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 meaningful 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, (one less than a triangular number. A pattern?)

Fifteen is a pronic or oblong number, the product of two primes. If you try any other pronic, you will get the same result as for 15, the sum of the cubes of the set of divisors (and the square of their sum) will be 81.

Numbers like 2*3*5=30 are called sphenic numbers. They should, like pronic numbers, all have the same Louiville sets for their divisors. The eight divisors of 30, (1, 2, 3, 5, 6, 10,15, 30) have a divisors set of (1, 2, 2, 2, 4, 4, 4, 8) . The sum of the set is 27 (also one less than a triangular number) so the sum of the cubes should be 27^2 = 729, and it is.

I haven't yet tried 210. Will return later unless one of you beats me to it.

Sets that have multiple factors, other than powers of a prime seem not to form such sets. Try 12 for instance.

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.

So for 210, we can multiply the Louisville set of 30, {(1, 2, 2, 2, 4, 4, 4, 8) by the set for 7 (1,2), so we just add in a double for all of these and for a product of four primes we get, in order (1, 2, 2, 2, 4, 4, 4, 8) conjoined with (2, 4, 4, 4, 8, 8, 8, 16) So I'm counting a 1, four 2's, six 4's, four 8's, and a 16 . 1^3 + 4*2^3 + 6*4^3 + 4*8^3 + 16^3 = 6561, and 1 + 4*2 + 6*4 + 4*8 +16 = 81 and 81^2 (drum roll please...da da da.. ) is 6561. (End of hoped for pattern of sum as one more than a triangular number, but I now see a binomial of the terms. For a semi-prime we had 1, 2, and 1 0f the numbers 1, 2, 4. For a product of three primes we got 1, 3, 6, 3, and 1 of the values 1, 2, 4, 6, 8, & 16.

Looking at the first four cubes we got 9, 81, 729 and 6561 which are the squares of 3, 9, 27, and 81; so hoping for 243 for the Liouville sum for this one, fingers crossed...

and on that I will predict the Liouville series for the product of five prime factors to have 1*1^3 + 5*2^3 + 10*4^3 + 10 * 8^3 + 5*16^3 + 1*32^3 = (Eureka, 59049 or 81^2.

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.

## 1 comment:

I discovered, by accident, the fact that the first n cubes is the square of the first n integers. I was able to prove it by deriving the formula for the first n cubes, but that provided no good insights into why.

I wrote a program that would find the formula for sum(x^a) for x = 1..n (call it f(n,a)), but could find no other instances where f(n,a) = f(n,b)^c. I convinced my self that a =3, b=1,c=2 was a special case but I don’t recall why.

This is a fascinating chapter in the story for me, I was completely unaware of this theorem by Louiville!

Post a Comment