# Sets of integers that turned out to be finite

In the spirit of the previous question "Conjectures that have been disproved with extremely large counterexamples?", and as an attempt to salvage this closed question, I'm interested in sets of natural numbers or integers such that it was historically an open question whether they contained infinitely many numbers, but have now been proven to be finite.

For example, it is still not known whether there are infinitely many perfect numbers. If it were proved that there are in fact only finitely many, that would make them a valid answer to this question.

I know that one can easily make up trivial examples ("I don't know whether $A=\{n:|n|<10^{1000}\}$ is finite. Oh wait, it is. Done!") but those would not count as historical open questions.

Note: This is not identical to the previous question on extremely large counterexamples, because one may have an extremely large counterexample while still having infinitely many larger non-counterexamples.

• Love this question! – goblin GONE Jul 13 '17 at 4:53
• Someone more informed might have the story for the monster group and such. Also dimensions in which done geometric property holds, those often end up smaller than I'd ever expect. – Artimis Fowl Jul 13 '17 at 6:02
• If you want to make it different to "extremely large counter-examples" and not to include every proven (or disproven) conjecture in number-theory, you need to say you are considering the original problem itself, not its many equivalent or stronger or weaker problems. For example the twin-prime conjecture is about the $n$ such that $n,n-2$ are primes, not about the $n$ such that there are more than $-1+\log n$ twin primes $< n$. – reuns Jul 13 '17 at 7:47

Many number-theoretic theorems are of the form: all sufficiently large $n$ have property $P(n)$. Thus until the theorem was proved, $\{n \in \mathbb N:\; \text{not }P(n)\}$ was not known to be finite, but now it is.

EDIT: E.g. for given positive integer $k$ and real number $\delta$ with $0 < \delta < 1$, Szemerédi's theorem says every sufficiently large $n$ has the property: every subset of $\{1, \ldots, n\}$ of size $\ge \delta n$ contains an arithmetic progression of length $k$.

• I thought this might be the case, but I don't know much about historically important number-theoretic theorems. Can you add one or two notable examples? – user856 Jul 13 '17 at 6:43

One important recent (2013) result: Let $p_n$ be the $n$th prime. Let $G(m)= \min \{p_{n+1}-p_n:n\geq m\}.$ The result was that $\{G(m):m\in \mathbb N\}$ is finite. In particular that $G(m)<7\cdot 10^7$ for all sufficiently large $m$. Further work has replaced the value $7\cdot 10^7$ with a value less than $300.$ (Sorry, I can't recall the details or names.) Equivalently, there exist infinitely many $n$ such that $p_{n+1}-p_n\leq 300.$

Here's an extreme one referenced in the article "On the greatest common divisor of the value of two polynomials" in the July 2017 American Mathematical Monthly.

The reference is to the Prime Glossary at http://primes.utm.edu/glossary/xpage/LawOfSmall.html

$gcd(n^{17}+9, (n+1)^{17}+9)$ seems to always be one. In fact, if you had your computer checking this for $n=1, 2, 3, . . .$ successively, it would never find a counter-example. That is because the first counter-example is the 52-digit number 8424432925592889329288197322308900672459420460792433.

• Unless I'm misunderstanding, shouldn't this be an answer to the previous question on extremely large counterexamples? Please see the clarification I added to the end of the question. – user856 Jul 13 '17 at 5:14
• The previous question was closed, so this is the best I could do. – marty cohen Jul 13 '17 at 5:24
• The previous question on extremely large counterexamples doesn't look closed to me. – user856 Jul 13 '17 at 5:27
• However, you could get your computer to find that counterexample in a fraction of a second if you were a bit cleverer. In Maple: f:= n^17+9: g:= (n+1)^17 + 9: Q:=resultant(f,g,n): {msolve(f, Q)} intersect {msolve(g,Q)}; – Robert Israel Jul 13 '17 at 5:40
• I long ago realized that I was not anywhere as clever as Robert Israel and that that was OK. – marty cohen Jul 13 '17 at 5:46

Probably the most famous finite set of integers, that is hard to prove to be finite, is the set $$B \subset \mathbb{N}$$ of integers $$n$$ satisfying the statement of Burnside conjecture («All groups with exponent $$n$$ are locally finite»). This set is finite due to Adyan-Novikov theorem («There exists an infinite finitely generated group of exponent $$n$$ for any odd $$n \geq 665$$») and Ivanov theorem («There exists an infinite finitely generated group of exponent $$n$$ for any even $$n \geq 2^{48}$$»). However, despite $$B$$ is finite, its exact size remains unknown. The only numbers that are known for sure to belong to $$B$$ are 1, 2, 3, 4 and 6.

If you want something more tangible, then there is Zivcovic Theorem, that states, that there are finitely many numbers $$n \in \mathbb{N}$$, such, that $$\Sigma_{k = 0}^{n - 1} {(-1)}^k (n - k)!$$ is prime. Such numbers form the OEIS sequence A001272.

Another example of such nontrivially finite sets of integers can be found in this MSE question: Equation $x = \tau(2^x - 1)$. The equation here appeared to have exactly 7 solutions: 1, 2, 4, 6, 8, 16, and 32.

And of course we should mention the set of all $$n \in \mathbb{N}$$, such that $$\exists x, y, z \in \mathbb{N} x^n + y^n = z^n$$. Actually, this set is two-element and contains only 1 and 2. This statement is called Fermat’s Last Theorem and was one of the most long-standing open problems (formulated by Fermat in 1637 and proved by Wiles in 1995)