Slashdot Mirror


Mathematician Claims Proof of Riemann Hypothesis

TheSync points to this press release about a Purdue University mathematician, Louis de Branges de Bourcia, who claims to have "proven the Riemann hypothesis, considered to be the greatest unsolved problem in mathematics. It states that all non-trivial zeros of the zeta function lie on the line 1/2 + it as t ranges over the real numbers. You can read his proof here. The Clay Mathematics Institute offers a $1 million prize to the first prover."

34 of 561 comments (clear)

  1. Googlized HTML version by Anonymous Coward · · Score: 3, Informative
  2. Failed proof by MobyDisk · · Score: 5, Informative

    Ha! They've already found an error in the proof! All that he posted was his apology! :-)

    Yes, I was actually confused at first. For the non-math geeks like myself, who are feeling stupid, look at definition 2a of apology.
  3. WTF? Mods? by Unnngh! · · Score: 5, Informative
    From reference.dictionary.com:

    Apology - 2: a formal written defense of something you believe in strongly

    This should at most have earned a "Funny", or is there something I'm missing here?

  4. Re:Apology by ssssmemyself · · Score: 4, Informative

    Note to mods: Mod parent funny, not interesting! This is a play off a quote from the beginning credits sequence in Monty Python and the Holy Grail. As for the pdf link, it's the first link in the purdue page referenced in the article. RTFA, people!

  5. There is no impact on crypto by Anonymous Coward · · Score: 4, Informative

    The Riemann Hypothesis, among other things, implies that the Prime Number Theorem is off in the distribution of primes by no more than O(sqrt(n)*log(n)). However even without the full result, we already had very good error bounds for the approximation of the prime number theorem for "small" numbers, including numbers far larger than any which come up in cryptography.

  6. Re:Impact on crypto? by mdrejhon · · Score: 4, Informative

    The art of cryptography can be summed up as: Easy to encode, but hard to decode.

    Prime numbers are easy to multiply together. Little CPU needed.

    But it's hard to do the reverse: Factor a big number into two separate prime numbers. Lots of CPU needed.

    It's based on that principle.

  7. Seems not-unlikely to be wrong by k98sven · · Score: 4, Informative

    Sorry to burst the bubble, but some usenetting shows:

    The same guy claimed to have solved the same problem at least 4 years ago.
    The guy has a reputation for sometimes getting it wrong.

    (Probably because he has published flawed proofs of other well-known problems.)

    He could be right, but I wouldn't get my hopes up.

    1. Re:Seems not-unlikely to be wrong by roll_w.it · · Score: 5, Informative

      otoh, he proved the Bieberbach conjecture in 84 and has been working on this since. Perhaps this is why he posted it before it is formally published in a journal.

    2. Re:Seems not-unlikely to be wrong by mrthoughtful · · Score: 5, Informative

      Well, he is reliably credited with solving the Bieberbach conjecture - the guy isn't a complete nut.

      However, a quick scan suggests that if his proof is indeed verified, it won't do what a lot of people want it to do: Quote from the article: "The proof of the Riemann hypothesis verifies a positivity condition only for those Dirichlet zeta functions which are associated with nonprincipal real characters. The classical zeta function does not satisfy a positivity condition since the condition is not compatible with the singularity of the function. But a weaker condition is satisfied which has the desired implication for zeros."

      So I may be wrong, but it looks like he may have found ground on a restricted interpretation of the GRH (or Generalized Riemann Hypothesis), -ie concerning Dirichlet zeta functions which are associated with nonprincipal real characters only.

      As for consequences, If GRH is indeed true, then e.g. the Miller-Rabin primality test is guaranteed to run in polynomial time.

