} \phi(d)$ — it is merely adding the sum numbers up in the opposite order. $\square$


In class we verified this by calculating the value of $\phi$ on divisors of 12. We found

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

We also verified that for each divisor d, the set $S_d$ had $\phi\left(\frac{12}{d}\right)$ elements. For instance, $S_3 = \{3,9\}$, and $\phi\left(\frac{12}{3}\right) = \phi(4) = 2.$ $\square$

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

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

One of the important properties that nearly all the arithmetic functions we'll deal with share is that they somehow respect multplication, 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

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

To prove either of these results, just use induction and the definition of multiplicativity (or complete multiplicativity).

An Interesting property of multiplicative functions

We'll finish our discussion of arithmetic functions in much the same way we finished our discussion of Euler's $\phi$ function.

Theorem: If f is a multiplicative function, then

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

is also a multiplicative function.

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, and I won't prove it here. This is good practice for the exam, so try proving it yourself!

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}


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