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Math The Almighty Buck

Banker Offers $1M To Solve Beal Conjecture 216

Posted by timothy
from the it's-no-fermat's-but-hey dept.
oxide7 writes "A Texas banker with a knack for numbers has offered $1 million for anyone who can solve a complex math equation that has stumped mathematicians since the 1980s. The Beal Conjecture states that the only solutions to the equation A^x + B^y = C^z, when A, B and C are positive integers, and x, y and z are positive integers greater than two, are those in which A, B and C have a common factor. Like most number theories, it's "easy to say but extremely difficult to prove.""
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Banker Offers $1M To Solve Beal Conjecture

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  • Fermat? (Score:2, Informative)

    by SleazyRidr (1563649)

    Is that not Fermat's last theorem? Wasn't that proved a few years ago? I'm a math enthusiast rather than an actual mathematician, so I imagine someone here can correct my incorrect assumptions.

    • Re:Fermat? (Score:5, Informative)

      by Samantha Wright (1324923) on Thursday June 06, 2013 @12:57PM (#43927007) Homepage Journal
      Fermat's last theorem requires x = y = z, and argues that there is no answer. This is a generalization.
      • Re: (Score:3, Informative)

        by Garridan (597129)

        Worse than 'a generalization': if this conjecture is true, FLT is a trivial consequence. That's a clue that Beal's conjecture is likely significantly harder than Fermat's.

        • by Joce640k (829181)

          Worse than 'a generalization': if this conjecture is true, FLT is a trivial consequence. That's a clue that Beal's conjecture is likely significantly harder than Fermat's.

          ...or easier. Beal removes a constraint from Fermat's theorum.

          All you need to do is find one exception and you've won. Removing a constraint makes that search easier.

          • I'm no mathematician but I could write a program to brute-force this puppy. Worth trying? :-P

          • All you need to do is find one exception and you've won. Removing a constraint makes that search easier.

            If you want to try and find a counterexample, you'll have to start with values greater than 1000 because the conjecture has already been verified by exhaustion for all values of all six variables and exponents up to 1000 which is a fancy way of saying that brute force computation failed to find an easy counter example. There may still be one out there, but it might be so large that it would be computationally impractical to find it. On the other hand, the conjecture may be true in which case a brute force m

        • Re:Fermat? (Score:5, Funny)

          by zrbyte (1666979) on Thursday June 06, 2013 @01:34PM (#43927405)

          Huh. For a second there my brain thought: "FTL? Faster Than Light (travel)? WTF!"

        • by suutar (1860506)
          I'm afraid I don't see how FLT would be a consequence. Beal's says that A, B, and C have to have a common factor for there to be x, y, and z > 2 that work; FLT says that if x == y == z even a common factor won't make it work. Can you describe (roughly :) how FLT would derive from Beal's?
          • Re:Fermat? (Score:5, Informative)

            by Anonymous Coward on Thursday June 06, 2013 @01:49PM (#43927585)

            Assume Beal's conjecture and you have a minimal counterexample to FLT where A^x + B^x = C^x. Then A, B, C have a common prime factor p, so (A/p)^x + (B/p)^y = (C/p)^z is a smaller counterexample to FLT, which contradicts our minimality assumption.

        • Maybe Fermat's actual "proof in the margin" was using Beal's conjecture, and FLT was just what was more aesthetic to him.

          Remember that the special case is not always solved before the more general one. Poincare was a famous unsolved problem, but Perelman actually solved the Thurston Geometrization Conjecture which was significantly more general. Poincare was just a special case of Thurston.
          • by semi-extrinsic (1997002) <asmunder&stud,ntnu,no> on Thursday June 06, 2013 @04:22PM (#43929431)
            We have a fairly good hunch what Fermat's actual "proof in the margin" was. I can't remember how it goes, but it falls apart because rings Z^n with n>13 are no longer Unique Factorization Domains (UFD: a ring where all numbers have a single unique prime factorization) (or something like that). The concept of something not being a UFD was unheard of at the time of Fermat. Disclaimer: it's a few years since I did Algebra, so there may be errors in this post.
        • That isn't immediately intuitive, since Beal would allow solutions with a common factor, so for everyone else who got confused by that:

          Suppose (a*k)^n+(b*k)^n=(c*k)^n for some k>1, where k is the greatest common divisor of ak,bk,ck. Then (a^n + b^n)* k^n = c^n * k^n, and a solution a^n+b^n=c^n without a common factor follows, contradicting the conjecture.

        • by arobatino (46791)

          Worse than 'a generalization': if this conjecture is true, FLT is a trivial consequence. That's a clue that Beal's conjecture is likely significantly harder than Fermat's.

          If a claim makes unnecessary assumptions, then the simplest proof for it doesn't necessarily make use of those assumptions. They may simply cause people to waste time trying to find proofs that incorporate them.

    • Re:Fermat? (Score:5, Informative)

      by Kjella (173770) on Thursday June 06, 2013 @01:00PM (#43927043) Homepage

      No, Fermat's last theorem says a^n + b^n = c^n for n > 2 has no solutions. Here's it says a^x + b^y = c^z for x,y,z > 2 only have solutions when they have a common factor. Example: 3^3 + 6^3 = 3^5.

      • by gr8_phk (621180)
        Fermat's last theorem is easy to prove if one takes this conjecture as true. I don't recall the logic ATM but my limited number theory was able to figure it out all by itself.
        • by TC Wilcox (954812)

          Fermat's last theorem is easy to prove if one takes this conjecture as true. I have discovered a truly marvelous proof of this, which this comment box is too small to contain.

          Fixed that for you.

    • Similar, but Fermat's Last Theorem states that no integers x,y,z can satisfy the equation x^n + y^n = z^n where n is an integer > 2
      This one allows every exponent to vary separately as well as the bases and is trying to make a guarantee rather than proving an impossibility (and, contrary to popular belief, proving a negative is not only possible but trivial to prove )

    • Re: (Score:2, Informative)

      by Anonymous Coward

      For a proof or counterexample published in a refereed journal, