I’ll just throw together this proof very quickly because I am very excited about it!! This problem has been plaguing my mind for months and the answer to it just absolutely came out of nowhere and was so so so excited to realize I could finally have some closure on it.

There exists a well known property that

,

where is Euler’s totient function, and we are summing over all positive divisors of .

I was playing around with some stuff in the middle of class a few months back (instead of paying attention), and I noticed a peculiar thing. It seemed to always be true that for any positive integer and ,

,

which some of you might recognize (though I didn’t at the time) as the Dirichlet Convolution of the totient function and an exponentiation function. Notice that if we set , we are only summing over the totient function, and we know that result will be . Since divides , there are no problems.

I tried a lot of things that wouldn’t work. The function wasn’t multiplicative so there was no luck there. I recently learned Mobius Inversion but couldn’t figure out how to get it to work. One day, I decided to watch a lecture on Mobius Inversion by Kevin Atienza. I paused the video to try and answer the second question myself, and all of a sudden I realized that it also answered my problem.

It was a combinatoric proof all along!

First, as a warmup, let’s try proving that . We will also prove this combinatorially.

First let’s answer the question: How many integers in the inclusive range have a gcd of with ?

Let the answer to this question be . First make the observation that the only valid would be divisors of ; can’t be a greatest common divisor of if it’s not even a divisor of in the first place! Next, we use the property of gcd’s that implies that ; conversely, implies that . These both follow from the definition of the greatest common divisor.

Therefore, there exists a bijection between every positive integer such that and the positive integers such that . But we already know how to count the former; it’s . Therefore

Now, consider the sequence of fractions . How many fractions are there in that sequence? Well, rather obviously, we can see that we enumerated fractions, since the denominator is fixed at and our numerators go from 1 to .

Let’s count it another way. Let us reduce all the fractions in that sequence to lowest terms, i.e. for , we would have the sequence

What are all the possible denominators that could show up in this lowest terms sequence? Well, each positive divisor would show up as a denominator—and these are the only possible denominators—but how many times each? Consider some numerator . If , then when reduced to lowest terms, the denominator of the fraction would be . Therefore, each divisor of would show up as a denominator exactly times, and from earlier we know that this is . To count the total number of fractions in the sequence, we sum this up over all possible values of .

Equating the two sides gives us

.

The fact that this was a combinatoric proof should’ve clued me in to the answer of my question!

The question from the video is: A pizza is cut into slices, and each slice can have one of different toppings. How many different pizzas can there be if rotations of a pizza are considered the same pizza?

This problem just screams that it’s a textbook application of Burnside’s Lemma . Let us define the group to be the set of all unique clockwise rotations of an -slice pizza (shift 1 place, … shift places, shift places), and the operation of this group is composition of these rotations. The group has elements.

Suppose we are shifting places to right, where is in the range . How many orbits are there when shifting the pizza spaces clockwise? Let’s label the slices 0 to and examine slice 0. In order for this pizza to look exactly the same after rotation, the topping of slice 0 should match the topping of slice , the spot that it lands on after rotation. Similarly, the topping of slice should match the topping of slice . In general, the topping of slice should be the same as the topping of slice . This continues on and on until we end up with a closed “loop”. Each slice on the pizza belongs to exactly one such loop, and every slice within a given loop should have the same topping.

I encourage you to try and draw out these loops for yourself for various of a given to see the pattern that we’ll discuss next.

How long will each of these loops be for a given ? You might observed that each loop has slices in it. In the sequence , , , … …, we need the first value of such that . Well, it will be equal to 0 when is divisible by , so since is fixed, we find that = .

Exercise: Prove that the loop “closes” and repeats itself at the 0th slice, and not on a later one.

If each loop has length , that means that there will be of these loops. Each of these loops must contain toppings of the same kind. Therefore, since there are loops whose toppings we have to decide and there are possible toppings per loop, the number of orbits for rotating slices clockwise is .

If we let be the number of pizzas with slices, made from toppings, where rotations aren’t distinct, then by Burnside’s Lemma,

.

However, again note that the only values can take will be divisors of . Let us define as the number of in the inclusive range such that . Therefore the above summation can be rephrased as,

.

However, we know that is actually just . Therefore,

.

Also, since each divisor corresponds to the divisor , the above sum is the same as

.

Finally, move the to the left side, and we have

.

Since is the answer to a counting problem, we know it must be an integer. Therefore, the right hand side, the sum we wanted, is divisible by . Proof complete!