Not especially famous, long-open problems which anyone can understand

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.

https://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.

• 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" Jun 21, 2012 at 20:53
• Nice question!! Jun 22, 2012 at 3:25
• 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. Jun 22, 2012 at 12:05
• More precisely, K-12 means anything up to high school (K = Kindergarten, 12 = 12th grade, and K-12 covers this range). Jun 22, 2012 at 13:05
• There seems to be a claimed proof of the union-closed sets conjecture by Blinovsky arxiv.org/abs/1507.01270 Oct 22, 2015 at 14:08

One problem which I think is mentioned in Guy's book is the integer block problem: does there exist a cuboid (aka "brick") where the width, height, breadth, length of diagonals on each face, and the length of the main diagonal are all integers?

update 2012-07-12 Since the question has returned to the front page, I'm taking the liberty to add some links that I found after Scott Carnahan's comments. (Scott deserves the credit, really, but I thought the links belonged in the answer rather than in the comments.)

• Because so much has been known about Pythagorean triples for so long, I'm shocked that this problem is open. Is there an intuitive explanation of why the problem is so hard? Jun 22, 2012 at 5:38
• I'm afraid I have no idea (mind you, I can think of no intuitive reason why it wouldn't be hard). Further details at mathworld.wolfram.com/PerfectCuboid.html Jun 22, 2012 at 5:49
• The solution space forms an algebraic surface in the projectivized space of box dimensions. The surface has rather high degree, and in fact van Luijk showed that it is of general type, (and therefore rather resistant to standard methods). Jun 22, 2012 at 6:03
• @Dal I am not a number theorist, but I believe that a solution here would require one to introduce new techniques or make significant improvements to existing techniques. Note that the original question did not ask for questions which were "important to current mathematical research", merely that "There should already exist at least a small community of mathematicians who will care if one of these problems gets solved." Dec 23, 2014 at 0:52
• What about a brick with sidelength 0? ;) Oct 26, 2015 at 4:35

The moving sofa problem: What rigid two-dimensional shape has the largest area $$A$$ that can be maneuvered through an L-shaped planar region with legs of unit width?

So far the best results are $$2.219531669\lt A\lt 2.37$$.

• The upper bound of $2\sqrt{2}$ was recently improved to 2.37 in this paper by Yoav Kallus and me. Sep 14, 2017 at 18:00

Can we cover a unit square with $\dfrac1k \times \dfrac1{k+1}$ rectangles, where $k \in \mathbb{N}$?

(Note that the areas sum to $1$ since $\displaystyle \sum_{k \in \mathbb{N}}\dfrac1{k(k+1)} = 1$)

Here is an MO thread discussing some of the progress on this problem.

This is the second time I've seen this question on MathOverflow and this will be the second time I've posted this answer.

Singmaster's conjecture says there is a finite upper bound on the number of times a number (other than the $$1$$s on the edge) can appear in Pascal's triangle. The upper bound may be as low as $$8$$. If so, then no number (besides those $$1$$s) appears more than eight times in Pascal's triangle. Only one number is known to appear that many times: $$\binom{3003}{1} = \binom{78}{2} = \binom{15}{5} = \binom{14}{6}$$

It has been proved that infinitely many numbers appear twice; similarly three times, four times, and six times. It is unknown whether any number appears five times or seven times.

Singmaster states that Erdős said the conjecture is probably true but probably difficult to prove.

• We don't really need Erdős to tell us it's probably true when we can do straightforward probabilistic estimates (plus some geometry of plane curves). A short computation shows that there are no numbers less than $10^{1000}$ that have odd multiplicity greater than 3, and heuristics suggest it is quite unlikely that such numbers exist. Jul 2, 2012 at 9:49
• @S.Carnahan : How did you do that "short computation"? Jul 6, 2012 at 21:49
• Odd multiplicity means you have a number of the form $\binom{2k}{k}$. It's not hard to check whether a number has the form of a binomial coefficient $\binom{m}{n}$ in SAGE, since you have a built-in function that estimates integer $n$-th roots. Jul 12, 2012 at 7:02
• I love this problem! Everything about it is simple and compelling, and it can be understood by anyone who knows how to add. Is there also a simple heuristic argument for why it should be true? @S. Carnahan , can you flesh out your heuristics a little more? What's this stuff about geometry of plane curves? Jul 22, 2012 at 18:44
• The conjecture has been proved in a large interior region in this paper by 5 coauthors arxiv.org/abs/2106.03335 Jul 20, 2021 at 8:51

The lonely runner conjecture. As Wikipedia puts it:

Consider $k + 1$ runners on a circular track of unit length. At $t = 0$, all runners are at the same position and start to run; the runners' speeds are pairwise distinct. A runner is said to be lonely if at distance of at least $1/(k + 1)$ from each other runner. The lonely runner conjecture states that every runner gets lonely at some time.

