In mathematics, is there any conjecture about the existence of an object that was proven to exist but that has not been explicitly constructed to this day? Here object could be any mathematical object, such as a number, function, algorithm, or even proof.

4The exact meaning of your question is hard to understand. You mentioned that english is not you mother tongue, so I recommend the following: extend your post by the questions asked in *your language* with the request for someone to translate it properly. – M. Winter Aug 12 '18 at 22:05

6I don't understand the question. What do you mean by a problem that is "solvable"? (Note that this term has vernacular meaning in English, as well as several precise mathematical meanings, depending on context.) As it stands, it reads to me like you are asking about the existence of [nonconstructive proofs](http://mathworld.wolfram.com/NonconstructiveProof.html). – Xander Henderson Aug 12 '18 at 22:05

6@M. Winter For example, if Galois theory was discovered before Ferrari's formula, We know that an equation of $4$ degrees can solvable but we did not know the formula yet.. – lone student Aug 12 '18 at 22:15

@Xander Henderson I tried to give an example for the understanding of the question... – lone student Aug 12 '18 at 22:17

4In that case, this question (although a bit specific) might be interesting [Proving the existence of a proof without actually giving a proof](https://math.stackexchange.com/questions/1642225/provingtheexistenceofaproofwithoutactuallygivingaproof), or perhaps [Are there any nonconstructive proofs for which an example was never constructed?](https://math.stackexchange.com/questions/575835/arethereanynonconstructiveproofsforwhichanexamplewasneverconstructed/) – Sil Aug 12 '18 at 22:22

@Sil I think you understand me correct..Do you think the question should be edited? – lone student Aug 12 '18 at 22:24

1@Student I tried to modify it a bit, please check it and update accordingly. Maybe I expanded the scope too much :) I am not native speaker as well, so I would consider it a try – Sil Aug 12 '18 at 22:29

@Sil Thank you very much! :) – lone student Aug 12 '18 at 22:32

4Another closely related question is [Is there any conjecture that we know is provable/disprovable but we haven't found a proof of yet?](https://math.stackexchange.com/q/2273525/26369) – Mark S. Aug 13 '18 at 01:24

@Mark S. Thank you so much for suggestion. – lone student Aug 13 '18 at 10:59

2This reminds me of studying ODEs. Very often we would prove both existence and uniqueness of a solution to the ODE but we would never actually solve the ODE itself. Though this doesn't actually mean it wasn't solvable, only that we proved that there was a solution without actually knowing said solution. – nurdyguy Aug 13 '18 at 21:06

1Another kind of example: With usual axioms, a [wellordering](https://en.wikipedia.org/wiki/Wellorder) of the set $\mathbb{R}$ (or just any set equipotent to $2^\mathbb{N}$) exists. However such a wellordering cannot be given explicitly. If I understand it correctly (which I probably do not), it is independent of the usually chosen axioms whether a ___definable___ wellordering of $\mathbb{R}$ exists. – Jeppe Stig Nielsen Aug 13 '18 at 21:51

Would the Busy Beaver function values qualify? (Given $k$, $BB(k)$ may be defined as the smallest natural number such that any twoletter Turing Machine, with $k$ internal states starting on an empty tape, which continues beyond $BB(k)$ steps will continue indefinitely). We know they exist for any $k\in \Bbb N$, but we only know the first four values, and we also now that, for instance, $BB(8000)$ [eludes standard (ZF) set theory](https://www.scottaaronson.com/blog/?p=2725). – Arthur Aug 17 '18 at 11:32
11 Answers
It is known that there is an even integer $n\le246$ such that there are infinitely many primes $p$ such that the next prime is $p+n$, but there is no specific $n$ which has been proved to work (although everyone believes that every even $n\ge2$ actually works).
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3You can state it three ways: There are infinitely many primes that are at most 246 apart (at most 2 would be twin primes). There is an integer n ≤ 246 such that infinitely many primes are exactly n apart (n = 246 would be twin primes). Of the twin prime conjecture, (n,n+4) prime conjecture, ..., (n, n+246) conjecture, at least one is true. – gnasher729 Aug 15 '18 at 22:55

4@jkd, if we could get that 246 down to 2, we'd have the twin prime conjecture. This is as close as anyone has been able to get to it. – Gerry Myerson Aug 16 '18 at 04:37

2It has been proven that there exists an $n \leq 246$, but has it been proven that this is provable for a specific $n$? – JiK Aug 16 '18 at 13:01

