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?

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Justin L.
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    You can give your friend this 'conjecture': All integers are smaller than n. Counter example n. Here n = something extremely large. ;) – Chao Xu Jul 22 '10 at 21:07
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    Did you really test up to 5.76 × 10^18, or are you quoting someone else's result? I'm assuming you're joking about computing it yourself, but, if you did, I'd like to know how you did it. I've done some interesting things w/ cloud computing, but never THAT interesting. –  Apr 03 '11 at 14:05
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    @barrycarter heh, this question was posted during the site's beta phase as my attempt to seed the site with more questions, and I wasn't being too serious. – Justin L. Apr 04 '11 at 02:56
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    For Reference: http://mathoverflow.net/questions/15444/the-phenomena-of-eventual-counterexamples – JavaMan Aug 29 '11 at 22:07
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    You may have been joking about doing though 10^18 via cloud computing; but an ongoing BOINC project has done an exhaustive search to 2.3*10^21. http://boinc.thesonntags.com/collatz/high_steppers.php – Dan Is Fiddling By Firelight Feb 21 '12 at 14:00
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    Here $\downarrow$ is an example of a fairly large counter-example to the OP's conjecture. https://math.stackexchange.com/questions/2673678/do-n-2m1-and-big2m-bmodm-cdot-n-big-in-n1-3n-1-imply-n-prime The counter-example is the very first comment :) – Mr Pie Mar 06 '18 at 10:00
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    @ChaoXu Pretty sure your conjecture is true as I have tested up to every $n-1$ and found no counterexamples so far. – samerivertwice Mar 25 '18 at 08:54

20 Answers20


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?" :-)

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    This one I like a lot, and it's pretty simple to explain :) I also appreciate the historical background offered. – Justin L. Jul 29 '10 at 05:20
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    On the other hand the numbers involved in the above examples are fairly small (compared to the millions of millions of millions). I find Justin's question interesting and I'm not sure I'm so convinced by those counterexamples which wouldn't take long to find using today's technology. – David Kohler Mar 31 '11 at 16:47
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    @David: 31858749840007945920321 is pretty large, and it took until 1988. Naïvely speaking, you have to try all triples of fourth powers, with the largest number going up to 414560. Finding this counterexample, even with today's technology, would take more than a year on a desktop computer. (Of course, there are presumably number-theoretic insights which make it faster to find counterexamples by computer, but as I saw it, the whole point of the question was that the "simply try as many numbers as you can on a computer" approach need not give the right answer.) – ShreevatsaR May 10 '11 at 07:19
  • @ShreevatsaR: I edited as this question has just been bumped up anyway. – George Lowther Aug 29 '11 at 23:03
  • @ShreevatsaR: no problem. Although, now, it displays badly on my iphone as the first double-dollar sign does not get interpreted as converting to tex. Think that might be because the long numbers are interpreted as phone numbers and not as tex. – George Lowther Aug 30 '11 at 09:17
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    It may be reasonable to mention Elkies explicitly here. – Mikhail Katz Aug 22 '17 at 13:05
  • @MikhailKatz Why do you say so? The counterexample found in 1966 was by Lander and Parkin ([link to paper](http://www.ams.org/journals/bull/1966-72-06/S0002-9904-1966-11654-3/S0002-9904-1966-11654-3.pdf)), published a couple of months before Elkies was born. The smallest $n=4$ example (not relevant to the disproving of the original conjecture; only included in this answer as additional information) found in 1988 was by Roger Frye. No doubt Elkies is relevant to the latter, but all this information is in the linked article. – ShreevatsaR Apr 19 '19 at 16:49

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 $n=8424432925592889329288197322308900672459420460792433$

