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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.

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.

David Feldman
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    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
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    Here is some collection of some other "collect open problems" quests. on MO: http://mathoverflow.net/questions/96202/open-problems-questions-in-representation-theory-and-around PS Nice question ! PSPS may be add tag "open-problems" – Alexander Chervov Jun 21 '12 at 20:53
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    Nice question!! – Suvrit Jun 22 '12 at 03:25
  • Does Traveling Salesman Problem count? (It's famous but can be explained easily to K-12 students.) – Sniper Clown Jun 22 '12 at 09:38
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    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
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    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
  • See also http://math.stackexchange.com/q/532544/18398 – JRN Oct 20 '13 at 23:49
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    Very easy to understand, hard to solve: to find an algorithm which answers for any 3 words A,B,C if consequent substitutions of word A with word B in word C can be performed infinitely or not: http://mathoverflow.net/questions/181057/whats-the-current-state-of-one-rule-semi-thue-system-termination-problem – TT_ stands with Russia Nov 07 '14 at 21:58
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    Chvatal conjecture ( see http://www.math.illinois.edu/~dwest/regs/chvatal.html) very manichean like Union Closed conjecture. – Jérôme JEAN-CHARLES Sep 22 '15 at 17:22
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    There seems to be a claimed proof of the union-closed sets conjecture by Blinovsky http://arxiv.org/abs/1507.01270 – Marco Oct 22 '15 at 14:08
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    What about famous problems, which are super hard to understand, but turned out to be very easy to solve? – Per Alexandersson Oct 22 '15 at 14:27

115 Answers115

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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.)

Yemon Choi
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    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? – Vectornaut Jun 22 '12 at 05:38
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    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 http://mathworld.wolfram.com/PerfectCuboid.html – Yemon Choi Jun 22 '12 at 05:49
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    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). – S. Carnahan Jun 22 '12 at 06:03
  • @YemonChoi, could I ask why this open problem is important to current mathematical research? –  Dec 23 '14 at 00:48
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    @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." – Yemon Choi Dec 23 '14 at 00:52
  • Dear @YemonChoi, thank you very much for the insight; my question was just out of curiosity. –  Dec 23 '14 at 00:54
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    Arguably, the brick violates the "outside mathematician" condition. I even tried to convince some crank (may one say this word here? :-) not to waste as many of his lifetime to the problem as I did. – Hauke Reddmann May 14 '15 at 14:49
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    What about a brick with sidelength 0? ;) – Sidharth Ghoshal Oct 26 '15 at 04:35
  • What do you say on following plan for proving that perfect cuboid does not exist. Let $a$ be odd edge being product of different primes. Let $b_n$ be set of integers forming Pythagorean pair with $a$. We should now prove that there are no Pyth. pair within $b_n$. –  Dec 12 '17 at 07:22
162

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 $2.219531669\lt A\lt 2.37$.

JRN
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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.

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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.

Michael Hardy
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    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. – S. Carnahan Jul 02 '12 at 09:49
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    @S.Carnahan : How did you do that "short computation"? – Michael Hardy Jul 06 '12 at 21:49
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    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. – S. Carnahan Jul 12 '12 at 07:02
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    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? – Vectornaut Jul 22 '12 at 18:44
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    "It has been proved that infinitely many numbers appear twice; similarly three times, four times".

    I guess most of this is easy: every prime number $p$ appears exactly twice, every number of the form $\binom{p}{2}$ ($p\ge 5$) appears exactly four times, and every number of the form $\binom{2k}{k}$ appears an odd and at least three number of times.

    – Francisco Santos Oct 17 '17 at 01:07
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    @FranciscoSantos : Certainly $\dbinom {2k} k$ appears an add number of times and that odd number is at least three. But it's been proved that infinitely many appear exactly three times. – Michael Hardy Oct 17 '17 at 03:26
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    A bit late to the party, but you might be interested in that Singmaster's Conjecture appeared on a Matt Parker video: https://youtu.be/tjJ2qL9uaz4 Timestamp 3:14 :) – Benjamin Wang Dec 03 '20 at 01:15
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    Here's a link to the other MO question that you alluded to. – Timothy Chow Mar 31 '21 at 00:59
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    The conjecture has been proved in a large interior region in this paper by 5 coauthors https://arxiv.org/abs/2106.03335 – Archie Jul 20 '21 at 08:51
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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.

Timothy Chow
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  • For k=1, this does not happen for very long. Gerhard "As( Me About System Design" Paseman, 2012.06.22 – Gerhard Paseman Jun 22 '12 at 18:43
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    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 – Gerhard Paseman Jun 22 '12 at 18:59
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    Observe that the human condition implies that everyone gets lonely at some time. In particular the runners get lonely. – Asaf Karagila Jun 22 '12 at 20:59
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    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.

    – Andrew Dudzik Jul 02 '12 at 02:14
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    Here's a post by Terence Tao on this conjecture: https://terrytao.wordpress.com/2015/05/13/a-remark-on-the-lonely-runner-conjecture/ – pageman Oct 05 '15 at 06:45
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    The conjecture was proved in the special case where the distance is long. See https://en.wikipedia.org/wiki/The_Loneliness_of_the_Long-Distance_Runner – PatrickT Feb 02 '19 at 13:14
  • Ten years on from Asaf's comment I would like to mention that I have never felt loneliness. Nor guilt or disgust, for that matter. But I'm not dead yet, so this of course doesn't actually disprove anything. – Harry Wilson Jul 08 '22 at 20:13
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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 $a\in k$.

Denis Serre
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  • $k$ is any field? – Joël Jun 22 '12 at 19:19
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    I guess $k$ must be of characteristic $0$ – Joël Jun 22 '12 at 19:35
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    @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. – Denis Serre Jun 22 '12 at 20:38
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    For those interested in this conjecture, here is what I believe the current state of knowledge on the conjecture : http://arxiv.org/abs/math/0605090 The first open case is $n=12$. Interestingly, the proofs in the known cases use scheme theory (over $\mathbf{Z}$). – François Brunault Jun 22 '12 at 22:57
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    @François Brunault. Some months ago I asked this question http://mathoverflow.net/questions/94838/emptyness-of-a-projective-variety with the Casas-Alvero conjecture in mind. It appeared from the answers that instead of the argument using scheme theory, the simpler Lefschetz principle ( http://www.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 '12 at 06:22
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    http://arxiv.org/abs/1504.00274 – Andreas Thom Jul 31 '15 at 07:34
  • @Andreas. Strange that this paper deals only with $k={\mathbb C}$. – Denis Serre Jul 31 '15 at 14:51
  • @DenisSerre. Did you check the proof? – Andreas Thom Jul 31 '15 at 16:33
  • @AndreasThom. Not yet. – Denis Serre Jul 31 '15 at 19:49
  • @DenisSerre: The title was changed to "Towards the Casas-Alvero Conjecture". – Andreas Thom Nov 13 '15 at 16:27
  • A proof of the conjecture is on arxiv http://arxiv.org/abs/1511.04932v1 – François Brunault Nov 19 '15 at 07:28
  • @FrançoisBrunault: This paper has been withdrawn today (see http://arxiv.org/abs/1511.04932v2) – Matheus Nov 20 '15 at 18:23
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    Whaaat? I thought this was a trivially true statement – Amr Apr 24 '16 at 13:29
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    Oh ok I misunderstood the conjecture the first time I read it – Amr Apr 24 '16 at 13:49
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    I think this squarely fails the 'anyone can understand it' requirement. –  Jul 31 '17 at 02:51
  • (1/2) Do you know if the conjecture can be paraphrased in terms of certain rational functions satisfying certain conditions? I'm not a professional mathematician, but I wondered if you @FrançoisBrunault can/want tell me if it is potentially interesting, after I known this nice post from the professor. I was trying to propose informal conditions for $d-1$ rational functions $R_k(x)=\frac{P_k(x)}{Q_k(x)}$, thus $1\leq k\leq d-1$ to evoke an equivalent form of Casas-Alvero conjecture. My toy model are these $\frac{f'(x)}{f(x)},\frac{f''(x)}{f'(x)},\ldots ,\frac{f^{(d-1)}(x)}{f^{(d-2)}(x)}$, – user142929 Jan 25 '20 at 23:29
  • (2/2) where $f(x)=(x-a)^d$ is the polynomial in Casas-Alvero conjecture and $a\in k$, and I try to evoke/transfer conditions to state the equivalent form as the statement A suitable set of conditions for the functions $R_k(x)$ implies (say) $Q_1(x)=(x-a)^d$. This can be a naive strategy, my informal reasonings were of the kind taking products or realize those keywords in the conjecture: monic polynomials and degrees; and I don't know if an equivalent formulation in terms of rational functions can be exploited as advantage. My apologizes and thanks to the OP. – user142929 Jan 25 '20 at 23:29
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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

