# Summary

We started class by going over the midterm grades. We then discussed $\sigma$, the function which adds up the divisors of a given integer *n*. We saw that $\sigma$ was multiplicative and we gave a formula for evaluating $\sigma(n)$ based on a prime factorization of *n*. We then collected together a few interesting results about this important function.

# Midterm results

In class today I handed back the first midterms, and I was generally quite impressed with everyone's scores. If you feel that your score doesn't accurately reflect your understanding in the course, don't despair! There is ample opportunity to pick up your grade if you start working on it now. In particular, make sure that you're studying on a day-to-day basis so that you can pick up the "easy points" on the next exam, thus leaving yourself more time for problems that seem more difficult. Here's the breakdown on scores

Raw Score | Curved Score | Students in Range (pre-bonus) |
---|---|---|

88-100 | 90 - 100 | 3 |

78-87 | 80-89 | 7 |

67 - 76 | 70-79 | 7 |

As usual, if you have any questions, don't be afraid to ask.

# The Sigma Function

Today in class we're going to introduce a new arithmetic function: the $\sigma$ function.

Definition: For a given integer

n, the $\sigma$ function is gives the sum of the divisors ofn.

For example, the divisors of 10 are $\{1,2,5,10\}$, and hence $\sigma(10) = 18$.

As with all the arithmetic functions we've considered before, the first thing we want to know about this new function is whether or not it is multiplicative.

Proposition: The $\sigma$ function is multiplicative.

Proof: We have a result from a while back that shows any multiplicative function *f* can be used to define another multiplicative function $F(n) = \sum_{d\mid n} f(d)$. Notice that the function $f(d)=d$ is multiplicative (you showed this in your homework), and hence we have

is also multiplicative. $\square$

This is a huge advantage when it comes to computing the value of $\sigma$ for a particular number, since we now know that $\sigma$'s value can be determined by examining what it does to prime powers. With that as our motivation, notice that since the divisors of a prime power $p^a$ are given by $\{1,p,p^2,\cdots,p^a\}$, we have

(2)This is a geometric series, so we can use the formula for the sum of a geometric series to compute

(3)Corollary: For a number $n = p_1^{a_1}\cdots p_k^{a_k}$, we have

$\displaystyle \sigma(n) = \prod_{i=1}^k \frac{p_i^{a_i+1}-1}{p_i-1}$.

#### Example: Summing divisors of a googol

Suppose we wanted to compute $\sigma(10^{100})$. Doing this by writing out all divisors would be IMPOSSIBLE, since there are way too many divisors to try adding up one at a time. Fortunately, we can use our formula above:

(4)$\square$

# Interesting Properties of the Sigma Function

Now that we know how to evaluate the $\sigma$ function, let's try to come up with a few interesting properties of the $\sigma$ function.

## Detecting Primality

The first such property will be a result that tells us the $\sigma$ function can be used to characterize prime numbers.

Lemma: A number

nis composite if and only if $\sigma(n) > n+\root\of{n}$.

Proof: First, assume that *n* is composite. An old result tells us that there is a prime divisor *p* of *n* which satisfies $p\leq \root\of{n}$. Hence the integer $\frac{n}{p}$ is a divisor of *n* with $\frac{n}{p}\geq \root\of{n}$.

Now if we want to compute $\sigma(n)$ we should add up all its positive divisors. Notice, however, that this sum of *all* divisors is at least as big as the sum of the 3 divisors [[$\{1,\frac{n}{p},n\}$ of *n* — after all, we're just throwing away the remaining divisors of *n*, so the sum of these three divisors is no bigger than the sum of all the divisors. Hence we have

So we have shown that when *n* is composite, $\sigma(n)$ is appropriately "big."

To prove the converse, we'll show that when *n* is prime, then $\sigma(n)$ is "small". In fact, it's easy to see that when *n* is prime, we have $\sigma(n) = n+1 < n+\root\of{n}$. $\square$

**Remark** It is worth noting that even though this theorem does give a *bona fide* way of testing whether an integer is prime, it doesn't give a very *practical* method for doing so. After all, the 2 ways we have of computing $\sigma(n)$ rely on either knowing all divisors of *n* (in which case you'll be able to tell pretty quickly whether the number is prime) or on the prime factorization of *n* (in which case, again, you will already be able to tell if the number is prime).

## Evenness and Oddness

As with the $\nu$ function, one might be curious to know which integers *n* have $\sigma(n)$ an odd number. For this, we note that a number $n=p_1^{a_1}\cdots p_k^{a_k}$ has

so that $\sigma(n)$ is odd if and only if all these factors of $\sigma(n)$ are odd.

First, if one of the $p_i=2$, then we have

(7)which is always an odd number (regardless of the value of $a_i$. Hence the power of 2 which appears in the prime factorization of *n* is of no consequence to this question.

Now suppose that $p_i>2$, so that $p_i$ is odd. Then we have

(8)Notice that each of the summands in this sum are odd numbers (since each is a power of an odd number, and you have proven that the product of odd numbers is odd). Hence this sum is odd if and only if the number of terms in the sum is odd, and we can see that the number of terms in the sum is $a_i+1$. Hence if we want this term to be odd, we need $a_i$ to be even. We have proven the following

Lemma: An integer

nhas $\sigma(n)$ odd if and only if $n=2^e\cdot s^2$, wheresis an odd integer.

# Summing Powers of Divisors

We finished class by discussing a generalization of the $\sigma$ function.

Definition: For a given integer

n, the $\sigma_k$ function is gives the sum of thek-th powers of the divisors ofn.

In equation form, the definition of $\sigma_k$ says

(9)As with all arithmetic functions, your first question is to find out whether this function is multiplicative. But notice that the function $f(d) = d^k$ is multiplicative (you proved this in your homework!), and so our old result on building new multiplicative functions out of old multiplicative functions gives that $\sigma_k$ is multiplicative.

Now that you know the function is multiplicative, you can give a formula for evaluating $\sigma_k(n)$ given a prime factorization of *n*. This is worth trying out, so give it a shot.

One final note: the function $\sigma_0(n)$ is defined as $\sigma_0(n) = \sum_{d\mid n} d^0 = \sum_{d \mid n} 1$, and hence we see that $\sigma_0 = \nu$.