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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 high school (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.

http://en.wikipedia.org/wiki/Union-closed_sets_conjecture

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

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    $\begingroup$ 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 $\endgroup$ – Gerhard Paseman Jun 21 '12 at 19:11
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    $\begingroup$ Here is some collection of some other "collect open problems" quests. on MO: mathoverflow.net/questions/96202/… PS Nice question ! PSPS may be add tag "open-problems" $\endgroup$ – Alexander Chervov Jun 21 '12 at 20:53
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    $\begingroup$ 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. $\endgroup$ – Emil Jeřábek supports Monica Jun 22 '12 at 12:05
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    $\begingroup$ More precisely, K-12 means anything up to high school (K = Kindergarten, 12 = 12th grade, and K-12 covers this range). $\endgroup$ – Henry Cohn Jun 22 '12 at 13:05
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    $\begingroup$ There seems to be a claimed proof of the union-closed sets conjecture by Blinovsky arxiv.org/abs/1507.01270 $\endgroup$ – Marco Oct 22 '15 at 14:08

105 Answers 105

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How many trees are there?

Let $T(n)$ be the number of trees on $n$ vertices up to graph isomorphism. There is no known closed formula for $T(n)$.

In 1947 Richard Otter proved[Source] the asymptotic result $$T(n) \sim A \cdot B^n \cdot n^{-\frac{5}{2}}$$ where $A \approx 0.535$ & $B \approx 2.996$.

By way of contrast, let $L(n)$ be the number of labelled trees, i.e. trees formed from vertices labelled $1,...,n$ where isomorphism additionally preserves the label. In 1889, Arthur Cayley showed[Source] that $$L(n)=n^{n-2}$$

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    $\begingroup$ What exactly is the problem here? $\endgroup$ – Harry Altman Jul 24 '12 at 22:03
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    $\begingroup$ The problem is to find a closed form for $T(n)$, I think. $\endgroup$ – I. J. Kennedy Oct 3 '13 at 16:14
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    $\begingroup$ Is there a good reason to suspect such a formula exists? It could equally be that the simplest description of $T(n)$ is the one just given. $\endgroup$ – Jakub Konieczny Jun 4 '15 at 19:41
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    $\begingroup$ “Closed formula” is a bit of a slippery goal — there are specific definitions, but they are mostly somewhat ad hoc. A more mathematically natural goal along similar lines might be to find a polynomial-time algorithm (or some other reasonable sense of “fast”) for computing T(n). $\endgroup$ – Peter LeFanu Lumsdaine Jan 24 '16 at 0:23
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Easy-to-Explain but Hard-to-Solve Problems About Convex Polytopes slides by Jes´us De Loera contains 7 open problems (Hirsch conjecture is also there so it is out-of-date).

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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 $0$.

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    $\begingroup$ >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. $\endgroup$ – David Feldman Jul 13 '12 at 5:36
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    $\begingroup$ @David just wanted to write the same :) mathoverflow.net/questions/101169/… $\endgroup$ – Alexander Chervov Jul 13 '12 at 6:18
  • $\begingroup$ 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 $\endgroup$ – Yemon Choi Jul 14 '12 at 12:34
  • $\begingroup$ @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. $\endgroup$ – Alexander Chervov Jul 14 '12 at 17:53
  • $\begingroup$ 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. $\endgroup$ – Daniel Moskovich Jul 14 '12 at 19:47
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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.

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  • $\begingroup$ Is it easy to describe an optimal tiling for a $a\times (a-1)$ rectangle? Is the exact asymptotic known for this case? $\endgroup$ – Harry Richman Oct 12 '17 at 18:36
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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 $0$?

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  • $\begingroup$ Hi! Does this problem have a name? $\endgroup$ – Dror Speiser Sep 15 '17 at 20:51
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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?

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Ron Graham [1,2] asked if there are infinitely many positive integers $n$ such that the central binomial coefficient $\binom{2n}{n}$ is coprime to $105 = 3 \times 5 \times 7$, and offered a prize of $1.000 for a proof/disproof.

Accordingly to some heuristics [3, §4], there should be infinitely many such $n$, but if instead of $105$ a product of four primes is taken, than only finitely many such $n$ are expected.

Furthermore, it has been proved [4] that $\binom{2n}{n}$ if coprime to $pq$ for infinitely many $n$, where $p$ and $q$ are two fixed odd primes.

[1] D. Berend and J. E. Harmse, On some arithmetic properties of middle binomial coefficients, Acta Arith. 84 (1998), 31–41.

[2] OEIS, https://oeis.org/A030979

[3] C. Pomerance, Divisors of the middle binomial coefficient, Amer. Math. Monthly, 112 (2015), 636-644.