There is no specific $n$ for which it has been proved. But I'm not sure I understand your question. – Gerry Myerson Aug 16 '18 at 21:28

@JiK "proven to be provable" and "proven to be true" is the same; and as stated in the answer, no. – user202729 Aug 17 '18 at 09:24

1@JiK I had the same concern, until I read the actual question post, not just the title. "is there any conjecture about the existence of an object that was proven to exist but that has not been explicitly constructed to this day?" We know such an $n$ exists, and that it is smaller than $246$, but we haven't explicitly constructed such a number. – Arthur Aug 17 '18 at 10:55

@user202729 That's not what I meant; I was asking whether it was proven that it's provable for the right $n$ without knowing what that $n$ is. For example, the proposition "the Ramsey number $R(5,5)$ is $x$" is certainly provable for some $x$, but we don't know what that $x$ is. – JiK Aug 17 '18 at 12:44

@JiK, "the right $n$" makes it sound like there's only one $n$ that works. But everyone believes that every even $n$ works. – Gerry Myerson Aug 17 '18 at 13:15
I am not sure this answers your question but it seems to come close. The Wikipedia article Skewes number states
In number theory, Skewes's number is any of several extremely large numbers used by the South African mathematician Stanley Skewes as upper bounds for the smallest natural number $x$ for which $\, \pi(x) > \textrm{li}(x) \,$ where $\pi$ is the primecounting function and $\, \textrm{li} \,$ is the logarithmic integral function.
It further states that
All numerical evidence then available seemed to suggest that $\, \pi(x) \,$ was always less than $\, \textrm{li}(x). \,$ Littlewood's proof did not, however, exhibit a concrete such number $x$.
The problem is that even though the existence of the number $x$ has been proven, we only know huge upper bounds on the first such number.
Perhaps a better example is from the Wikipedia article Ramsey theory where
Problems in Ramsey theory typically ask a question of the form: "how many elements of some structure must there be to guarantee that a particular property will hold?"
These Ramsey numbers have the property that
Results in Ramsey theory typically have two primary characteristics. Firstly, they are nonconstructive: they may show that some structure exists, but they give no process for finding this structure (other than bruteforce search). For instance, the pigeonhole principle is of this form. Secondly, while Ramsey theory results do say that sufficiently large objects must necessarily contain a given structure, often the proof of these results requires these objects to be enormously large – bounds that grow exponentially, or even as fast as the Ackermann function are not uncommon.
Thus, in theory, some of these enumerative numbers exist but they are too huge to write explicitly. In other cases, there exist small bounds but it is very hard to narrow the bounds.
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5See also the idea of the minimax value or "God's Number" for combinatorial problems like puzzle games. Often the exact value is not known, but an upper bound is. E.g. until 2010 is was not known that the maximum number of steps to solve any Rubik's Cube is 20; the first known upper bound for this value was 277. https://en.wikipedia.org/wiki/Optimal_solutions_for_Rubik%27s_Cube – m69 is disappointed in SE Aug 12 '18 at 23:27

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2@m69: Yep. In general, finding the minimum natural number that satisfies some complicated computable property (after proving nonconstructively that one exists) is explicitly solvable but whose solution might take a long time to find, especially if the property has combinatorial explosion. The minimum number of moves to solve a 3x3x3 Rubik's cube is known today, but surely we will never find the minimum number of moves to solve a 7x7x7 Rubik's cube! – user21820 Aug 13 '18 at 10:18
It's known that no matter the size of the board, the first player has a winning strategy in Chomp, but an explicit winning strategy is only known for smallish boards.
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21One can apply Zermelo's theorem to chess, Go, and other finite games. They are all determined, but we don't know who has the winning strategy (or at least, nonlosing one) – Asaf Karagila Aug 13 '18 at 10:13

5or any other ultraweakly solved game https://en.wikipedia.org/wiki/Solved_game – Charon ME Aug 15 '18 at 06:42
Well, this might be an example of what you are asking for:
Take the question "are there irrational numbers $a,b$ such that $a^b$ is rational?"
A quick way to see that there are is to consider $\alpha =\sqrt 2^ {\sqrt 2}$.
Either $\alpha$ is rational or irrational. If it is rational, then we are done. If it is irrational then consider $\alpha^{\sqrt 2}=2$. Either way, we can find an example.
To be sure, it is possible (though quite difficult) to show that $\alpha$ is irrational and there are other ways to get direct examples (such as $e^{\ln 2}$) that settle the problem . But this indirect proof is so simple it is, I think, worth study.
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11@lhf Yes, I pointed that out in my post. Still though, the simplicity of this argument has always appealed to me. Even an expression like $e^{\ln 2}$ needs hard work...unless I am missing something, you need the transcendence of $e$ to show that $\ln 2 $ is irrational. (though I suppose an example like $(\sqrt 3)^{\log_{3} 4}=2$ is simple enough). – lulu Aug 13 '18 at 16:32

