# Fundamental problems whose solution seems completely out of reach [closed]

In many areas of mathematics there are fundamental problems that are embarrasingly natural or simple to state, but whose solution seem so out of reach that they are barely mentioned in the literature even though most practitioners know about them. I'm specifically looking for open problems of the sort that when one first hears of them, the first reaction is to say: that's not known ??!! As examples, I'll mention three problems in geometry that I think fall in this category and I hope that people will pitch in either more problems of this type, or direct me to the literature where these problems are studied.

The first two problems are "holy grails" of systolic geometry---the study of inequalities involving the volume of a Riemannian manifold and the length of its shortest periodic geodesic---, the third problem is one of the Busemann-Petty problems and, to my mind, one of the prettiest open problems in affine convex geometry.

Systolic geometry of simply-connected manifolds. Does there exist a constant $C > 0$ so that for every Riemannian metric $g$ on the three-sphere, the volume of $(S^3,g)$ is bounded below by the cube of the length of its shortest periodic geodesic times the constant $C$?

• For the two-sphere this is a theorem of Croke.
• Another basic test for studying this problem is $S^1 \times S^2$. In this case the fundamental group is non-trivial, but in some sense it is small (i.e., the manifold is not essential in the sense of Gromov).
• There is a very timid hint to this problem in Gromov's Filling Riemannian manifods.

Sharp systolic inequality for real projective space. If a Riemannian metric in projective three-space has the same volume as the canonical metric, but is not isometric to it, does it carry a (non-contractible) periodic geodesic of length smaller than $\pi$?

• For the real projective plane this is Pu's theorem.
• In his Panoramic view of Riemannian geometry, Berger hesitates in conjecturing that this is the case (he says it is not clear that this is the right way to bet).
• In a recent preprint with Florent Balacheff, I studied a parametric version of this problem. The results suggest that the formulation above is the right way to bet.

Isoperimetry of metric balls. For what three-dimensional normed spaces are metric balls solutions of the isoperimetric inequality?

• In two dimensions this problem was studied by Radon. There are plenty of norms on the plane for which metric discs are solutions of the isoperimetric problem. For example, the normed plane for which the disc is a regular hexagon.

• This is one of the Busemann-Petty problems.

• The volume and area are defined using the Hausdorff $2$ and $3$-dimensional measure.
• I have not seen any partial solution, even of the most modest kind, to this problem.
• Busemann and Petty gave a beautiful elementary interpretation of this problem:

Take a convex body symmetric about the origin and a plane supporting it at some point $x$. Translate the plane to the origin, intersect it with the body, and consider the solid cone formed by this central section and the point $x$. The conjecture is that if the volume of all cones formed in this way is always the same, then the body is an ellipsoid.

Additional problem: I had forgotten another beautiful problem from the paper of Busemann and Petty: Problems on convex bodies, Mathematica Scandinavica 4: 88–94.

Minimality of flats in normed spaces. Given a closed $k$-dimensional polyhedron in an $n$-dimensional normed space with $n > k$, is it true that the area (taken as $k$-dimensional Hausdorff measure) of any facet does not exceed the sum of the areas of the remaining facets?

• When $n = k + 1$ this is a celebrated theorem of Busemann, which convex geometers are more likely to recognize in the following form: the intersection body of a centrally symmetric convex body is convex. A nice proof and a deep extension of this theorem was given by G. Berck in Convexity of Lp-intersection bodies, Adv. Math. 222 (2009), 920-936.
• When $k = 2$ this has "just" been proved by D. Burago and S. Ivanov: http://front.math.ucdavis.edu/1204.1543
• It is not true that totally geodesic submanifolds of a Finsler space (or a length metric space) are minimal for the Hausdorff measure. Berck and I gave a counter-example in What is wrong with the Hausdorff measure in Finsler spaces, Advances in Mathematics, vol. 204, no. 2, pp. 647-663, 2006.
• are you asking for problems in any area or just geometry? – Michael Bächtold Jun 30 '12 at 19:44
• As this question has no correct answer, it should be community wiki. – HJRW Jun 30 '12 at 20:00
• The big problems always seem out of reach, until they are solved. – Angelo Jun 30 '12 at 22:01
• @François: well, we really don't know whether their ARE completely out of reach ... Mathematicians do not always look for solutions to problems in the place where the problems are, but in the place where other solutions are. A bit like searching for your lost car keys next to the lamp-post instead of where you lost them, because there is more light around the lamp-post ... I don't know how to put this in a title! – alvarezpaiva Jul 1 '12 at 8:45
• I voted to reopen. Unlike most recent soft questions, this one attracted good answers from which I learned something. – Andy Putman Jul 3 '12 at 22:18

A proof of this conjecture of Erdos would certainly turn heads, raise eyebrows, and garner the attention of the Fields Medal committee.