Joe K
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    And $n^{19} + 6$ and $(n+1)^{16}+9$ are relatively prime until $n = 1578270389554680057141787800241971645032008710129107338825798$ (61 digits), where the two numbers have gcd equal to $5299875888670549565548724808121659894902032916925752559262837$. – KCd Apr 18 '13 at 22:19
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    @KCd: These numbers (for that $n$) have $\gcd=2$. And the smallest value of $n$ with $\gcd>1$ is much much smaller. You intended $n^{19}+6$ and $(n+1)^{19}+6$. Do you know (or Joe K) any reference about this 'conjecture'? – P.. Jun 05 '13 at 09:44
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    This phenomenon is related to the resultant of the polynomials $x^{17}+9$ and $(x+1)^{17}+9$, or of $x^{19}+6$ and $(x+1)^{19}+6$, or of $x^{29}+14$ and $(x+1)^{29}+14$. Polynomials with even a moderately large degree, even if they have small coefficients, can have a truly gigantic resultant. – KCd Jun 05 '13 at 13:59
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    I don't think anyone actually ever conjectured these numbers to be relatively prime. I think they were concocted specifically to illustrate that statements could be plausible, and verifiable out to large $n$, yet still be false. – Gerry Myerson Jan 12 '15 at 08:12
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    Guys, if the numbers are flying off the screen in the comments... then perhaps a link would be better. – Simply Beautiful Art Apr 08 '17 at 21:30
  • @GerryMyerson Why do you say so? Some time back it might have been a good way to get sets of relatively prime numbers. – DS R Jan 27 '18 at 14:44
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    @DSR, seems unlikely to me. There are much simpler ways to get pairs of relatively prime numbers. E.g., $n$ and $n+1$. – Gerry Myerson Jan 27 '18 at 22:45
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    @KCd Can I have some more explanation on the connection to the resultant? – alphacapture Apr 14 '18 at 22:27

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.

Larry Wang
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    There's a nice article about "prime number races" on the [MAA Writing Awards](http://mathdl.maa.org/mathDL/22/?pa=content&sa=viewDocument&nodeId=3147) website. – ShreevatsaR Jul 29 '10 at 01:30
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    Also we can disprove this conjecture without finding a counter-example, using [this](https://math.stackexchange.com/questions/2345525/does-the-mertens-function-have-an-infinite-number-of-integer-zeros/2345591#2345591) argument applied to [$\sum_{p^k} \chi(p^k) p^{-sk} = -\frac{L'}{L}(s,\chi)$](https://en.wikipedia.org/wiki/Dirichlet_L-function) once we have shown $L(s,\chi)$ doesn't vanish on $(0,1)$ @ShreevatsaR – reuns Jul 13 '17 at 05:33

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!

[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.


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.

Eric O. Korman
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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:

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

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

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

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

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

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

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

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

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


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 $4$


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

Yuriy S
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    A somewhat arbitrary but perhaps nevertheless interesting association: The sum of the reciprocals of all known primes is about $4$, whereas the sum of the reciprocals of all primes is unbounded. – joriki Feb 17 '20 at 16:10

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 2018

Daniel Fischer
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Charles Stewart
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    This one is the most accessible/easy to explain of them all...but 257 seems like an arbitrary place to stop the conjecture at so I'm not sure it's as pretty. – Justin L. Jul 23 '10 at 19:21
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    @Justin: Well, Mersenne had an intuition. Sometimes intuitions are sound, and sometimes they're not. – Charles Stewart Jul 23 '10 at 20:02
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    I doubt you mean Euclid, since he lived more than a thousand years before Mersenne. Euler, I presume? – Pete L. Clark Jul 29 '10 at 07:07
  • @Pete: Quite so. – Charles Stewart Aug 08 '10 at 07:59
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    As of November 2014 there are 48 known Mersenne primes with M{57,885,161} added in January 2013. – Mr. Llama Nov 11 '14 at 16:14
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    How did people back then prove that such large numbers are prime? Surely not by exhaustive calculation..? – Epiousios Nov 01 '17 at 08:32
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    @Epiousios the largest calculation ever done by hand was proving that a certain mersenne number was indeed prime. I can’t remember the details, but I think it was by a mathematician with a name *Lucas*. – Mr Pie Feb 19 '18 at 14:19
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    Also, there are now $50$ mersenne primes (with a number below $M_{50}$ that was conjectured to be prime at first). – Mr Pie Feb 19 '18 at 14:20
  • https://primes.utm.edu/notes/proofs/MerDiv.html –  Mar 01 '19 at 00:22
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    I just read the Wikipedia article and I think this post can lead readers to the wrong conclusion (see Justins comment). 1. Mersenne did not stop at 257, but his conjecture was "for n <= 257, there are only these 11 Primes" and the counter-examples to his conjectures are thus all numbers for n <= 257 - for example M257 is actually a composite number and not a prime. 2. How did Euler refute the conjecture by proving M31 is prime, when Mersenne originally stated M31 is prime? – Falco Jul 31 '19 at 10:50
  • @Falco: your second point is certainly valid; according to [this Wikipedia article](https://en.wikipedia.org/wiki/Mersenne_prime), Euler proved that $M_{31}$ is prime; that is a support not a refutation of Mersenne's conjecture; Mersenne made a mistake by not including $M_{61}$ though, as per an earlier version of this answer (though it was not Euler who proved $M_{61}$ prime, so the earlier version of this answer was not entirely correct either) – J. W. Tanner May 22 '20 at 15:57
  • Still no 52nd Mersenne prime :( – HelloWorld May 29 '22 at 12:33