• Also, I suspect this is equivalent to the lonely starting post conjecture, which is the conjecture above except that one of the runners has speed 0 and the statement is that he/she gets lonely. Gerhard "Ask Me About Going Slow" Paseman, 2012.06.22 Jun 22, 2012 at 18:59
• Observe that the human condition implies that everyone gets lonely at some time. In particular the runners get lonely. Jun 22, 2012 at 20:59
• This is open for $k\geq 7$. The proof for $k=6$ was done by Barajas and Serra using elaborate computer-assisted casework, and many simplifications that rely on the fact that $6+1$ is prime. It is worth noting that when the ratio of two speeds is irrational, the problem is made easier by density arguments, so the essentially hardest case is when all the speeds are integers. Therefore this is a combinatorial number theory question disguised as basic calculus. Jul 2, 2012 at 2:14
• Here's a post by Terence Tao on this conjecture: terrytao.wordpress.com/2015/05/13/… Oct 5, 2015 at 6:45
• The conjecture was proved in the special case where the distance is long. See en.wikipedia.org/wiki/… Feb 2, 2019 at 13:14

The Casas-Alvero conjecture: let the characteristic of the field $k$ be $0$. If a monic polynomial $f\in k[X]$ of degree $n$ has a common root with each of its derivatives $f',\ldots,f^{(n-1)}$, then $f(X)=(X-a)^n$ for some $a\in k$.

• I guess $k$ must be of characteristic $0$
– Joël
Jun 22, 2012 at 19:35
• @Joel. Right! If $k$ is of finite characteristic $p$, then $X^{2p}+X^p$ does share a root with every derivative, but is not a monomial. Jun 22, 2012 at 20:38
• For those interested in this conjecture, here is what I believe the current state of knowledge on the conjecture : arxiv.org/abs/math/0605090 The first open case is $n=12$. Interestingly, the proofs in the known cases use scheme theory (over $\mathbf{Z}$). Jun 22, 2012 at 22:57
• @François Brunault. Some months ago I asked this question mathoverflow.net/questions/94838/… with the Casas-Alvero conjecture in mind. It appeared from the answers that instead of the argument using scheme theory, the simpler Lefschetz principle ( proofwiki.org/wiki/Lefschetz_Principle_(First-Order) ) can be used. (answering to my question, Qiaochu Yuan also indicated an ultraproduct construction which is even simpler than the Lefschetz Principle, since no completeness result is used).
– js21
Jun 25, 2012 at 6:22
• I think this squarely fails the 'anyone can understand it' requirement.
– user49512
Jul 31, 2017 at 2:51

Gourevitch's conjecture1: $$\sum_{n=0}^\infty \frac{1+14n+76n^2+168n^3}{2^{20n}}\binom{2n}{n}^7 = \frac{32}{\pi^3}.$$

1Jesús Guillera: About a New Kind of Ramanujan-Type Series; Experimental Mathematics (2003), Volume: 12, Issue: 4, page 507-510; DOI: 10.1080/10586458.2003.10504518, eudml

• wow, at first look it seems hard to believe that this is still a conjecture! Jun 22, 2012 at 3:26
• As I understand it, this kind of identity is amenable in principle to automatic theorem-proving methods, but (using known techniques) is out of reach of current computers. Jun 22, 2012 at 14:40
• Tim, there is also an example, from December 2011, for $1/\pi^4$ due to Jim Cullen (members.bex.net/jtcullen515), another mathematics amateur; I cannot easily fine it online though. Aug 25, 2012 at 11:13
• What is the theory that led to such formula? It's hard to believe that it could have been conjectured from playing with random expressions. Dec 3, 2014 at 0:41
• @Michael : Try following the link. There are similar-looking formulas that have been proved. One can then hypothesize the existence of other formulas having a similar general form, and search for them numerically using a linear-relation-finding algorithm such as PSLQ. Dec 3, 2014 at 21:48