Timothy Chow
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    wow, at first look it seems hard to believe that this is still a conjecture! – Suvrit Jun 22 '12 at 03:26
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    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. – Timothy Chow Jun 22 '12 at 14:40
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    Tim, there is also an example, from December 2011, for $1/\pi^4$ due to Jim Cullen (http://members.bex.net/jtcullen515/), another mathematics amateur; I cannot easily fine it online though. – Wadim Zudilin Aug 25 '12 at 11:13
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    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. – Michael Dec 03 '14 at 00:41
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    @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. – Timothy Chow Dec 03 '14 at 21:48
  • @WadimZudilin Can you post a reference/link to the $1/ \pi^4$ identity, or post the identity here? Thanks in advance? – MathGod Oct 28 '16 at 13:07
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    @TimothyChow I found the link to Cullen's $1/ \pi^4$ conjecture. – MathGod Oct 30 '16 at 00:04
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    @Michael this is in fact part of experimental mathematics. – Anixx Oct 31 '16 at 09:52
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    Just out of curiosity, I plugged this one in to Mathematica - It gives a fairly complicated answer using whatever HypergeometricPFQ is, but if you turn it into a decimal, it's equal to 32/Pi^3 as a decimal to at least 500 digits. Doesn't prove anything, but it's interesting nonetheless. – numbermaniac Sep 28 '17 at 03:46
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    @MathGod it looks like bex.net is now dead. heres wayback machine, is this about how it should look? https://web.archive.org/web/20210226213502/http://members.bex.net/jtcullen515/Math.htm – Sidharth Ghoshal Sep 26 '22 at 01:17
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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 $$\sum_{k=1}^{p-1} k^{p-1} \equiv -1 \pmod{p}$$

Xarles
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  • P would be at least a carmichael number or fermat pseudoprime. – Takahiro Waki Jan 03 '17 at 02:39
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    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? – feralin Feb 09 '17 at 00:58
  • @feralin arXiv $\text{}$ – Brevan Ellefsen Mar 18 '18 at 20:26
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    @feralin So please tell me, Did you actually prove it? Did you publish it somewhere? – Reader Manifold Sep 30 '19 at 20:09
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    @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 '19 at 03:34
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    @DeepakMS for an update on the situation: I made a dumb mistake in my "proof", which I caught shortly after. – feralin Dec 27 '19 at 17:39
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Is $e+\pi $ rational?

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    Too famous? Most mathematicians have heard of this one, haven't they? – Timothy Chow Jun 22 '12 at 14:41
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    I'm not so sure. Which popular books, widespread textbooks or articles do you know where it is stated? – Georges Elencwajg Jun 22 '12 at 17:12
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    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. – Timothy Chow Jun 22 '12 at 18:17
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    Well, Timothy, I think not and it seems to be difficult to know what "most professional mathematicians are familiar with". Anyway, I think this is not important at all: the question is extremely soft and I take in the spirit of an amusing break from actual mathematics. And my only reason for answering it was that I found it amusing to write an open, extremely elementary question with 14 typographical characters ... (By the way, I have just upvoted your four fine answers) – Georges Elencwajg Jun 22 '12 at 19:31
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    I have never heard this before... – Filippo Alberto Edoardo Jul 03 '12 at 14:53
  • That's a very nice question, but is it elementary enough? (Not the word 'rational', but the definition of $e$ and $\pi$, which should be precise enough for the question to make sense ...) – Vincent Beffara Jul 23 '12 at 09:59
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    Even though I've seen this one at many different places, what I don't know is: why is this question so exceedingly tricky? – Suvrit Jul 27 '12 at 20:59
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    @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. – Timothy Chow Aug 27 '12 at 03:19
  • @TimothyChow: can we then interpret this phenomenon as part of a larger concern, where we can say stuff with high probability, but turning the statement into a deterministic truth is out of scope? – Suvrit Apr 01 '15 at 01:37
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    @Suvrit : The irrationality of $e+\pi$ would follow from a far more general statement called Schanuel's conjecture, which in its full generality seems to be out of reach. For example Baker won a Fields medal for a partial result. – Timothy Chow Apr 01 '15 at 15:21
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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".

Ian Morris
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    An excellent reference is the recent book Distribution modulo one and Diophantine approximation, by Yann Bugeaud. – Andrés E. Caicedo Jan 06 '14 at 19:35
  • In my youth I did a modest contribution to the related Mahler 3/2 problem (http://dx.doi.org/10.5802/jtnb.588). – Wadim Zudilin Nov 11 '15 at 03:51
  • I think this squarely fails the 'anyone can understand it' requirement. –  Jul 31 '17 at 02:51
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    @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? – Ian Morris Aug 01 '17 at 17:06
  • So put that in the answer then. –  Aug 01 '17 at 23:13
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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
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    More info can be found at http://www.adnu.edu.ph/bmc2012/Noche.pdf – JRN Jun 22 '12 at 05:24
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    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. – Alex Becker Jul 03 '12 at 04:08
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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.

Ronak
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Ng Yong Hao
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    Is there something special about 33?! – Vectornaut Jun 22 '12 at 05:42
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    For small numbers (<100), 33, 42 and 74 are still unresolved. See this: http://www.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 $(\sqrt[3]{33/3},\sqrt[3]{33/3},\sqrt[3]{33/3})$. – Ng Yong Hao Jun 22 '12 at 13:45
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    But I feel like this is not a good introduction to what actual research, at least for a beginning researcher, is like. Usually, you are taught fairly advanced methods and some result that was achieved using those methods and then are asked to modify it a little bit to see what you can do. – David Corwin Jun 24 '12 at 18:30
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    Related: http://math.stackexchange.com/questions/88709/complex-math-problem-that-is-easy-to-solve/88776#88776 – KConrad Oct 22 '13 at 01:38
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  • Tim Browning gave a nice introduction to this problem on Numberphile – Mark S Feb 25 '18 at 17:10
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    @MarkS Your link states it was Andrew Booker from Bristol who found the solution, and links the paper. –  Mar 11 '19 at 22:00
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    $33=8866128975287528^3+(−8778405442862239)^3+(−2736111468807040)^3$ - see https://gilkalai.wordpress.com/2019/03/09/8866128975287528%C2%B3-8778405442862239%C2%B3-2736111468807040%C2%B3/ – Mark S Mar 11 '19 at 22:09
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    @MarkS Very nice, so I guess $42$ is the only one left. Maybe it will be found be the same method eventually. – Ng Yong Hao Mar 12 '19 at 02:50
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    @NgYongHao queue the Numberphile follow-up video on $33$ - $42$ is on the same list. – Mark S Mar 13 '19 at 11:55
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    $42=(-80538738812075974)^3 + 80435758145817515^3 + 12602123297335631^3$ is now solved (by Andrew Booker and Andrew Sutherland). – Simon Pepin Lehalleur Oct 24 '19 at 17:30
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    Related: https://math.stackexchange.com/a/1565427 – Rosie F Apr 26 '20 at 07:39
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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.