3I think a better example would be that at least one of `e+pi` and `e*pi` is irrational, since it's not yet known which is irrational, or if both are. – BallpointBen Aug 15 '18 at 19:32

1@BallpointBen Ok, though both underlying theorems there ("there exist irrationals whose sum is irrational, and there exist irrationals whose product is irrational") are extremely easy to prove. – lulu Aug 15 '18 at 19:44
Define the following number:
$$K = \begin{cases} 1, & \text{if Riemann Hypothesis is true} \\ 0, & \text{if it is false} \end{cases}$$
Since the Riemann Hypothesis is either true or false, the above constant $K$ is welldefined, mathematically. It is one specific number. But no one knows whether it is 0 or 1 (because knowing it means knowing the answer to the hypothesis).
I know this example is very artificial, but nevertheless I thought it was interesting to share  basically any unsolved problem in mathematics can be converted in an object that certainly exists but knowing what it is becomes equivalent to solving the problem itself in the first place.
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5Is this correct? Can we say any hypothesis is true? Isn't the correct wording "can be proven within an axiom system"? And if that's so, what if it is like the continuum hypothesis? – chx Aug 13 '18 at 19:44

45This fails the "the problem was proved to be solvable" requirement. – Reinstate Monica Aug 13 '18 at 20:06

5@Solomonoff'sSecret: Trivially fixable. Given the definition of K, consider `K*K = K`. Obviously true, but we don't know whether the proof is `1*1=1` or `0*0=0`. – MSalters Aug 14 '18 at 15:21

6@MSalters I think that still doesn't qualify. You give a certain positive proof that $K^2 = K$, not a proof that something or its negation can be proven without knowing which. – Reinstate Monica Aug 14 '18 at 16:25

I secretly wish the Riemann Hypothesis is independent of ZFC. That would be an interesting earthquake to witness. – Aug 14 '18 at 20:49

@RobertWolfe We already know that the Continuum Hypothesis is independent of ZFC, would it be more shocking if RH was as well? Also, it seems that if it is false then it is provably so. https://mathoverflow.net/questions/79685/cantheriemannhypothesisbeundecidable – badjohn Aug 15 '18 at 20:46
There are sentences, called Parikh sentences for which there are short proofs that the sentence is provable but all proofs are very long. My discussion is based on page 17 of Noson Yanofsky's arXiv article A Universal Approach to SelfReferential Paradoxes, Incompleteness and Fixed Points. He constructs a sentence $\mathcal{C}_{n}$ that says, "I do not have a proof of myself that is shorter than $n$. Then choose a large $n$, say $P!$, where $P$ is a reasonable estimate of the number of electrons in the observable universe.
In the above article Yanofsky states his discussion is based on R. Parikh's Existence and Feasiblity in Arithmetic.
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But there *is* an algorithm to construct this sentence, right? Just try all natural numbers until you find the correct one. – Federico Poloni Aug 12 '18 at 23:38


5Where does the quoted sentence end? (Or is it intended that all subsequent text, including this comment, is part of the sentence?) – Eric Towers Aug 14 '18 at 16:21
Not every mathematical object can be explicitly constructed. For example, there are uncountably many real numbers, but only countably many finite strings in a given finite alphabet (let's say ASCII). An explicit construction of a real number is a finite ASCII string, and by definition it can only construct at most one real number, so that leaves uncountably many real numbers that can't be explicitly constructed.
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5How do you reconcile this with models of set theory where *every* set is definable (including every real number)? – Asaf Karagila Aug 13 '18 at 09:23

1@AsafKaragila _you don't_, that's the point: such models are “unrealistic” in the sense that not every set they call “definable” could actually be specified by a human. – leftaroundabout Aug 13 '18 at 14:22


The catch is that these models are "really" countable, but an enumeration of the reals can't be defined within the model. – Robert Israel Aug 13 '18 at 16:45