• If $\sum_{a \in A} \frac 1a$ diverges and $A\subseteq {\mathbb N}_{>0}$, then $A$ contains a 3-term arithmetic progression.

Probably "diverges" can be replaced with "is bigger than 4".

• wow, in my ignorance, I find it hard to believe that this question is within the "completely out of reach" category! really awesome question though! – Suvrit Nov 7 '13 at 4:31
• @Kevin: What doing mean by saying $A$ contains a 3-term arithmetic progession? – crskhr May 8 '18 at 13:38

Is every algebraic curve in $\mathbb P^3$ the set-theoretic intersection of two algebraic surfaces ? Not known!

• This is a nice example! – François Brunault Jul 1 '12 at 8:14
• Great! Exactly what I was looking for. – alvarezpaiva Jul 1 '12 at 8:45
• Can you give a reference where this problem is (partially) analyzed? – Martin Brandenburg Jul 2 '12 at 10:20
• @Martin: Here is a thesis devoted to the theme of complete intersections and which is a nice synthesis of the subject. The very last sentence of that document states the open problem I mention! – Georges Elencwajg Jul 2 '12 at 12:23
• Is this known in the complex case? – Steven Gubkin Jul 2 '12 at 18:33

This comes up in Waring's Problem, but it is so freakishly simple that it has taken on a life of its own. Let $\{ x \} = x \mod 1 = x-\lfloor x \rfloor$ be the fractional part of $x$.

• Say anything about the sequence $\{ (3/2)^n \}.$

Computations support the thought that the sequence should uniformly distributed in $[0,1]$, as for almost all $x$ the sequence $\{x^n\}$ is u.d. But with $x=3/2$, there is no value known to be a limit point, nor any value known to not be a limit point, it's unknown if there are two limit points, unknown if the sequence is infinitely often in $[0,1/2)$, or that it is infinitely often not in $[0,1/2)$. Really, nothing is known.

As a final comment on this problem, the golden ratio is special. With $x=\phi=(1+\sqrt 5)/2$, for every $\epsilon>0$ there are only finitely many $n$ with $$\epsilon< \{\phi^n \} < 1-\epsilon.$$

• I'd done some special visualizations of this problem, but due to lack of background in my NT-experience I#ve not been able to proceed in a meaningful way then. But I find it still an interesting approach to display such a property as tried in that link as part of my earlier Collatz-discussion: go.helms-net.de/math/collatz/aboutloop/… The golden ratio, for instance, shows this significant behave, that you mention, in a surprising regularity. – Gottfried Helms Jul 1 '12 at 15:17
• Actually, it is known that the sequence $\\{ (3/2)^n \\}$ has infinitely many limit points. More generally, Vijayaraghavan proved in the 40s ("On the fractional parts of the powers of a number. II.") that if $\theta>1$ is algebraic and is not Pisot, then $\\{ \theta^n \\}$ has infinitely many limit points. – Pablo Shmerkin Jul 2 '12 at 1:16
• Sorry about the bad formatting... I meant the sequence of fractional parts of $(3/2)^n$ and $\theta^n$. – Pablo Shmerkin Jul 2 '12 at 1:18
• @Pablo: I didn't know that! I'll read up and fix my answer. – Kevin O'Bryant Jul 2 '12 at 3:07
• @Kevin: I learned this from a draft of Yann Bugeaud's soon to be published book "Distribution modulo one and Diophantine approximation". There it is mentioned that the proof that there are infinitely many points was found independently by Vijayaraghavan, by Pisot ("La repartition modulo 1 et les nombres algebriques") and by Rédei ("Zu einem Approximationssatz von Koksma"). All of these are from around 1940 and it seems there has been no progress since. – Pablo Shmerkin Jul 2 '12 at 8:29