There is a lot of number theory elementary conjectures, but one that is especially elementary is the so called Giuga Conjecture (or Agoh-Giuga Conjecture), from the 1950: a positive integer $$p>1$$ is prime if and only if $$\sum_{k=1}^{p-1} k^{p-1} \equiv -1 \pmod{p}$$

• P would be at least a carmichael number or fermat pseudoprime. Jan 3, 2017 at 2:39
• I think I may have actually solved this problem, unless I've done something terrible in my proof. I didn't even realize this was an open problem; a classmate posed the question to me as a kind of converse to Fermat's Little Theorem, and I spent about an hour to find a proof. Is there someplace I can put my "proof" to see if it is correct? Feb 9, 2017 at 0:58
• @feralin So please tell me, Did you actually prove it? Did you publish it somewhere? Sep 30, 2019 at 20:09
• @feralin Since you mention Fermat's little Theorem it sounds like you proved that the formula works if p is prime. But the main question you would need to answer is whether or not there is a composite number that would also satisfy this formula.
– Ryan
Oct 24, 2019 at 3:34
• @DeepakMS for an update on the situation: I made a dumb mistake in my "proof", which I caught shortly after. Dec 27, 2019 at 17:39

Is $e+\pi$ rational?

• Too famous? Most mathematicians have heard of this one, haven't they? Jun 22, 2012 at 14:41
• Popular books, I don't know, but David Feldman wrote, "I'm looking for problems that, with high probability, a mathematician working outside the particular area has never encountered." (Emphasis mine.) It feels to me that most professional mathematicians, even those not working in transcendental number theory, are familiar with this one. Jun 22, 2012 at 18:17
• I have never heard this before... Jul 3, 2012 at 14:53
• Even though I've seen this one at many different places, what I don't know is: why is this question so exceedingly tricky? Jul 27, 2012 at 20:59
• @Suvrit: The way I think of it, it's not so much that this particular question is exceedingly tricky; it's that we don't know that many ways to prove that a specific number is irrational. For "most" numbers that one can name, irrationality is unknown. This just happens to be one of the simplest examples. Similarly, it's not hard to write down a simple Diophantine equation or PDE whose solvability is unknown. Aug 27, 2012 at 3:19

Is the sequence $(3/2)^n \mod 1$ dense in the unit interval?

In the other direction, Mahler's 3/2 problem:

Do all elements of this sequence with large enough index $n$ lie in the interval $(0,1/2)$?

It is known that $\beta^n$ is uniformly distributed modulo one for almost all $\beta>1$, but explicit examples of $\beta$ for which density holds are not known. This question seems to originate in work of Weyl and Koksma on uniform distribution.

Update: Since posting this answer I've attempted to find some references with which to flesh it out, with only modest success. The earlier paper I have identified which deals with this question directly is T. Vijayaraghavan's 1940 article On the fractional parts of the powers of a number, in which it is shown that the sequence $(3/2)^n \mod 1$ has infinitely many limit points. Mahler conjectured in 1968 that the answer to his question is negative. Jeffrey Lagarias' 1985 survey on the Collatz problem, The 3x + 1 Problem and Its Generalizations, includes a one-page overview of the literature on the distribution of this sequence. Flatto, Lagarias and Pollington subsequently proved that the diameter of the set of accumulation points is at least 1/3; Mahler's question would be answered in the negative if this is improved to "at least 1/2".

• An excellent reference is the recent book Distribution modulo one and Diophantine approximation, by Yann Bugeaud. Jan 6, 2014 at 19:35
• In my youth I did a modest contribution to the related Mahler 3/2 problem (dx.doi.org/10.5802/jtnb.588). Nov 11, 2015 at 3:51
• I think this squarely fails the 'anyone can understand it' requirement.
– user49512
Jul 31, 2017 at 2:51
• @MilesRout, I have to disagree. The requirement is that it has understandable formulations. Suppose you multiply $3/2$ by itself $n$ times. What's the first digit after the decimal point? Can it be anything you want (for at least one value of $n$) or are there forbidden values? What about the first two digits, can you make them be anything you like by choosing $n$ correctly? What about the first 100 digits, can we make those be anything we want by choosing the right $n$? If $N \geq 1$ is given, can we fix the first $N$ digits after the decimal point to be anything we want them to be? Aug 1, 2017 at 17:06
• So put that in the answer then.
– user49512
Aug 1, 2017 at 23:13

