(Off Topic) Absurd and Adsurd

by admin on March 30, 2018

With apologies, I am going to go off topic today and talk about math. Some chess players like math, so I hope you’ll be interested in this post!

Big news! Today, for the first time in 23 years, I’ve had an article published in a mathematical research journal. If you want to read it, go to the journal Integers, Volume 18, and scroll down the page to paper #A33. It’s right after paper #A32 by Doron Zeilberger, who is, like, a famous number theorist, so I am in great company.

Here’s a friendly, non-technical explanation of what the article is about. It starts with a puzzle by a 19th century popular math writer, August de Morgan, who asked: “At one point in my life, the square of my age was the same as the year. When was I born?”

At first it seems that there cannot possibly be enough information to answer the question. But then, especially if you’ve had some high school algebra, you might think of letting X be his age in the year X^2. Then he had to be born in the year X^2 – X. And here’s the other piece of information you need to know: he wrote this puzzle in 1864. So X = 44 is too big; he would have to be born in 1892, and that year hadn’t happened yet. And X = 42 is too small; he would have to be born in 1722, which would make him more than 140 years old when he wrote these words. So the only reasonable solution is X = 43, and he was born in 43^2 – 43 = 1806.

Of course, we can do this same trick with powers higher than squares if we want, and it’s interesting to make a list of the next few “de Morgan years”: 1892, 1980, 2046, 2070, 2162, 2184, 2184, …

Wait, what’s that? Was that a typo? No, it wasn’t. In fact, 2184 is a de Morgan year in two different ways. It’s equal to 3^7 – 3 and it’s also equal to 13^3 – 13. Or to put it another way, people born in 2184 will get two chances at glory: in 2187 their age will be the seventh root of the year, and in case they miss this great occasion, ten years later their age will be the cube root of the year.

I’m not the first person to notice this; in fact, Matt Parker made a YouTube video about it, called Why 1980 was a great year to be born… but 2184 will be better. Check it out, and all of his Numberphile videos! (By the way, Matt told me that he was born in 1980, which is what gave him the idea for the video.)

When I realized that 2184 had this amazing property, there were two things I had to know: are there any other numbers like this? And is there some explanation for it?

Short answer: Yes, and yes. That’s what my article is about.

For the long answer, first we need a better word than “numbers like this.” I decided to call numbers of the form X^n – X “absurd numbers.” This is a pseudo-Latin combination of the prefix “ab-” meaning “away, from,” and the word “surd,” which is an old-fashioned term for “n-th root.” (It’s used mostly in typography as a name for the square-root symbol.) So “absurd” means a number minus its n-th root. Perfect! I also coined the word “adsurd,” which is a number plus its surd; for example, 2^7 + 2 = 130 is an adsurd number.

So our question is whether there are any other numbers that are absurd in two different ways, or doubly absurd. In fact, there are eight doubly absurd numbers known:

  • 6 = 2^3 – 2 = 3^2 – 3
  • 30 = 2^5 – 2 = 6^2 – 6
  • 210 = 6^3 – 6 = 15^2 – 15
  • 240 = 3^5 – 3 = 16^2 – 16
  • 2184 = 3^7 – 3 = 13^3 – 13
  • 8190 = 2^13 – 2 = 91^2 – 91
  • 78120 = 5^7 – 5 = 280^2 – 280
  • 24299970 = 30^5 – 30 = 4930^2 – 4930.

I strongly suspect (or to use a mathematician’s word, “conjecture”) that these are the only ones. But I also think that proving this is far beyond human abilities at present.

Notice that 2184 is special, even among the numbers on this list, because it is the only one where both powers are cubes or higher. (In every other example, one of the absurd expressions has the form X^2 – X.) When I see this, I think, “This cannot be an accident.” Some of the earlier equations on this list could just be coincidences. The great number theorist Richard Guy once postulated the semi-humorous Law of Small Numbers, which says that there are not enough small numbers to do all of the things that we expect of them. So there are bound to be coincidences. But to me, 2184 isn’t small. Anything under 100 is small, and anything under 1000 is small-ish, but 2184 demands an explanation.

And that’s what I found! Not a complete explanation, perhaps, but a good enough one to satisfy me. I proved that 2184 is the only integer N that solves the equation

(*) N = X^odd – X = prime^3 – prime.

The way I proved it was to reduce the problem using inequalities to a classical problem that was solved not too long ago, called Catalan’s conjecture. Catalan was a 19th century mathematician who noticed that 8 and 9 are the only two consecutive numbers that are pure powers (8 = 2^3, 9 = 3^2). He couldn’t prove it, of course. That is the problem with number theory, it is so easy to notice stuff, but so hard to prove anything.

However, in the century and a half after Catalan, number theory has advanced a little bit, and in 2002 the Romanian mathematician Preda Mihailescu proved Catalan’s conjecture using techniques that are too advanced for me to understand. (Well, if I really applied myself for a year I might be able to sort of understand his proof, but I’m not a professional mathematician any more — I don’t have time to do that.) But fortunately, for my purposes I don’t need to understand the proof; I can just use his theorem as a “black box” to help with my problem.

The most amazing and unexpected thing to me was when I figured out how to generate solutions to Catalan’s problem using doubly absurd numbers. In fact, any solution N to the equation (*) must give rise to a solution to Catalan’s problem. The number N = 2184 reduces to the solution 2^3 + 1 = 3^2 of Catalan’s problem. And we know that there are no other solutions to Catalan’s problem, thanks to Mihailescu and his “black box.” Therefore there are no other solutions to the above equation. QED!

This was really fun for me because it was the first time in my life that I was able to make any progress at all on a number theory problem that hasn’t been solved before. I want to emphasize that it’s really a tiny, tiny, TINY bit of progress. And my article does not involve any deep, subtle or revolutionary ideas. The deep ideas were all Mihailescu’s, and all I’m doing is pointing out that once you have his result, here is another cute fact that comes along for the ride. Nevertheless, it’s fun to feel like a researcher for the first time in 23 years. I’ve spent the last two decades writing about other people who are contributing to the advance of human knowledge, but this time I’m doing it myself.

By the way, you might wonder, “What about doubly adsurd numbers?” I’ve left this question completely open — maybe someone else will get interested in it? The only two examples known are:

  • 30 = 3^3 + 3 = 5^2 + 5
  • 130 = 2^7 + 2 = 5^3 + 5

There are no other examples less than 2 billion. The trouble is, I think that these might just be coincidences, according to Guy’s Law of Small Numbers. (As I said before, 130 is small-ish.) But one person’s coincidence might be another person’s inspiration!


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{ 1 comment… read it below or add one }

Larry Smith March 31, 2018 at 5:21 pm

Congratulations! I like the old-timey feel of the publication screenshot.

Bringing a chess angle back into this, I dare say (conjecture?) this is the best theoretical novelty (TN) in the Catalan so far in 2018!

I wonder why you suspect that there are only 8 doubly absurd numbers. As a layperson, it seems to me that since the highest one known is “only” 8 digits long, there ought to be more, maybe even an infinite quantity. I’m not a mathematician, though, so this is only FWIW.

Whenever I hear the obscure word “surd” I think of Charles Dodgson’s little rhyme:

Yet what are all such gaieties to me
Whose thoughts are full of indices and surds?
x2 + 7x + 53
= 11 / 3.


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