It is still not known whether the problem of determining whether a linear integer recurrence (of which the Fibonacci recurrence $F_n = F_{n-1}+F_{n-2}$, $F_1=F_0=1$ is the most well known) contains a zero is decidable or not. Even the case of recurrences of depth 6 is currently open. (I discussed this problem at http://terrytao.wordpress.com/2007/05/25/open-question-effective-skolem-mahler-lech-theorem/ .) We do have the famous Skolem-Mahler-Lech theorem that gives a simple criterion as to when the number of zeroes is finite, but nobody knows how to get from that to deciding when there is a zero at all. (This is perhaps the simplest example of a large family of results in number theory in which one has an ineffective finiteness theorem for the number of solutions to a certain number-theoretic problem (in this case, an exponential Diophantine problem), but no way to determine if a solution exists at all. Other famous examples include Faltings' theorem and Siegel's theorem.)

EDIT: See also this survey of Halava-Harju-Hirvensalo-Karhumäki from 2005 on this problem: http://tucs.fi/publications/view/?id=tHaHaHiKa05a

Artin's Conjecture: There are infinitely many primes $p$ for which 2 is a primitive root, i.e., 2 generates the multiplicative group of ${\mathbb Z}/p{\mathbb Z}$.

The conjecture is actually a bit more general, but we should at least be able to say what happens with 2! The OEIS lists the first several such primes.

• Dear Kevin, As you probably know, Hooley proved this contingent on a GRH. Thus it reduces to a more-well known open problem! It could be worth adding this to your answer. Regards, Matthew – Emerton Jul 2 '12 at 12:03
• Typo: "on a GRH" should just read "on GRH". (More precisely, he needs the RH for the Dedekind zeta-functions of a certain infinite collection of number fields.) – Emerton Jul 2 '12 at 12:04
• Follows from, but doesn't imply. This is (I suspect) much easier than GRH, especially just asking about $2$. – Kevin O'Bryant Jul 2 '12 at 16:52
• Dear Kevin, From the point of algebraic number theory (or more specifically, my understanding of Hooley's proof!), $2$ doesn't look so different from other numbers. But maybe from other points of view that's less true. Also, I agree that this seems easier (or, at least, much more specialized) than GRH, but on the other hand, Hooley's proof is quite natural, so viewing it through the GRH lens doesn't seem unreasonable. (E.g. the same method applies to prove many variants --- again dependent on GRH.) Regards, Matthew – Emerton Jul 2 '12 at 22:46

Can we exactly calculate Ramsey numbers? Erdős once famously remarked:

"Suppose aliens invade the earth and threaten to obliterate it in a year's time unless human beings can find the Ramsey number for red five and blue five. We could marshal the world's best minds and fastest computers, and within a year we could probably calculate the value. If the aliens demanded the Ramsey number for red six and blue six, however, we would have no choice but to launch a preemptive attack."

• There seems to be something wrong with the link. – alvarezpaiva Jul 2 '12 at 12:04
• Great load of finding precise answers to problems in Ramsey theory would qualify as well. – Wojowu Dec 25 '15 at 13:58

Is there an algebraic irrational number in the Cantor set?

More generally: Are algebraic irrational numbers normal in all bases?

Normality of numbers. Is Pi normal?

Here are two from ergodic theory:

• (The problem of smooth realizations) Let $X$ be a Lebesgue space with measure $\mu$, and let $T:X\to X$ be a transformation preserving the measure $\mu$. If the entropy $h_\mu(T)$ is finite, is $(X,T,\mu)$ always measurably isomorphic to a smooth system $(M,f,v)$, where $M$ is a compact manifold, $f$ is a diffeomorphism of $M$ and $v$ is a smooth volume?

• (Furstenberg's $\times 2 \times 3$ problem) Does there exist a Borel probability measure $\mu$ on the unit circle $\mathbb{R}/\mathbb{Z}$, which is neither discrete nor Haar measure, and which is invariant under both $x\to 2x \bmod 1$ and $x\to 3x\bmod 1$?

For the first problem, as far as I know there has been no significant progress.

