Conjectures that have been disproved with extremely large counterexamples?

Question

I just came back from my Number Theory course, and during the lecture there was mention of the Collatz Conjecture.

I'm sure that everyone here is familiar with it; it describes an operation on a natural number – $n/2$ if it is even, $3n+1$ if it is odd.

The conjecture states that if this operation is repeated, all numbers will eventually wind up at $1$ (or rather, in an infinite loop of $1-4-2-1-4-2-1$).

I fired up Python and ran a quick test on this for all numbers up to $5.76 \times 10^{18}$ (using the powers of cloud computing and dynamic programming magic). Which is millions of millions of millions. And all of them eventually ended up at $1$.

Surely I am close to testing every natural number? How many natural numbers could there be? Surely not much more than millions of millions of millions. (I kid.)

I explained this to my friend, who told me, "Why would numbers suddenly get different at a certain point? Wouldn't they all be expected to behave the same?"

To which I said, "No, you are wrong! In fact, I am sure there are many conjectures which have been disproved by counterexamples that are extremely large!"

And he said, "It is my conjecture that there are none! (and if any, they are rare)".

Please help me, smart math people. Can you provide a counterexample to his conjecture? Perhaps, more convincingly, several? I've only managed to find one! (Polya's conjecture). One, out of the many thousands (I presume) of conjectures. It's also one that is hard to explain the finer points to the layman. Are there any more famous or accessible examples?

Answer 1

Another example: Euler's sum of powers conjecture, a generalization of Fermat's Last Theorem. It states:
If the equation $\sum_{i=1}^kx_i^n=z^n$ has a solution in positive integers, then $n \leq k$ (unless $k=1$). Fermat's Last Theorem is the $k=2$ case of this conjecture.

A counterexample for $n=5$ was found in 1966: it's
$$ 61917364224=27^5+84^5+110^5+133^5=144^5 $$ The smallest counterexample for $n=4$ was found in 1988:
$$ 31858749840007945920321 = 95800^4+217519^4+414560^4=422481^4 $$ This example used to be even more useful in the days before FLT was proved, as an answer to the question "Why do we need to prove FLT if it has been verified for thousands of numbers?" :-)

Answer 2

My favorite example, which I'm surprised hasn't been posted yet, is the conjecture:

$n^{17}+9 \text{ and } (n+1)^{17}+9 \text{ are relatively prime}$

The first counterexample is $n=8424432925592889329288197322308900672459420460792433$

Answer 3

A famous example that is not quite as large as these others is the prime race.

The conjecture states, roughly: Consider the first n primes, not counting 2 or 3. Divide them into two groups: A contains all of those primes congruent to 1 modulo 3 and B contains those primes congruent to 2 modulo 3. A will never contain more numbers than B. The smallest value of n for which this is false is 23338590792.

Answer 4

I heard this story from Professor Estie Arkin at Stony Brook (sorry, I don't know what conjecture she was talking about):

For weeks we tried to prove the conjecture (without success) while we left a computer running looking for counter-examples. One morning we came in to find the computer screen flashing: "Counter-example found". We all thought that there must have been a bug in the algorithm, but sure enough, it was a valid counter-example.

I tell this story to my students to emphasize that "proof by lack of counter-example" is not a proof at all!

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[Edit] Here was the response from Estie:

It is mentioned in our paper:
Hamiltonian Triangulations for Fast Rendering
_{E.M. Arkin, M. Held, J.S.B. Mitchell, S.S. Skiena (1994). Algorithms -- ESA'94, Springer-Verlag, LNCS 855, J. van Leeuwen (ed.), pp. 36-47; Utrecht, The Netherlands, Sep 26-28, 1994.}

Specifically section 4 of the paper, that gives an example of a set of points that does not have a so-called "sequential triangulation".

The person who wrote the code I talked about is Martin Held.

Answer 5

The wikipedia article on the Collatz conjecture gives these three examples of conjectures that were disproved with large numbers:

Polya conjecture.

Mertens conjecture.

Skewes number.

Answer 6

In this paper http://arxiv.org/abs/math/0602498 a sequence of integers is proposed, which, when started with $1$ begins like this:

$$1, 1, 2, 1, 1, 2, 2, 2, 3, 1, 1, 2, 1, 1, 2, 2, 2, 3, 2, \dots$$

This is also the sequence A090822 at OEIS. The description there is somewhat better:

Gijswijt's sequence: $a(1) = 1$; for $n>1$, $a(n) =$ largest integer $k$ such that the word $a(1)a(2)...a(n-1)$ is of the form $xy^k$ for words $x$ and $y$ (where $y$ has positive length), i.e. the maximal number of repeating blocks at the end of the sequence so far.

