MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Question: I'm asking for a big list of not especially famous, long open problems that anyone can understand. Community wiki, so one problem per answer, please.

Motivation: I plan to use this list in my teaching, to motivate general education undergraduates, and early year majors, suggesting to them an idea of what research mathematicians do.

Meaning of "not too famous" Examples of problems that are too famous might be the Goldbach conjecture, the $3x+1$-problem, the twin-prime conjecture, or the chromatic number of the unit-distance graph on ${\Bbb R}^2$. Roughly, if there exists a whole monograph already dedicated to the problem (or narrow circle of problems), no need to mention it again here. I'm looking for problems that, with high probability, a mathematician working outside the particular area has never encountered.

Meaning of: anyone can understand The statement (in some appropriate, but reasonably terse formulation) shouldn't involve concepts beyond (American) K-12 mathematics. For example, if it weren't already too famous, I would say that the conjecture that "finite projective planes have prime power order" does have barely acceptable articulations.

Meaning of: long open The problem should occur in the literature or have a solid history as folklore. So I do not mean to call here for the invention of new problems or to collect everybody's laundry list of private-research-impeding unproved elementary technical lemmas. There should already exist at least of small community of mathematicians who will care if one of these problems gets solved.

I hope I have reduced subjectivity to a minimum, but I can't eliminate all fuzziness -- so if in doubt please don't hesitate to post!

To get started, here's a problem that I only learned of recently and that I've actually enjoyed describing to general education students.

Edit: I'm primarily interested in conjectures - yes-no questions, rather than classification problems, quests for algorithms, etc.

share|cite|improve this question
You might get more success if you sampled certain open problem lists and indicated which ones fit your list and which ones did not. I could mention various combinatorial problems such as integer complexity, determinant spectrum, covering design optimization, but I can't tell from your description if they would be suitable for you. Gerhard "They Are Suitable For Me" Paseman, 2012.06.21 – Gerhard Paseman Jun 21 '12 at 19:11
Here is some collection of some other "collect open problems" quests. on MO:… PS Nice question ! PSPS may be add tag "open-problems" – Alexander Chervov Jun 21 '12 at 20:53
To save the search for explanation of cryptic acronyms for those of us outside US, K-12 means high school. @Mahmud: You are using a wrong meaning of the word “problem”. The TSP is not an unproved mathematical statement, it is a computational task. – Emil Jeřábek Jun 22 '12 at 12:05
More precisely, K-12 means anything up to high school (K = Kindergarten, 12 = 12th grade, and K-12 covers this range). – Henry Cohn Jun 22 '12 at 13:05
There seems to be a claimed proof of the union-closed sets conjecture by Blinovsky – Marco Oct 22 '15 at 14:08

99 Answers 99

What is the largest possible volume of the convex hull of a space curve having unit length?

share|cite|improve this answer
It was discussed here:… – Vladimir Reshetnikov Jul 28 '12 at 21:10

I think you could give an accessible K-12 formulation of the definition of a group (as a group of permutations, for instance) and of an integral group ring. The Zero Divisor Conjecture (Kaplansky, 1940) then states, in one version, that if $G$ is a torsion-free group then the group ring $\mathbb{Z}[G]$ has no zero divisors besides the number $0$.

share|cite|improve this answer
>I think you could give an accessible K-12 formulation of the definition of a group ... In any case this is perfect for my follow-up question which the net gods decided to close for the time being. – David Feldman Jul 13 '12 at 5:36
@David just wanted to write the same :)… – Alexander Chervov Jul 13 '12 at 6:18
I agree that one could just about formulate the conjecture for "higher mathematics beginners" -- though exactly where would be an issue, IMHO -- but I am not so sure that one can get people to appreciate the conjecture – Yemon Choi Jul 14 '12 at 12:34
@Yemon Choi it might be useful for those who wish to become mathematicians, may be they will not appreciate it immediately but will be aware of it. It does not mean that will work on it some future, there are some indirect uses like - creating some taste what is good what is bad, etc. – Alexander Chervov Jul 14 '12 at 17:53
Speaking for myself, when I first encountered this, I found it very surprising it wasn't known, and deeply intriguing on its own merits. My context was low dimensional topology (for me, G was the fundamental group of a 3-manifold). But even without getting into applications, it's clear that without answering this we're ignorant about how to work with integral group rings. – Daniel Moskovich Jul 14 '12 at 19:47

Do there exist five positive integers such that the product of any two of them increased by 1 is a perfect square?