2Robert, I'm not entirely sure what you mean. Yes, these models are countable, but it just might be that the universe of mathematics [that you are working in, if you prefer to avoid Platonism,] is such model (of course, we have no way of proving something like that, but the point is that there is no way of disproving it either). – Asaf Karagila Aug 13 '18 at 18:16

There's an important difference here between a proof saying that something "exists" and having a "construction" of a specific object. Maybe part of the trouble is that "define"/"definition"/etc. could be used in either sense, depending on context. – aschepler Aug 14 '18 at 10:35

I found this answer very relatable, and it also made me realize that the title of the original question does not match the body of the question: proofs v. constructions. – Seth Aug 14 '18 at 15:35
One could say that for many problems that are NP hard, it is often known that the solution exists, but (in many cases) we don't have the computational power to evaluate it exactly, so we use algorithms that we hope are close to the optimal solution.
Probably one of the most commonly known NP hard problems is the traveling salesman. In this problem, we want to find the shortest route connecting a set of nodes. Clearly, a solution does exist; in fact, it's really easy to consider a finite set of candidate solutions, one of which must have the shortest route.
The problem is that generally speaking, this finite set we need to consider is still very large, and we must evaluate the length on the routes for each element in these sets to obtain the answer. Simplistic approaches result in a set of size $O(n!)$. Reading through the wikipedia page on the problem, it seems that there are methods that reduce this to $O(n^2 2^n)$. Note that this means the human race does not have enough computational power to solve this problem for a couple hundred nodes within our lifetime [citation needed].
But a solution must exist!
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yes, pointing to the NP hard class is a really good hint. And there are many other problems. NP complete problems fit in too, like the Clique, Map Colorability and Packing problems. We know there is a solution, we are just not able to find it yet. All we have are approximation algorithms (which I dont like because they are not exact) – undefined Aug 14 '18 at 11:08

1As a computer scientist I would say we do know how to solve TSP, it just takes a huge amount of time to do so exhaustively. Thus the "direct solution" mentioned in the title is known, although very impractical. – Andrea Lazzarotto Aug 14 '18 at 21:41

@AndreaLazzarotto do you have a link or something to a working (solution) algorithm for TSP? – undefined Aug 15 '18 at 06:39

4@undefined, every instance of TSP is a finite problem, so, in principle, you can just look at every single one of the possible paths through all the points, and pick out the shortest one. That's a guaranteedtowork algorithm for TSP. It's just a tiny bit slow, that's all. – Gerry Myerson Aug 15 '18 at 07:23

4@undefined, enumerate all possible solutions, compute their weight, chose the best one. – Andrea Lazzarotto Aug 15 '18 at 10:12
Maybe the Picard–Lindelöf theorem is what you are looking for. It is an existence theorem for solutions of differential equations and it uses the Banach fixed point theorem. It basically says, under certain conditions differential equations have a unique solution. But to find the solution for any given differential equation is rather hard (and not alway possible).
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4What do you mean by "find the solution"? The solution is a certain function, which can be approximated to arbitrary accuracy by numerical methods. It might not have a "closed form" expression, though. Well, not all functions have closed form expressions, just as not all real numbers are rational. – Robert Israel Aug 14 '18 at 15:32

1@RobertIsrael True. Do you know if there is a full classification of differential equations if they have closed form solution or dont admit one. How would you interpret the original question instead? – lalala Aug 15 '18 at 10:34

"Closed form" is not precisely defined. There are algorithms in some cases, e.g. Kovacic's for determining whether a secondorder linear homogeneous d.e. with rational coefficients has Liouvillian solutions, but AFAIK there is no "full classification". – Robert Israel Aug 15 '18 at 15:59
It is known that the game of Hex cannot end in a draw (visually this makes sense, either there is a path side to side or up and down). This implies that the first player has a winning strategy since the first player can always 'steal' the second player's strategy. However, an explicit strategy is only known for very small boards.
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stackzebra posted an answer, which has been deleted by a moderator. It's a very simple answer, but, I think, a good one, so I'm going to post it (in a modified form) here:
It has been proved that there is a prime greater than $10^{10^{100}}$ – indeed, Euclid proved that there are infinitely many such primes – but no example has ever been given. The largest known primes are as infinitesimals compared to this number.
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Indeed, if $n=10^{10^{100}}$, it has been proved that there is a prime between $n$ and $2n$, and of course it's just a finite computation to find one, but no one has ever carried out that computation. – Gerry Myerson Dec 08 '19 at 23:05