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What are best examples of questions in mathematics that are not interesting until one knows the answers, whose answers themselves are what is interesting?

The thing that prompts me to post this is just one example. I've seen others, but they escape me at the moment. Here it is:

A torus is embedded in just the usual way in $\mathbb R^3$. It has parallels of latitude and meridians of longitude. A curve that meets every parallel of latitude at the same angle, or, equivalently, meets every meridian of longitude at the same angle, is a loxodrome. Suppose that angle is so chosen, given the shape of the particular torus, that the loxodrome goes through all $360^\circ$ of longitude in just the length it takes to go through all $360^\circ$ of latitude, returning there to its starting point. (There must be some conventional terminology for describing these windings, but I don't know it.) The question is: What are the curvature and torsion at the various points along this curve? Doubtless some will consider this question interesting, but to me, and, I suspect, to many, the answer, because it is so unexpected, is where this starts to get interesting. The answer is that the curvature is constant --- the same at all points on the curve --- and the torsion is everywhere $0$. (And it's really easy to deduce from that the precise value of the curvature.) I believe this was discovered in the 1890s and is stronger than the celebrated theorem of Villarceau, published in 1848. Villarceau's theorem says that a plane bitangent to a torus intersects the torus in two circles. This proposition does not assume as a hypothesis, but rather has as a (trivial corollary of its) conclusion, that the curve lies in a plane.

Michael Hardy
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    Somewhat subjective, isn't it? I find the Whitehead problem to be very uninteresting, but the answer itself is quite interesting. – Asaf Karagila Mar 24 '15 at 07:39
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    Can you be more specific as to what you're trying to get at by "best examples"? I could surely write down a long list of uninteresting generators and relations and say, "what is this structure?" or "what is its order?", and the group turns out to be the Monster Group. – Kimball Mar 24 '15 at 09:05
  • Reminds me of RDRR – Mikey Mouse Mar 24 '15 at 09:55
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    I think this is what the Reversal badge on MathOverflow is supposed to be for: http://mathoverflow.net/help/badges/50/reversal. Doesn't look like there are many! – Chris Heunen Mar 24 '15 at 10:27
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    @ChrisHeunen Then should we downvote this question a lot so the question can be an answer to itself? – Kimball Mar 24 '15 at 12:53
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    I suspect that any interesting mathematical result can be made the answer of some uninteresting mathematical question, or at least, quite less interesting than the answer. For instance: how little is the minimum algebraic sum $\pm a^2\pm b^2\pm c^2$ for a right triangle with edges $a,b,c$? etc. – Pietro Majer Mar 25 '15 at 21:36
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    I have to say I am shocked that this question has received so many upvotes and has not yet been closed. This question is very subjective and MO is not for collecting bits of mathematical trivia. This question could perhaps have some value if it were geared toward understanding how it is that uninteresting questions have interesting answers (see Timothy Chow's answer, for instance), but in its current form I don't see the point of it and it is attracting a lot of dubious answers. Pietro Majer's comment is also illustrative of the arbitrariness of this question. – Eric Wofsey Mar 25 '15 at 23:28
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    I agree with Eric that this question is too subjective, as people often can't seem to agree on which questions are "uninteresting". I'm casting the fifth vote to close. – Todd Trimble Mar 26 '15 at 11:13
  • @Todd, you could always let the voters decide which questions are uninteresting. – Gerry Myerson Mar 26 '15 at 11:53
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    @GerryMyerson I don't understand. I am acting as an ordinary voter (i.e., I'm not exercising any moderator powers; I waited until four votes were cast), and I join four others to vote to close this as "primarily opinion-based". If you disagree, I invite you to post at meta. – Todd Trimble Mar 26 '15 at 13:51
  • @Todd, what I meant was, by our upvotes and downvotes on answers, users are deciding deciding which questions are uninteresting. "How many primes are there" is currently at minus 2; the voters seem to agree that that question is not uninteresting. Joe Silverman's answer is at plus 33; the voters seem to agree that that question is uninteresting. It might be somewhat subjective, but there's also a fair bit of agreement here. – Gerry Myerson Mar 26 '15 at 22:48
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    @GerryMyerson I see. Thanks for clarifying. Still, I'm voting by my conscience that this is not a good question for MO. It's also not clear to me that upvotes really reflect both interest in the answer and disinterest in the question; maybe people just think the answers are cool. Little can be deduced, I'm afraid, except maybe from comments from a few users. – Todd Trimble Mar 26 '15 at 23:57

