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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 (American) K-12 mathematics. For example, if it weren't already too famous, I would say that the conjecture that "finite projective planes have prime power order" does have barely acceptable articulations.

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

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

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

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

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

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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: mathoverflow.net/questions/96202/… 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 3:25
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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

79 Answers 79

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.)

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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 5:38
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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 mathworld.wolfram.com/PerfectCuboid.html –  Yemon Choi Jun 22 '12 at 5:49
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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 6:03

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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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 $2.219531669\lt A\lt 2.8284$ (approximately).

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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 $a\in k$.

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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 : 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 mathoverflow.net/questions/94838/… with the Casas-Alvero conjecture in mind. It appeared from the answers that instead of the argument using scheme theory, the simpler Lefschetz principle ( proofwiki.org/wiki/Lefschetz_Principle_(First-Order) ) can be used. (answering to my question, Qiaochu Yuan also indicated an ultraproduct construction which is even simpler than the Lefschetz Principle, since no completeness result is used). –  js21 Jun 25 '12 at 6:22

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

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wow, at first look it seems hard to believe that this is still a conjecture! –  Suvrit Jun 22 '12 at 3:26
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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 (members.bex.net/jtcullen515), another mathematics amateur; I cannot easily fine it online though. –  Wadim Zudilin Aug 25 '12 at 11:13

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.

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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 2 '12 at 2:14

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 $$\sum_{i=1}^{p-1} i^{p-1} \equiv -1 \pmod{p}$$

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for all $i$? For some $i$? Quantifier needed. –  temp Jun 25 '12 at 6:26
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@temp: $i$ is the summation variable. –  Emil Jeřábek Jun 25 '12 at 10:25
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Thank you, Emil. I was under the impression that $i^2=-1$, I couldn't help it. –  Włodzimierz Holsztyński May 4 '13 at 3:51

This is the second time I've seen this question on mathoverflow and this will be the second time I'vve 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.

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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 2 '12 at 9:49
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@S.Carnahan : How did you do that "short computation"? –  Michael Hardy Jul 6 '12 at 21:49
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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

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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I have never heard this before... –  Filippo Alberto Edoardo Jul 3 '12 at 14:53
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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

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".

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An excellent reference is the recent book Distribution modulo one and Diophantine approximation, by Yann Bugeaud. –  Andres Caicedo Jan 6 at 19:35

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

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More info can be found at adnu.edu.ph/bmc2012/Noche.pdf –  Joel Reyes Noche Jun 22 '12 at 5:24
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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 3 '12 at 4:08

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 $k$. en.wikipedia.org/wiki/Hadamard_matrix#The_Hadamard_conjecture –  François Brunault Jun 22 '12 at 9:19
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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

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.

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Is there something special about 33?! –  Vectornaut Jun 22 '12 at 5:42
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For small numbers (<100), 33, 42 and 74 are still unresolved. See this: asahi-net.or.jp/~kc2h-msm/mathland/math04/matb0100.htm . @Vectornaut when I saw your comment the first thing I thought off was the irrational solution $(\sqrt[3]{33/3},\sqrt[3]{33/3},\sqrt[3]{33/3})$. –  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: math.stackexchange.com/questions/88709/… –  KConrad Oct 22 '13 at 1:38

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.

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I think Artin's primitive root conjecture counts as pretty well known. –  John Pardon Jun 22 '12 at 1:51
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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

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.

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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 4:05
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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 6 '12 at 4:11
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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 6 at 19:28

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 $f$.

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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).
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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. –  Chandrasekhar Jul 1 '12 at 7:40

Here are a few others:

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

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

  3. The largest integer that cannot be written in the form $xy+xz+yz$, where $x,y,z$ are positive integers, is 462. It is known that there exists at most one such integer $n>462$, which must be greater than $2\cdot 10^{11}$. See J. Borwein and K.-K. S. Choi, On the representations of $xy+yz+xz$, Experiment. Math. 9 (2000), 153-158; http://projecteuclid.org/Dienst/UI/1.0/Summarize/euclid.em/1046889597.

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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
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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 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

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.

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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.

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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.

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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 2:05
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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 6:44
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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 3:40
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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

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

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

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and, in my answer at math.SE which you link here, I refer to the mathoverflow question mathoverflow.net/questions/24579. –  George Lowther Jun 24 '12 at 0:35

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?

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Here is a link to Henriksen's paper related to this question. google.com/… –  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 0:04

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

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This is the congruent number problem and leads to the Birch-Swinnerton-Dyer conjecture... 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 3:09

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.
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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 $3$.
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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 Erdos-Straus Conjecture [ http://planetmath.org/ErdHosStrausConjecture.html ], 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 [ http://en.wikipedia.org/wiki/Abc_conjecture ], or following the simple proof of the corresponding result for polynomials [ http://en.wikipedia.org/wiki/Mason%E2%80%93Stothers_theorem ]

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6  
Also: one problem per answer –  Yemon Choi Jun 23 '12 at 9:42

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.

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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.

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