      --
      This comment was written with the intention to opt out of advertising.
  8. Re:Impact on crypto? by Mahrtian · · Score: 3, Informative

    The magic of PKI occurs through the use of extremely long prime numbers, called keys. Two keys are involved - a private key, which only you have access to, and a public key, which can be accessed by anyone. The two keys work together, so a message scrambled with the private key can only be unscrambled with the public key and vice versa. The more digits in these keys, the more secure the process. --Public-key encryption for dummies

    Not the best explanation, I prefer this

    --

    --
  9. The Problem by Anonymous Coward · · Score: 5, Informative

    The problem is simple enough to understand, assuming you know some math basics. As most of you know, any function f(X) where f(Xo)=0 is said to have a zero at Xo. For functions of complex numbers f(z) where z=x+iy and x,y are real numbers, you obviously have the function taking on different values for every x and y, so the zeros can be anywhere on the x-y plane. For the zeta function, "trivial zeros" occur at the negative even integers (z=-2+i0,-4+i0,...) and also at points on the line x=1/2 (i.e 1/2 +iy for certain y).The Riemann Hypothesis says that all zeros that aren't negative even integers lie on this line.

    Most of you have who have taken basic calculus courses have probably seen a simplified definition of the zeta function for real intergers greater than 1. when z=n, a natural number, the zeta function reduces to the infinite series Zeta(n)= SUM (k=1-->inf) 1/k^n

  10. Re:What are the consequences for cryptography? by Anonymous Coward · · Score: 5, Informative

    Nope, probably not.
    Most mathematicians felt that the Riemann Hypothesis was true so that this view has been taken into consideration into mathematics for a long time. Perhaps if he developed some new methods for playing with numbers in the proof, but it doesn't seem like it to me.
    There's a ton of math papers that begin with "Assume the extended riemann hypothesis...".

    At least that's my guess.

  11. Re:Impact on crypto? by cperciva · · Score: 3, Informative

    One of the fallout corollaries from a proof of the Riemann hypothesis is that there exists a simple algorithm for factorization (read: p-time).

    No. GRH implies that isprime() is in P (by bounding the cost of a strong pseudoprime test); but we already knew that, thanks to AKS.

  12. Re:Impact on crypto? by susano_otter · · Score: 5, Informative
    This theorem is a theory of how prime numbers are distributed...

    It's actually a little more complex than that.

    Riemann was investigating the distribution of prime numbers. Along the way he devised (discovered?) the Zeta Function, which describes with considerable accuracy the distribution of prime numbers. While working with the Zeta Function, he discovered an interesting property: It appeared that all the non-trivial zeroes of the function had a real part of one-half. However, since this property of the function was not related to the prime-distribution work he was doing, he did not bother to prove this apparent property, which came to be known as the "Riemann Hypothesis" (presumably, once it is proven it will be known as the Riemann Theorem, or some such).

    Thus, the Riemann Hypothesis is in fact tangential to (and possibly unrelated to) the distribution of prime numbers. Riemann's notes on the Zeta Function, regarding his work on prime distribution, are pretty explicit about this.

    --

    Any sufficiently well-organized community is indistinguishable from Government.

  13. Re:Impact on crypto? by cperciva · · Score: 4, Informative

    does it's proof have any impact on crypto?

    No. Almost all mathematicians have assumed for years that GRH is true anyway; proving it would mean that all those footnotes ([1] Under the assumption of the Riemann Hypothesis) could be removed, but that's the only practical effect it would have.

  14. Usage of "apology" by mhotas · · Score: 3, Informative

    This usage of "apology" is fashionable in math circles; a prime example is the title of G. H. Hardy's memoir : A Mathematician's Apology.

  15. actual paper by Anonymous Coward · · Score: 5, Informative

    The 23 page "apology" is not the actual purported proof, contrary to what the article and press release say. The actual proof is the manuscript "Riemann zeta functions", the third link on de Branges' home page, which weighs in at 124 pages!

    So if his "proof" isn't obviously wrong, it'll likely take quite a while for the experts to verify.