It is currently unknown if all triangles have a periodic billiard path. (See, for example, https://en.wikipedia.org/wiki/Outer_billiards#Existence_of_periodic_orbits)

– JRN
Jun 22, 2012 at 5:24
• Additional info: The best known result is that all triangles of maximum angle 100 degrees admit a periodic orbit. It is also known that all triangles (in fact, all polygons) with angles that are rational multiples of $\pi$ admit periodic orbits. Jul 3, 2012 at 4:08

From "An Invitation to Mathematics":

Are there any integer solutions to $$x^3 + y^3 + z^3 = 33$$?

I thought this might be a good candidate since that book was meant as a bridge from competitive Mathematics to research. There are a few other examples, but I am quoting only one here due to your requirement. Edit: Such integers x, y and z have been found.

• Is there something special about 33?! Jun 22, 2012 at 5:42
• For small numbers (<100), 33, 42 and 74 are still unresolved. See this: asahi-net.or.jp/~kc2h-msm/mathland/math04/matb0100.htm . @Vectornaut when I saw your comment the first thing I thought off was the irrational solution $(\sqrt[3]{33/3},\sqrt[3]{33/3},\sqrt[3]{33/3})$. Jun 22, 2012 at 13:45
• @NgYongHao 74 is solved.
– orlp
Oct 4, 2016 at 19:17
• $33=8866128975287528^3+(−8778405442862239)^3+(−2736111468807040)^3$ - see gilkalai.wordpress.com/2019/03/09/… Mar 11, 2019 at 22:09
• $42=(-80538738812075974)^3 + 80435758145817515^3 + 12602123297335631^3$ is now solved (by Andrew Booker and Andrew Sutherland). Oct 24, 2019 at 17:30

I always enjoyed telling people about the Inscribed square problem :

Does every (Jordan) curve in the plane contain all four vertices of some square?

Update: Here is a variation due to Helge Tverberg: Does every (polygonal) curve in the plane outside of the unit circle, contain all four vertices of some square with side length >0.1? This version implies the original problem and lacks disadvantages pointed out by Tim Chow and Henry Cohn. See Ville H. Pettersson, Helge A. Tverberg, and Patric R.J. Östergård, "A Note on Toeplitz' Conjecture," Discrete Comput. Geom. 51 (2014), 722–738.

• This is a nice problem but it's only open in the case where the curve is pathologically ugly, in a way that perhaps not "anyone can understand." Jun 22, 2012 at 2:05
• I do think that anyone can understand whats an injective, continuous map from the circle to the plane. Jun 22, 2012 at 6:44
• Actually, I disagree that anyone can (quickly, easily) understand what such a map is for the purposes of this problem, since the maps for which it's not known are of a sort even mathematicians didn't realize existed until well into the 19th century. One can still state the problem, but it's likely to lead to conversations of the following sort. "Wow, so you mean nobody knows in advance if this curve [draws a curve] has a square in it?" "Well, actually we know that case, or really any curve you can draw, but mathematicians have discovered exotic curves for which we don't know the answer." Jun 22, 2012 at 13:14
• The issue here is that intuitive "definitions" of continuous tend to be wrong. "You can draw it without lifting your pencil" really means at least piecewise smooth. Jun 24, 2012 at 3:40
• Well, not if you shake your hand fast enough (or with enough brownian motion) Aug 24, 2012 at 22:48

There are infinitely many primes $p$ such that the repeating part of the decimal expansion of $1/p$ has length $p-1$.

First explicitly asked by Gauss, now generally thought of as a corollary of Artin's primitive root conjecture.

• I think Artin's primitive root conjecture counts as pretty well known. Jun 22, 2012 at 1:51
• @unknown: That's a fair comment. Still, if the goal is to find conjectures that are accessible to the general math-loving public that they may not have heard of before, I think the decimal expansion problem counts. Perhaps David Feldman can clarify whether he really means that 90% of non-number theorists haven't heard of the conjecture of which this happens to be a corollary, or whether he means something weaker than that. Jun 22, 2012 at 14:37

Problem: The partition function $$p(n)$$ is even (resp. odd) half of the time.