[4] P. Erdős, R. L. Graham, I. Z. Russa and E. G. Straus, On the prime factors of C(2n,n), Math. Comp. 29 (1975), 83-92.

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Some pages:
Open Problem Garden
The Open Problems Project

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    $\begingroup$ [Removed the link to my open-problem page, which is more than a decade old; most of those problems are now solved.] $\endgroup$ – JeffE Jun 22 '12 at 12:06
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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

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    $\begingroup$ The link given to the Diophantine m-tiples pages now says that an article posted on arXiv in October 2016 (arXiv:1610.04020 [math.NT]) claims to answer the first of these two questions in the negative. $\endgroup$ – paul Monsky Nov 4 '16 at 5:26
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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.

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

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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 $n\geq 5$.

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Waring's problem inequality

One of the oldest (Since 1770) and famous open problem in number theory is Waring's problem. It has been conjectured that if

$$ \left\{\left(\frac{3}{2}\right)^n\right\} \le 1 - \left(\frac{3}{4}\right)^n. $$

(where $\{ \cdot \}$ denotes the fractional part) true then, the general solution of Waring's problem is

$$ g(n) = 2^n + \left\lfloor{\left(\frac{3}{2}\right)^n}\right\rfloor - 2. $$

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    $\begingroup$ If I understand correctly, the statement given here has actually been proved in work by Dickson, Pillai, Rubugunday, and Niven. (This is stated in section 6.2.7 of Bombieri and Gubler's book, Heights in Diophantine Geometry.) The conjecture, which coupled to this result would complete the solution of Waring's problem, is that the the stated diophantine inequality (on the fractional part of $(3/2)^n$), should hold true for all $n$. It is an ineffective consequence of Roth's theorem (as extended by Mahler to several places) that it holds for $n \gg 0$. $\endgroup$ – Vesselin Dimitrov Dec 17 '14 at 17:35
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    $\begingroup$ The post is misstated: the inequality is known to be true eventually, thanks to K. Mahler. See the related dx.doi.org/10.5802/jtnb.588 for a historical account and recent novelties. $\endgroup$ – Wadim Zudilin Nov 11 '15 at 4:01
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Grundy's game is a two-player mathematical game of strategy.

The starting configuration is a single heap of objects, and the two players take turn splitting a single heap into two heaps of different sizes. The game ends when only heaps of size two and smaller remain, none of which can be split unequally.

Whether the sequence of nim-values of Grundy's game ever becomes periodic is an unsolved problem. Elwyn Berlekamp, John Horton Conway and Richard Guy have conjectured that the sequence does become periodic eventually, but despite the calculation of the first $2^{35}$ values by Achim Flammenkamp, the question has not been resolved.

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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.)

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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.)

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    $\begingroup$ Where are functions of exponential growth used in the classical proof, or the Euler product + divergence of harmonic series? $\endgroup$ – Douglas Zare Jun 24 '12 at 14:06
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    $\begingroup$ In the classical proof, the exponential growth is in the product of the known primes. See mathoverflow.net/questions/59262/… 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. $\endgroup$ – Henry Cohn Jun 24 '12 at 16:49
  • $\begingroup$ 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. $\endgroup$ – Douglas Zare Jun 24 '12 at 18:01
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    $\begingroup$ See also mathoverflow.net/questions/76058 . 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. $\endgroup$ – Emil Jeřábek supports Monica Jun 25 '12 at 11:35
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    $\begingroup$ 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... $\endgroup$ – Emil Jeřábek supports Monica Jun 25 '12 at 11:52
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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).

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    $\begingroup$ Cool ! what are the references for current state of art ? $\endgroup$ – Alexander Chervov Jul 24 '12 at 9:57
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    $\begingroup$ 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; math.sunysb.edu/Videos/AGNES/video.php?f=04-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: front.math.ucdavis.edu/0808.3351. $\endgroup$ – Arend Bayer Jul 24 '12 at 22:03
  • $\begingroup$ K-12??? $\textrm{}$ $\endgroup$ – Victor Protsak Jan 6 '14 at 20:12
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  • 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?
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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 $1/2$.)

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  • $\begingroup$ Is this the same problem as in mathoverflow.net/q/3307/5340 "Can a discrete set of the plane of uniform density intersect all large triangles?" $\endgroup$ – Zsbán Ambrus Nov 14 '16 at 13:27
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A list of fundamental geometry problems with simple intuitive statements involving curves and surfaces in Euclidean space is maintained at:

Open Problems in Geometry of Curves and Surfaces

Five of these problems are listed below in no particular order (see the paper for references, background, and more precise statements):

Durer's Unfolding Problem: Can every convex polyhedron be cut along a collection of its edges and developed into the plane in a single non-overlapping piece?