The rules are better explained by demonstration:

$$\color{blue}{1} \to 1$$

$$\color{blue}{1} \color{red}{1} \to 2$$

$$11 \color{blue}{2} \to 1$$

$$112 \color{blue}{1} \to 1$$

$$112 \color{blue}{1}\color{red}{1} \to 2$$

$$ \color{blue}{112}\color{red}{112} \to 2$$

$$11211 \color{blue}{2}\color{red}{2} \to 2$$

$$11211 \color{blue}{2}\color{red}{2}\color{green}{2} \to 3$$

$$11211222 \color{blue}{3} \to 1$$

etc.

What's really surprising:

• $4$ appears for the first time in position $220$
• $5$ appears for the first time in approximately position $10^{10^{23}}$ (sic !)
• The sequence is unbounded

To clarify, this fits the question like this: If someone tried to check this sequence for large numbers experimentally they would most likely conclude that it's bounded, and has no numbers larger than $4$

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Edit

Curiously, this paper explicitly states that the authors initially thought that no number greater than $4$ appears in the sequence.

Answer 7

For an old example, Mersenne made the following conjecture in 1644:

The Mersenne numbers, $M_n=2^n − 1$, are prime for n = 2, 3, 5, 7, 13, 17, 19, 31, 67, 127 and 257, and no others.

Pervushin observed that the Mersenne number at $M_{61}$ is prime, so refuting the conjecture.

$M_{61}$ is quite large by the standards of the day: 2 305 843 009 213 693 951.

According to Wikipedia, there are 51 known Mersenne primes as of 2023.

Answer 8

The first example which came to my mind is the Skewes' number, that is the smallest natural number n for which π(n) > li(n). Wikipedia states that now the limit is near e^727.952, but the first estimation was much higher.

Answer 9

Another class of examples arise from diophantine equations with huge minimal solutions. Thus the conjecture that such an equation is unsolvable in integers has only huge counterexamples. Well-known examples arise from Pell equations, e.g. the smallest solution to the classic Archimedes Cattle problem has 206545 decimal digits, namely 77602714 ... 55081800.

Answer 10

In combinatorics there are quite many such disproven conjectures. The most famous of them are:

1. Tait conjecture:

Any 3-vertex connected planar cubic graph is Hamiltonian

The first counterexample found has 46 vertices. The "least" counterexample known has 38 vertices.

2. Tutte conjecture:

Any bipartite cubic graph is Hamiltonian

The first counterexample found has 96 vertices. The "least" counterexample known has 54 vertices.

3. Heidetniemi conjecture

Chromatic number of tensor product of finite undirected simple graph is equal to the least of chromatic numbers of those graphs.

The first known counterexample to this conjecture has more than $4^{10000}$ vertices

4. Thom conjecture

If two finite undirected simple graphs have conjugate adjacency matrices over $\mathbb{Z}$, then they are isomorphic.

The least known counterexample pair is formed by two trees with 11 vertices.

5. Borsuk conjecture:

Every bounded subset $E$ of $\mathbb{R}^n$can be partitioned into $n+1$ sets, each of which has a smaller diameter, than $E$

In the first counterexample found $n = 1325$. In the "least" counterexample known $n = 64$.

6. Danzer-Gruenbaum conjecture:

If $A \subset \mathbb{R}^n$ and $\forall u, v, w \in A$ $(u - w, v - w) > 0,$ then $|A| \leq 2n - 1$

This statement is not true for any $n \geq 35$

7. The Boolean Pythagorean Triple Conjecture:

There exists $S \subset \mathbb{N}$, such that neither $S$, nor $\mathbb{N} \setminus S$ contain Pythagorean triples

This conjecture was disproved by M. Heule, O. Kullman and V. Marek. They proved, that there do exist such $S \subset \{n \in \mathbb{N}| n \leq k\}$, such that neither $S$, nor $\{n \in \mathbb{N}| n \leq k\} \setminus S$ contain Pythagorean triples, for all $k \leq 7824$, but not for $k = 7825$

8. Burnside conjecture:

Every finitely generated group with period n is finite

This statement is not true for any odd $n \geq 667$ (proved by Adyan and Novikov).

9. Otto Shmidt conjecture:

If all proper subgroups of a group $G$ are isomorphic to $C_p$, where $p$ is a fixed prime number, then $G$ is finite.

Alexander Olshanskii proved, that there are continuum many non-isomorphic counterexamples to this conjecture for any $p > 10^{75}$.

10. Von Neuman conjecture

Any non-amenable group has a free subgroup of rank 2

The least known finitely presented counterexample has 3 generators and 9 relators

11. Word problem conjecture:

Word problem is solvable for any finitely generated group

The "least" counterexample known has 12 generators.

12. Leinster conjecture:

Any Leinster group has even order

The least counterexample known has order 355433039577.

