Like many teachers at the upper level high school math classes, over the years I've presented the sum of the Cubes of the natural numbers formula above many dozens of times. Then, perhaps like many others, I would point out how nice it is that it turns out to be the square of the nth triangular number, a happy coincidence that would make it easier to remember. Usually then, we would challenge them to extend the idea to fourth powers and see if they could do the induction proof, even though there was no really nice simplification (to my knowledge) of the sums of fourth powers.

But then I reread a book that has been in my library for about six years, and realized that many of those teachers may have known a different approach to sums of cubes equaling square of sums that I had been completely unaware of. In case there are other teachers who somehow also didn't know, I share my newfound ancient knowledge.

The book was David Wells', Games and Mathematics, Subtle Connections, and as I reread it, I came across a theorem by Joseph Liouville (1809,1882) It is most easily understood if you illustrate with an example I think, so, as Wells did in his book, I'll use the number 15, although you can you any integer and it's still true.

If you write down all the divisors of the number, including 1 and the number, you see that there are four such divisors, 1, 3, 5, and 15. Now the unusual step, we want to write down the number of divisors each of the divisors has, so one has only 1 divisor, three has 2 divisors, five has 2 divisors, and fifteen has 4 divisors. Now let's make a set of these four new numbers, {1,2,2,4} What Liouville discovered was that the sum of the cubes of these four numbers, \( (1^3 + 2^3 + 2^3 + 4^3) = 81\(; is equal to the square of their sum. (1 +2+2+4)

^{2}.
And then, he goes on to explain why that is meanigful in relation to the natural numbers, the WHY of the fact I thought was just a coincidence. Take any prime to a power, for example 64 = 2

^{6}. Now the divisors of 64 are 1, 2, 4, 8, 16, 32, and 64. When we take the number of divisors of each of these divisors, we get 1, 2, 3, 4, 5, 6, 7,
When I played around with the idea, I realized that all the strings of natural numbers (powers of a prime) summed to a triangular number (64 above sums to 28, the 7th triangular number), but no others. Semiprimes always produce {1,2,2,4} which has a sum of nine,

I also discovered that the sums of sets with this Liouville property had a multiplication rule similar to the ones for finding the number of divisors....

Ok, I need to explain that since it is important for younger students if they've never tried to find the number of divisors for larger numbers, which gets tedious by hand. First, Primes to any power (even zero power) always have one more divisor than their exponent, so 3

^{2}has three divisors, 1, 3, and 9. , so if you wonder about numbers like 18, which is 3^{2}* 2, you can multiply the three divisors of 9, by the two divisors of 2 , to find that 18 has six total divisors.
Now the same kind of properties work with the sums of the Liouville sets. The set for 9 is {1,2,2,4) which sums to nine, and the sum of the set for 3 is {1,2} which sums to three. As long as the two numbers have no common factor you can multiply their sums, so the sum for the set for 18 will be 2*9 = 18. In fact any number that is a square times a prime will have 18 for it's sum.

There is another sort of multiplication of the sets that I found on another paper on line (now lost it seems) that allows you to take any two sets and make another that also has the Liouville property. For example the set for three is {1,2} and the set for 10 is (1, 2, 2, 4} . You can form a type of cross product by multiplying each number of one set by each number of the other set independently, so we could get {1*1, 1*2, 1*2, 1*4, 2*1, 2*2, 2*2, 2*4} or {1,2,2,4,2,4,4,8} or in preferred order, {1,2,2,2,4,4,4,8} . The cubes sum to 729 and the sum (27) also squares to 729. And if you walk through the process of producing the Liouville set for 3*10 = 30, you will find it matches....but then so would 105, or in general, any sphenic number, the product of three distinct primes.

The open question remaining is, Are there any sets that have this property that are NOT derived from divisors of a natural number? I am right where David Wells left me on that question. I would really be surprised if there were, but then I was surprised when I found out all the above.