The same question for seven distinct nonzero rationals.

Diophantine m-tuples pages

share|cite|improve this answer

Let $R(x)=P(x)/Q(x)$ where $P(x)$ and $Q(x)$ are polynomials with integer coefficients and $Q(0)\neq 0$. Is there an algorithm that given $P(x)$ and $Q(x)$ as an input always halts and decides if the Taylor series of $R(x)$ at $x=0$ has a coefficient $0$?

share|cite|improve this answer

The Polya--Szego conjecture for polygonal drums: among the polygonal drums with $n$ sides and given area, the regular one has the slowest vibration (and therefore the lowest tone).

As far as I know, this remains open for $n\geq 5$.

share|cite|improve this answer

Imre Ruzsa conjectured in 1971 (Mat. Lapok 22, in hungarian) that a congruence-preserving mapping $f : \mathbb{N} \to \mathbb{Z}$ is a polynomial as soon as the power series $A(t) := \sum_{n \in \mathbb{N}} f(n)t^n \in \mathbb{Z}[[t]]$ has radius of convergence $> 1/e$. (Congruence-preserving simply means $n-m \mid f(n)-f(m)$.)

This is still an open problem, although A. Perelli and U. Zannier have shown that the power series $A(t)$ must be $D$-finite ("On recurrent mod $p$ sequences," J. reine angew. Mat. 348, 1984). The best result on Ruzsa's problem is due to U. Zannier ("On periodic mod $p$ sequences and G-functions," Manuscripta math. 90, 1996).

share|cite|improve this answer

The Alon-Tarsi Conjecture:

A latin square of order $n$ is a filling of an $n\times n$ matrix with the numbers $1, 2,\ldots,n$ such that each row or column gives a permutation of $1,2,\ldots,n$. Take the product of the signs of these $2n$ permutations and call it the sign of the latin square. Let $EVEN(n)$ be the number of latin squares with sign $+1$ and let $ODD(n)$ be the number of latin squares with sign $-1$. The conjecture says:

If $n$ is even then $EVEN(n)\neq ODD(n)$.

The original reference is: N. Alon and M. Tarsi, Colorings and orientations of graphs, Combinatorica 12 (1992), 125-134. See also this preprint by Landsberg and Kumar for a recent update.

share|cite|improve this answer

Can one prove the infinitude of the primes without employing any functions of super-polynomial growth?

(Of course I confess I have in mind Paris and Wilkie's more precise and sophisticated question concerning primes in the theory of bounded induction, but I think a high school student could think about looking for a positive answer without that background.)

share|cite|improve this answer
Where are functions of exponential growth used in the classical proof, or the Euler product + divergence of harmonic series? – Douglas Zare Jun 24 '12 at 14:06
In the classical proof, the exponential growth is in the product of the known primes. See… for a more precise discussion. In the Euler product proof, I imagine the growth occurs in the Chinese-remainder-theorem-based coding tricks necessary to express this proof in the language of first-order arithmetic, but I haven't thought this through carefully so maybe I'm totally off base. – Henry Cohn Jun 24 '12 at 16:49
Thanks. The second answer to that question says in part, "functions of exponential growth aren't necessary, but we are still using a function of super-polynomial growth." That seems to contradict the statement that this is open. – Douglas Zare Jun 24 '12 at 18:01
See also . David, contrary to what you write, it is possible to define in bounded arithmetic a function computing rational approximations of logarithm. This does not imply that exponentiation is total, since there may be numbers greater that all values of logarithm. As for divergence of harmonic series, the problem is even to express this statement in bounded arithmetic: you cannot in general define $\sum_{n\le x}f(n)$ by a bounded formula, unless $x$ is logarithmic. – Emil Jeřábek Jun 25 '12 at 11:35
Even if you restrict attention to small $x$, there is the question how do you formulate “divergence”. Bounded arithmetic certainly cannot prove that for every $y$, there exists $x$ such that $\sum_{n\le x}n^{-1}$ is defined and larger than $y$. However, I think that with appropriate formulations, it can prove that for every $y$, either there exists such an $x$, or all sums $\sum_{n\le x}n^{-1}$ that are defined have value less than, say, $y/2$ (which means the sum does not converge to $y$). Anyway, this is largely irrelevant. The real show-stopper in the proof using the Euler product is... – Emil Jeřábek Jun 25 '12 at 11:52

From Rick Kenyon's open problem list:

What are the minimal number of squares needed to tile an $a \times b$ rectangle?