13. Rotman conjecture:

Automorphism groups of all finite groups not isomorphic to the trivial group or $C_2$ have even order

The first counterexample found has order 78125. The least counterexample has order 2187. It is the automorphism group of a group with order 729.

14. Rose conjecture:

Any nontrivial complete finite group has even order

The least counterexample known has order 788953370457.

15. Hilton conjecture

Automorphism group of a non-abelian group is non-abelian

The least counterexample known has order 64.

16)Hughes conjecture:

Suppose $G$ is a finite group and $p$ is a prime number. Then $[G : \langle\{g \in G| g^p \neq e\}\rangle] \in \{1, p, |G|\}$

The least known counterexample has order 142108547152020037174224853515625.

17)$\frac{p-1}{p^2}$-conjecture:

Suppose $p$ is a prime. Then, any finite group $G$ with more than $\frac{p-1}{p^2}|G|$ elements of order $p$ has exponent $p$.

The least counterexample known has order 142108547152020037174224853515625 and $p = 5$. It is the same group that serves counterexample to the Hughes conjecture. Note, that for $p = 2$ and $p = 3$ the statement was proved to be true.

18. Moreto conjecture:

Let $S$ be a finite simple group and $p$ the largest prime divisor of $|S|$. If $G$ is a finite group with the same number of elements of order $p$ as $S$ and $|G| = |S|$, then $G \cong S$

The first counterexample pair constructed is formed by groups of order 20160 (those groups are $A_8$ and $L_3(4)$)

19. This false statement is not a conjecture, but rather a popular mistake done by many people, who have just started learning group theory:

All elements of the commutant of any finite group are commutators

The least counterexample has order 96.

If the numbers mentioned in this text do not impress you, please, do not feel disappointed: there are complex combinatorial objects "hidden" behind them.

Answer 11

It is well known that Goldbach's conjecture is one of the oldest unsolved problems in mathematics. A counterexample if it exists it will be a number greater than $4\cdot10^{18}$.

What is not well-known is that Goldbach made another conjecture which turned out to be false. The conjecture was

All odd numbers are either prime, or can be expressed as the sum of a prime and twice a square.

The first of only two known counterexamples is $5777$ (The second being $5993$).

This number is not "extremely large" for today's data but surely it was on 1752 when Goldbach proposed this conjecture in a letter to Euler who failed to find the counterexample. It was found a century later in 1856 by Moritz Abraham Stern (see this). The prime numbers that cannot be written as a sum of a (smaller) prime and twice a square are called Stern primes. It is believed that there are only finitely many Stern primes.

Answer 12

I was so pissed after testing one of my own conjectures that I remembered this question and wanted to post it here.

I conjectured after numerical observations that for every prime p, and integers $k \ge 1, n \ge 1$, that $$ p^k \, || 2^n-1 \quad \Longleftrightarrow \quad p^{k-1} \, || \, n \quad and \quad O(2,p) \, |\, n, $$ where $O(2,p)$ is the least integer $m$ such that $2^m \equiv 1 \pmod p$, and $||$ stands for exact division (i.e. $a^k \, | \, m$ but $a^{k+1} \, \nmid \, m$ is written $p^k \, || \, m$). This conjecture happens to be true for the first $180$ primes and the first $3000$ multiples of $O(2,p)$ (when $n$ is not a multiple of $O(2,p)$ we already know what happens). But it so happens that $1093$ is prime, that $O(2,1093) = 364$ and $2^{364} \equiv 1 \pmod {1093^2}$, so that the statement above is not true when $k=1$, $n = 364$ and $p=1093$ because the division on the LHS is not exact.

Answer 13

I don't know if I would consider this accessible or 'large', but the counterexample of Adyan to the famous General Burnside Problem in group theory requires an odd exponent greater than or equal to 665. The "shorter" counterexample (proof) due to Olshanskii requires an exponent greater than $10^{10}$. The reason for the large number in the latter proof is essentially due to 'large scale' consequences of Gauss-Bonnet theorem for certain planar graphs expressing relations in groups. It may be that a finer analysis can show that a counterexample can occur at exponent as low as 5, but this is still not known.

This is probably essentially different than what you are asking, since we aren't forced to consider 665 because the cases 1-664 are known to be true. I thought it may be fun to point out, here, though!

Answer 14

Further counterexamples can be found here: https://mathoverflow.net/questions/15444/the-phenomena-of-eventual-counterexamples

Answer 15

Let’s take the number $12$. This number is not prime. It is a composite number, equal to $2^2\times 3$. Also, $121$ is not prime either. It is equal to $11^2$. And $1211$ is not prime as well. It is equal to $7\times 173$. Now you might notice a pattern here.

