# Open mathematical questions for which we really, really have no idea what the answer is

There is no shortage of open problems in mathematics. While a formal proof for any of them remains elusive, with the "yes/no" questions among them mathematicians are typically not working in both directions but rather have a pretty clear idea of what the answer should be. Famous conjectures such as Riemann and Collatz are supported by some very convincing heuristics, leading mathematicians to believe in their validity so strongly that they write papers based on the assumption that they are true. For other wide open problems such as $P$ vs. $NP$, one side ($P=NP$ in this case) is usually considered so unlikely to be true that almost nobody seriously works on it. Of course, whenever a "conjecture" is attached to an open question that already implies that one answer is preferred over the other – people don't conjecture $A$ and $\neg A$ simultaneously.

Are there any open mathematical questions with a yes/no answer for which we have no good reason to assume one or the other, for which we really have absolutely no idea what the true answer might be?

• Maybe are there infinitely many Fermat primes. – André Nicolas Oct 27 '14 at 5:18
• From Wikipedia: "The following heuristic argument suggests there are only finitely many Fermat primes..." – user139000 Oct 27 '14 at 5:19
• @DavidH: My money is on the "not" side. – user139000 Oct 27 '14 at 5:21
• What about the values of sufficiently large Ramsey numbers? See, e.g., the Erdős quote about how we're better off trying to destroy the omnicidal aliens who want to know $R(6,6)$ than trying to compute it for them... – Micah Oct 27 '14 at 5:22
• This is outside of any expertise I have, so I'll put it as a comment rather than an answer. It is not known whether the Burnside group $B(2,5)$ is finite. Some groups $B(m,n)$ are, some aren't. I don't know whether the experts have a consensus on $B(2,5)$. For some details, see www-history.mcs.st-and.ac.uk/HistTopics/Burnside_problem.html – Gerry Myerson Oct 27 '14 at 6:15

I believe whether or not the Thompson group $F$ is amenable is such question. The paper/article "WHAT is... Thompson's Group" mentions that at a conference devoted to the group there was a poll in which 12 said it was and 12 said it was not. There are in fact papers claiming (at least at the time) to have proofs for both sides. Here are some posts to get an idea of the "controversy": 1, 2,3.

• It seems to me that whenever someone utters the words "Thompson's group" (no matter which one they are actually discussing!) someone else will ask them about the amenability of $F$. I think this problem is so interesting simply because of the number of false attacks on it (including by Justin Moore, who spoke at the ICM), and because those attacks claim opposite results. – user1729 Nov 11 '14 at 10:54
• Accepted, as this answer not only gives an open problem, but also evidence that either side could be true, based on current knowledge. – user139000 Feb 27 '15 at 10:31

In 4 dimensions, it is an open question as to whether there are any exotic smooth structures on the 4-sphere.

• In this vein, is there a complex structure on $S^6$? – user98602 Nov 9 '14 at 6:01
• @MikeMiller If this paper is correct math.bme.hu/~etesi/s6-spontan.pdf – Loreno Heer Nov 9 '14 at 21:28
• There are a great many papers that claim to have solved the problem (either positively or negatively), and a great many have been shown to have errors. My impression is that the problem is still generally accepted to be open. – user98602 Nov 9 '14 at 22:22
• Indeed, this is a great example. I don't think I've heard any experts in the field even guess whether there is one, finitely many or infinitely many exotic structures on $S^4$. – Cheerful Parsnip Nov 10 '14 at 21:19
• Scorpan mentions that there are no compact simply-connected smooth $4$-manifolds known to admit only finitely many smooth structures. – Jesse Madnick Nov 11 '14 at 7:30

A more or less elementary example I'm quite fond of is the Erdős conjecture on arithmetic progressions, which asserts the following:

If for some set $S\subseteq \mathbb{N}$ the sum $$\sum_{s\in S}\frac{1}s$$ diverges, then $S$ contains arbitrarily long arithmetic progressions.

I've never seen a heuristic argument one way or the other - I believe the strongest known result, as of now is Szemerédi's theorem, which, more or less, states that if the lower asymptotic density of $S$ is positive (i.e. there are infinitely many $n$ such that $|[1,n]\cap S|>n\varepsilon$), then it contains arbitrarily long arithmetic progressions. There's also the Green-Tao theorem which is a special case of the conjecture, giving that the primes have arbitrarily long arithmetic progressions (and, indeed, establishes the fact for a larger class of sets as well).