Kenyon showed the correct order is $\log a$ assuming $a/b$ is bounded with $b \leq a$. However, there is plenty of room for improvement in the constant factor, and an exact formula seems far, far away.

share|cite|improve this answer

Is there such $n\in\mathbb{N}$ that ${^n\pi}\in\mathbb{N}$? (see tetration)

share|cite|improve this answer

The following conjecture by Carsten Thomassen:

If $G$ is a 3-connected graph, every longest cycle in $G$ has a chord.

Thomassen has proven the conjecture true for 3-connected cubic graphs.

share|cite|improve this answer

Is there a positive integer which is both triangular and factorial except these obvious examples: $1, 6, 120$? (Tomaszewski conjecture,

share|cite|improve this answer
  • Is Hilbert's tenth problem for Diophantine equations in rational numbers decidable?
  • Is Hilbert's tenth problem for Diophantine equations of power $3$ decidable?
  • Is there a universal Diophantine equation of power $3$?
  • Is there a universal Diophantine equation containing less than $9$ variables? If so, what is the minimal number of variables? What minimal power can be achieved for that number of variables?
  • Is there a universal Diophantine equation that can be written using less than $100$ arithmetic operations (additions or multiplications)? If so, what is the minimal number of operations?
share|cite|improve this answer

Is the density of $1$s in the Kolakoski sequence $122112122122112112212112\dots$ (Wikipedia, OEIS) equal to $1/2$? Also, does every consecutive block, which occurs at all in the Kolakoski sequence, occur infinitely often?

share|cite|improve this answer

The easy-to-understand "equal sums of like powers" problem, which generalizes Pythagorean triples:

$$3^2+4^2 = 5^2$$

$$3^3+4^3+5^3 = 6^3$$

In general, does,

$$x_1^k+x_2^k+\dots +x_k^k=z^k$$

have a non-zero integer solution for all positive integer $k$?

So far, integer solutions are known for all $k\leq9$, except $k=6$.

(Unfortunately, the Eulernet search for $k=6$ has been stopped since the mid-2000s. With today's computers, and with a distributed search, it may be feasible to find it now.)

share|cite|improve this answer

The following problem is very well-known among algebraic geometers:

Does there exist a cubic 4-fold that is not rational?

It's probably not well-known outside of algebraic geometry, even though it can easily be explained in every elementary terms:

Does there exist a polynomial equation $F$ of degree three in five variables with the following property: Let $X \subset \mathbb C^5$ be the solution set of $F = 0$. Then there exists no chart $U \subset \mathbb C^4, \phi \colon U \to X$ such that $\phi$ is defined by rational functions (i.e., quotients of polynomials).

share|cite|improve this answer
Cool ! what are the references for current state of art ? – Alexander Chervov Jul 24 '12 at 9:57
Basically nothing is known. On the other hand, it is easy to construct cubics that are rationals - e.g. cubics containing two planes. There are two conjectural descriptions of the locus of rational cubics inside the moduli space of cubics: one is due to Hassett and Harris; is video of a talk by Harris on the question; for Hassett's related results search "Hassett cubic fourfolds" on google scholar. Kuznetsov has a conjecture in terms of derived categories, the reference is: – Arend Bayer Jul 24 '12 at 22:03
K-12??? $\textrm{}$ – Victor Protsak Jan 6 '14 at 20:12

The Kurepa problem: Show that for all primes $p>3$ we have that $$ 0!+1!+2!+\dots+(p-1)! $$ is not divisible by $p$. Kurepa posed this problem in 1971. For an overview see the article by Ivic and Mijajlovic (

share|cite|improve this answer
Already on this list: see – Yemon Choi Dec 17 '14 at 16:18
Is there a particular reason to believe that this problem has a positive answer? -- Naively, for large enough $n$ I would expect about $\ln(\ln(n))$ counterexamples less than $n$. But maybe there is a particular reason why this heuristics is not applicable here? – Stefan Kohl Dec 17 '14 at 16:30
@StefanKohl you can try search for n=exp(exp(3)) if you find this large enough. – joro Dec 17 '14 at 16:53
@Yemon Choi: Sorry, I didn't see that there is a second page. – Jan-Christoph Schlage-Puchta Dec 18 '14 at 14:05
@Stefan Kohl: Ivic told me that the Barsky-Benzaghou-proof is philosophically correct in the sense that although it does not prove the conjecture, it gives a reason why the conjecture should be true. I haven't looked at the paper itself, though. – Jan-Christoph Schlage-Puchta Dec 18 '14 at 14:08