For Furstenberg's conjecture, Furstenberg himself solved the analog question for sets (answer is negative), and Rudolph proved that the answer is negative under an extra positive entropy assumption. While there has been a huge amount of progress in the positive entropy case since, the zero entropy case remains untractable despite the simplicity of the statement.

Does $H^\infty(D)$, the Banach space of all bounded holomorphic functions on the unit disc, have Grothendieck's approximation property?

Here is an old question by Borel: is there any a priori growth restriction on entire functions $f(z)$ satisfying polynomial differential equations $P(z,f(z),\dots,f^n(z))=0$ where $P$ is a polynomial with complex coefficients in $n+2$ variables?

• @fedja: $P$ seems to be polynomial in $n+1$ variables, isn't it? – shuhalo Jul 2 '12 at 13:30
• @Martin: No, we have $z$ and the 0th-nth derivative of $f(z)$. – temp Jul 3 '12 at 1:36

Chromatic Number of the Plane (Hadwiger-Nelson Problem): What is the minimum number of colors required to color the plane so that no two points which are unit distance apart are the same color? Let $\chi$ denote this number. The current bounds on $\chi$ are

$$4\leq \chi \leq 7$$.

• I know this thread is closed, but for posterity, thanks to de Grey we now know $5\le\chi\le7$. – Mark S Apr 25 '18 at 22:44

The Bunyakovsky conjecture (or Bouniakowsky conjecture) stated in 1857 by the Russian mathematician Viktor Bunyakovsky, claims that

• an irreducible polynomial of degree two or higher with integer coefficients generates for natural arguments either an infinite set of numbers with greatest common divisor (gcd) exceeding unity, or infinitely many prime numbers.
• Often referred to as "Schinzel's hypothesis" nowadays, though Schinzel is somewhat more general than this. – JSE Jul 2 '12 at 6:04
• @JSE: Yes I truly agree. – crskhr Jul 2 '12 at 6:43

Every finite abelian group is (isomorphic to) the class group of the ring of algebraic integers of some number field.

For Dedekind domains this is well-known (even for any abelian group); due to Claborn and Pete L. Clark has an alternate proof/a refinement.

Also a 'geometric analog' is known (Perret, 1999).

And every finite ablian group is at least a subgroup of a classgroup (even for a cyclotomic field).

It can also be shown that, for a fixed prime $p$, every finite abelian $p$-group is the $p$-Sylow of the class group of the ring of algebraic integers of some number field (by Yahagi, Tokyo J. of Math 1978) and that every finite $p$-group is the Galois group of the maximal unramified $p$-extension of a number field (Ozaki, Inventiones 2011); note that this Galois group coincides with the class group if one adds the condition that it be abelian, by Class Field Theory.

ps. Not sure this passes all (or any) of the criteria; I'll let you decide :)

ps2. Searching for a reference, I found this math.SE question on exactly this https://math.stackexchange.com/questions/10949/finite-abelian-groups-as-class-groups

• I don't think that this problem qualifies. On the other hand, the question is quite vague ... – Martin Brandenburg Jul 1 '12 at 7:36
• Also, every finite abelian group is an $S$-class group of a number field. – user19475 Jul 1 '12 at 8:05
• @Martin Brandenburg: in particular, since there are several criteria to be met, could you elabortate, which one(s) is/are not met in your opinion. – user9072 Jul 1 '12 at 8:33

Are there infinitely many regular primes? We know there are infinitely many irregular ones, and that their percentage should be much smaller than the regular ones, still it is unproven that the latter are infinite.

Let me recall that a prime $p$ is irregular if it divides the class number of $\mathbb{Q}(\zeta_p)$, the cyclotomic field.

Similarly, we cannot prove that there are infinitely many real quadratic fields of class number $1$.

An obvious problem in algebraic topology would be the computation of the homotopy groups of spheres.