8 Answers8

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Uninteresting question: find $$\int_0^1{x^4(1-x)^4\over1+x^2}\,dx$$ Interesting answer: $${22\over7}-\pi$$

Gerry Myerson
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    That one actually crossed my mind while I pondered this question, but at the moment of posting it I had forgotten it. ${}\qquad{}$ – Michael Hardy Mar 24 '15 at 06:01
  • Hm... When do the approximants for $\pi$ appear in such way? – Ilya Bogdanov Mar 24 '15 at 08:40
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    @Ilya, there has been work on finding integrals that involve other approximations to $\pi$. For example, see http://www.austms.org.au/Publ/Gazette/2005/Sep05/Lucas.pdf – Gerry Myerson Mar 24 '15 at 10:07
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    Is this the easiest proof that $\pi < 22/7$? :) – Bill Johnson Mar 24 '15 at 16:06
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    @IlyaBogdanov you might be interested in an question on the history and context of that integral –  Mar 24 '15 at 16:37
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    @Bill Johnson, it depends on what "easiest" means. Archimedes proved $\pi < \frac{22}{7}$, but I doubt he could have evaluated that integral. – Nathan Mar 24 '15 at 16:50
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    There is also a Wikipedia article, originally created by me: http://en.wikipedia.org/wiki/Proof_that_22/7_exceeds_%CF%80 ${}\qquad{}$ – Michael Hardy Mar 24 '15 at 19:41
  • @BillJohnson I'd say Machin's formula (http://en.wikipedia.org/wiki/Machin-like_formula) gets you there pretty quickly, with a hand-held calculator. We have $\pi = 16\arctan(1/5) - 4\arctan(1/239)$ where three terms of $16\arctan(1/5)$ minus two terms of $4\arctan(1/239)$ gets you comfortably between $\pi$ and $22/7$. – Todd Trimble Mar 26 '15 at 02:59
  • Another answer for the same question is $$\frac{16}{21}\sum_{k=1}^\infty \frac{B(4k+1,8)}{B(k+1,2)}$$ where $B$ is the Beta integral, so $$\frac{22}{7}-\frac{16}{21}\sum_{k=1}^n \frac{B(4k+1,8)}{B(k+1,2)}$$ gives fractions closer to $\pi$ as $n$ grows (@IlyaBogdanov) http://mathoverflow.net/a/231250/85758 – Jaume Oliver Lafont Feb 18 '16 at 11:24
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Another classical geometry problem with a similar flavor to Michael's example:

Let $C$ be a smooth convex plane curve, let $L$ be a small line segment, let $P$ be the mid-point of $L$. Slide $L$ around the curve, keeping the endpoints on $C$, so $P$ traces out a curve $C'$ inside of $C$. What is the area between $C$ and $C'$? For any particular curve $C$, this seems (to me) to be a rather uninteresting calculus problem. What's interesting, of course, is that the answer is independent of $C$ and depends only on the length of $L$.

(Actually, one can mark any point $P$ on $L$, and then the area depends on the lengths of the two sub-segments of $L$.)