Alexandrov's Conjecture on Intrinsic Diameter: Of all convex surfaces with a fixed intrinsic diameter (the distance between the farthest two points as measured within the surface), the one with the greatest area is the doubled disk (the degenerate convex surface obtained by gluing a pair of disks along their boundaries).

Zalgallers's Problems on Width and Inradius: What are the shorted closed curves in Euclidean 3-space with a given width or inradius? (The width is the smallest distance between all pairs of parallel planes which contain the curve in between them, and the inradius is the radius of the largest ball which is contained in the convex hull of the curve and is disjoint from the curve; see the post by Joseph O'Rourke).

Euler-Maxwell Rigidity Problem: Can one continuously deform a smooth closed surface in Euclidean 3-space without changing the distances between its points, as measured within the surface?

Converse of the Archimedes Hatbox Theorem: Is the sphere the only convex surface such that whenever it is cut by a pair of parallel planes separated by a fixed distance, the area of the portion of the surface trapped between the planes is always the same? See this earlier post.

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http://en.wikipedia.org/wiki/Keller%27s_conjecture

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.

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  • $\begingroup$ Apparently the final case of Keller's conjecture was resolved recently (in the affirmative) by a computer assisted proof: arxiv.org/abs/1910.03740 $\endgroup$ – Terry Tao Oct 12 at 1:22
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Is there such $n\in\mathbb{N}$ that ${^n\pi}\in\mathbb{N}$? (see tetration)

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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).

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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?

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This requires some multivariable calculus, so maybe it is not strictly speaking to "everyone", but you could still use it when teaching undergraduates: the unique continuation for the $p$-Laplace equation.

Let $\Omega \in \mathbb R^3$ be an open domain. Suppose $u \in C^2(\Omega)$ and $$ \nabla \cdot (|\nabla u|^{p-2}\nabla u) = 0 \ \ \text{in}\ \Omega, \quad 1 < p \neq 2, $$ with $u\equiv 0$ in some open ball $B \Subset \Omega$. Then the question is to show that necessarily $u \equiv 0$ in all of $\Omega$.

The real open problem is to show this for all weak solutions (which are known to be $C^{1,\alpha}$), but I think this is open also for $C^2$-functions; so posing this makes sense also without any knowledge above multivariable calculus.

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  • $\begingroup$ $p\neq2$ I guess. $\endgroup$ – username Aug 1 '15 at 14:55
  • $\begingroup$ Yes, of course. :) $\endgroup$ – Juhana Siljander Aug 1 '15 at 17:45
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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.

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Given $n\in\mathbb N$, what is the smallest $k\in\mathbb N$ such that the harmonic number $H_k>n$?
It has been conjectured that for all $n$ the answer is $\lfloor\exp(n-\gamma)-1/2\rfloor$. See A002387.

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Does every nonseparating planar continuum have the fixed point property?

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    $\begingroup$ For those of us whose general topology is rusty: what is the K-12 level formulation of this question? $\endgroup$ – Yemon Choi Jun 21 '12 at 22:50
  • $\begingroup$ I'm happy to learn that this unknown, even if doesn't meet the stipulations. So it got my up vote. $\endgroup$ – David Feldman Jun 22 '12 at 3:02
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    $\begingroup$ Uniform continuity is a K-12 concept now? $\endgroup$ – Douglas Zare Jun 22 '12 at 14:25
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    $\begingroup$ 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: $\endgroup$ – Paul Fabel Jun 22 '12 at 17:56
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    $\begingroup$ 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). $\endgroup$ – Paul Fabel Jun 22 '12 at 17:56
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Is there any odd perfect number?

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    $\begingroup$ Methinks this one is both pretty famous and long open... $\endgroup$ – José Hdz. Stgo. Jun 14 '13 at 15:21
  • $\begingroup$ Well, maybe too famous, but I was not sure. $\endgroup$ – The User Jun 14 '13 at 23:05
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    $\begingroup$ answer all mathematicians are both odd and perfect at the same time $\endgroup$ – Riemann-bitcoin. Apr 18 '17 at 14:19
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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 (http://arxiv.org/abs/math/0312202).

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    $\begingroup$ Already on this list: see mathoverflow.net/a/114639/763 $\endgroup$ – Yemon Choi Dec 17 '14 at 16:18
  • $\begingroup$ 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? $\endgroup$ – Stefan Kohl Dec 17 '14 at 16:30
  • $\begingroup$ @StefanKohl you can try search for n=exp(exp(3)) if you find this large enough. $\endgroup$ – joro Dec 17 '14 at 16:53
  • $\begingroup$ @Yemon Choi: Sorry, I didn't see that there is a second page. $\endgroup$ – Jan-Christoph Schlage-Puchta Dec 18 '14 at 14:05
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    $\begingroup$ @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. $\endgroup$ – Jan-Christoph Schlage-Puchta Dec 18 '14 at 14:08

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