Of course you need to explain to a general audience what the partition function is, but that's not hard, my daughter in K1 got as an assignment to compute $$p(n)$$ for $$n$$ up to 4. You also need to explain "half of the time", which means that the number of $$n < x$$ such that $$p(n)$$ is even, divided by $$x$$, has limit 1/2 when $$x$$ goes to infinity, so you need the notion of limit of a sequence, which is in K12, isn't it ?

The problem is certainly famous among specialists, but not too famous. I don't think there are books on it, for instance. It is old (formulated as a conjecture during the 50th), with an history going back to Ramanajunan. And I like it very much.

UPDATE (28/2/2015) Here is a useful reference:
Ken Ono, The parity of the partition function, Electronic Res. Ann. (1995)

• The notion of limit of a sequence is not usually taught in the US until a real analysis course, which is usually taken only by students in mathematics and frequently not until the third (or even last) year of university. (But I think this case is concrete enough that the necessary ideas here could be explained to a high school student.) Jun 22, 2012 at 4:05
• Ah, that is bad. Isn't there an option for seniors in good high school to learn some calculus ?
– Joël
Jun 22, 2012 at 12:21
• Sequences are taught before real analysis, usually in Calc 2 along with infinite series. And the more basic material is suitable for high school, even a decent precalculus class. These are only sequences of reals so it isn't very general, and while they are taught, students might not really "understand" them until later. Jun 22, 2012 at 12:51
• Yes, there is an option for seniors in a good high school to learn some calculus, but most calculus courses in the United States no longer give a rigourous definition of a limit. Without a rigourous definition, there are some subtle possibilities for what might go wrong that won't be appreciated. (Of course, very few students at that level have the mathematical maturity to understand a rigourous definition well enough to appreciate the subtle possibilities anyway, which is why the rigourous definition isn't taught anymore.) Sep 6, 2012 at 4:11
• Also, "half of the time" can be restated in probabilistic terms. In other words, instead of framing it as a real analysis question, appeal to probabilistic intuition. Alexander Woo's remarks about subtle possibilities notwithstanding, vastly larger numbers of students learn elementary probability and statistics than calculus. Jan 6, 2014 at 19:28

The circulant Hadamard matrix conjecture, first stated in print by Ryser in 1963. It can be stated as follows. If $n>4$, then there does not exist a sequence $(a_1,a_2,\dots,a_n)$ of $\pm 1$'s satisfying $$\sum_{i=1}^n a_i a_{i+k}=0,\ 1\leq k\leq n-1,$$ where the subscript $i+k$ is taken modulo $n$.

• Related to this, the Hadamard conjecture : there exist Hadamard matrices of order $4k$ for every $k$. en.wikipedia.org/wiki/Hadamard_matrix#The_Hadamard_conjecture Jun 22, 2012 at 9:19
• Further related: Let m be the largest integer such that the integer interval (-m,m) is contained in the set D_n, the set of determinants of order n 0-1 matrices. What function of n are very good bounds for approximating m? Cf determinant spectrum on Will Orrick's maxdet site. Gerhard "Ask Me About Binary Matrices" Paseman, 2012.06.22 Jun 22, 2012 at 19:08
• It seems the circulant Hadamard matrix conjecture has been solved on 4 pages: arxiv.org/abs/2302.08346
– sdd
Feb 6 at 3:49

At the risk of stretching my own rule, please allow that I could define "ring" for a high school senior. Then I'd proffer this question I heard years ago from Melvin Henriksen:

Must a non-commutative ring (with identity) contain a non-zero-divisor aside from the identity?

• Here is a link to Henriksen's paper related to this question. google.com/… Jun 23, 2012 at 18:52
• I've known ring theory for a while, and it never even occurred to me that that was difficult (let alone possibly true). Jul 22, 2012 at 20:35
• Mel, to whom I will be eternally grateful for my low Erdos number and much else, was a master of the uniquely mathematical game of one downsmanship: "You don't if ..., we I don't even know if ...!!! Jul 23, 2012 at 0:04
• This looks like a fun way to spend some time in $\mathbb{Z}/(2)$-vector spaces... Feb 6, 2017 at 5:05
• That's a very interesting problem ... note that a finite non-commutative ring with unity has more than one units ( see mathoverflow.net/questions/284684/… ) ; so that only leaves for infinite rings to be considered .
– user111524
Oct 29, 2017 at 8:42

Sendov's Conjecture

For a polynomial $$f(z) = (z-r_{1}) \cdot (z-r_{2}) \cdots (z-r_{n}) \quad \text{for} \ \ \ \ n \geq 2$$ with all roots $$r_{1}, ..., r_{n}$$ inside the closed unit disk $$|z| \leq 1$$, each of the $$n$$ roots is at a distance no more than $$1$$ from at least one critical point of $$f$$.

