# The Summer 2017 Mathematics A To Z: Sárközy’s Theorem

Gaurish, of For the love of Mathematics, gives me another chance to talk number theory today. Let’s see how that turns out.

# Sárközy’s Theorem.

I have two pieces to assemble for this. One is in factors. We can take any counting number, a positive whole number, and write it as the product of prime numbers. 2038 is equal to the prime 2 times the prime 1019. 4312 is equal to 2 raised to the third power times 7 raised to the second times 11. 1040 is 2 to the fourth power times 5 times 13. 455 is 5 times 7 times 13.

There are many ways to divide up numbers like this. Here’s one. Is there a square number among its factors? 2038 and 455 don’t have any. They’re each a product of prime numbers that are never repeated. 1040 has a square among its factors. 2 times 2 divides into 1040. 4312, similarly, has a square: we can write it as 2 squared times 2 times 7 squared times 11. So that is my first piece. We can divide counting numbers into squarefree and not-squarefree.

The other piece is in binomial coefficients. These are numbers, often quite big numbers, that get dumped on the high school algebra student as she tries to work with some expression like $(a + b)^n$. They’re also dumped on the poor student in calculus, as something about Newton’s binomial coefficient theorem. Which we hear is something really important. In my experience it wasn’t explained why this should rank up there with, like, the differential calculus. (Spoiler: it’s because of polynomials.) But it’s got some great stuff to it.

Binomial coefficients are among those utility players in mathematics. They turn up in weird places. In dealing with polynomials, of course. They also turn up in combinatorics, and through that, probability. If you run, for example, 10 experiments each of which could succeed or fail, the chance you’ll get exactly five successes is going to be proportional to one of these binomial coefficients. That they touch on polynomials and probability is a sign we’re looking at a thing woven into the whole universe of mathematics. We saw them some in talking, last A-To-Z around, about Yang Hui’s Triangle. That’s also known as Pascal’s Triangle. It has more names too, since it’s been found many times over.

The theorem under discussion is about central binomial coefficients. These are one specific coefficient in a row. The ones that appear, in the triangle, along the line of symmetry. They’re easy to describe in formulas. for a whole number ‘n’ that’s greater than or equal to zero, evaluate what we call 2n choose n:

${{2n} \choose{n}} = \frac{(2n)!}{(n!)^2}$

If ‘n’ is zero, this number is $\frac{0!}{(0!)^2}$ or 1. If ‘n’ is 1, this number is $\frac{2!}{(1!)^2}$ or 2. If ‘n’ is 2, this number is $\frac{4!}{(2!)^2}$ 6. If ‘n’ is 3, this number is (sparing the formula) 20. The numbers keep growing. 70, 252, 924, 3432, 12870, and so on.

So. 1 and 2 and 6 are squarefree numbers. Not much arguing that. But 20? That’s 2 squared times 5. 70? 2 times 5 times 7. 252? 2 squared times 3 squared times 7. 924? That’s 2 squared times 3 times 7 times 11. 3432? 2 cubed times 3 times 11 times 13; there’s a 2 squared in there. 12870? 2 times 3 squared times it doesn’t matter anymore. It’s not a squarefree number.

There’s a bunch of not-squarefree numbers in there. The question: do we ever stop seeing squarefree numbers here?

So here’s Sárközy’s Theorem. It says that this central binomial coefficient ${{2n} \choose{n}}$ is never squarefree as long as ‘n’ is big enough. András Sárközy showed in 1985 that this was true. How big is big enough? … We have a bound, at least, for this theorem. If ‘n’ is larger than the number $2^{8000}$ then the corresponding coefficient can’t be squarefree. It might not surprise you that the formulas involved here feature the Riemann Zeta function. That always seems to turn up for questions about large prime numbers.

That’s a common state of affairs for number theory problems. Very often we can show that something is true for big enough numbers. I’m not sure there’s a clear reason why. When numbers get large enough it can be more convenient to deal with their logarithms, I suppose. And those look more like the real numbers than the integers. And real numbers are typically easier to prove stuff about. Maybe that’s it. This is vague, yes. But to ask ‘why’ some things are easy and some are hard to prove is a hard question. What is a satisfying ’cause’ here?

It’s tempting to say that since we know this is true for all ‘n’ above a bound, we’re done. We can just test all the numbers below that bound, and the rest is done. You can do a satisfying proof this way: show that eventually the statement is true, and show all the special little cases before it is. This particular result is kind of useless, though. $2^{8000}$ is a number that’s something like 241 digits long. For comparison, the total number of things in the universe is something like a number about 80 digits long. Certainly not more than 90. It’d take too long to test all those cases.

That’s all right. Since Sárközy’s proof in 1985 there’ve been other breakthroughs. In 1988 P Goetgheluck proved it was true for a big range of numbers: every ‘n’ that’s larger than 4 and less than $2^{42,205,184}$. That’s a number something more than 12 million digits long. In 1991 I Vardi proved we had no squarefree central binomial coefficients for ‘n’ greater than 4 and less than $2^{774,840,978}$, which is a number about 233 million digits long. And then in 1996 Andrew Granville and Olivier Ramare showed directly that this was so for all ‘n’ larger than 4.

So that 70 that turned up just a few lines in is the last squarefree one of these coefficients.

Is this surprising? Maybe, maybe not. I’ll bet most of you didn’t have an opinion on this topic twenty minutes ago. Let me share something that did surprise me, and continues to surprise me. In 1974 David Singmaster proved that any integer divides almost all the binomial coefficients out there. “Almost all” is here a term of art, but it means just about what you’d expect. Imagine the giant list of all the numbers that can be binomial coefficients. Then pick any positive integer you like. The number you picked will divide into so many of the giant list that the exceptions won’t be noticeable. So that square numbers like 4 and 9 and 16 and 25 should divide into most binomial coefficients? … That’s to be expected, suddenly. Into the central binomial coefficients? That’s not so obvious to me. But then so much of number theory is strange and surprising and not so obvious.

## Author: Joseph Nebus

I was born 198 years to the day after Johnny Appleseed. The differences between us do not end there. He/him.

## 5 thoughts on “The Summer 2017 Mathematics A To Z: Sárközy’s Theorem”

1. Thank you. And I’m not sure how I overlooked that, since Bertrand’s Postulate is such a nice, easy-to-understand result. (The Postulate, which can be proven, is that there is always at least one prime between a whole number ‘n’ and its double, ‘2n’. With Sarkozy’s Theorem you can show this has to be true for numbers larger than 468. For the numbers from 1 up to 468, you can just check each case. It’s time-consuming but not hard.)

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