Joe Silverman
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    That looks very surprising. Is anyone aware of any 'fundamental' reason which could explain this invariance phenomenon ? – Hachino Mar 24 '15 at 11:57
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    Interesting, indeed. Though I know it might not matter much to folks here, this has a real-world application, too. If the line segment represents a cutting tool, the area is the material removed by the tool motion. Understanding rates of material removal is very important in manufacturing. References, please?? – bubba Mar 24 '15 at 12:17
  • There seem to be lots of problems in geometry, including this one, and the one in my posted question and the napkin ring problem, and others that escape me at the moment, that seem essentially the same as each other in all respects except the specifics. ${}\qquad{}$ – Michael Hardy Mar 24 '15 at 16:03
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    I am guessing that $C$ is supposed to be closed? – Willie Wong Mar 24 '15 at 16:04
  • Is there a formulation involving signed area for nonconvex curves, or is the convexity really essential? – Robin Saunders Mar 24 '15 at 16:17
  • @WillieWong Yes, $C$ is supposed to be closed. – Joe Silverman Mar 24 '15 at 16:44
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    @RobinSaunders Probably it's fine for non-convex curves with signed area, as you say. It's also okay for piecewise smooth curves. In any case, the proof is a lovely application of Green's theorem, and from the proof one should be able to see how to formulate it more generally than I did in the answer that I gave. But note, for example, if $L$ is too long and $C$ is too small, it may not be possible to slide $L$ around inside $C$, so some restrictions are necessary. – Joe Silverman Mar 24 '15 at 16:46
  • What are some references for this result? – Wojowu Mar 24 '15 at 16:55
  • @Wojowu It was assigned as a homework problem when I took undergrad differential geometry class, at which time I knew who it was named after. But I'm afraid I've forgotten even that piece of information. And I don't know any references. Sorry. – Joe Silverman Mar 24 '15 at 17:24
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    This looks like Mamikon's "visual calculus": https://www.its.caltech.edu/~mamikon/VisualCalc.html – zeb Mar 25 '15 at 00:04
  • @Wojowu: Step 1: The result for $C$ being a circle follows directly from Pythagoras Theorem: Basically, $L/2$ is one side of the triangle, the ray from the center of $C$ to the middle of $L$ is the other side, and the radius is the hypotenuse. Step 2: "morally conclude" the case of a general closed curve $C$ by considering at every point $P$ of $C$ the tangent circle $C'$ of the same curvature as $C$. The infinitesimal amount of area swept by $L$ at $P$ is virtually the same for $C$ and for $C'$. Since that amount does not depend on the radius of $C'$ it's straightforward to conclude the rest – Michael Mar 25 '15 at 20:30
  • @bubba, Carpenters call it kerf and they must account for it to build things. Sawmills do the same to determine board-feet of a log. The volume of this kerf is the sawdust. – Fred Daniel Kline Mar 26 '15 at 08:52
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According to Gauss, Fermat's Last Theorem is an example.

Pietro Majer
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  • Not sure if this example qualifies. – Pietro Majer Mar 25 '15 at 21:40
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    on the contrary, I think this is the ultimate example. – shadowtalker Mar 26 '15 at 00:17
  • How did Gauss formulate this sentiment? (I guess he hadn't access to this MO question) – Vincent Feb 16 '16 at 16:09
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    It's a rather popular quotation. Here it is:

    I confess that Fermat's Theorem as an isolated proposition has very little interest for me, because I could easily lay down a multitude of such propositions, which one could neither prove nor dispose of.

    [A reply to Olbers' attempt in 1816 to entice him to work on Fermat's Theorem.]

    Quoted in J. R. Newman, The World of Mathematics (New York 1956).

     – Pietro Majer Feb 22 '16 at 15:00
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Gerry Myerson's integral and Joe Silverman's geometry problem fall into the category of problems that seem uninteresting at first because they can be answered by a straightforward calculation that is not expected to yield any insight after the answer is obtained.

Another potential category consists of problems that seem hopelessly difficult but that turn out to be tractable. As an example, I propose the question, "What are all the finite simple groups?" Superficially, this might seem (almost) as hopeless, and therefore as uninteresting, as the question, "What are all the finite groups?" Only when you know that there is a nice answer that can actually be proved does the question reveal itself to be extremely interesting.

Timothy Chow
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How many lines in $\mathbb{CP}^3$ meet four lines in general position?

Given the `linear' nature of the problem one may be tempted to guess $0,1$ or $\infty$. But the answer is actually 2.