A conjecture arising from Waring problem: a number of solutions of the equation $$x_1^3+x_2^3+x_3^3=y_1^3+y_2^3+y_3^3,\qquad|x_i|,|y_i|<P,$$ is $O(P^{3+\varepsilon})$. The best known estimate is $O(P^{7/2+\varepsilon})$ (Hua Loo Keng).

share|cite|improve this answer

In number theory, the Odd greedy expansion problem concerns a method for forming Egyptian fractions in which all denominators are odd.

Stein, Selfridge, Graham, and others have posed the question of

whether the odd greedy algorithm terminates with a finite expansion for every $x/y$ with $y$ odd?

share|cite|improve this answer

More than ten years ago I posed the following problem in a couple of math-related mailing lists:

Let $G_n$ be the graph with vertex set $\{1, 2, \dots, 2n\}$ such that $\{i,j\}$ is an edge if and only if $i+j$ is a prime number. Is it true that $G_n$ is eulerian for every $n \geq 2$?

It is a simple consequence of Bertrand's Postulate (there is always a prime between $k$ and $2k$) that $G_n$ is connected and has a perfect matching for every $n$.

The problem turned out to be an old one. I believe that some variation of it appears in Richard K. Guy's "Unsolved Problems in Number Theory" and according to this article, it was originally posed in the Journal of Recreational Mathematics in 1982.

Michael A. Jones and Leslie Cheteyan, "Two observation on unsolved problem #1046 on prime circles of $\{1, 2, . . . , 2m\}$", J. Recreational Mathematics Vol.35(1) (2006), 15--19.

share|cite|improve this answer

Does every nonseparating planar continuum have the fixed point property?

share|cite|improve this answer
For those of us whose general topology is rusty: what is the K-12 level formulation of this question? – Yemon Choi Jun 21 '12 at 22:50
I'm happy to learn that this unknown, even if doesn't meet the stipulations. So it got my up vote. – David Feldman Jun 22 '12 at 3:02
Uniform continuity is a K-12 concept now? – Douglas Zare Jun 22 '12 at 14:25
Thanks Yemon. I agree with Doug that uniform continuity is not a stand alone K-12 concept. I mentioned it in order to avoid trying to decode its meaning in K-12 terms in this context, as follows: – Paul Fabel Jun 22 '12 at 17:56
f is a set of 4-tuples (x1,y1,x2,y2) so that each point (x1,y,1) in K appears precisely once as the 1st two coordinates of some point in f, and we also require that the last two coordinates of each point of f is a point of K. For each positive radius R we can find a smaller positive radius r(R) so that if (x1,y1,x2,y2) is in f, then if (w1,z1) is both in K and also in the disk of radius r(R) centered at (x1,y1), and if (w1,z1,w2,z2) is in f, then (w2,z2) is in the disk of radius R centered at (x2,y2). – Paul Fabel Jun 22 '12 at 17:56

Alexander's Conjecture, and by extension a lot of open problems about combinatorial subdivision, are as easy to understand as they are maddening. To quote Melikhov:

Alexander's 80-year old problem of whether any two triangulations of a [3-dimensional] polyhedron have a common iterated-stellar subdivision. They are known to be related by a sequence of stellar subdivisions and inverse operations (Alexander), and to have a common subdivision (Whitehead). However the notion of an arbitrary subdivision is an affine, and not a purely combinatorial notion. It would be great if one could show at least that for some family of subdivisions definable in purely combinatorial terms (e.g. replacing a simplex by a simplicially collapsible or consructible ball), common subdivisions exist...

Stellar subdivision (and arbitrary subdivisions) can be explained to a K-12 student with a picture. For a stellar subdivision, choose a face F, take its midpoint, and connect it to all vertices of tetrahedra of which F is a face. For arbitrary subdivision, invent some silly triangulation of a simplex, and just plug it inside. refining heighbouring simplexes as needed.

share|cite|improve this answer

What is the least $V$ such that any convex body of unit volume can be fit into a tetrahedron of volume $V$? It is known that $V \ge 9/2$ and conjectured that $V = 9/2$.

share|cite|improve this answer

Is there any odd perfect number?