Malik Younsi
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    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." – Timothy Chow Jun 22 '12 at 02:05
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    I do think that anyone can understand whats an injective, continuous map from the circle to the plane. – Fernando Muro Jun 22 '12 at 06:44
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    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." – Henry Cohn Jun 22 '12 at 13:14
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    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. – Noah Snyder Jun 24 '12 at 03:40
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    Well, not if you shake your hand fast enough (or with enough brownian motion) – Feldmann Denis Aug 24 '12 at 22:48
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    3Blue1Brown did a video on this problem and a weakening of it (the inscribed rectangle problem) which was recently solved, that I feel is rather understandable to a layman. – Zemyla Sep 14 '18 at 21:56
  • For reference, this was asked by Toeplitz in 1911. Here is a text by Étienne Ghys (in French) with various historical facts about it. – YCor Oct 24 '19 at 14:32
  • @NoahSnyder also if you draw with the pencil the thickness of the line is not zero. Actually I think that people beyond mathematicians are able to conceive by themselves that the mathematical definition is an extrapolation of the intuitive definition, rather than a literal interpretation. – YCor Oct 24 '19 at 14:36
  • @YCor: I think there's a huge difference between how easy it is to imagine the line being infinitely thin, and how easy it is to imagine continuous functions that are not piecewise smooth. I think the average calculus student thinks the word function means piecewise smooth function (possibly with isolated non-continuous points). – Noah Snyder Oct 24 '19 at 15:40
  • @NoahSnyder I know people who have an intuition of what is a "curve in the plane" and do not know what a derivative is. What is their intuition of what it is, and whether it involves smoothness, I wouldn't decide for them. – YCor Oct 24 '19 at 16:12
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    I'm very happy to bet if you ask 100 non-mathematicians to draw a curve in the plane that at least 99 of them will draw something piecewise smooth! – Noah Snyder Oct 24 '19 at 19:02
  • I would make the same bet for mathematicians to be honest, especially since you’ve already pitched it as “drawing” a curve.... I don’t think that addresses YCor’s point. – Ryan Jun 25 '20 at 14:56
69

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.

Timothy Chow
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    I think Artin's primitive root conjecture counts as pretty well known. – John Pardon Jun 22 '12 at 01:51
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    @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. – Timothy Chow Jun 22 '12 at 14:37
65

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)

Joël
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    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.) – Alexander Woo Jun 22 '12 at 04:05
  • Ah, that is bad. Isn't there an option for seniors in good high school to learn some calculus ? – Joël Jun 22 '12 at 12:21
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    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. – Francis Adams Jun 22 '12 at 12:51
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    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.) – Alexander Woo Sep 06 '12 at 04:11
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    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. – Victor Protsak Jan 06 '14 at 19:28
61

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 $n$.

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    Related to this, the Hadamard conjecture : there exist Hadamard matrices of order $4k$ for every $k$. http://en.wikipedia.org/wiki/Hadamard_matrix#The_Hadamard_conjecture – François Brunault Jun 22 '12 at 09:19
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    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 – Gerhard Paseman Jun 22 '12 at 19:08
  • It seems the circulant Hadamard matrix conjecture has been solved on 4 pages: https://arxiv.org/abs/2302.08346 – sdd Feb 06 '24 at 03:49
60

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?

David Feldman
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  • That's interesting. Do you have some further pointers? – Pasha Zusmanovich Jun 23 '12 at 15:04
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    Here is a link to Henriksen's paper related to this question.

    http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=4&ved=0CI4BEBYwAw&url=http%3A%2F%2Fwww.math.hmc.edu%2F~henriksen%2Fpublications%2F1989_Henriksen_Rings_with_a_unique_regular_element.pdf&ei=BxDmT5iUKoPA0AHUh6yICg&usg=AFQjCNF_N1lOOPd3hADCA0yxGEET-cNLvQ&sig2=sXA9eJqSmyGXGaHbk9K7zw

    – David Feldman Jun 23 '12 at 18:52
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    I've known ring theory for a while, and it never even occurred to me that that was difficult (let alone possibly true). – David Corwin Jul 22 '12 at 20:35
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    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 ...!!! – David Feldman Jul 23 '12 at 00:04
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    This looks like a fun way to spend some time in $\mathbb{Z}/(2)$-vector spaces... – DavidLHarden Feb 06 '17 at 05:05
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    That's a very interesting problem ... note that a finite non-commutative ring with unity has more than one units ( see https://mathoverflow.net/questions/284684/does-there-exist-a-finite-non-commutative-ring-with-unity-having-exactly-one-inv ) ; so that only leaves for infinite rings to be considered . –  Oct 29 '17 at 08:42
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    The original of the Henriksen paper is now gone. Here is the Wayback Machine link: https://web.archive.org/web/20100813070706/https://math.hmc.edu/~henriksen/publications/1989_Henriksen_Rings_with_a_unique_regular_element.pdf As an aside, it's shocking to me how little universities care about preserving the scholarship of their own researchers. – arsmath Jun 24 '20 at 10:20
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    The answer to Henriksen's question is yes not only for finite rings (as noted by user111524 in his comment above), but also for right (or left) artinian rings (see https://mathoverflow.net/a/395778/16537 for further details). – Salvo Tringali Jun 20 '21 at 08:52
53

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 $f$.

C.S.
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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.

Ian Agol
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    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] (http://dx.doi.org/10.1007/s00208-003-0420-2). – Wadim Zudilin Nov 11 '15 at 03:55
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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)

the L
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    and, in my answer at math.SE which you link here, I refer to the mathoverflow question http://mathoverflow.net/questions/24579. – George Lowther Jun 24 '12 at 00:35
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    This is really cool! Everyone should take a look at the actual math.SE answer, which explains why this question is hard. – Vectornaut Jul 22 '12 at 18:53
  • For comparison, see also this question: https://mathoverflow.net/questions/282259/is-the-series-sum-n-sin-nn-n-convergent – Timothy Chow Dec 10 '17 at 22:08
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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.

C.S.
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  • cured link to Sendov's conj. Great answer ! – Alexander Chervov Jun 22 '12 at 12:07
  • @alexander: thanks for fixing the broken link. – C.S. Jun 22 '12 at 17:20
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    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. – Emil Jeřábek Jun 25 '12 at 10:34
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    @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 01 '12 at 07:40
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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? – Noam D. Elkies Jun 22 '12 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. – Richard Stanley Jun 22 '12 at 20:31
  • Yes, I understand why it's difficult... And it seems that the third one that you added since then is another incarnation of the "numeri idonei" problem. – Noam D. Elkies Jun 23 '12 at 00:47
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    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? – David Feldman Jun 23 '12 at 19:17
  • For #3, you say that 462 is the largest integer with this property, yet that there exists at most one such integer $n<462$. Do you mean to say that 462 is the largest integer with this property other than one possibly $n>462$? – David Corwin Jul 22 '12 at 20:24
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    @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. – Owen Biesel Jul 27 '12 at 21:15
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    Question #2 was addressed in the paper http://www.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. – Marc Chamberland Aug 19 '12 at 18:40
45

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.

Timothy Chow
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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 $n=3$.

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    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... – Richard Montgomery Jan 28 '15 at 04:55
  • I do not understand what is the relation between the problems. Can you explain? In Maxwell problem $x_j$ are FIXED. – Alexandre Eremenko Jan 28 '15 at 14:48
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    I just saw this claim for a proof for the case $n=3$ with equal $a_j$: http://www.sciencedirect.com/science/article/pii/S0167278915001347 – Suvrit Oct 12 '15 at 01:53
  • @Suvrit: Interesting! Thanks for the reference. – Alexandre Eremenko Oct 12 '15 at 10:41
40

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.
39

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

39

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.