• Now proved by Terence Tao for n large enough, see arxiv.org/abs/2012.04125 Jul 20, 2021 at 8:35
• @Archie Thanks for the update :)
– C.S.
Jul 21, 2021 at 3:37

The irrationality of Catalan's constant $$G=1-1/3^2+1/5^2-1/7^2+\cdots$$.

Remarks: Although Catalan's constant is certainly well-known, the irrationality is the tip of the iceberg of a related conjecture of Milnor about the linear independence over the rationals of volumes of certain hyperbolic 3-manifolds (which is a special case of a conjecture of Ramakrishnan). The irrationality of Catalan's constant would imply that the volume of the unique hyperbolic structure on the Whitehead link complement is irrational. To this date, it is not known that any hyperbolic 3-manifold has irrational volume.

• There are some partial results towards the irrationality: [T. Rivoal and W. Zudilin, Diophantine properties of numbers related to Catalan's constant, Math. Ann. 326:4 (2003), 705–721] (dx.doi.org/10.1007/s00208-003-0420-2). Nov 11, 2015 at 3:55

Does the series $\sum_{n=1}^{\infty} \frac{1}{n^3 \sin^2 n}$ converge?

(Taken from https://math.stackexchange.com/questions/20555/are-there-any-series-whose-convergence-is-unknown where there are more such examples)

• and, in my answer at math.SE which you link here, I refer to the mathoverflow question mathoverflow.net/questions/24579. Jun 24, 2012 at 0:35
• This is really cool! Everyone should take a look at the actual math.SE answer, which explains why this question is hard. Jul 22, 2012 at 18:53
• For comparison, see also this question: mathoverflow.net/questions/282259/… Dec 10, 2017 at 22:08

Here is one which I found at this MO link:

$$\frac{24}{7\sqrt{7}} \int_{\pi/3}^{\pi/2} \log \left| \frac{\tan(t)+\sqrt{7}}{\tan(t)-\sqrt{7}}\right|\ dt = \sum_{n\geq 1} \left(\frac n7\right)\frac{1}{n^2},$$ where $$\displaystyle\left(\frac n7\right)$$ denotes the Legendre symbol. Not really my favorite identity, but it has the interesting feature that it is a conjecture! It is a rare example of a conjectured explicit identity between real numbers that can be checked to arbitrary accuracy. This identity has been verified to over 20,000 decimal places. See J. M. Borwein and D. H. Bailey, Mathematics by Experiment: Plausible Reasoning in the 21st Century, A K Peters, Natick, MA, 2004 (pages 90-91).

P.S. This problem was resolved before this post was placed in Section 5 of [D.H. Bailey, J.M. Borwein, D. Broadhurst and W. Zudilin, Experimental mathematics and mathematical physics, in "Gems in Experimental Mathematics", T. Amdeberhan, L.A. Medina, and V.H. Moll (eds.), Contemp. Math. 517 (2010), Amer. Math. Soc., 41–58]. In fact, the problem was solved even before its mentioning in the 2004 book; the details of the story can be found in the article.

• cured link to Sendov's conj. Great answer ! Jun 22, 2012 at 12:07
• @alexander: thanks for fixing the broken link.
– C.S.
Jun 22, 2012 at 17:20
• It was a good idea to split the two conjectures to two answers, but you should have done it the other way around. I venture to guess that most people, like me, originally upvoted this answer because of Sendov’s conjecture, not because of an obscure integral equality which I couldn’t explain to any high school student I now of. Jun 25, 2012 at 10:34
• @Emil: Emil, The answers were split because of an user requesting me to do so. Otherwise I would have kept it here itself.
– C.S.
Jul 1, 2012 at 7:40

Here are a few others:

1. Let $$H_n=\sum_{j=1}^n 1/j$$. Then for all $$n\geq 1$$, $$\sum_{d|n}d\leq H_n+(\log H_n)e^{H_n}.$$ Jeff Lagarias showed that this is equivalent to the Riemann hypothesis!

