Divisor Sums in Number Theory

students, this lesson introduces divisor sums, one of the most useful ideas in arithmetic functions and a key part of your Midterm 1 review 📘. The main goal is to understand how adding up divisors of integers reveals patterns, helps classify numbers, and connects to broader ideas in number theory.

By the end of this lesson, you should be able to:

explain what divisor sums are and what the common notation means,
compute divisor sums for specific integers,
use number theory reasoning to find patterns in divisor sums,
connect divisor sums to multiplicative functions and the study of arithmetic functions,
summarize why divisor sums matter in Midterm 1 and beyond.

A divisor sum is a function built from the divisors of a number. It shows up in problems about perfect numbers, abundant numbers, and multiplicative functions. Think of it like taking a number and asking, “What story do its factors tell?” 🧩

What Is a Divisor Sum?

For a positive integer $n$, the divisors of $n$ are the positive integers that divide $n$ with no remainder. For example, the divisors of $12$ are $1, 2, 3, 4, 6,$ and $12.

The most common divisor sum is the function $\sigma(n)$, defined by

$$\sigma(n)=\sum_{d\mid n} d,$$

where $d\mid n$ means “$d$ divides $n$.” This notation tells us to add up all positive divisors of $n$.

For $12$,

$$\sigma(12)=1+2+3+4+6+12=28.$$

Another important function is the sum of proper divisors, which adds all divisors except the number itself. It is often written as

$$s(n)=\sum_{d\mid n,\ d<n} d,$$

so for $12$,

$$s(12)=1+2+3+4+6=16.$$

These two functions are closely related because

$$\sigma(n)=s(n)+n.$$

This simple relationship is useful all over number theory. When you know one, you can find the other.

How to Compute Divisor Sums

There are two main ways to compute divisor sums: listing divisors directly or using prime factorization. For small numbers, listing works well. For larger numbers, prime factorization is much faster.

Method 1: Listing divisors

Suppose you want $\sigma(18)$. First list the divisors of $18$:

$$1,2,3,6,9,18.$$

Then add them:

$$\sigma(18)=1+2+3+6+9+18=39.$$

This method is easy, but it can become slow if $n$ has many divisors.

Method 2: Using prime factorization

A very important formula in divisor sums is based on the prime factorization of $n$. If

$$n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k},$$

then

$$\sigma(n)=\prod_{i=1}^k \left(1+p_i+p_i^2+\cdots+p_i^{a_i}\right).$$

This works because every divisor of $n$ is formed by choosing powers of each prime from $0$ up to $a_i$.

For example, let

$$n=12=2^2\cdot 3^1.$$

Then

$$\sigma(12)=(1+2+2^2)(1+3)=(1+2+4)(1+3)=7\cdot 4=28.$$

This matches the direct list method.

Here is another example:

$$n=36=2^2\cdot 3^2.$$

$$\sigma(36)=(1+2+4)(1+3+9)=7\cdot 13=91.$$

Using factorization is especially helpful in exam problems because it saves time and reduces arithmetic mistakes. ✅

Why Divisor Sums Are Important

Divisor sums are not just a computation trick. They help classify numbers and reveal deeper structure.

Perfect numbers

A number $n$ is perfect if the sum of its proper divisors equals the number itself:

$$s(n)=n.$$

Since $s(n)=\sigma(n)-n$, this is equivalent to

$$\sigma(n)=2n.$$

The number $6$ is perfect because its proper divisors are $1,2,3, and

$$1+2+3=6.$$

Also,

$$\sigma(6)=1+2+3+6=12=2\cdot 6.$$

Another perfect number is $28$:

$$1+2+4+7+14=28.$$

Perfect numbers are famous in number theory because they connect divisor sums with special arithmetic patterns.

Abundant and deficient numbers

A number is abundant if its proper divisor sum is bigger than the number:

$$s(n)>n.$$

A number is deficient if

$$s(n)<n.$$

For example, $12$ is abundant because

$$s(12)=16>12.$$

The number $8$ is deficient because its proper divisors are $1,2,4, and

$$1+2+4=7<8.$$

These categories are based directly on divisor sums, so the function $\sigma(n)$ gives a quick way to study the “size” of a number’s divisor structure.