Jim Belk
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  • 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. – The Masked Avenger Jun 07 '15 at 19:19
35

Schinzel-Sierpinski Conjecture

Taken from this MathOverflow link.

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 $3$.
C.S.
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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.

mrtaurho
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paul Monsky
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31

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.)

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Ramanujan's conjecture [*] If $2^x$ and $3^x$ are both rational (hereafter assumed) integers for some non-zero $x$ then $x$ is an integer.

[*] I think that is the accepted name for this problem. He certainly proved the weaker corresponding result with $2^x$, $3^x$, and $5^x$ all assumed to be integers.

Unlike some of the other fascinating conjectures already listed here, this one seems "obviously" true. Yet I gather little progress has been made on it. It must be hard to find a foothold, so to speak, or know where to start.

Another easily understood example is the Erdős-Straus Conjecture which asserts that for every integer $n > 1$, there is at least one set of positive integers $x, y, z$ with $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{4}{n}$. The result is trivially true if negative integers are also allowed.

In this case, by contrast, it's easy(ish) to "almost" prove it, and with patience and ingenuity one can proceed (apparently) ever closer to a solution. But a few annoying special cases always seem to slip through the net!

One more example - I think a high school kid would have little difficulty understanding the abc conjecture, or following the simple proof of the corresponding result for polynomials Mason-Stothers theorem.

  • @John: Dear John, this is a well known conjecture. See the question once again. – C.S. Jun 23 '12 at 09:13
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    Also: one problem per answer – Yemon Choi Jun 23 '12 at 09:42
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    I was under the impression that the result attributed here to Ramanujan was in fact first proved as an application of the Six Exponentials Theorem (https://en.wikipedia.org/wiki/Six_exponentials_theorem) in the 1960s. Do you have a reference for Ramanujan's work? – Gerry Myerson Oct 23 '15 at 06:17
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    Am I right to think your Ramanujan's conjecture would follow from Schanuel's Conjecture? Writing $n=2^x $and $m=3^x$ would give $\ln(n)\ln(3)=\ln(m)\ln(2)$, so an algebraic relation between $\ln(2),\ln(3),\ln(m)$ and $\ln(n)$. But the exponentials of the logs are the rationals $2,3,m$ and $n$, so Schanuel would expect a linear relation over the integers between the logs, and then exponentiation and unique factorization would finish the job. – David Feldman Apr 25 '23 at 06:59
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This is basically copied from my answer on this question, which I've now updated some.

Let's let $\|n\|$ denote the smallest number of 1's needed to write n using an arbitrary combination of addition and multiplication. For instance, ||11||=8, because $11=(1+1)(1+1+1+1+1)+1$, and there's no shorter way. This is sequence A005245.

Then we can ask: For n>0, is $\|2^n\|=2n$?

Since it is known that for m>0, $\|3^m\|=3m$, we can ask more generally: For n, m not both zero, is $\|2^n 3^m\|=2n+3m$?

Attempting to throw in powers of 5 will not work; ||5||=5, but $\|5^6\|=29<30$. (Possibly it could hold that $\|a^n\|=n\|a\|$ for some yet higher choices of a, but I don't see any reason why those should be any easier.)

Jānis Iraids has checked by computer that this is true for $2^n 3^m\le 10^{12}$ (in particular, for $2^n$ with n≤39), and Joshua Zelinsky and I have shown that so long as $n\le 21$, it is true for all m. (Fixed powers of 2 and arbitrary powers of 3 are much easier than arbitrary powers of 2!) In fact, using an algorithmic version of the method in the linked preprint, I have computed that so long as $n\le 41$, it is true for all $m$, though I'm afraid it will be some time before I get to writing that up...

That seems to be the best known.

Harry Altman
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The Cerny conjecture says that if X is a collection of mappings on an n element set such that some iterated composition (repetitions allowed) of elements of X is a constant map then there is a composition of at most $(n-1)^2$ mappings from X which is a constant mapping. This comes from automata theory. See https://en.wikipedia.org/wiki/Synchronizing_word.

28

I think nobody pointed this problem, if it is repeated, please say me to delete it. This problem killed me for three weeks, when I was a young student in high school. So, I want to recall it again.

Problem: Find all right triangles with rational sides, where the area of these triangles are integer?

I think it is still open problem and if somebody can solve it, I will give 100$ as a small award.

After I searched, I found these two interesting sources. I hope it will be helpful.

1) N.Koblitz, Introduction to elliptic curves and modular forms, volume 97 of Graduate Texts in Mathematics. Springer-Verlag, New York, second edition, 1993.

2) Washington, Lawrence C., Elliptic Curves : Number Theory and Cryptography, CRC Press Series On Discrete Mathematics and Its Applications

Shahrooz
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    This is the congruent number problem and leads to the Birch-Swinnerton-Dyer conjecture... http://www.math.jussieu.fr/~colmez/congruents.pdf – François Brunault Jun 22 '12 at 23:13
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    That's presumably the intention, though the problem as stated looks simpler... (The "congrent number problem" amounts to asking which integers are the areas of right triangles all of whose sides are rational.) – Noam D. Elkies Jun 23 '12 at 03:09
  • I am not sure that the problems I posted under numbers 223484, 237438, and 216320 will be suitable for you, but the are definitely not very famous, some of them were stated about 10 years ago, they are formulated in a very elementary way, and I think that they are not trivial since some of very good mathematicians (not me, but really good) could not solve them quickly. And the answer will be interesting to many people. Will be glad if this information is useful. – Deepti Oct 31 '16 at 07:15
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Proving the Inequality of the Means by fitting boxes into a cube. From Berlekamp, Conway and Guy's Winning Ways for Your Mathematical Plays, Academic Press, New York 1983. See the discussion of this problem on Dror Bar-Natan's webpage for details, pictures, etc.

Question: Is it possible to pack $n^n$ rectangular n-dimensional boxes whose sides are $a_1, a_2,\ldots, a_n$ inside one big n-dimensional cube whose side is $a_1+a_2+\cdots+a_n$?

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The Kurepa conjecture : For every odd prime $p$, one has $$ 0!+1!+\cdots+(p-1)!\not\equiv0\pmod p $$ A proof was claimed and published in 2004 but the claim was withdrawn in 2011. See also my comment on the accepted answer to MO24265.

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Are there eight points on the plane, no three on a line, no four on a circle, with integer pairwise distances?

The analogous question for seven points was posed by Paul Erdős and answered positively by Kreisel, Kurz 2008, who have then asked this question.

In general, problems by Paul Erdős are worth to check if you want to find problems you are asking for here.

Tobias Kreisel, Sascha Kurz, There Are Integral Heptagons, no Three Points on a Line, no Four on a Circle, Discrete & Computational Geometry 39/4 (2008), 786-790. (Wayback Machine)

23

The Graph Reconstruction Conjecture:

Let $G, H$ be finite, simple, loopless graphs such that $|V(G)|$ and $|V(H)|$ are at least $4$. If there is a bijection $\varphi:V(G)\to V(H)$ such that for all $v\in V(G)$ the graphs $G\setminus \{v\}$ and $H\setminus \{\varphi(v)\}$ are isomorphic, then $G\cong H$.

20

An open problem I find surprising, the PAC (Perimeter to Area Conjecture) due to Keleti (1998):

The perimeter to area ratio of the union of finitely many unit squares in the plane does not exceed 4.

See for example Bounded - Yes, but 4? and references therein.