  16. Re:The Reimann Hypothesis by Phragmen-Lindelof · · Score: 3, Informative

    I think you are going a bit overboard here. The Riemann hypothesis is the greatest open problem in mathematics right now and solving it would be HUGE :-). However, famous open problems usually do not advance mathematics that much and I suspect that a proof of the Riemann hypothesis would not introduce new techniques which would have wide (or even slightly wide) use in math. Look at some of the Fields metal papers (e.g. restricted Burnside problem - Zelmanov - 1994 metal) and tell me how they changed mathematics.
    For influences on math, consider Dirac (crazy British scientist who predicted to existence of the positron) whose ideas led L. Schwartz, L. to write "Théorie des distributions. Tome I,II"; distribution theory has had a huge influence on analysis.

  17. "Apology" an old paper? by jim3e8 · · Score: 3, Informative

    The 23-page "Apology" referred to in the press release is also apparently mentioned in this 1996 Usenet post. So is there a new proof? No one seems to know yet.

  18. Cool background material by Anonymous Coward · · Score: 5, Informative

    A cool overview of why this is such an interesting hypothesis.

    If nothing else check out the animation.

    mind-boggling

  19. The question, explained by SamSim · · Score: 5, Informative

    First: complex numbers, explained. You may have heard the question asked, "what is the square root of minus one?" Well, maths has an answer and we call it i. i*i = -1. If the real number line ...-4, -3, -2, -1, 0, 1, 2, 3, 4... is represented as a horizontal line, then the numbers ...-4i, -3i, -2i, -i, 0, i, 2i, 3i, 4i... can be thought of as the *vertical* axis on this diagram. The whole plane taken together is then called the complex plane. This is a two-dimensional set of numbers. Every number can be represented in the form a+bi. For real numbers, b=0.

    Right. Now the Riemann Zeta Function is a function/map (like f(x)=x^2 is a function) on the complex plane. For any number a+bi, zeta(a+bi)will be another complex number, c+di.

    Now, a zero of a function is (pretty obviously) a point a+bi where f(a+bi)=0. If f(x)=x^2 then the only zero is obviously at 0, where f(0)=0. For the Riemann Zeta Function this is more complicated. It basically has two types of zeros: the "trivial" zeroes, that occur at all negative even integers, that is, -2, -4, -6, -8... and the "nontrivial" zeroes, which are all the OTHER ones.

    As far as we know, *all* the nontrivial zeroes occur at 1/2 + bi for some b. No others have been found in a lot of looking... but are they ALL like that? The Riemann Hypothesis suggests that they are... but until today nobody has been able to prove it.

    1. Re:The question, explained by h4rm0ny · · Score: 5, Informative
      It seems like the answer (well, we'll call it "i") has been proposed before anyone has shown if can really happen.

      Great Cthulhu help me, but I'm going to try and answer this for you.

      We have natural numbers - 1,2,3, ... - and people are happy with this. It's an abstract way of representing a real property. I have five oranges, I owe you four oranges. Natural.

      And then we have Zero and once upon a time this disturbed people. You grew up with it, you're happy with it; but we can see that it was less intuitive than 1,2,3, ... because it developed so much later and the greeks managed without it for quite a long time. It's not an abstraction in the same way that these other numbers are. People used to ask questions such as, 'how can something exist and yet be nothing?' 'How can zero x zero = zero since that means you have no zero's?' Can you prove that it does mathematically, right now? *

      And yet, the discovery (or creation ;) of Zero allowed people to abstract in new ways that produced real world results. The same can be said of Negative numbers which are even less intuitive. If I give you those four oranges mentioned earlier (not bloody likely since I'm writing this before breakfast), then that leaves me with one. But suppose I owe you six oranges? We can't carry out that operation with oranges, but the operation is useful in many other areas, the most obvious is probably money. You can be overdrawn for example - that's applied negative numbers. Is there really anti-money in your account? Well, yes, why not? It's just numbers, and numbers are an abstraction, a model of something if you like. It's perfectly normal to represent some properties as negatives. Try basic Newtonian physics - two bodies moving in opposite directions towards each other. You treat the momentum of one of them as negative and the other positive which lets you work out which direction they're going in after collision.

      Now perhaps at this point, you're nodding and saying 'yes, yes, I know that already.' If so, then good, because you've just understood the principle of a complex number. It's another abstraction that can't easily be represented in the real world (nuclear physicists shut up, please). And yet, it has very real use in making calculations.