Yet, neither of these suggests that the result holds in general. It's tempting to believe it's true, because it'd be such a beautiful theorem, but there's not much to support that - it's really unclear why the sum of the reciprocals diverging would have anything to do with arithmetic progressions. Still, there's no obvious examples of where it fails, so it's hard to make an argument against it either.

• "...it's really unclear why the sum of the reciprocals diverging would have anything to do with arithmetic progressions" seems too defeatist. The intuition is clear: if a collection of natural numbers is large enough then it cannot avoid certain patterns. That is one of the basic kinds of lessons you learn in Ramsey theory, and both Szemeredi's theorem and the Green-Tao theorem are excellent examples of that philosophy at work. The sum of the reciprocals diverging is a decent candidate for "large enough" - surely the example of the primes is what motivated Erdos to conjecture the result... – Qiaochu Yuan Nov 8 '14 at 5:27
• @Qiaochu What I mean is that there is a range of functions which might satisfy "$\sum_{s\in S}f(s)$ is large implies $S$ contains arbitrarily long AP" - the Szemeredi theorem provides a non-trivial $f$ which works, which suffices for an upper bound. And, yes, the sum of reciprocals of primes diverge - but so does the sum of $\frac{1}{n\log(\log(n))}$, and, to my knowledge, we have no reason to suppose that this is a less reasonable lower bound that $\frac{1}n$. What distinguishes $\frac{1}n$ is that it is, more or less, the slowest decreasing function for which the statement might hold... – Milo Brandt Nov 8 '14 at 15:35
• ...since summing over $\frac{1}{n^{1-\varepsilon}}$ is almost surely going to give sets which have diverging sums, but no long enough AP (since the sets only need $O(x^{\frac{1}{1-\varepsilon}})$ elements). So, $\frac{1}n$ is left on the boundary - meaning the conjecture is a very reasonable question to ask, since $\frac{1}n$ is the first open case we have. But "reasonable question" is far from "the answer is yes" – Milo Brandt Nov 8 '14 at 15:39
• The formulation with the sum was just a catchy way to phrase things. Anyway this is a non-answer as there are pretty good reasons to believe it. The case for 3-AP is in some sense quite close to being solved by recent results of Tom Sanders. – quid Nov 14 '14 at 22:02
• See this talk by Gowers renyi.hu/conferences/erdos100/slides/gowers.pdf – quid Nov 14 '14 at 22:51

Hilbert's 10th problem over over $\mathbb{Q}$/Mazur's conjecture. These are two open problems that point in opposite directions, and I think experts really aren't sure which way to guess.

Hilbert's 10th problem over $\mathbb{Q}$ Is there an algorithm which, given a collection of polynomial equations with rational coefficients, do they have a rational solution?

The problem is open. Here are heuristics each way. For "no".

• Individual diophantine equations are really hard. Think of how many mathematicians worked to prove $x^n+y^n=1$ has no solutions other than $(0,1)$ and $(1,0)$ for various values of $n$. Is it really plausible that all of their work could be reduced to running an algorithm?

• There is no such algorithm over $\mathbb{Z}$. (Matiyasevich-Robinson-Davis-Putnam)

For "yes":

• There are powerful theorems and conjectures about diophantine equations having finitely many solutions. For example, Mordell's Conjecture (now Falting's Theorem) immediately tells us that there are finitely many rational points on $x^n+y^n=1$ for any given $n$. If the Bombieri-Lang conjecture were proved, which doesn't seem impossible, we'd have much more powerful tools. And while finite is not the same as zero, we have developed a lot of tools to find those finitely many solutions in many cases. See Bjorn Poonen's course notes for a survey.

But here is the really frustrating thing. Suppose you believe that the answer is "no". Then you probably want to prove that you can encode the halting problem as a question about diophantine equations (this was how MDRP was proved), or else encode solving diophantine equations over $\mathbb{Z}$ into solving diophantine equations over $\mathbb{Q}$. In order to do this, you'd presumably write down some diophantine equation whose solutions looked like the states of a universal Turing machine, or looked like $\mathbb{Z}$. In any case, it would probably have infinitely many solutions, spread out discretely. And thus runs you into

Mazur's conjecture Given any collection of polynomial equations over $\mathbb{Q}$ in $n$-variables, let $X(\mathbb{Q})$ be the set of their solutions and let $\overline{X(\mathbb{Q})}$ be the topological closure of $X(\mathbb{Q})$ in $\mathbb{R}^n$. Then $\overline{X(\mathbb{Q})}$ has finitely many connected components.

