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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    $\begingroup$ You can give your friend this 'conjecture': All integers are smaller than n. Counter example n. Here n = something extremely large. ;) $\endgroup$
    – Chao Xu
    Commented Jul 22, 2010 at 21:07
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    $\begingroup$ 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. $\endgroup$
    – user2469
    Commented Apr 3, 2011 at 14:05
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    $\begingroup$ @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. $\endgroup$
    – Justin L.
    Commented Apr 4, 2011 at 2:56
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    $\begingroup$ 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. boinc.thesonntags.com/collatz/high_steppers.php $\endgroup$ Commented Feb 21, 2012 at 14:00
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    $\begingroup$ @ChaoXu Pretty sure your conjecture is true as I have tested up to every $n-1$ and found no counterexamples so far. $\endgroup$ Commented Mar 25, 2018 at 8:54

20 Answers 20


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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    $\begingroup$ This one I like a lot, and it's pretty simple to explain :) I also appreciate the historical background offered. $\endgroup$
    – Justin L.
    Commented Jul 29, 2010 at 5:20
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    $\begingroup$ 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. $\endgroup$ Commented Mar 31, 2011 at 16:47
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    $\begingroup$ @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.) $\endgroup$ Commented May 10, 2011 at 7:19
  • $\begingroup$ @ShreevatsaR: I edited as this question has just been bumped up anyway. $\endgroup$ Commented Aug 29, 2011 at 23:03
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    $\begingroup$ It may be reasonable to mention Elkies explicitly here. $\endgroup$ Commented Aug 22, 2017 at 13:05

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 $n=8424432925592889329288197322308900672459420460792433$

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    $\begingroup$ 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 $5299875888670549565548724808121659894902032916925752559262837$. $\endgroup$
    – KCd
    Commented Apr 18, 2013 at 22:19
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    $\begingroup$ 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. $\endgroup$ Commented Jan 12, 2015 at 8:12
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    $\begingroup$ Guys, if the numbers are flying off the screen in the comments... then perhaps a link would be better. $\endgroup$ Commented Apr 8, 2017 at 21:30
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    $\begingroup$ @DSR, seems unlikely to me. There are much simpler ways to get pairs of relatively prime numbers. E.g., $n$ and $n+1$. $\endgroup$ Commented Jan 27, 2018 at 22:45
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    $\begingroup$ @KCd Can I have some more explanation on the connection to the resultant? $\endgroup$ Commented Apr 14, 2018 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.

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    $\begingroup$ There's a nice article about "prime number races" on the MAA Writing Awards website. $\endgroup$ Commented Jul 29, 2010 at 1:30
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    $\begingroup$ Also we can disprove this conjecture without finding a counter-example, using this argument applied to $\sum_{p^k} \chi(p^k) p^{-sk} = -\frac{L'}{L}(s,\chi)$ once we have shown $L(s,\chi)$ doesn't vanish on $(0,1)$ @ShreevatsaR $\endgroup$
    – reuns
    Commented Jul 13, 2017 at 5:33

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.

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    $\begingroup$ Please ask Estie Arkin to identify the conjecture. It's a great story, but would be more useful if we knew the precise counterexample. $\endgroup$ Commented Aug 6, 2010 at 18:44
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    $\begingroup$ @Joseph: See edit $\endgroup$ Commented Mar 31, 2011 at 16:33
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    $\begingroup$ Thanks for asking the source! $\endgroup$ Commented May 10, 2011 at 7:21
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    $\begingroup$ Interesting statement. It highlights the difference between Physics and Mathematics: In Physics, ALL proofs are proofs by lack of counter-example! :) $\endgroup$
    – Wildcard
    Commented Aug 23, 2017 at 4:59

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.

  • $\begingroup$ Numberphile did episodes on two of these: Merten's Conjecture, Skewes' Number $\endgroup$ Commented May 15, 2020 at 7:59
  • $\begingroup$ The Pólya conjecture seems like a bad example because it has counterexamples under $10^9$; they just weren't the first to be found. Skewes's number was also just an upper bound, and the best lower bound for that problem seems to be just $10^{14}$ (per MathWorld). $\endgroup$
    – benrg
    Commented Jun 5, 2023 at 16:47

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.

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    $\begingroup$ 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. $\endgroup$
    – joriki
    Commented Feb 17, 2020 at 16:10
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    $\begingroup$ This reminds me of the inverse ackermann function. I wonder if it has a similar complexity to it $\endgroup$
    – gist076923
    Commented Feb 8, 2023 at 19:53
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    $\begingroup$ @gist076923 We have to compare the inverse Ackermann function to the sequence of the highest number reached so far, since this sequence is not monotonic. However, once we do, this sequence is estimated to contain element $t$ for the first time at around position $2^{3^{4^{5^{\dots^{t-1}}}}}$, which leads to an inverse-power-tower rate of growth: slow, but much faster than inverse Ackermann. Most numbers that are practical to write down will appear between the $A(4,4)$-th and $A(5,5)$-th positions. $\endgroup$ Commented Sep 9, 2023 at 19:22

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.