share|cite|improve this answer
Methinks this one is both pretty famous and long open... – José Hdz. Stgo. Jun 14 '13 at 15:21
Well, maybe too famous, but I was not sure. – The User Jun 14 '13 at 23:05

I like the Montesinos-Nakanishi 3-move conjecture from knot theory. A 3-move on a link is the replacement of two parallel strands by three half twists. The conjecture is that any link can be turned into the trivial link by a sequence of such moves. (If you think of this as a conjecture on diagrams, then you also need to allow Reidemeister moves.) According to an Encyclopedia of Mathematics article:

The conjecture has been proved for links up to 12 crossings, 4-bridge links and five-braid links except one family represented by the square of the centre of the 5-braid group. This link, which can be reduced by 3-moves to a 20-crossings link, is the smallest known link for which the conjecture is open (as of 2001).

share|cite|improve this answer

One I saw in a talk yesterday by Faustin Adiceam (I hope I have remembered this correctly):

Danzer's problem (in dimension 2): Is there a subset $S$ of $\mathbb{R}^2$ of finite density (the number of points at distance $\le r$ from the origin is $O(r^2)$) that hits every rectangle of unit area?

There is also a version where instead of positive density, a stronger condition is imposed: there is $\delta > 0$ such that any two points in $S$ are at least distance $\delta$ apart.

(Both versions are usually stated for convex sets, but the rectangle versions are equivalent, as any convex set of area $1$ is contained in a rectangle of area $2$ and contains a rectangle of area $1/2$.)

share|cite|improve this answer

I get the feeling that you will enjoy reading about the Simonyi and Chvatal conjectures described here by some guy called Gil Kalai. Anyone know who that is? ;)

share|cite|improve this answer
Yes, I would say that a lot of people know who he is :-) For example he is active on MO: – Daniel Soltész Dec 7 '14 at 17:36

The continuum hypothesis. Of course it's extremely famous, but everyone thinks it's resolved. I was astonished to find out that some serious set theorists apparently consider it (I mean in the present, decades past Cohen's proof) to be an important open problem that people should be working on solving (for some meaning of "solve").

P. Koellner ( ) describes some current approaches.

share|cite|improve this answer
You seem to allude that it has not been resolved yet. But in that case you should explain what you mean by that. On the other hand, the continuum hypothesis is so well-known that it does not fit to the question. – Martin Brandenburg Jul 19 '12 at 12:17
+1 because the note you linked is really interesting! Although the "problem" is rather open-ended, I think it could be stated in a more well-defined way. Here's a stab at it: "If we want to settle the continuum hypothesis, which axioms should we add to ZFC in order to do it?" – Vectornaut Jul 22 '12 at 18:24
By the way, I think your statement that "everyone thinks it's resolved" is a little misleading. Maybe it would be better to say it this way: "most people think there's nothing left to say about it, because it's been proven independent of ZFC." – Vectornaut Jul 22 '12 at 18:26

From Wikipedia:

Keller's conjecture is the conjecture introduced by Ott-Heinrich Keller (1930) that in any tiling of Euclidean space by identical hypercubes there are two cubes that meet face to face.

Keller's original cube-tiling conjecture remains open in dimension 7.

Conjecture was shown to be true in dimensions at most 6 by Perron (1940a, 1940b). However, for higher dimensions it is false, as was shown in dimensions at least 10 by Lagarias and Shor (1992) and in dimensions at least 8 by Mackey (2002), using a reformulation of the problem in terms of the clique number of certain graphs now known as Keller graphs. Although this graph-theoretic version of the conjecture is now resolved for all dimensions, Keller's original cube-tiling conjecture remains open in dimension 7.

The related Minkowski lattice cube-tiling conjecture states that, whenever a tiling of space by identical cubes has the additional property that the cube centers form a lattice, some cubes must meet face to face. It was proved by György Hajós in 1942.

Szabó (1993), Shor (2004), and Zong (2005) give surveys of work on Keller's conjecture and related problems.

share|cite|improve this answer

The list coloring conjecture: A list of colors is assigned to each edge of a finite graph $G$. A "list coloring" of $G$ is an edge-coloring such that (1) each edge is colored with a color from its list, and (2) edges that meet at a vertex have different colors. Suppose the graph $G$ admits a list coloring when the list $\{1,2,\dots,n\}$ is assigned to every edge; does it still admit a list coloring when an arbitrary list of $n$ colors is assigned to each edge?

share|cite|improve this answer

protected by Community Oct 26 '15 at 4:22

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.