So the general difficulty of diophantine equations leads one to imagine that the problem is unsolvable, but Mazur's conjecture blocks the most plausible route to proving it is unsolvable. Of course, one can imagine that diophantine equations over $\mathbb{Q}$ are unsolvable, and yet they can't encode a Turing machine. I think it is unclear whether most unsolvable problems are unsolvable because they are equivalent to the Halting problem, or whether that is (essentially) the only kind of problem we know how to prove is unsolvable.

• This is really compelling! +1 – Kevin Arlin Nov 14 '14 at 23:31
• "Is it really plausible that all of their work could be reduced to running an algorithm?" Because a proof was eventually found, it would have been found by an algorithm trying and checking all possible proofs, right? – JiK Dec 8 '14 at 22:27
• "I think it is unclear whether most unsolvable problems are unsolvable because they are equivalent to the Halting problem, or whether that is (essentially) the only kind of problem we know how to prove is unsolvable." - It is my understanding that you can prove the busy beaver function noncomputable without using the halting problem, but I could be mistaken, and I know there are proofs that use it. – Kevin Feb 3 '16 at 3:09
• @Kevin I actually know a bunch about this. I'm leaving a comment so I can find this thread later when I have time to type something up. – Stella Biderman Apr 4 '17 at 13:32
• I forget whether it's all of them, some of them, or just a few of them, but I do remember that it's very possible for unsolvable problems to also have the agonizing property that even their unsolvability cannot be proven. – Trevor Dec 3 '19 at 15:22

In the theory of dynamical systems, problems involve limit cycles in general are always very difficult. The second part of Hilbert's sixteenth problem is my personal "favorite". The upper bound for the number of limit cycles of planar polynomial vector fields of degree $n$ remains unsolved for any $n>1$. For example, can quadratic plane vector fields ($n=2$) have more than four limit cycles? It may be extremely tricky to find a quadratic system with five limit cycles, but we really have absolutely no idea. In 1950s, mathematicians claimed quadratics systems have maximal three limit cycles and had multiple other mathematicians conformed, but it was shown wrong when a quadratic system with four limit cycles was found. For details, you can check this article.

The smooth Poincare conjecture in dimension 4 has already been mentioned, so I'll mention the smooth Schönflies Problem in that dimension. The question is whether there is a diffeomorphism of $S^4$ taking any smoothly embedded copy of $S^3$ in $S^4$ to the standard equatorial $S^3\subset S^4$. This is true in all other dimensions, but $4$ is such an unusual dimension that it's difficult to speculate what the answer is in this case.

From a number theoretic perspective, there are a few famous problems related to ranks of elliptic curves, which a lot of modern research in the area is geared towards solving. For example, Manjul Bhargava recently received the Fields medal partly for his work on bounding average ranks of elliptic curves (and proving that the Birch and Swinnerton Dyer conjecture is true for ever-increasing percentages of elliptic curves).

To describe some of the results: an elliptic curve over $\mathbb{Q}$ is a rational smooth genus 1 projective curve with a rational point, or in less scary terms, the set of solutions to an equation that looks like $$E(\mathbb{Q}) = \{(x,y) \in \mathbb{Q}^2: y^2 = x^3 + ax + b\}$$ where $a, b \in \mathbb{Q}$. It's a fact that any such set forms a finitely generated abelian group, so by the structure theorem for such objects the group of rational points is $$E(\mathbb{Q}) \cong \mathbb{Z}^r + \Delta,$$ where $\Delta$ is some finite group. Now, we have complete descriptions of what this group $\Delta$ can be - a Theorem of Mazur limits it to a small finite list of finite groups of size less than 12. However the values of $r$ are much more mysterious. We define the rank of $E$ to be this $r = r(E)$.

