# Divisor Functions

Definition. The sum of divisors function is given by

As usual, the notation " " as the range for a sum or product means that d ranges over the positive divisors of n.

The number of divisors function is given by

For example, the positive divisors of 15 are 1, 3, 5, and 15. So

I want to find formulas for and in terms of the prime factorization of n. This will be easy if I can show that and are multiplicative. I can do most of the work in the following theorem.

Theorem. The divisor sum of a multiplicative function is multiplicative.

Proof. Suppose f is multiplicative, and let be the divisor sum of f. Suppose . Then

Then

Now , so if and , then . Therefore, multiplicativity of f implies

Now every divisor d of can be written as , where and . Going the other way, if and then . So I may set , where , and replace the double sum with a single sum:

This proves that is multiplicative.

Theorem. (a) The sum of divisors function is multiplicative.

(b) The number of divisors function is multiplicative.

Proof. (a) The identity function is multiplicative: for all m, n, so obviously it's true for . Therefore, the divisor sum of is multiplicative. But

Hence, the sum of divisors function is multiplicative.

(b) The constant function is multiplicative: for all m, n, so obviously it's true for . Therefore, the divisor sum of I is multiplicative. But

Hence, the number of divisors function is multiplicative.

I'll use multiplicativity to obtain formulas for and in terms of their prime factorizations (as I did with ). First, I'll get the formulas in the case where n is a power of a prime.

Lemma. Let p be prime.

(a) .

(b) .

Proof. The divisors of are 1, p, , ..., . So the sum of the divisors is

And since the divisors of are 1, p, , ..., , there are of them, and

Theorem. Let , where the p's are distinct primes and for all i. Then:

Proof. These results follow from the preceding lemma, the fact that and are multiplicative, and the fact that the prime power factors are pairwise relatively prime.

Here is a graph of for .

Note that if p is prime, . This gives the point , which lies on the line . This is the line that you see bounding the dots below.

For each n, there are only finitely many numbers k whose divisor sum is equal to n: that is, such that . For k divides itself, so

This says that k must be less than n. So if I'm looking for numbers whose divisors sum to n, I only need to look at numbers less than n. For example, if I want to find all numbers whose divisors sum to 42, I only need to look at .

Here is a graph of for .

If p is prime, . Thus, repeatedly returns to the horizontal line , which you can see bounding the dots below.

The formulas given in the theorem allow us to compute and by hand for at least small values of n. For example, , so

Example. Find all positive integers n such that .

Since doesn't work, I can assume .

I have

In other words, is a sum of distinct positive integers other than 1 and n that is equal to 7. I have to consider all possible ways of doing this. I'll consider cases according to the largest element of this sum, which is the largest divisor d of n other than 1 and n.

Suppose .

Then the only divisor of n other than 1 and n is 7. Since , I know for . But if , then 49 would be a divisor of n other than 1 and n. Hence, , and this is a solution.

Suppose . Then the expression must have the form , which contradicts the assumption that the sum does not include 1.

Suppose . Then the expression must have the form . In this case, .

Suppose . Then

But if , then . So must have the form , contradicting the assumption that the sum does not include 1.

Suppose . Then

However, can't include 1, and can't use 2 twice. Hence, this isn't possible.

Suppose . Then the remaining terms in must sum to 5 and can only use 1, which is excluded by assumption. Hence, this isn't possible.

Therefore, or .

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