      If you're a programmer, think about how much code there is behind the scenes of a program to produce the result you want from it. Suppose that your program counts how many oranges people have given you. Maybe it has the line
      for (i=0; i &lt oranges_owed; i++) {}

      Well i isn't physically real, it doesn't represent a physical aspect of what you are modelling (the oranges) but it's useful. And in the same way, i is also useful, even if it's just part of a intellectual model.

      For a mathematician: I think therefore i is.

      The only thing remaining is to give you an example of how it is useful. Easily done - Quantum Physics. All of it. ;)

      Hope this helps, IASNAM (I Am Surprisingly Not...)


      * Proof that 0x0=0:
      0=1x0
      0=(0+1)x0
      0=0x0+1x0
      0=0x0+0
      0=0+0x0
      0=0x0
      --

      Aide-toi, le Ciel t'aidera - Jeanne D'Arc.
  20. Re:Impact on crypto? by NonSequor · · Score: 4, Informative
    Along the way he devised (discovered?) the Zeta Function, which describes with considerable accuracy the distribution of prime numbers.


    Actually, as with most things Euler was the first to study it. The zeta function is also the simplest of a class of functions that Dirichlet studied Dirichlet L-series. There is also a Generalized Riemann Hypothesis that states that no Dirichlet L-series has zero with real part greater than 1/2.

    The Riemann Hypothesis is more than tangential to the study of the distribution of primes. There is a function derived from the distribution of the primes that can be expressed in terms of the non-trivial zeros of the zeta function. The Prime Number Theorem is also equivalent to the statement that the zeta function has no zeros with real part 1. The Generalized Riemann Hypothesis implies the weak form of Goldbach's conjecture (i.e. that any odd number greater than 7 can be expressed as the sum of three odd primes).
    --
    My only political goal is to see to it that no political party achieves its goals.
  21. Re:If there's one thing I know by PeeCee · · Score: 5, Informative
    Next he'll be solving problems that are NP-Complete. We'll have to re-write all our textbooks!

    Not to spoil your joke or anything, but actually, AFAIK, NP-complete problems are perfectly solvable. The problem is how long it takes to solve them in general (a certain instance of a problem could prove easy). They cannot be solved deterministically in polynomial time (i.e., quickly).

  22. Why people haven't believed him so far by This+is+outrageous! · · Score: 5, Informative
    As others mentioned, de Branges has been claiming a proof along the same lines for years. He's hard to dismiss because he actually proved the Bieberbach conjecture -- a startling exception in the series of wrong proofs he's been famous for, before and since.

    The reasons why most specialists doubt that his approach can ever yield the result are well described in this paper from 1998:

    In this note, we shall (...) give examples showing that de Branges' positivity conditions, which imply the generalized Riemann hypothesis, are not satisfied by defining functions of reproducing kernel Hilbert spaces associated with the Riemann zeta function zeta(s)
    (i.e., despite the name, the "generalized RH" proved by de Branges actually did not include the standard RH as a special case.)
    --
    This is...

    O
    U
    T
    R
    A
    G
    E
    O
    U
    S

    !

  23. Riemann Hypothesis Interview by qpacberty · · Score: 3, Informative

    Berkeley Groks has an interview that aired today with John Derbyshire discussing the Riemann Hypothesis. He states that after talking with many mathematicians in the field, the prospects for a solution any time soon are quite low.

  24. Probably a hoax: by usmcpanzer · · Score: 4, Informative

    mathworld.wolfram.com
    Riemann Hypothesis "Proof" Much Ado About Noithing A June 8 Purdue University news release reports a proof of the Riemann Hypothesis by L. de Branges. However, both the 23-page preprint cited in the release (which is actually from 2003) and a longer preprint from 2004 on de Branges's home page seem to lack an actual proof. Furthermore, a counterexample to de Branges's approach due to Conrey and Li has been known since 1998. The media coverage therefore appears to be much ado about nothing

  25. Much ado about nothing? by Scorillo47 · · Score: 5, Informative

    The proof (or, better said, the sketch of the proof) actually starts at the end of page 21, very close to the last page. The original work is actually pretty hard to find since it is buried in so many unrelated side notes.