Now, we know quite a lot about $r$ - for example, in "100%" of cases the rank is $0$ or $1$ (where here "100%" is used in the probablistic sense, not to mean that every elliptic curve has rank $0$ or $1$!). There is also the Birch and Swinnerton Dyer Conjecture (BSD), which is one of the very open problems that you mention that nobody has any idea how to prove, but which most people believe. It relates the rank of the elliptic curve to the order of vanishing of its $L$-function at 1. Perhaps the strongest heuristic for it is that it's been proved in certain special cases, as well as Bhargava's work. So much of modern number theory research goes towards BSD, and it's one of the famous Millenium problems.

However, what we don't have much intuition with is:

Question: Are the ranks of elliptic curves over $\mathbb{Q}$ bounded? That is, is there some $R$ such that for any elliptic curve $E/\mathbb{Q}$, we have $r(E) \leq R$?

As of last year, it was very open - there were loose heuristics both ways. The largest rank we've found so far is a curve with rank at least 28, due to Elkies, which has been the record-holder for a long time now. As I mentioned before, Bhargava has proved the average rank is bounded by at least 1.5, and this was enough to win a Fields medal.

However, having said all that, I think there has been some excitement recently with some stronger heuristics that lean towards the rank being bounded. I don't know enough about these heuristics to comment any further, but there's more information here: http://quomodocumque.wordpress.com/2014/07/20/are-ranks-bounded/

• Is there any connection here to ECC? (In other words, are there any implications about elliptic curves over $\mathbb{F}_q$?) – Alexander Gruber Nov 11 '14 at 18:45
• Short answer: I'm not sure. Longer answer: I asked in my department group meeting whether there were any results that might follow from this, and the consensus seemed not; it's a result of a more standalone nature, and having a definite answer wouldn't further our understanding that much with current methods! Of course, as nobody knows what the answer might be, it's unlikely there has been any work done assuming ranks are bounded (or unbounded), so it might be hard to tell what might lead from a proof either way. – Chris Williams Nov 12 '14 at 12:31

I don't think anyone has a clear idea whether there exist classical solutions to the Navier-Stokes equation. http://www.claymath.org/millenium-problems/navier%E2%80%93stokes-equation

Most attempts have focused on trying to prove it is true. However Leray gave a suggestion for looking for a counterexample. It was later shown that his proposed counterexample would never work: J. Nečas, M. Růžička, and V. Šverák, On Leray's self-similar solutions of the Navier-Stokes equations, Acta Mathematica, 1996, Volume 176, Issue 2, pp 283-294. However the fact that counterexamples have been proposed does suggest that it is reasonable to think the conjecture is false.

• My (naive, non-specialist) viewpoint. As I know, N-S equations are deduced from a false hypothesis: that the fluid is a continuum medium. Is really reasonable expecting physically realistic solutions in any case? Analogous example: singularities in classical gravity. – Martín-Blas Pérez Pinilla Nov 15 '14 at 14:59
• I think there is no compelling reason why N-S should model realistic situations. But I did hear from someone who did comparisons of experiment and numerics that it does surprisingly well, even when there is a tremendous amount of turbulence. (He was considering air flow over cars, and found strange phenomena that occurred in both the experiments AND the numerics.) – Stephen Montgomery-Smith Nov 15 '14 at 16:51
• If the regularity of the N-S turns out to be true, I do think it would strongly suggest that the continuum hypothesis is a valid approximation to the particles of the fluid being very small. – Stephen Montgomery-Smith Nov 15 '14 at 16:53

An easy-to-understand open problem involves the first counterexample to Euler's Sum of Powers conjecture:

Q: Does $x_1^5+x_2^5+x_3^5+x_4^5+x_5^5=0$ have infinitely many primitive non-zero integer solutions?

(Primitive being the $x_i$ have no common factor.) Only three are known so far and nobody has given a good heuristic argument that the list is finite, or if there are infinitely many. There are interesting congruential constraints on the $x_i$.

More generally,

Q: For odd $k>3$, does $x_1^k+x_2^k+\dots+x_{k}^k = 0$ have infinitely many primitive non-zero integer solutions?

From what I can tell, neither the existence nor non-existence of the Moore Graph of degree 57 and diameter 2 is strongly attested. Most of the work to date on the subject revolves around the various properties such a graph (should it exist) must or must not possess, but none of these seem to give a strong indication to lean one way or the other. Also, the respondents to a poll on this blog post from 2009 seem to be split pretty evenly.