Seirios
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Here's another Birch Swinnerton-Dyer related problem. Sylvester conjectured that every prime that is 4,7 or 8 mod 9 is a sum of two rational cubes. Elkies (unpublished?) settled the first two cases. As far as I know, the third is still open.

paul Monsky
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    This conjecture of Sylvester is indeed not so widely known and the case $p=8 \mod{9}$ is still open. For some informations on Elkies's construction, see http://www.math.harvard.edu/~elkies/sel_p.html For published results, see Dasgupta-Voight's article http://people.ucsc.edu/~sdasgup2/clay.pdf – François Brunault Jun 23 '12 at 12:51
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    Yes, still unpublished alas. When I was working on it I looked up Sylvester's work on $x^3+y^3=a$ and didn't find any evidence that he actually conjectured this, though he did make some speculations about the case $p \equiv 1 \bmod 9$, which is the one case where $a$ is prime and the rank might be as high as $2$. For $p \equiv 4, 7, 8 \bmod 9$ the earliest statement of the conjecture that I found is Birch-Stephens (Topology 1966), prefigured by Selmer (Acta Math. 1951). It is a special case of the parity conjecture for the rank of elliptic curves. – Noam D. Elkies Jun 24 '12 at 15:09
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The complexity of matrix multiplication (i.e. the asymptotic number of steps required to multiply two n-by-n matrices).

This is an important problem in CS theory, but is non-famous enough in other fields that a mathematician (Andrew Stothers) made a significant advance in it in 2010 (beating a 20-year-old bound of Coppersmith and Winograd), and wrote up the result on page 71 of his PhD thesis without bothering to state it as a theorem or otherwise call attention to it. Word of it only got around a year or so later, when a computer scientist (Virginia Vassilevska Williams) independently made a further improvement.

The obvious multiplication algorithm takes $O(n^3)$ steps, and a well-known Karatsuba-like rearrangement gets the exponent $\omega$ down to about 2.8. There is a simple proof that the smallest possible $\omega$ is $\ge 2$. Coppersmith and Winograd got an exponent of 2.376 and the more recent results have it at 2.373. Apparently nobody has even shown that the minimum is not equal to 2: there are some who believe there's an algorithm faster than $O(n^{2+\epsilon})$ for any $\epsilon>0$ but not an $O(n^2)$ algorithm, but this is not known.

More info is in this blog post of Scott Aaronson: https://scottaaronson.blog/?p=839

none
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Is there an upper bound of quotients in the continued fraction representation of $\sqrt[3]{2}=[ 1; 3, 1, 5, 1, 1, \dots]$?

19

For integers $a_1, a_2, \ldots$, let $[a_1,a_2,\ldots]$ denote a continued fraction expansion $a_1 + \frac{1}{a_2 + \frac{1}{\ddots}}$.

Zaremba conjectured in 1972 that there must exist some constant $A > 1$ such that given any integer $d > 1$, you can always find some coprime $b$ such that if we write out the continued fraction expansion $b/d = [a_1, a_2, \ldots, a_k]$, all of the coefficients $a_i \leq A$.

Everyone seems to believe that $A = 5$, but we still don't know if such a constant exists. The current best known result (of Bourgain and Kontorovich) is that almost all integers $d$ permit a $b$ such that $b/d$ has the desired property (in their paper, they took $A = 50$, but I think that has been improved slightly since).

18

The Littlewood conjecture:

For any $\alpha, \beta \in \mathbb{R}$ we have $$\lim\textrm{inf}_{n\to\infty} (n\cdot||n\alpha||\cdot||n\beta||) = 0$$

where $||\cdot||$ denotes the distance to the nearest integer.

17

Can a disk be dissected into two or more congruent pieces, with its centre lying within one of the pieces?

maproom
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3D Version Of Blaschke-Lebesgue(1914) Theorem

The planar, compact.convex set of constant width, say 1, of minimal area is the Reuleaux triangle: Blaschk-Lebesgue(1914). The 3D set of constant width and minimal volume is unknown.

16

Bonnessen—Fenchel conjecture: Which convex body of constant width has the least volume? Is it Meissner's tetrahedron?

16

The following is a conjecture of Wlodzimierz Kuperberg:

Every convex planar set of area 1 is contained in a quadrilateral of area $1+\frac{4}{5}\tan\frac{\pi}{5}\sin\frac{\pi}{5}$.

In other words, such a set is contained in a quadrilateral of area less that $\sqrt{2}$, and the minimum is obtained for the minimum area quadrilateral containing a regular pentagon.

The conjecture involved only elementary plane geometry, and can be found in:

W. Kuperberg, On minimum area quadrilaterals and triangles circumscribed about convex plane regions, Elem. Math. 38 (1983), no. 3, 57–61, MR0703939 (85a:52009)

It is presented as a challenge to the MO community here:

Small quadrilaterals containing a given convex region


It is easy to prove that

(*) Every convex planar set of area 1 is contained in a quadrilateral of area 2.

It is also easy to see that statement (*) remains true if the constant 2 is replaced with a somewhat smaller one. Contest: Find such a constant, the smaller the better.

Update:

Reaching $\sqrt{2}$ and even a strictly smaller value was proved by Chakerian (1973) and Kuperberg (1983) and the research challenge offered is to improve it even further, and perhaps even to verify the conjecture that the minimum is attained by a regular pentagon. But any nice arguments for bounds below 2 are welcome.

15

Is it true that any word of length $n$ contains less than $n$ squares?

(A square is a factor of the form $uu$ for a non-empty word $u$.)

15

The Feit-Thompson conjecture is not too famous and it is still open: there are no prime numbers $p\neq q$ such that $$\frac{p^q-1}{p-1}\text{ divides }\frac{q^p-1}{q-1}.$$

14

Here is a nice question due to John Conway. In a magical 4x4 square, show that the XOR composition of the four numbers, written in base 2, in every row and in every column is zero. This applies to a square in which the numbers 0 to 15 are used (rather than 1 to 16).

For instance, a typical row might be 0 15 14 1, which in binary is 0000 1111 1110 0001, and in each of the four positions there happen to be two entries 0 and two entries 1, so the binary sum is zero.

Of course there are only finitely many possible magic 4x4 squares, and you can give proof by "complete inspection" (aka brute force). In fact, that has been done, so the result is true. But neither he nor I know a conceptual proof. Should be easy to understand about a classical problem -- and yet seems not obvious. Try it!

(Incidentally, the binary sum along the diagonals need not always be zero; that's not part of the question.)

Dierk
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  • Is it still true if the constraint on the sum of the diagonals is removed? What about replacing 4 by other powers of 2? – Harry Altman Jul 11 '12 at 21:57
  • Consider the following transform: f(x)= x div 2 + 8(x mod 2). If you have four positive numbers sum to 30, and exactly two of those numbers are odd, then the transformed sum is also 30. Now if one can find an algebraic proof that there is no magic square with four even numbers in a row or column, then f is an action on 4x4 magic squares, and there are exactly two ones in any slice off our such numbers. But Conway probably knows this already, and f may not be an action on magic squares. Gerhard "If Only Wishes Were Proofs" Paseman, 2012.07.12 – Gerhard Paseman Jul 12 '12 at 15:52
  • According to a website by H. B Meyer, the stronger statement in my comment above about slices is true for order 4 magic squares. Thus I suspect the problem is not open. If I find an algebraic proof that no row or column can have all even numbers, I will post a reference here. Gerhard "Ask Me About Right Shifts" Paseman, 2012.07.12 – Gerhard Paseman Jul 12 '12 at 19:59
  • Of course, I got the argument backwards. Consider the inverse of f. It will leave sums of 30 invariant, provided exactly two of the four numbers are 8 or greater. But no row (column) can have three or more numbers greater than 7, as then one row would have one or fewer numbers greater than 7. I claim this problem can now be solved, and is no longer open. Gerhard "Ask Me About System Design" Paseman, 2012.07.16 – Gerhard Paseman Jul 17 '12 at 01:23
  • After sleeping on it, I find the flaw in the above reasoning. I can use parity to restrict the number of low order bit ones that occur in a row, and I can use size considerations for the hhigh order bits, but neither case gives me exactly what is needed, and I don't see a way to combine the two (If bit string reversal preserved the sum, that would do it. I'm not there yet.) Gerhard "Tried It, Now Moving On" Paseman, 2012.07.17 – Gerhard Paseman Jul 17 '12 at 17:52
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    The 2017 book "The Mathematics of Various Entertaining Subjects", volume 2, at chapter 5 "Frenicle’s 880 Magic Squares", by Conway, Norton and Ryba, describes this problem as "Nimm0" property. It also states in note 1 that Friedrich Fitting and Oliviero Giordano Cassani solved it independently. However I wasn't able to find any paper online regarding the proof. – Fabius Wiesner Aug 31 '19 at 08:06
  • The paper by Fitting is "Rein mathematische Behandlung des Problems der magischen Quadrate von 16 und von 64 Feldern," Jahresbericht der Deutschen Mathematiker-Vereingung 40 (1931), 177-199. – Timothy Chow Jan 07 '24 at 20:06
14