Divisor Sums and Multiplicative Functions

Divisor sums fit into the larger chapter of arithmetic functions. An arithmetic function is a function whose input is a positive integer and whose output is usually a number related to that integer.

The divisor sum function $\sigma(n)$ is a classic example. It is also multiplicative, meaning that if $\gcd(m,n)=1$, then

$$\sigma(mn)=\sigma(m)\sigma(n).$$

This property is extremely useful. It lets us break a hard problem into smaller pieces.

For example, since $8$ and $9$ are relatively prime,

$$\sigma(72)=\sigma(8\cdot 9)=\sigma(8)\sigma(9).$$

Now compute each part:

$$\sigma(8)=1+2+4+8=15,$$

and

$$\sigma(9)=1+3+9=13.$$

$$\sigma(72)=15\cdot 13=195.$$

This is much faster than listing all divisors of $72$ by hand.

A useful related fact is that many divisor-based functions are built by summing values over divisors. For example, if $f(n)$ is another arithmetic function, then the divisor-sum transform

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

creates a new arithmetic function from an old one. This idea appears often in more advanced number theory because it links local divisor data to global behavior.

Working Through More Examples

Let’s practice with a few examples to strengthen your understanding.

Example 1: Find $\sigma(15)$

The divisors of $15$ are

$$1,3,5,15.$$

$$\sigma(15)=1+3+5+15=24.$$

Since $s(15)=24-15=9$, the proper divisor sum is $9$.

Example 2: Find $\sigma(30)$ using factorization

First factor $30$:

$$30=2\cdot 3\cdot 5.$$

Then use the product formula:

$$\sigma(30)=(1+2)(1+3)(1+5)=3\cdot 4\cdot 6=72.$$

$$s(30)=72-30=42.$$

Since $42>30$, the number $30$ is abundant.

Example 3: Identify the type of number

Suppose you know that

$$\sigma(n)=2n.$$

Then

$$s(n)=\sigma(n)-n=2n-n=n,$$

so $n$ is perfect.

Suppose instead that

$$\sigma(n)<2n.$$

Then

$$s(n)<n,$$

so $n$ is deficient.

$$\sigma(n)>2n,$$

then $n$ is abundant.

These comparisons are a quick way to classify numbers using divisor sums.

Common Mistakes to Avoid

When working with divisor sums, students, a few mistakes show up often:

forgetting to include $1$ and $n$ among the divisors,
mixing up $\sigma(n)$ with $s(n)$,
using the product formula without first writing the prime factorization correctly,
assuming $\sigma(mn)=\sigma(m)\sigma(n)$ for all $m,n$ even when $\gcd(m,n)\neq 1$.

For example, the multiplicative rule works for relatively prime numbers, not for every pair of integers. That detail matters a lot on exams.

A good habit is to check your answer by listing divisors for small examples. This builds confidence and helps you spot errors early. 🔍

Conclusion

Divisor sums are a central topic in arithmetic functions and a major part of your Midterm 1 preparation. The function $\sigma(n)$ adds all positive divisors of $n$, while the proper divisor sum $s(n)$ leaves out the number itself. These ideas help classify numbers as perfect, abundant, or deficient and connect directly to multiplicative functions.

The key lesson is that divisor sums are both practical and theoretical. They give you efficient computation methods through prime factorization, and they reveal deep patterns in the structure of integers. If you understand $\sigma(n)$, you are building strong foundations for the rest of number theory 📚.

Study Notes

The divisor sum function is $\sigma(n)=\sum_{d\mid n} d$.
The proper divisor sum is $s(n)=\sum_{d\mid n,\ d<n} d$.
The relationship is $\sigma(n)=s(n)+n$.
If $n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$, then

$$\sigma(n)=\prod_{i=1}^k \left(1+p_i+p_i^2+\cdots+p_i^{a_i}\right).$$

If $\gcd(m,n)=1$, then $\sigma(mn)=\sigma(m)\sigma(n)$.
A number is perfect if $\sigma(n)=2n$.
A number is abundant if $\sigma(n)>2n$.
A number is deficient if $\sigma(n)<2n$.
Divisor sums are part of arithmetic functions and help explain number patterns in Midterm 1.
Always check whether a problem asks for $\sigma(n)$ or $s(n)$ before calculating.