Let $$12\,\|\, \underbrace{1\,\|\, 1\,\| 1\,\|\cdots}_{k\text{ times}}\tag{$\star$}$$ such that $a\,\|\, b = \left\{10^na + b : b \text{ has $n$ digits}\right\}$. Then, for all $k\in\mathbb{Z}_{>1}$, Eq. $(\star)$ will never be prime and always remain composite ($k > 1$ since $121$ is trivially not prime).

This conjecture sounds reasonable, but the first counter-example is obtained when $k = 136$, for which Eq. $(\star)$ is finally a prime number.

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Let’s also look at the equation, $$\frac{a}{b+c} + \frac{b}{a + c} + \frac{c}{a + b} = 4.\tag{$\star\star$}$$ It was conjectured that if we set $\{a, b, c\}\subset \mathbb{Z}^+$ then there do not exist such values of $a$, $b$, and $c$ to satisfy Eq. $(\star\star)$.

However, there exists a counter-example, and the following equation shows the smallest values of $a$, $b$, and $c$. $$(a, b, c)= (\ldots,\ldots,\ldots)$$ such that $$a =$$ $$154476802108746166441951315019919837485664325669565431700026634898253202035277999.$$ $$b =$$ $$36875131794129999827197811565225474825492979968971970996283137471637224634055579.$$ $$c =$$ $$4373612677928697257861252602371390152816537558161613618621437993378423467772036.$$

Edit:

A link on the $MSE$ discussing this particular example can be found here, and a similar link on the $MO$ can be found here (credit to @B.Mehta).

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So yeah, to sum it all up, there are tons of conjectures disproven by large (and very large) counter-examples :)

Answer 16

I had a conjecture that for any two natural numbers with the same least prime factor, there must be at least one number in between them with a higher least prime factor. It seemed extremely robust for small numbers and gave every indication via empirical trends that it would hold for arbitrarily large numbers as well.

Just this morning, I discovered a counterexample at 724968762211953720363081773921156853174119094876349. While this may not be the smallest one possible, it's easy to show that any counterexample that does exist can't be too much smaller. I was amazed to see such a big number pop out of a relatively simple problem statement.

Answer 17

Here's a recent one I didn't see on either page. The following are true statements:

$\displaystyle{ \int_0^\infty \frac{\sin t}{t} \, dt = \frac{\pi}{2} }$
$\displaystyle{ \int_0^\infty \frac{\sin t}{t} \, \frac{\sin \left(\frac{t}{101}\right)}{\frac{t}{101}} \, dt = \frac{\pi}{2} }$
$\displaystyle{ \int_0^\infty \frac{\sin t}{t} \, \frac{\sin \left(\frac{t}{101}\right)}{\frac{t}{101}} \, \frac{\sin \left(\frac{t}{201}\right)}{\frac{t}{201}} \, dt = \frac{\pi}{2} }$
$\displaystyle{ \int_0^\infty \frac{\sin t}{t} \, \frac{\sin \left(\frac{t}{101}\right)}{\frac{t}{101}} \, \frac{\sin \left(\frac{t}{201}\right)}{\frac{t}{201}} \, \frac{\sin \left(\frac{t}{301}\right)}{\frac{t}{301}} \, dt = \frac{\pi}{2} }$

Is it true that

$\forall n,\displaystyle{ \int_0^\infty \prod_0^n\frac{\sin \left(\frac{t}{100n+1}\right)}{\frac{t}{100n+1}} \, dt = \frac{\pi}{2} }$

?

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No, it's not. However, you could have a computer calculating this for the rest of your life and never find a counter-example. The first counter-example for n has 43 digits.

I found this example here. It was specially constructed to have a large counter-example - the page gives a way to construct similar statements with arbitrarily large counter-examples.

Answer 18

A case where you can "dial in" a large counterexample involves this theorem:

"A natural number is square if and only if it is a $p$-adic square for all primes $p$."

A $p$-adic square is a number $n$ for which $x^2\equiv n\bmod p^k$ has a solution for any positive $k$.

If we include all primes $p$ we have no counterexamples, but suppose we are computationally testing for squares and we have only space and time to include finitely many primes. How high we can go before we get a non-square number that slips through the sieve depends on how many primes we include in our test.

With $p=2$ as the only prime base, the first non-square we miss is $17$. Putting $p=3$ in addition to $p=2$ raises that threshold to $73$. Using $2,3,5,7$ gives $1009$ as the first "false positive". The numbers appear to be growing fast enough to make the first counterexample large with a fairly modest number of primes. See http://oeis.org/A002189 for more details.

Answer 19

Fermat conjectured that $F_n=2^{2^n}+1$ is prime for all $n$,

but Euler showed that $ F_{5}=2^{2^{5}}+1=2^{32}+1=4294967297=641\times 6700417.$

Answer 20

One of Euler conjectures was disproven with a huge counterexample.