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, DOI: 10.1515/crll.1984.348.135, eudml). The best result on Ruzsa's problem is due to U. Zannier ("On periodic mod $p$ sequences and G-functions," Manuscripta math. 90, 1996, eudml, DOI: 10.1007/BF02568314).

14

Are there infinitely many $n\ge1$ such that $$ \gcd(2^n-1,3^n-1)=1 ? $$ Ailon and Rudnick conjectured that the answer is affirmative around 2000. What I like about this problem is that it could appear in Euclid, yet wasn't asked until fairly recently. There are obvious generalizations, and the analogue with $\mathbb Z$ replaced by $\mathbb C[T]$, is proven in the Ailon-Rudnick paper. There are also some (deep) results of Bugeaud, Corvaja, and Zannier giving the following related results:

  • There is a constant $C>0$ and infinitely many $n\ge1$ such that $$ \gcd(2^n-1,3^n-1) \ge \exp(C n/\log\log n) . $$
  • For every $\epsilon>0$ there is a $C_\epsilon$ such that $$ \gcd(2^n-1,3^n-1) \le C_\epsilon\exp(\epsilon n) \quad\text{for all $n\ge1$.} $$
Joe Silverman
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This problem is open (to my knowledge), surely accessible to anybody, and not too famous. As for the "Long open" requirement, there is at least a sizable group of people interested in it, as evidenced by this MO post: Factoring 0-1 polynomials

The problem is:

Is it true that if a polynomial $p(x)$ whose coefficients are in $\{0,1\}$ factors as $p(x)=p_1(x)p_2(x)$, where $p_1$ and $p_2$ are monic polynomials with non-negative real coefficients, then in fact also $p_1$ and $p_2$ have coefficients in $\{0,1\}$?

Valerio
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In an oriented graph, is there always a vertex from which there are at least as many vertices that one can access by moving along exactly two edges, than there are vertices that one can access by moving along one edge?

This is known as Seymour's second neighborhood conjecture, and might be on the verge to being too famous (but it seems few of my colleagues know it).

13

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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The Happy Ending Problem

  • Says that any set of five points in the plane in general position has a subset of four points that form the vertices of a convex quadrilateral. More generally, Erdös and Szekeres proved that for any positive integer $N$, there is a minimal integer $f(N)$ such that any set of $f(N)$ points in the plane in general position has a subset of $N$ points that form the vertices of a convex polygon, and it is known that $f(N)$ is at least $1+2^{N-2}$.

An open question is: does $f(N)=1+2^{N-2}$ hold?. Taken from this MO link.

C.S.
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    This recent preprint http://arxiv.org/abs/1604.08657 claims to "nearly settle" this conjecture.... – Suvrit May 02 '16 at 01:13
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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.

Zach H
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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?

bof
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Here is another problem on equilibrium points of potentials: suppose that we have infinitely many point masses in $R^3$ (the points do not accumulate). Must there exist a point where the gravitational force created by these masses is zero?

If the masses $m_k>0$ are placed at $x_k\to\infty$ then the force is $$\sum_{k=1}^\infty m_k\frac{x-x_k}{|x-x_k|^3},\quad \mbox{where}\quad\sum_{k=1}^\infty m_k|x_k|^{-2}<\infty.$$ Does every such function have a zero?

This is a version of the problem proposed by Lee Rubel in 1980-th. For some partial results see https://www.math.purdue.edu/~eremenko/dvi/equil.pdf This question can be easily modified for any dimension $n\geq2$, using Newtonian ($n\geq 3$) or logarithmic potential ($n=2$). The question is substantially easier in dimension $2$, but even for $n=2$ it is not solved in full generality.

  • Can we simplify the problem giving all points an equal mass 1? Would that make it any easier to understand for students etc. in your opinion? P.s. 20th Necromancer badge! You'd deserve a gold badge. :) – Nemo May 02 '15 at 07:10
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    @Nemo: If you look at the file which I cited you find some partial results. In particular, in $R^3$, if $m_k\geq c>0$, and $\sum_k m_k/|x_k|<\infty$, there are infinitely many zeros. But convergence condition here is more restrictive than conjectured. – Alexandre Eremenko May 02 '15 at 22:33
12

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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A meta-answer: I recommend Guy's Unsolved Problems in Number Theory and perhaps some of his others (Unsolved Problems in Geometry, Unsolved Problems in Combinatorial Games), which have many unsolved problems (both well-known and obscure), grouped into categories. Many of these are of attackable difficulty.

Charles
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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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    What exactly is the problem here? – Harry Altman Jul 24 '12 at 22:03
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    The problem is to find a closed form for $T(n)$, I think. – I. J. Kennedy Oct 03 '13 at 16:14
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    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. – Jakub Konieczny Jun 04 '15 at 19:41
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    “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). – Peter LeFanu Lumsdaine Jan 24 '16 at 00:23
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Enumeration of meanders. (See also meander).

Problem is to find some formula for the number of meanders or at least some good asymptotic.

As far as I understand the attention to it has been attracted by V.I. Arnold. The problem is so "everyone can understand" that there is an article by him in the math. journal for shool-children "Quant" (sorry it is in Russian. I remember it from my school years): djvu file from the site.

There are plenty papers in arXiv on the problem.

E.g. https://arxiv.org/abs/cond-mat/0003008 Exact Meander Asymptotics: a Numerical Check Philippe Di Francesco, Emmanuel Guitter (SPHT-Saclay), Jesper Lykke Jacobsen (LPTMS-Orsay)

As far as I understand from the nice book (or) by S. Lando and A. Zvonkin the problem is still open.

11

Is there a positive integer which is both triangular and factorial except these obvious examples: $1, 6, 120$? (Tomaszewski conjecture, http://oeis.org/A000217)

11

Is there a rectangle that can be cut into $3$ congruent connected non-rectangular parts?

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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 $0$?

11

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.

11

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.

10

Some pages:
Open Problem Garden
The Open Problems Project edited by Erik D. Demaine, Joseph S. B. Mitchell, Joseph O’Rourke

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

Arend Bayer
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    Cool ! what are the references for current state of art ? – Alexander Chervov Jul 24 '12 at 09:57
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    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; http://www.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: http://front.math.ucdavis.edu/0808.3351.

    – Arend Bayer Jul 24 '12 at 22:03
  • K-12??? $\textrm{}$ – Victor Protsak Jan 06 '14 at 20:12
10

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 $n\geq 5$.

Federico
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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.)