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    $\begingroup$ 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. $\endgroup$
    – Justin L.
    Commented Jul 23, 2010 at 19:21
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    $\begingroup$ I doubt you mean Euclid, since he lived more than a thousand years before Mersenne. Euler, I presume? $\endgroup$ Commented Jul 29, 2010 at 7:07
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    $\begingroup$ As of November 2014 there are 48 known Mersenne primes with M{57,885,161} added in January 2013. $\endgroup$
    – Mr. Llama
    Commented Nov 11, 2014 at 16:14
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    $\begingroup$ @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. $\endgroup$
    – Mr Pie
    Commented Feb 19, 2018 at 14:19
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    $\begingroup$ 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? $\endgroup$
    – Falco
    Commented Jul 31, 2019 at 10:50

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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    $\begingroup$ Without RH, large Skewe's number is $10^{10^{10^{964}}}$. $\endgroup$ Commented Apr 8, 2017 at 21:31
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    $\begingroup$ @SimplyBeautifulArt: well, that paper gives the upper boundary of $\exp(727.951336108)$ withour RH and $\exp(727.951335622)$ with RH (Lemma 9.6), if I understand it correctly. $\endgroup$ Commented Jun 16, 2017 at 14:53
  • $\begingroup$ @colt_browning I'm just copying from the Wikipedia linked in this answer. $\endgroup$ Commented Jun 16, 2017 at 14:59
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    $\begingroup$ @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. $\endgroup$ Commented Jun 16, 2017 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.

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    $\begingroup$ 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 $\alpha$. $\endgroup$
    – yo'
    Commented Aug 23, 2017 at 7:08

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.

  1. Tutte conjecture:

Any bipartite cubic graph is Hamiltonian

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

  1. 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

  1. 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.

  1. 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 $n = 64$.

  1. 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 $n \geq 35$

  1. 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 $k = 7825$

  1. 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).

  1. 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 $p > 10^{75}$.

  1. 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

  1. Word problem conjecture:

Word problem is solvable for any finitely generated group

The "least" counterexample known has 12 generators.

  1. Leinster conjecture:

Any Leinster group has even order

The least counterexample known has order 355433039577.

  1. 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.

  1. Rose conjecture:

Any nontrivial complete finite group has even order

The least counterexample known has order 788953370457.

  1. 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.

  1. 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 $L_3(4)$)

  1. 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.

  • $\begingroup$ Multiple sources suggest that the minimal order of a counterexample to (11) is 96, e.g. math.stackexchange.com/questions/7811/…. Do you have a source for the Rotman conjecture? I can't find any information about it elsewhere. $\endgroup$ Commented Apr 21, 2019 at 1:17
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    $\begingroup$ @RavenclawPrefect, the sources of all mentioned counterexamples to Rotman conjecture are listed in answers to this question: math.stackexchange.com/q/2165784/407165 $\endgroup$ Commented Apr 21, 2019 at 12:09
  • $\begingroup$ 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). $\endgroup$ Commented Aug 7, 2019 at 0:16
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    $\begingroup$ 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. $\endgroup$ Commented Aug 7, 2019 at 0:20
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    $\begingroup$ 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/ $\endgroup$ Commented Aug 20, 2019 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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    $\begingroup$ Why isn’t 1 a counterexample? $\endgroup$
    – DonielF
    Commented Aug 21, 2017 at 3:12
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    $\begingroup$ @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'. $\endgroup$
    – P..
    Commented Aug 21, 2017 at 3:46

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.

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    $\begingroup$ @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. $\endgroup$ Commented Jul 31, 2012 at 7:57
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    $\begingroup$ Maybe you can just exclud'em: Let $p$ not be a Wiefereich prime, then$\color{green}{.}\color{goldenrod}{.}\color{red}{.}$ $\endgroup$
    – draks ...
    Commented Jul 31, 2012 at 8:01
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    $\begingroup$ @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. $\endgroup$ Commented Jul 31, 2012 at 8:05
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    $\begingroup$ so you won't like Wolstenholme primes as well...anyway, good luck for research. $\endgroup$
    – draks ...
    Commented Jul 31, 2012 at 8:08

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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    $\begingroup$ 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. $\endgroup$ Commented Aug 6, 2019 at 23:59

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


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 :)

  • $\begingroup$ What’s the name of the (⋆⋆) conjecture? Also the a, b, c provided don’t seem to satisfy the equation. $\endgroup$ Commented Mar 25, 2018 at 20:42
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    $\begingroup$ @user477343 With the substitutions provided, I get a value of $3.9997650454...$ on the left hand side. $\endgroup$
    – B. Mehta
    Commented May 8, 2018 at 21:02
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    $\begingroup$ It also looks like you made a transposition error for $c$: this link has $z=43736...$ while you have $c=43763...$. Similarly you can see here $\endgroup$
    – B. Mehta
    Commented May 8, 2018 at 21:08
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    $\begingroup$ @B.Mehta oh, hahah. It was just a typo. Thank you very much for that :) $\endgroup$
    – Mr Pie
    Commented May 8, 2018 at 23:26
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    $\begingroup$ 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. $\endgroup$
    – Rosie F
    Commented Feb 24, 2019 at 19:24

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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    $\begingroup$ What?! This is the reason why I stay away from prime numbers. $\endgroup$ Commented Nov 29, 2023 at 21:12
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    $\begingroup$ Shouldn't the counterexample be two numbers..? $\endgroup$ Commented Feb 18 at 10:01

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.

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    $\begingroup$ 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 oeis.org/A002189 for a tabulation and links and references. $\endgroup$ Commented Aug 7, 2019 at 0:10
  • $\begingroup$ I clicked on the link you included in your edit, and it worked fine for me. $\endgroup$ Commented Aug 7, 2019 at 2:12

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.$


One of Euler conjectures was disproven with a huge counterexample.enter image description here

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    $\begingroup$ Euler's sum of powers conjecture is already mentioned in this answer $\endgroup$
    – Sil
    Commented Jun 1, 2023 at 21:39

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