2. Let $$x_0=2$$, $$x_{n+1}=x_n-\frac{1}{x_n}$$ for $$n\geq 0$$. Then $$x_n$$ is unbounded.

3. The largest integer that cannot be written in the form $$xy+xz+yz$$, where $$x,y,z$$ are positive integers, is 462. It is known that there exists at most one such integer $$n>462$$, which must be greater than $$2\cdot 10^{11}$$. See J. Borwein and K.-K. S. Choi, On the representations of $$xy+yz+xz$$, Experiment. Math. 9 (2000), 153-158; https://projecteuclid.org/journals/experimental-mathematics/volume-9/issue-1/On-the-representations-of-xyyzzx/em/1046889597.full.

• Where does the second of these come from? Jun 22, 2012 at 20:24
• @Noam: I can't recall. Someone mentioned this to me many years ago. A little thought will show why this is so difficult. Jun 22, 2012 at 20:31
• I'm wondering if I "get" #2. I see an implicit map from $S^1$ to $S^1$ of index 2, so yes, it seems generally hard to understand the dynamical fate of a given starting value. A similar question might ask if the binary expansion of $\sqrt{2}$ contains strings of 0's of arbitrary length. But is #2 specifically conjugate to something more familiar? Jun 23, 2012 at 19:17
• @Davidac897: I think the conjecture part of #3 is the first sentence: "The largest integer... is 462." If I'm reading the rest correctly, it's known that if the conjecture is false, it's only because of a single counterexample that must be greater than 200 billion. Jul 27, 2012 at 21:15
• Question #2 was addressed in the paper math.grinnell.edu/~chamberl/papers/mario_digits.pdf The real problem concerns the initial value $x_0=2$. It can be shown that the set of initial values which produce an unbounded sequence $\{x_n\}$ has full measure, so from a probabilistic perspective, one expects the statement in question 2 to hold. Aug 19, 2012 at 18:40

The Kneser–Poulsen conjecture in dimension 3: An arrangement of (possibly overlapping) unit balls in space is tighter than a second arrangement of the same balls if, for all $$i$$ and $$j$$, the distance between the centers of ball $$i$$ and ball $$j$$ in the first arrangement is less than or equal to the distance between the centers of ball $$i$$ and ball $$j$$ in the second arrangement. The conjecture is that a tighter arrangement always has equal or smaller total volume. True in the plane, open in higher dimensions.

Here is another easy to state problem which is 140 years old but not very famous. Consider the potential of finitely many positive charges: $$u(x)=\sum_{j=1}^n\frac{a_j}{|x-x_j|},\quad x,x_j\in R^3,\quad a_j>0$$ How many equilibrium points can this potential have? Equilibrium points are solutions of $\nabla u(x)=0$.

First conjecture: it is always finite.

Second conjecture: when finite, it is at most $(n-1)^2$. This estimate is stated by Maxwell in his Treatease on Electricity and Magnetism, vol. I, section 113, as something known. The editor (J. J. Thomson) wrote a footnote that he "could not find any place where this result is proved".

Nobody could find this place to this time. This is even unknown in the simplest case when all $a_j=1$ and $n=3$.

• This is quite similar to the problem regarding finiteness of the number of central configurations for the n-body problem which made it into Smale's problem list, known for centuries for n=3, proven for n=4 [Moeckel and Hampton] and with substantial results for n=5 [Albouy and Kaloshin]. To get to Maxwell's from the n-body one, formally take n bodies and place them at the $x_j$, with masses $a_j$, and then add a n+1st body, at variable x, with mass $\epsilon$ and let $\epsilon \to 0$. Hmmm... Jan 28, 2015 at 4:55
• I do not understand what is the relation between the problems. Can you explain? In Maxwell problem $x_j$ are FIXED. Jan 28, 2015 at 14:48
• I just saw this claim for a proof for the case $n=3$ with equal $a_j$: sciencedirect.com/science/article/pii/S0167278915001347 Oct 12, 2015 at 1:53
• @Suvrit: Interesting! Thanks for the reference. Oct 12, 2015 at 10:41

Let $${^n a}$$ denote tetration: $${^0 a}=1, {^{n+1} a}=a^{({^n a})}$$.