• I'd like the statement of problem a bit more if it was more concrete: sort of what's $\pi_5(S^2)$ ? Actually, what are the smallest $k$ and $n$ for which $\pi_k(S^n)$ is not known ? – alvarezpaiva Jul 2 '12 at 18:25
• What does "compute" here mean precisely? Do these answers not count: mathoverflow.net/questions/31004 ? – Michael Bächtold Jul 2 '12 at 19:01
• Dear Michael, In this context, "compute" means something like "give a closed form expression for". Regards, – Emerton Jul 2 '12 at 22:39
• Dear Emerton, why is there hope that such closed form expression exists for homotopy groups of spheres? – Michael Bächtold Jul 2 '12 at 23:54
• Well, in contrast, it is known by work of Jie Wu that the homotopy groups of $S^2$ are given by the centres of a sequence of combinatorially described groups (math.nus.edu.sg/~matwujie/newnewpis_3.pdf from 2001). I don't know how far we are now from having a combinatorial or algorithmic description of these centres though. – David Roberts Jul 3 '12 at 1:01

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 $f$.

From the Overview of the Royal Danish Sciences Institution's work and its members' work in the year 1882.

In the notes from a meeting on March 9th 1877, after discussing papers by Legendre, J. W. L. Glaisher, and Meissel, Oppermann stated:

At the same occasion, I made people aware of the not yet proven conjecture, that when $n$ is a whole number $>1$, at least one prime number lies between $n(n-1)$ and $n^2$ and also between $n^2$ and $n(n+1)$.

A solution to Oppermann's Conjecture leads to simple solutions to Legendre's, Brocard's, and Andrica's Conjectures.

Proof of Andrica assuming Oppermann

• That's a neat one! – alvarezpaiva Jul 2 '12 at 12:02
• Or, to ask a stronger conjecture, is it true that for any positive real $\epsilon$, there is a natural number $N$ such that if $N < n$ then there is a prime between $n$ and $n + n^\epsilon$. – Zsbán Ambrus Jul 2 '12 at 16:40

Schinzel-Sierpinski Conjecture

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

• A conjecture of Schinzel and Sierpinski asserts that every positive rational number $x$ can be represented as a quotient of shifted primes, that $x=\frac{p+1}{q+1}$ for primes $p$ and $q$. It is known that the set of shifted primes, generates a subgroup of the multiplicative group of rational numbers of index at most $3$.

The Whitehead asphericity conjecture. Let $X$ be a $2$-dimensional aspherical simplicial complex and let $Y \subset X$ be a connected subcomplex. The conjecture then is that $Y$ is aspherical.

Very little is known about this, but a deep theorem of Bestvina and Brady says that the Eilenberg-Ganea conjecture and the Whitehead asphericity conjecture cannot both be true.

• This is just in two dimensions or are you giving us the simplest unkown case ? – alvarezpaiva Jul 3 '12 at 5:12
• @alvarezpaiva : It's something special about dimension $2$. It's easy to come up with counterexamples in higher dimensions; for instance, you can triangulate $\mathbb{R}^3$ so that it contains $S^2$ as a subcomplex. – Andy Putman Jul 3 '12 at 16:02
• The complex is finiite? – Alexander Chervov Jul 10 '12 at 4:59
• @Alexander Chernov : No. – Andy Putman Jul 14 '12 at 5:51

I'm not sure what your threshold for "barely mentioned in the literature" is, since some of the highly-voted answers seem rather well known to me, but here's one that is certainly fundamental, seemingly out of reach, and perhaps not so well known except to complexity theorists.

Describe explicitly a Boolean function whose minimum circuit size is superlinear.

A simple counting argument shows that almost all Boolean functions require exponentially large circuits to express. However, giving explicit examples is another matter. Here, "explicit" is a bit vague, but let's say for example that it means that the truth table can be computed in time polynomial in the size of the truth table. Thus NP-complete Boolean functions count as "explicit," and proving superpolynomial circuit lower bounds for them would separate P from NP, but even if we weaken the requirement to a superlinear lower bound on any explicit function, nobody seems to have any clue.

The Eilenberg-Ganea conjecture. Recall that the cohomological dimension $\text{cd}(G)$ of a discrete group $G$ is the maximal $n$ such that there exists a $G$-module $M$ with $H^n(G;M) \neq 0$. The geometric dimension $\text{gd}(G)$ of $G$ is the smallest $n$ such that $G$ has a $K(G,1)$ which is an $n$-dimensional CW complex. It is elementary that $\text{cd}(G) \leq \text{gd}(G)$. Moreover, if $\text{cd}(G) \neq 2$, then it is classical that $\text{cd}(G) = \text{gd}(G)$. The Eilenberg-Ganea conjecture says that this also holds if $\text{cd}(G)=2$. It is known, by the way, that if $\text{cd}(G)=2$ then $2 \leq \text{gd}(G) \leq 3$.