I think a proper answer to this question are examples of questions where numerical evidence is extremely difficult to obtain. So for example, we don't know anything interesting about the Collatz conjecture but at least we know that it's true for a huge number of cases.

As an example of something we don't know at all, consider $S_n$ the symmetric group, and define $s_i$ to be the adjacent transposition $(i,i+1)$. Then for a permutation $\pi\in S_n$, defined a reduced decomposition of $\pi$ to be a minimal length product of transpositions that gives you $\pi$. For example if $\pi=4321$ then $w=s_1s_2s_1s_3s_2s_1$ is a reduced decomposition of $\pi$.

It's easy to see the minimal length of the reduced decomposition is the number of inversions in $\pi$. On the other hand, the question of how many distinct reduced decompositions $R(\pi)$ there are of $\pi$ is a ridiculously complicated question. We know the answer for permutations that have particular (lack of) patterns such as vexellary permutations, in particular the reverse permutation $(n,n-1,\cdots,1)$, which has $f_n$ reduced decompositions, where $f_n$ is the number of staircase shaped Young tableaux of shape $(n-1,n-2,\cdots,1)$. For any other non-trivial permutation there is essentially nothing known. For $n=7$, the number of reduced decompositions for the reverse permutation exceeds the number of atoms in the known universe. A similar difficulty arises for pretty much any non-trivial permutation. One can't even obtain an order of magnitude estimate.

Do all compact smooth manifolds of dimension $\geq 5$ admit Einstein metrics?

(An Einstein metric is a Riemannian metric with constant Ricci curvature.)

A list of fundamental open problems in differential geometry and geometric analysis can be found at the end of Yau's excellent survey, Review of Geometry and Analysis. It was written in 2000, so is very current.

Aside: Some problems which don't fit the bill (in that there's evidence one way or another or in that I hear the same guesses), but are interesting anyway are:

• Does the 6-sphere admit an integrable almost complex structure?

• Hopf Conjecture: Does $\mathbb{S}^2 \times \mathbb{S}^2$ admit a metric with positive sectional curvature?

• Chern Conjecture: Does every compact affine manifold have vanishing Euler characteristic?

• What are the guesses/evidence for your last three? I haven't heard guesses either way before. – user98602 Nov 11 '14 at 16:57
• @MikeMiller: (1) Any almost complex structure on $\mathbb{S}^6$ would have to be incompatible with both the standard metric (LeBrun) and the standard symplectic structure (Bryant-Chern). (3) The Chern conjecture is true for complete affine manifolds. (2) I remember hearing the same guesses for the Hopf conjecture at one point (by a couple experts), but I honestly don't recall them and won't venture one myself. – Jesse Madnick Nov 12 '14 at 5:54

Does there exist a finitely presented, infinite torsion group?

A torsion group is a group where every element has finite order. Burnside's problem (1902) asked if there exists a finitely generated, infinite torsion group. Such a group of unbounded exponent was constructed by Golod and Shafarevich in 1964, while Novikov and Adian did it for bounded exponent in 1968. Ol'shanskii constructed finitely generated infinite groups, all of whose proper, non-trivial subgroups are cyclic of order a fixed prime $p$ ("Tarski monster" groups). However, all these examples are finitely generated but not finitely presentable. The question of whether or not there exists a finitely presented example is still wide open. Apparently, Rips gave a possible method of constructing such a group, but Ol'shanskii and Sapir turned his handle to no avail (reference).

It is worth mentioning that Efim Zelmanov was awarded a fields medal for a related problem, called the restricted Burnside problem.