• It is unknown if $${^5 e}$$ is an integer.
• It is unknown if there is a non-integer rational $$q$$ and a positive integer $$n$$ such that $${^n q}$$ is an integer.
• It is unknown if the positive root of the equation $${^4 x}=2$$ is rational (ditto for all equations of the form $${^n x}=2$$ with integer $$n>3$$)
• It is unknown if the positive root of the equation $${^3 x}=2$$ is algebraic.
• Relevant OEIS sequences: A199550 (${^3 x}=2$) and A225134 (${^4 x}=2$). Feb 10, 2017 at 19:08

Is there a dense subset of a plane having only rational distances between its points?

• This problem is given in Klee, Victor; Wagon, Stan (1991), "Problem 10 Does the plane contain a dense rational set?", Old and New Unsolved Problems in Plane Geometry and Number Theory, Dolciani mathematical expositions, 11, Cambridge University Press, pp. 132–135, ISBN 978-0-88385-315-3, and attributed to Ulam and Erdős. Jul 25, 2012 at 13:39
• It was also discussed here: mathoverflow.net/questions/19127/… Jul 28, 2012 at 20:30

Does there exist a point in the unit square whose distance to each of the four corners is rational?

This is sometimes called the rational distance problem, although that name often refers to a more general class of similar problems. It's discussed by Richard Guy in Unsolved Problems in Number Theory and in the following paper:

Guy, Richard K. "Tiling the square with rational triangles." Number theory and applications 265 (1989): 45-101.

It's also open whether there's a point outside the square whose distance to each of the four corners is rational, although it is known that no point on the edge of the square has this property.

• I think there is a rational transformation which shows the outside version reduces to the inside. The relevant picture may e en be in Guy's book. Jun 7, 2015 at 19:19

Schinzel-Sierpinski Conjecture

Melvyn Nathanson, in his book Elementary Methods in Number Theory (Chapter 8: Prime Numbers) states the following:

• A conjecture of Schinzel and Sierpinski asserts that every positive rational number $x$ can be represented as a quotient of shifted primes, that $x=\frac{p+1}{q+1}$ for primes $p$ and $q$. It is known that the set of shifted primes, generates a subgroup of the multiplicative group of rational numbers of index at most $3$.
• That's a nice one!
– user9072
Jun 23, 2012 at 9:46
• @quid: I agree with your suggestion and hence I have deleted my answer.
– C.S.
Jun 23, 2012 at 9:51
• @Chandrasekhar: thanks for letting me know. [To avoid confusion for others, this conversation refers to a different answer.]
– user9072
Jun 23, 2012 at 9:57
• @quid: No problem. The deleted answer does seem to appear in a red box.
– C.S.
Jun 23, 2012 at 10:12
• @Chandrasekhar: yes, but only for you (the owner), 10k+ users, and moderators.
– user9072
Jun 23, 2012 at 10:16

A few decades ago Sherman Stein asked whether a trapezoid whose parallel sides are in the ratio $$1:\sqrt2$$ can be dissected into triangles, all of the same area. This remains open--it's a mystery which trapezoids admit such dissections.

Erdos's problem on the length of lemniscates (it is somewhat famous in certain narrow circles). Let $P$ be a polynomial, and consider the set $E=\{ z:|P(z)|=1\}$ in the complex plane.

What is the maximum length of $E$ over all monic polynomials of degree $d$?

Erdos conjectured that an extremal $P$ is $P_0(z)=z^d+1$.

It is known that the asymptotic of maximal length is $2d+o(d).$ It is known that $P_0$ gives a local maximum. It is also known that for every extremal polynomial, all critical points lie on $E$, so $E$ must be connected.

However the conjecture is not established even for $d=3$.

After Erdos's death, I offered a $200 prize for the first solution. (Erdos had offered the same, but I do not know whether one can collect his prize.) • According to en.wikipedia.org/wiki/Paul_Erdős#Erd.C5.91s.27_problems offers Erdos made will be honored. Jan 29, 2015 at 2:18 • Thanks, this is nice to know. So you can collect$400 total for this problem. Jan 29, 2015 at 18:27