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 e727.952, but the first estimation was much higher.

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    Without RH, large Skewe's number is $10^{10^{10^{964}}}$. – Simply Beautiful Art Apr 08 '17 at 21:31
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    @SimplyBeautifulArt: well, [that paper](https://dx.doi.org/10.1090%2FS0025-5718-2015-02930-5) gives the upper boundary of $\exp(727.951336108)$ withour RH and $\exp(727.951335622)$ with RH (Lemma 9.6), if I understand it correctly. – colt_browning Jun 16 '17 at 14:53
  • @colt_browning I'm just copying [from the Wikipedia linked in this answer](https://en.wikipedia.org/wiki/Skewes%27_number#Skewes.27_numbers). – Simply Beautiful Art Jun 16 '17 at 14:59
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    @SimplyBeautifulArt: the paper referenced by me is in that article, too (and there is even an improved result): see the next section, first paragraph, last three sentences. $10^{10^{10^{964}}}$ is the historical result. – colt_browning Jun 16 '17 at 15:11

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.

Bill Dubuque
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    But from our mathematics point of view, we know that each Pell equation has a solution, and you can actually prescribe a lower bound $\alpha$ and I can give you a squarefree $D$ such that the fundamental unit of $\mathbb{Q}(\sqrt D)$ is greater than $\alpha$. – yo' Aug 23 '17 at 07:08

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 $C_2$ 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.


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.

Chain Markov
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  • Multiple sources suggest that the minimal order of a counterexample to (11) is 96, e.g. https://math.stackexchange.com/questions/7811/derived-subgroup-where-not-every-element-is-a-commutator. Do you have a source for the Rotman conjecture? I can't find any information about it elsewhere. – RavenclawPrefect Apr 21 '19 at 01:17
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    @RavenclawPrefect, the sources of all mentioned counterexamples to Rotman conjecture are listed in answers to this question: https://math.stackexchange.com/q/2165784/407165 – Chain Markov Apr 21 '19 at 12:09
  • Edits have affected the pointers in the preceding comments. I think that what @Raven calls (11) is currently (17), while Yanior's comment refers to (12). – Gerry Myerson Aug 07 '19 at 00:16
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    I also note that this answer begins, "In combinatorics there are quite many such disproven conjectures. The most famous of them are...." but many of the examples cited are from group theory rather than combinatorics. – Gerry Myerson Aug 07 '19 at 00:20
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    For Hedetniemi's conjecture in graph theory its first known counterexample is a graph with more than $4^{10000}$ vertices, see www.quantamagazine.org/mathematician-disproves-hedetniemis-graph-theory-conjecture-20190617/ – Tommy R. Jensen Aug 20 '19 at 18:29

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.

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    Why isn’t 1 a counterexample? – DonielF Aug 21 '17 at 03:12
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    @DonielF Yes sure you are right, but that time 1 was considered a prime number. It was considered a prime number until the beginning of 20th century. So nowadays the conjecture should be stated as 'All odd numbers greater than 1 are either prime, or can be expressed as the sum of a prime and twice a square'. – P.. Aug 21 '17 at 03:46

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.