9

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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    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 05:36
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    @David just wanted to write the same :) http://mathoverflow.net/questions/101169/not-especially-famous-long-open-problems-which-higher-mathematics-beginners-can – Alexander Chervov Jul 13 '12 at 06:18
  • 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
  • @Daniel: oh, I agree it's a fascinating question (I bumped into this when hearing about this thing called Baum-Connes that implied this thing called Kadison-Kaplanaksy) and indeed it can stated quite accessibly to people with relatively little background. I was just being pessimistic about getting people at lower levels - which I appreciate was not exactly David's original question - to realise why the question is not an isolated or arcane curiosity – Yemon Choi Jul 15 '12 at 00:09
  • (And I can usually spell Kaplansky properly...) – Yemon Choi Jul 15 '12 at 00:10
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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

duje
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    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. – paul Monsky Nov 04 '16 at 05:26
9

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.

utdiscant
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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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    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 $n \gg 0$. – Vesselin Dimitrov Dec 17 '14 at 17:35
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    The post is misstated: the inequality is known to be true eventually, thanks to K. Mahler. See the related http://dx.doi.org/10.5802/jtnb.588 for a historical account and recent novelties. – Wadim Zudilin Nov 11 '15 at 04:01
9

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 $1/2$.)

Colin Reid
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  • Is this the same problem as in http://mathoverflow.net/q/3307/5340 "Can a discrete set of the plane of uniform density intersect all large triangles?" – Zsbán Ambrus Nov 14 '16 at 13:27
9

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.

9

Farideh Firoozbakht conjecture states that the sequence $P_n^{1/n}$ is strictly decreasing, where $P_n$ is the $n-$th prime number. This conjecture can also give simple proofs to many theorems related to Prime numbers.

It is not proved yet.

8

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.)

David Feldman
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    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
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    In the classical proof, the exponential growth is in the product of the known primes. See http://mathoverflow.net/questions/59262/induction-the-infinitude-of-the-primes-and-workaday-number-theory 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
  • $n$ goes to $1+\prod_{i=1}^n p_i$ certainly grows super-exponentially.

    If you're formalizing your proof in a theory of the integers, employing a logarithm means employing the inverse of exponential function, so you can't make the Euler product + divergence of harmonic series proof fly either. In his thesis (I have a copy), Alan Woods analyzes many well-known proofs. He pays particular attention to a proof of Sylvester's, but can't make it work. See here for more http://mathoverflow.net/questions/59262/induction-the-infinitude-of-the-primes-and-workaday-number-theory. Sorry, I'm no expert.

    – David Feldman Jun 24 '12 at 18:26
  • Corrected. Thank you Douglas Zare. – David Feldman Jun 24 '12 at 18:28
  • Is it open whether you can prove that the harmonic series diverges without using functions of superpolynomial growth? – Douglas Zare Jun 24 '12 at 18:38
  • From the summation function of the harmonic series you can define a function that grows like logarithm, and then define a function that grows like $e^x$. The theory of bounded arithmetic will only allow you to define functions of polynomial growth. – David Feldman Jun 24 '12 at 19:46
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    See also http://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. – Emil Jeřábek Jun 25 '12 at 11:35
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    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
  • ... the Euler product itself: just as with sums, you can’t define products like $\prod_{p\le x}(1-p^{-1})^{-1}$ by a bounded formula. – Emil Jeřábek Jun 25 '12 at 11:55
8

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

8

It is not possible to find three corners of a square that form an equilateral triangle. (Left as an exercise for the reader.)

It is, however, possible to find four corners of a cube that form a regular tetrahedron:

                                          enter image description here

We could ask: in which dimensions $n>1$ is it possible to do such a thing - find a subset of the corners of the $n$-cube which form an $n$-simplex?

The answer: it is conjectured to be the case if and only if $n\cong 3\pmod{4}$, but remains open! The congruence condition is easily seen to be necessary, but existence of solutions for such $n$ has only been shown up through $n=663$ as of this writing.

(When you express the vertices of the $n$-cube as the set of length-$n$ $(0,1)$-tuples and do a little algebraic reshuffling, it becomes clear that this is equivalent to the Hadamard conjecture, a problem which was mentioned in an earlier comment but which I think becomes much more accessible under this framing.)

7

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 constructible 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.

7

https://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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    Apparently the final case of Keller's conjecture was resolved recently (in the affirmative) by a computer assisted proof: https://arxiv.org/abs/1910.03740 – Terry Tao Oct 12 '19 at 01:22
7

What is the least $S$ (if any) such that any subset of a plane of area $S$ contains $3$ vertices of a triangle of unit area?

7
  • 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?
7

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?

bof
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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 (https://arxiv.org/abs/math/0312202).

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    Already on this list: see http://mathoverflow.net/a/114639/763 – 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
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    @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
  • @StefanKohl According to the article by Ivić and Mijajlović, the conjecture is verified for $n<10^6$. – Alex Ravsky Oct 13 '22 at 00:08
7

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).

Jim Conant
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7

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?

7

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.

7

I really like this problem of Busemann and Petty (Problems on convex bodies, Math. Scand., 1956):

Given a $0$-symmetric convex body $K$ and one of its support hyperplanes, construct the solid cone whose apex is a point at which the hyperplane touches the body and whose base is the central section of $K$ that is parallel to the hyperplane. Clearly the volume of this cone will be independent of the choice of contact point (in the case the support hyperplane intersects the body in more than one point). Now assume that no matter what support hyperplane you choose, the volume of the cone is always the same. Is $K$ an ellipsoid?

As far as I know there are not even meaningful partial solutions to this problem (e.g., assume $K$ is three-dimensional and the boundary of $K$ is real analytic and strictly convex).

alvarezpaiva
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6

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 $V = 9/2$.

6

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 Hamiltonian 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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    Surely you don’t meen Eulerian? For example, $G_4$ isn’t Eulerian as it has many odd degree vertices. – OHO Mar 03 '20 at 07:36
  • I meant "Hamiltonian". Thanks for pointing that out. – Santi Spadaro Mar 23 '20 at 21:53
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    I can't find it in Guy's book. A recent paper is Chen, Fu, and Guo, From a consequence of Bertrand's postulate to Hamiltonian cycles, https://arxiv.org/abs/1804.07104 The authors prove the conjecture holds for infinitely many $n$. They seem to be unaware of previous discussions of the conjecture. – Gerry Myerson Mar 23 '20 at 23:47
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    Also, the question has been discussed here on MO: https://mathoverflow.net/questions/241569/for-which-n-is-there-a-permutation-such-that-the-sum-of-any-two-adjacent-eleme where Zhi-Wei Sun gives the reference Antonio Filz [J. Recreational Math. 14(1982), p.64] and Max Alekseyev gives the reference http://oeis.org/A051252 – oh, and it is in Guy's book, in the section (C1) on Goldbach's conjecture, with the attribution to Filz. – Gerry Myerson Mar 23 '20 at 23:59
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    Another reference is Stan Wagon's Problem of the week 1218, http://mathforum.org/wagon/current_solutions/s1218.html – Gerry Myerson Mar 24 '20 at 00:11
  • Thanks for the additional references, Gerry! – Santi Spadaro Mar 27 '23 at 15:11
6

Is there any odd perfect number?

The User
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6

Assuming that the definitions of a graph, its diameter and girth are something anyone can understand*, whether a graph with diameter $2$, girth $5$ and degree $57$ exists or not is a long standing famous open problem. See this or this.

There are several such (not especially famous) open problems regarding existence/uniqueness in the theory of bipartite Moore graphs (known as generalized polygons), i.e., graphs with diameter $d$ and girth $2d$, the prime power conjecture for finite projective planes that OP has mentioned being one of them. For example, is there a unique $4$-regular bipartite graph with diameter $6$, girth $12$, the so called generalized hexagon of order (3,3)?

*I have never had a problem with explaining this problem to people who have no math background beyond high school.

Anurag
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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).

6

I like Moser's worm problem: https://en.wikipedia.org/wiki/Moser%27s_worm_problem

I'm not sure how famous it is, but I believe less than it deserves. In short: what is the area of the smallest planar region such that every curve of unit length can be placed into it?