• I have been under the impression that most think that there should be such a group, just that with current tech we don't have much of an idea to construct such a group (or actually carry out the ideas that could construct such a group). Are there group theorists that seriously believe that there is no such group, or some reason to believe there is none? Are their opinions the same for both bounded and unbounded exponent case? – user29123 May 24 '15 at 17:44
• My impression, like that of @PaulPlummer, was that it's pretty commonly believed that such a group exists; it's just that we don't know how to construct one, and the methods we do have are rather delicate and complicated. (I remember one such existence proof from a seminar that involved checking a laundry list of over a hundred properties.) – anomaly Sep 27 '15 at 0:18
• @PaulPlummer If I remember correctly, my point was based on the sentence "Apparently, Rips gave a possible method of constructing such a group, but Ol'shanskii and Sapir turned his handle to no avail" and the associated link. The link reaffirms your impression that "most think there should be such a group", and gives a "general idea" for constructing such a group. However, from memory, between that post (2011) and my post (2014) Ol'shanskii and Sapir made this idea "work" but, as the sentence points out, it wasn't good enough. However, my dates may be incorrect (I haven't linked to the paper). – user1729 Sep 29 '15 at 15:57
• I have had a quick search for the paper of Ol'shanskii and Sapir, and it seems that they haven't written a paper together since 2006 source. So...I am not precisely sure what I was thinking. Sorry. – user1729 Sep 29 '15 at 16:15

The existence of projective finite planes. All the known examples have order prime power. Quote:

The existence of finite projective planes of other orders is an open question. The only general restriction known on the order is the Bruck-Ryser-Chowla theorem that if the order $N$ is congruent to 1 or 2 $\text{mod}$ 4, it must be the sum of two squares. This rules out $N = 6$. The next case $N = 10$ has been ruled out by massive computer calculations. Nothing more is known; in particular, the question of whether there exists a finite projective plane of order $N = 12$ is still open.
• This is a really cool problem but I'm not sure it applies. Don't most people think that this will turn out to be false? (i.e. that all projective planes have prime power order?) – Alexander Gruber Nov 11 '14 at 18:41
• @AlexanderGruber, there is some heuristic argument favoring the conjecture? – Martín-Blas Pérez Pinilla Nov 12 '14 at 14:13
• @AlexanderGruber: I have never heard a compelling heuristic argument either for or against existence of projective planes of non-prime-power order. Perhaps the intuition against is similar to that in Chris Godsil's answer to a question I once asked about biplanes, which are symmetric designs with $\lambda=2$ (projective planes are $\lambda=1$). – Will Orrick Nov 16 '14 at 13:02
• There is a related conjecture about projective planes. Desarguesian planes are those in which Desargues' theorem holds, which are those derived by identifying points with lines through the origin in a three-dimensional space over a finite field. For every prime order, the only plane currently known is the Desarguesian one and there is a conjecture that this is the only one that exists. In other prime power orders, many non-Desarguesian planes are known. If the non-Desarguesian plane are somehow deformations of the Desarguesian ones, and if such deformation is not possible in prime ... – Will Orrick Nov 16 '14 at 13:08
• ... order, that would explain what is known so far, but I have never heard any argument along these lines. From another point of view, one could take the existence of non-Desarguesian planes as reason to be optimistic about existence of planes in non-prime-power orders. – Will Orrick Nov 16 '14 at 13:09

The first proof that many people learn is that there are infinitely many primes. (If not the first, then it's often second to the fact that $\sqrt 2$ is irrational).

A natural generalization of this was considered by Dirichlet, who showed that as long as the arithmetic progression $a, a+d, a+2d, a+3d, ...$ doesn't have a trivial reason for not having many primes, then in fact it contains infinitely many primes. This is known as Dirichlet's Theorem on Primes in Arithmetic Progressions. Remarkably, if there are infinitely many primes, then it is also known that the sequence has asymptotically $1/\varphi(d)$ of all primes, where $\varphi(d)$ is the number of numbers up to $d$ that are relatively prime to $d$. In other words, every nontrivial arithmetic progression has the exact same percentage of primes, a sort of equidistribution theorem.

The next natural generalization is to consider higher polynomials, such as quadratic polynomials. (For the moment, call a polynomial quadratic if it's of the form $ax^2 + bx + c$ with $a \neq 0$). Is the analogue true? Can we predict distribution? In fact, we have not found a single quadratic polynomial that takes infinitely many primes (nor any polynomial of degree > 1). We have not even been successful showing that $x^2 + 1$ takes infinitely many primes, nor do we have any idea how.

Going a bit deeper, it is possible to conjecture densities using the circle method or its variants, even for higher degree polynomials. But we have no idea how to prove them.

In short, is $x^2 + 1$ prime infinitely often?