The proofs are also pretty interesting:

(1) Degenerate into two pairs of intersecting lines; the two lines are the intersection of the two planes containing them, and the line passing through the intersection point. Then infer that the number is an intersection number (so topological), and therefore independent of the configuration. Making this last claim rigorous was one of Hilbert's problems.

(2) Use the fact that three general lines determine a quadric surface in $\mathbb{CP}^3$. The fourth line will intersect the quadric in two more points. Now draw the two lines as rulings on that quadric in those points. Here it is also easy to see that there cannot be more than 2 lines with this property.

gsvr
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Watson's integral. Seemingly uninteresting question: calculate $$W_S=\frac{1}{\pi^3}\int\limits_0^\pi\int\limits_0^\pi\int\limits_0^\pi \frac{dx\,dy\,dz}{3-\cos{x}-\cos{y}-\cos{z}},$$ produces truly amazing answer: $$W_S=\frac{\sqrt{6}}{96\pi^3}\Gamma\left(\frac{1}{24}\right)\Gamma\left(\frac{5}{24}\right)\Gamma\left(\frac{7}{24}\right)\Gamma\left(\frac{11}{24}\right)= \frac{\sqrt{3}-1}{96\pi^3}\left[\Gamma\left(\frac{1}{24}\right) \Gamma\left(\frac{11}{24}\right)\right]^2.$$ See http://link.springer.com/article/10.1007%2Fs10955-011-0273-0 (70+ Years of the Watson Integrals, by I. J. Zucker).

Zurab Silagadze
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  • Rather subjective statement about the question... I would say the question is even VERY interesting, and consider it as an additional gadget that it has this amazing answer. Which in turn provokes the definitely interesting question: What about the $n$-fold such integral? Sadly I cannot access the paper. – Wolfgang Mar 26 '15 at 08:07
  • @Wolfgang Try this link http://www.inp.nsk.su/~silagadz/Watson_Integral.pdf – Zurab Silagadze Mar 26 '15 at 08:32
  • Sorry - why is this amazing? – gsvr Mar 26 '15 at 11:47
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    This is, of course, rather subjective to call it amazing. However, for me, these forms of the answer indicate that there is a deeper story behind them which allows to interrelate these funny expressions. By the way Watson's original result was $$W_S=(18+12\sqrt{2}-10\sqrt{3}-7\sqrt{6})\left[\frac{2}{\pi}K(k_6)\right]^2,$$ where $k_6=(2-\sqrt{3})(\sqrt{3}-\sqrt{2})$ and $K$ is the complete elliptic integral of the first kind. – Zurab Silagadze Mar 26 '15 at 13:23
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I hereby propose as one of innumerable possible answers to this question: Hilbert's 10th problem.

Doubtless it's an interesting problem, to those who are interested in that sort of thing; otherwise Hilbert would not have included it in his list. But to me, and again I suspect, to many, the answer is a lot more interesting than the question, partly, but not only, because it is surprising.

The problem is this: Is there an algorithm that given any polynomial in any finite number of variables with coefficients in $\mathbb Z$, correctly answers the question: is at least one tuple of integers a zero of this polynomial?

To understand the answer, let's establish some defintions:

  • A set $S$ of members of $\mathbb Z^n$ is diophantine if there is some $m\in\mathbb Z^+$ and some polynomial function $f$ in $m+n$ variables $y_1,\ldots,y_m,x_1,\ldots,x_n$ with coefficients in $\mathbb Z$ such that $(x_1,\ldots,x_n)\in S$ if and only if $\exists y_1,\ldots,y_m\in\mathbb Z\ f(y_1,\ldots,y_m,x_1,\ldots,x_n)=0$.
  • A set $S$ of members of $\mathbb Z^n$ is decidable if there is some algorithm that, given a member of $\mathbb Z^n$ correctly answers the question: Is this a member of $S$?
  • A set $S$ of members of $\mathbb Z^n$ is semi-decidable if there is some algorithm that, given a member of $\mathbb Z^n$, runs forever if the input is not a member of $S$, and ultimately halts if it is a member of $S$.