Patrick Da Silva
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    @draks : I didn't even know such primes existed. God I am damned. This was actually relevant to my research and I expected everything to work out nicely, and now I fall on some famous counter example. Gosh. – Patrick Da Silva Jul 31 '12 at 07:57
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    Maybe you can just exclud'em: *Let $p$ not be a Wiefereich prime, then$\color{green}{.}\color{goldenrod}{.}\color{red}{.}$* – draks ... Jul 31 '12 at 08:01
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    @draks : It's actually not that cool to exclude them given the context where I want to use them.. actually the reason for me to prove this statement is to establish a possible theorem/analogy between primes and irreducible polynomials over $\mathbb Z[x]$ (because there are lots, this one would be one more), so yeah, Wieferich primes... are not cool. – Patrick Da Silva Jul 31 '12 at 08:05
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    so you won't like [Wolstenholme primes](http://en.wikipedia.org/wiki/Wolstenholme_prime) as well...anyway, good luck for research. – draks ... Jul 31 '12 at 08:08

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!

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Jon Bannon
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    American Civil War buffs will be disappointed to learn that the General Burnside Problem was not named after General Ambrose E Burnside, who led the Union troops at the battle of Antietam. – Gerry Myerson Aug 06 '19 at 23:59

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

Doug Chatham
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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.

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 $$a =$$ $$154476802108746166441951315019919837485664325669565431700026634898253202035277999.$$ $$b =$$ $$36875131794129999827197811565225474825492979968971970996283137471637224634055579.$$ $$c =$$ $$4373612677928697257861252602371390152816537558161613618621437993378423467772036.$$


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).

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

Mr Pie
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  • What’s the name of the (⋆⋆) conjecture? Also the a, b, c provided don’t seem to satisfy the equation. – Roman Odaisky Mar 25 '18 at 20:42
  • @RomanOdaisky Well we know that $121 = 11^2$ so I have to include that $k > 1$. I also got the values of $a, b$ and $c$ from a site, as well as the former example. If I can remember, I can show you ..... I might have looked here? $\longrightarrow$ https://mathoverflow.net/questions/15444/examples-of-eventual-counterexamples If you know the values $a, b$ and $c$ don't satisfy the equation, what does the equation actually equal instead when given the substitutions? – Mr Pie Mar 25 '18 at 22:29
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    @user477343 With the substitutions provided, I get a value of $3.9997650454...$ on the left hand side. – B. Mehta May 08 '18 at 21:02
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    It also looks like you made a transposition error for $c$: [this link](https://math.stackexchange.com/questions/2192461/find-answer-of-fracxyz-fracyxz-fraczyx-4) has $z=43736...$ while you have $c=43763...$. Similarly you can see [here](https://mathoverflow.net/questions/227713/estimating-the-size-of-solutions-of-a-diophantine-equation) – B. Mehta May 08 '18 at 21:08
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    @B.Mehta oh, hahah. It was just a typo. Thank you very much for that :) – Mr Pie May 08 '18 at 23:26
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    Your numbers are $(10^k\times109-1)/9$. Similarly, for a given $n$, if all $2^k n-1$ are composite, $n$ is a Riesel number; if all $2^k n+1$ are composite, $n$ is a Sierpinski number. $n=10223$ was proved non-Sierpinski only with the discovery in 2016 that $2^{31172165}n+1$ is prime. – Rosie F Feb 24 '19 at 19:24
  • @RosieF Nice, thanks for the fact! It sound very interesting :P – Mr Pie Feb 24 '19 at 20:37

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.

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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

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


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.


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.

Oscar Lanzi
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    I take it "ccx" is one of those typos that slipped through. I think the numbers you are writing about are known as "pseudosquares", and Hugh Williams did calculations to find the first few dozen terms, and made estimates of the growth rate. See http://oeis.org/A002189 for a tabulation and links and references. – Gerry Myerson Aug 07 '19 at 00:10
  • I clicked on the link you included in your edit, and it worked fine for me. – Gerry Myerson Aug 07 '19 at 02:12

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.$

J. W. Tanner
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I have a conjecture that can be maybe disproved by an extremely huge counterexample:

Consider sequence https://oeis.org/A301806

I conjecture that if n divides a(n), then a(n) +1 is prime.

Enzo Creti
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