    Here is the general outline:
    1) At the end of page 19 he mentions that "The positivity condition which is introduced implies the Riemann hypothesis if it applies to Dirichlet zeta functions."
    2) After some introduction of the quantum gamma functions that lasts two pages, the actual proof starts at the end of page 21 with the phrase "A quantum gamma function is obtained when is nonnegative. A proof of positivity is given from properties of the Laplace transformation."
    3) The proof ends in the middle of page 23 with the a verification that W(z) is a quantum gamma function with quantum q = exp(-2*pi), obtained from a spectral theory of the shift operator.

    Overall this is just a very brief sketch of the whole proof.

    BTW, to add gas on fire, here is an exceprt from mathworld.com, which surprisingly was missed by /. until now :-)

    http://mathworld.wolfram.com

    Riemann Hypothesis "Proof" Much Ado About Noithing (sic)
    A June 8 Purdue University news release reports a proof of the Riemann Hypothesis by L. de Branges. However, both the 23-page preprint cited in the release (which is actually from 2003) and a longer preprint from 2004 on de Branges's home page seem to lack an actual proof. Furthermore, a counterexample to de Branges's approach due to Conrey and Li has been known since 1998. The media coverage therefore appears to be much ado about nothing.

    The counterexample to Brangles approach can be reached here: http://arxiv.org/abs/math.NT/9812166

    --
    Don't try to use the force. Do or do not, there is no try.
  26. Re:If there's one thing I know by smallfries · · Score: 4, Informative

    It would appear that mathworld.com agrees with you...

    ----------------

    Riemann Hypothesis "Proof" Much Ado About Noithing
    A June 8 Purdue University news release reports a proof of the Riemann Hypothesis by L. de Branges. However, both the 23-page preprint cited in the release (which is actually from 2003) and a longer preprint from 2004 on de Branges's home page seem to lack an actual proof. Furthermore, a counterexample to de Branges's approach due to Conrey and Li has been known since 1998. The media coverage therefore appears to be much ado about nothing.

    --
    Slashdot: where don knuth is an idiot because he cant grasp the awesome power of php
  27. Re:If there's one thing I know by PastaLover · · Score: 3, Informative

    NP-Complete problems are by definition problems that can't be solved in polynomial time(at least not by a Turing machine?). However, most problems that are considered NP-Complete are not mathematically proven to be so. Some are though, and the thing with NP-Complete problems is that you can always translate one NP-Complete problem to another NP-Complete problem.

    So in practice, NP-complete problems can be solved (you can solve just about anything by just trying every single solution) but for very big instances you will need several times the age of the universe etc.

    Several other posters in this thread seem to be mistaken about what the term actually means, but they were being so vague I thought I'd write this up. :)

  28. Re:If there's one thing I know by Deliberate_Bastard · · Score: 4, Informative

    The definition of NP-Complete has nothing to do with whether the problem can be solved in polynomial time.

    Not quite correct, because:

    The set NP is defined as all problems whose solutions can be verified in polynomial time

    ...is one of *two* equivalent definitions of the class NP. The other is:

    "Set of all problems which can be solved in polynomial time by a nondeterministic Turing Machine."

    So the question "Does P equal NP?" is also the question "Does an NTM have the same computional power(in time) as a TM, or does it have more?" (It is already known that as far as decidiblity is concerned, TMs and NTM are equivalent.)

    Because of the way the proofs have been constructed, if you can solve any of the NP-Complete problems in polynomial time, you can solve all of them in polynomial time.

    Some more detail here:

    An NP-hard problem is a problem to which any problem in NP can be reduced in polynomial time.

    (Essentially, it can be used as a subroutine for any NP problem, with only a polynomial number of calls. Thus a solution to it is a solution to any problem in NP.)