Obviously a set is decidable if and only if both the set and its complement are semi-decidable. The existence of semidecidable sets that are not decidable was discovered in the 1930s by several people working independently (I think including Stephen Kleene, Alan Turing, Alonzo Church and maybe others?) and some of them are noteworthy sets, e.g. the set of all satisfiable formulas in first-order logic.

Obviously all diophantine sets are semi-decidable.

The result that laid Hilbert's 10th problem to rest is Matiyasevich's theorem:

All semi-decidable sets are diophantine.

An immediate corollary is that no algorithm of the kind sought by Hilbert can exist.

In 1970, Yuri Matiyasevich finished off the proof, which had been worked on over a couple of decades by Julia Robinson, Martin Davis, and Hillary Putnam.

Michael Hardy
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    I would say that the problem is interesting and natural. Hilbert almost surely thought there would be an algorithm and was undoubtedly aware that many natural algorithmic problems are naturally (as opposed to the Halting problem) diophantine problems. – Benjamin Steinberg Mar 24 '15 at 17:04
  • @BenjaminSteinberg : OK, so your point is simply to agree with my second paragraph? – Michael Hardy Mar 24 '15 at 17:55
  • For semi-decidable, you may want to change it to say "may run forever if the input is not a member" because all decidable sets are semi-decidable (the current wording does not make it seem so). – Ryan Dougherty Mar 24 '15 at 18:18
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    I think it should not be referred to as just Matiyasevich's theorem, since (as you say) Davis, Robinson, and Putnam spent years and years on preliminary groundwork. Technically what Matiyasevich did in 1970 was prove the Fibonacci sequence is Diophantine; this by itself might not seem terribly exciting, but due to the work of the other three, all four knew that this technical result would imply the J.R. (Julia Robinson) hypothesis which in turn settled Hilbert's 10th. The theorem is often referred to as "the MRDP theorem". – Todd Trimble Mar 24 '15 at 18:49
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    My point is to disagree with the question being uninteresting. I think the question of which diophantine problems have solutions goes back to the earliest days of mathematics and is as natural as any mathematical problem I can imagine and I am not a number theorist. So I really don't think the problem was uninteresting until the answer was known. – Benjamin Steinberg Mar 24 '15 at 20:35
  • @Ryan : I don't think it's a precise definition with "may". And I state explictly in my posting that a set is decidable if and only if both it and its complement are semi-decidable, so that makes it clear if it weren't already that every decidable set is semi-decidable. That all decidable sets are semi-decidable is an easy exercise. ${}\qquad{}$ – Michael Hardy Mar 24 '15 at 22:39
  • @ToddTrimble : My posting says, in the first sentence in the second paragraph "Doubtless it's an interesting problem, to those who are interested in that sort of thing; otherwise Hilbert would not have included it in his list." That says it. But I was saying that even those who take no particular interest in that sort of thing can find Matiyasevich's theorem interesting because of the surprising nature of the result: that there are no semi-decidable sets except diophantine sets. – Michael Hardy Mar 24 '15 at 22:41
  • @MichaelHardy Did you mean to address Benjamin Steinberg? My comment was about what the theorem should be called. – Todd Trimble Mar 24 '15 at 23:20
  • Probably. @BenjaminSteinberg ${}\qquad{}$ – Michael Hardy Mar 25 '15 at 01:16
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One question which I think might be fitting is:

How many prime numbers are there?

After you learn about infinitude of primes you might think that there is nothing really much to say about the topic. But if we try to consider how many primes there are asymptotically, we reach a very interesting field of research, which I believe gave rise to the analytic number theory.

Wojowu
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    I understand that this might not be precisely in the flavor that OP was asking for, but I'd gladly hear what the person who down-voted has to say. – Wojowu Mar 25 '15 at 21:44
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    Most mathematicians (including Euclid) might feel that the question is immensely interesting -- that's why it doesn't fit the category of "questions that are initially uninteresting" – Dominic van der Zypen Mar 26 '15 at 07:57
  • @DominicvanderZypen I see your point. Thanks. – Wojowu Mar 26 '15 at 14:40