The only progress that I know of concerning this is a deep theorem of Bestvina and Brady that says that at the Eilenberg-Ganea conjecture and the Whitehead asphericity conjecture cannot both be true.

Gelfand's problem: can you find a closed, proper unital subalgebra A of C[0,1] such that the natural map from [0,1] to the character space of A is bijective?

(See for instance these notes of Feinstein )

Is every complemented subspace of $C[0.1]$ isomorphic to $C(K)$ for some compact metric space $K$?

Is every infinite dimensional complemented subspace of $L_1[0.1]$ isomorphic either to $L_1[0.1]$ or to $\ell_1$?

Is every finite lattice a congruence lattice of a finite (universal) algebra?

Astonishingly, by Pálfy and Pudlák, this question is equivalent to a question in group theory: is every finite lattice isomorphic to an interval of the subgroup lattice of a finite group?

Here is a variation of Georges Elencwajg's question, due to Gennady Lyubeznik. Is every closed point (of arbitrary degree over $\mathbb{Q}$) in $\mathbb{P}^2_{\mathbb{Q}}$ set-theoretically the intersection of two curves?

Let $G$ be a finite group. We define $r(G)$ to be the smallest number of relations possible in a presentation of $G$ with the minimal number of generators. If $G$ is a $p$-group, we can also consider "pro-$p$ presentations" of $G$ (using the free objects in the category of pro-$p$ groups); we write $r_p(G)$ for the smallest number of relations possible in a pro-$p$ presentation with the minimal number of generators.

Does $r(G) = r_p(G)$?

P vs NP

According to Leonid Levin (via Scott Aaronson), Richard Feynman could not be convinced that this was actually an open problem.

• Thanks JeffE, but this is well-known and documented problem with a million dollar check attached to it. I'm looking for problems that are almost embarrasingly natural and that are not often mentioned in the literature. – alvarezpaiva Jul 1 '12 at 8:50

What about Goldbach conjecure asking if every even natural number is the sum of two primes?

Another quite famous problem is Collatz' conjecture (also known as $3n+1$ problem), see http://en.wikipedia.org/wiki/Collatz_conjecture: consider the algorithm taking $n\in\mathbb{N}$ and sending it to $n/2$ if $n$ is even, and to $3n+1$ if $n$ is odd, iteratively. The question is whether the algorithm always ends up producing the loop $1\mapsto 3\cdot 1+1=4\mapsto 2\mapsto 1\mapsto 4\dots$ regardless of the initial input $n$.

• This is a well-known problem (by everybody), with a bit of literature, including a novel, behind it. I'm looking for the sort of problem that experts in some field would know, that may even consitute some sort of holy grail in the field, but that has not been very publicized in the literature because it is so hard that there are few if any partial solutions. – alvarezpaiva Jul 2 '12 at 5:14
• Ok, I am sorry, at first glance I did not notice that you were asking for problems which were barely unkown. I apologize for the bad answer. I edited it adding another problem, quite well-known in elementary number theory but may be not so well-known to everybody. – Filippo Alberto Edoardo Jul 2 '12 at 9:18

Can one transform a particular closed knotted piece of rope into another one?

• Do you mean the study of knot types when the length and thickness of the knot must be kept constant during the isotopy? Any reference for this problem or at least for a hint of it? – alvarezpaiva Jul 1 '12 at 19:20
• I mean a classification of knots under the isotopy. Just formulated it in physical setting to show its simple nature. A physical type problem is also unsolved. There are some development here, uncluding algorithms, depending on the thickness of a knot. – Andrew Jul 1 '12 at 20:26
• It's still not clear to me what question you are asking. – Douglas Zare Jul 2 '12 at 5:50
• I'm confused. Haken's algorithm determines whether two knot-complements are homeomorphic and hence, by the Gordon--Luecke Theorem, whether the two knots are isotopic (modulo orientation, admittedly). – HJRW Jul 2 '12 at 10:13
• I still don't know which problem is being suggested, but I think that with all of the progress on knot theory and geometrization, it would be very strange to say, "The solution seems completely out of reach." – Douglas Zare Jul 3 '12 at 0:49