    An NP-complete problem is one that is:

    1. NP-hard
    2. In the set NP.

    Thus if a polynomial-time solution exists to an NP-complete problem, then P=NP, because a polynomial number of calls to a function that terminates in polynomial time is O({polynomial}*{polynomial}) = O({polynomial}) .

    Please note, however, that not all NP-hard problems are NP-complete.

    --
    NOTICE: This notice will appear at the bottom of all my slashdot posts.
  29. Re:If there's one thing I know by Stephan+Schulz · · Score: 3, Informative
    NP-Complete problems are by definition problems that can't be solved in polynomial time(at least not by a Turing machine?). However, most problems that are considered NP-Complete are not mathematically proven to be so. Some are though, and the thing with NP-Complete problems is that you can always translate one NP-Complete problem to another NP-Complete problem.
    Sorry, but you got things mixed up.

    NP problems are problems that can be solved by a Nondeterministic Turing machine in Polynomial time. NP-Complete problems are the class of "hardest" problems in NP. All the usual suspects (Traveling Salesman, 3-SAT, SAT, ...) are proven to be in NPC.

    We know that we can solve NPC problems in exponential time (as we can simulate a non-deterministic Turing maschine on deterministic hardware with exponential overhead). What we do not know is if there is any smarter way. That is the P=?=NP question.

    --

    Stephan

  30. Well, not exactly by mrgeometry · · Score: 3, Informative

    Sorry, but...

    It is not proved; he is not at the top of his field; this "paper" will be quickly forgotten among professional mathematicians; and I doubt any professional mathematician is going over the proof with any sort of comb.

    L. de Branges first achieved fame for proving the Bieberbach conjecture. His proof went through strange and abstract methods. He went on the road to present his proof at various seminars in France, Russia, etc; IIRC a bunch of Russian students got very excited and basically rewrote his proof. Their new proof was much shorter and avoided the use of strange methods. Nowadays, their proof is remembered and his is not, but the proof still bears his name, since after all he was the first to come up with *some* kind of proof, and their proof did more or less come out of his.

    So he deserves credit for that, and it was quite an achievement to prove the Bieberbach conjecture. But even then he was using unwieldy proofs with unnecessarily abstract methods.

    For many years he has been claiming to have a proof of the Riemann Hypothesis. Professional mathematicians stopped listening a long time ago.

    This guy is washed-up.

    I whole-heartedly agree that this short article is hilarious, but I would like to add the adjective condescending. What kind of asshole apologizes for solving a problem? Does he think he lives on some higher plane, and therefore must take direct, personal responsibility for every aspect of our lives?

    Look at how G. Perelman submitted his ideas on proving the Poincare conjecture just a little while ago. He didn't waste anyone's time by rehashing the already-available history of the problem or its wider context in mathematics. Nor did he apologize for having an idea. Rather, he submitted his ideas for consideration, with the full awareness that there may have been a mistake. .... Now, this is where I admit that I do not really understand that area of math, and have not been closely following the status of (alliteration alert) Perelman's proposed proof. Still, Perelman is a real mathematician, and even if the proof is (was?) wrong, it has real ideas of value in it.

    de Branges is so full of crap, it makes me sick.

    zach

  31. Re:If there's one thing I know by saforrest · · Score: 3, Informative

    NP-Complete problems are by definition problems that can't be solved in polynomial time(at least not by a Turing machine?).

    No, you're wrong. NP problems are, by definition, problems that can be solved in polynomial time by a nondeterministic Turing machine .

    Essentially this means that a Turing machine could solve the problem in polynomial time, if it had some magic 'oracle' which instructed it on the right computational path to follow for a given input.

    Obviously there are problems out there that would require exponential time for even a nondeterministic Turing machine to solve. An example from the Wikipedia link I provided is finding the best move in a chess or Go game.

    Such problems are not in NP, and proving P=NP would not suddenly give us algorithms for solving these problems deterministically in polynomial time.

    However, most problems that are considered NP-Complete are not mathematically proven to be so.

    What? Sorry, if there ain't a proof, it ain't NP-complete. There are a lot of problems that are described as "believed to be NP-complete", but that's different.