• Is there anyone who really believes that it's not prime infinitely often? As you said, there are reasonable conjectures based on known number-theoretic techniques - this would seem to meet OP's definition of 'convincing heuristics'. – Steven Stadnicki Nov 11 '14 at 1:10
• Note also that the computational data we have gathered matches the heuristic prediction beautifully. So I think we have very good reason to believe this conjecture. – Greg Martin Nov 11 '14 at 8:38
• This is a special case of the following conjecture: Let $P(x)$ be a polynomial with integer coefficients. Then it is prime infinitely often unless it has a stupid reason not to be. ("Stupid reasons": $P$ is reducible; the coefficients aren't coprime; etc.) This general conjecture might be a better example of what OP is looking for. – Yoni Rozenshein Nov 11 '14 at 12:15
• I would think most number theorist would be extremely surprised if this were not prime infinitely often. – quid Nov 14 '14 at 22:10

Existence of rectangular cuboid with all edges, all faces' diagonals, and the main diagonal being integers.

Feasibility of reformulating all of math in only well-defined ultrafinitistic terms

From the point of view of physics there is something strange about the way math is used. By the Church–Turing–Deutsch principle all physical processes have (quantum) computable descriptions, but the way we do math invokes non-computable concepts such as the uncountable reals. What happens if we use math in practice is that the uncomputability of any concepts will always stay hidden in the intermediary parts, they will never arise in the final results.

This suggests that you don't need to invoke uncomputable concepts in the first place, but so far there has not been a lot of progress made by the advocates of ultrafinitism.

The problem of asymptotic behavior of the maximal cardinality of a cap sets in $\mathbb{Z}/3\mathbb{Z}^r$ as $r$ to infinity gives rise to the following yes/no question that is open.

A cap set, here, is a set with no three points on an affine line. This is equivalent to the existence of $x,y,z$ such that $x+y+z = 0$, or the existence of a 3-term arithmetic progression (see a related answer).

Is the maximal cardinality of a cap sets in $\mathbb{Z}/3\mathbb{Z}^r$ a $O((3- \delta)^{r })$ for some $\delta > 0$?

See a blog post of Terry Tao from a couple of years ago where he expresses an opinion on the matter but also acknowledges the dissent of a good friend of his.

• By now that's actually known. The answer is "yes" – quid Feb 8 '19 at 15:05

There is some "natural" axiom that added to ZFC decides CH? Interesting discussion in Introduction to Set Theory, Third Edition, Revised and Expanded.

• What is CH? (extra characters added for comment length) – DanielV Nov 14 '14 at 22:51
• @DanielV, Continuum Hypothesis. – Martín-Blas Pérez Pinilla Nov 15 '14 at 8:31

The Higman Conjecture concerns the number of conjugacy classes of $UT_n(\mathbb{F}_q)$, the group of unipotent upper-triangular matrices with entries in a finite field with $q$ elements. The conjecture is that for a fixed $n$ the number of conjugacy classes of $UT_n(\mathbb{F}_q)$ is given by a polynomial in $q$. This has been proven up to $n=13$, but beyond that it's unknown. The difficulty might be related to the fact that $UT_n(\mathbb{F}_q)$ has wild representation type. I know of a number of failed attempts at proof, and it seems that most of the people thinking about this conjecture believe it to be true. At the same time, there is a collection of subgroups of $UT_n(\mathbb{F}_q)$, known as "pattern groups," for which an analogous conjecture is known to be false.

## Kaplansky's Zero-Divisor Conjecture

Let $K$ be a field and $G$ a torsion free group, then is the group ring $KG$ a domain?

All the research up until now have been affirmative. This problem has been dealt in the book "The algebraic structure of group rings" by D. Passman.

It is one of the toughest and least approachable problem in the whole field of Algebra.

Latest I know is that it has been proved for torsion free solvable groups. Still a long way to go.

Another natural question after this (if the above is true) is, if we replace $K$ by any domain $D$, say $\Bbb{Z}$

What is the answer to the problem to which Graham's number is an upper bound?

• Graham and Rotschild proved that $6 \leq N \leq f(f(f(f(f(f(f(12)))))))$, where $f(x)=2 \uparrow^x 3$ and $\uparrow$ denotes up-arrow notation.
• Currently, the best bound known is: $13 \leq N < 2 \uparrow\uparrow\uparrow 6$.
• Mathematicians thought that the answer was $6$, until the lower bound of $11$ was proven. Now many have no idea where to expect $N$.