  • Is it even clear that there's a lower bound in the non-convex case? I feel like if I had to bet on a number I'd probably pick 0 by analogy with Kakeya... – Steven Stadnicki Aug 18 '21 at 19:32
  • I faintly remember that the problem appeared in the German version of Scientific American where it was called "mother worms" problem – Manfred Weis Nov 09 '22 at 16:13
5

Are there infinitely many partition numbers divisible by $3$? See A000041.

  • Could you perhaps expand a bit, i.e. for example what makes this a difficult problem, and what is known about the same question for divisors other than $3$? – Stefan Kohl Nov 09 '15 at 10:12
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    For example, it is known that there are infinitely many divisible by 2. It is hard for me to estimate how difficult the problem is, but it was publicly raised at OEIS more than 10 years ago (and I'm sure it had been discussed before), it is still open, and it is about a famous sequence that received much attention (and new interesting results) during last decade. The sequence graph suggests that the distribution of such numbers is quite irregular, they do not seem to to lie nicely on a smooth curve. – Vladimir Reshetnikov Nov 09 '15 at 17:23
5

(From Alon and Tarsi)

Let $A$ be a nonsingular $n$ by $n$ matrix over the finite field $\bf{F}_q$, $q>4$, then there exists a vector $x$ in $\bf{F}_q^n$ such that both $x$ and $Ax$ have no zero component.

5

If $2^x$ and $3^x$ are integers for some positive real number $x$, does this imply that $x\in\mathbb{N}$?

4

Does every nonseparating planar continuum have the fixed point property?

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    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 03:02
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    For those rusty on general topology, if K is a closed and bounded subset of the (x,y) plane, if K has connected and simply connected complement U, and if f:K-->K is continuous, must K have a fixed point?

    To aim at K-12, PL arcs in U and `uniform continuity' are adequate to reformulate the respective properties of U and f in an elementary manner.

    – Paul Fabel Jun 22 '12 at 13:05
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    Uniform continuity is a K-12 concept now? – Douglas Zare Jun 22 '12 at 14:25
  • Thanks, Paul. The question makes a lot more sense to me now. – Yemon Choi Jun 22 '12 at 16:32
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    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
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    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
  • @PaulFabel: How can a compact subset of the plane have simply connected complement? – Qfwfq Sep 15 '17 at 00:21
4

Can a square with integer sides contain a point whose distance from each corner is an integer?

3

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? ;)

Vidit Nanda
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    Yes, I would say that a lot of people know who he is :-) For example he is active on MO: http://mathoverflow.net/users/1532/gil-kalai – Daniel Soltész Dec 07 '14 at 17:36
3

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 ( http://logic.harvard.edu/EFI_CH.pdf, Wayback Machine ) describes some current approaches.

none
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    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
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    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
  • this is not an open problem. It has been settled (like many impossibility results) a long time ago. – andrey bovykin Sep 10 '19 at 04:08
  • I have added a Wayback Machine link. It seems that the same file can be found in some other places - of course, it's hard to say whether some of those links will remain stable. – Martin Sleziak Nov 09 '22 at 10:55
2

N. M. Katz: "Simple Things we don't know": https://web.math.princeton.edu/~nmk/pisa16.pdf

Jon Bannon
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Thomas Riepe
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  • These will be a better fit here:

    http://mathoverflow.net/questions/101169/not-especially-famous-long-open-problems-which-higher-mathematics-beginners-can

    when the collective net gods deem it time to reopen the question.

    – David Feldman Jul 14 '12 at 19:07
  • These will be a better fit here: mathoverflow.net/questions/101169/ when the collective net gods deem it time to reopen the question. Every vote helps, thanks. – David Feldman Jul 14 '12 at 19:08
2

Does the decimal expansion of $2^n$ contain the digit $7$ for all sufficiently large $n$?

hbjj
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1

$P(x)=x^2+1$. $\mathbb P$ the set of prime number.

  1. Is it true that $P(\mathbb N)\cap \mathbb P$ is an infinite set ?

  2. Can you find a $Q \in \mathbb Z[x]$, with $\deg(Q)\geq 2$ and $Q(\mathbb N) \cap \mathbb P$ is an infinite set ?

Dattier
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    This seems like a famous open problem since it is the 4th of Landau´s four problemshttps://en.wikipedia.org/wiki/Landau%27s_problems . – JoshuaZ Mar 05 '23 at 13:44
1

"Which regular polygons are constructible with a marked straightedge/ruler and compasses?"

This was posed by Baragar in 2002 after demonstrating that some (but not necessarily all) quintic equations can be solved with the above tools [1]. It follows that constructions for regular $n$-gons exist not only if the Euler totient of $n$ has a maximum prime factor of $2$ or $3$ such as $n=5$ or $n=13$ (with $2$ as the maximum prime factor, construction is also possible without marks on the straightedge), but maybe if the Euler totient has a maximum factor of $5$ like $n=11$ or $n=25$. The problem has since been solved (the answer is "yes") for $n=11$ (and thus also for $11$ times numbers with Euler-totient prime factors $\le3$), but not for $25,31$, or any other "quintic" cases.

Reference

  1. Baragar, A. (2002). "Constructions Using a Compass and Twice-Notched Straightedge". The American Mathematical Monthly, 109(2), 151–164. https://doi.org/10.2307/2695327
Oscar Lanzi
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Here are 2 problems related to cyclotomic polynomials that are unsolved :

Let $n,m>1$ and $$A(x) = \Phi_n(x) + \Phi_m(x)$$

  1. If $n,m$ are distinct odd primes then $A(x)$ is irreducible.

  2. If $A(x)$ factors then one of its factors is a cyclotomic polynomial.

Example :

$$\Phi_7(x) + \Phi_{22}(x) = \Phi_4(x) b(x)$$

$$\Phi_4(x) = x^2 +1 $$ $$ b(x) = x^8 - x^7 + 2 x^4 + 2$$

mick
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  • This is interesting, but do you have a reference to somewhere these conjectures are discussed? – Sam Hopkins Jan 02 '24 at 23:09
  • @SamHopkins Apparently it has been checked up to $m,n < 151$ in the year $2000$ by some "Nicol ". I am unaware of theoretical ideas or numerical tests after the year $2000$. I have not seen good arguments for or against. – mick Jan 02 '24 at 23:20
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    When it comes to cyclotomic polynomials, that's not a very big range. See https://mathoverflow.net/a/15506/25028. – Sam Hopkins Jan 02 '24 at 23:21
  • @SamHopkins I know, but I am unaware of anything beyond it. That does not mean it does not exist. Nobody has every found a counterexample. And $2000$ is $24$ years ago. Larger numbers have been checked, but I have no source. – mick Jan 02 '24 at 23:24
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    Without more references I fear this one fails the requirement that "[t]he problem should occur in the literature or have a solid history as folklore." – Sam Hopkins Jan 02 '24 at 23:43
  • @SamHopkins I understand but that somewhat contradicts " not especially famous " imo. Many people tried to find counterexamples and cylclotomic polynomials are well studied objects which anyone can understand. Irreducible polynomials is also a big topic. – mick Jan 04 '24 at 23:38
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Ore's odd Harmonic number conjecture.

Timothy Chow
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Gorka
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Denote by $\sigma(m)$ be the sum of divisors of $m$. When is $ \sigma(n!-1) $ a perfect square?

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    Is this a long-open problem? Does it have a history? – Gerry Myerson Sep 15 '17 at 23:02
  • Numbers k such that σ(k) is a square are tabulated at oeis.org/A006532 – it is not even known that there are infinitely many of them (although this would follow from standard conjectures) – zeraoulia rafik Oct 17 '17 at 19:17
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    Isn't https://www.jstor.org/stable/10.4169/amer.math.monthly.119.05.373?seq=1#page_scan_tab_contents proving that there are infinitely many? – Dima Pasechnik Jan 12 '19 at 14:23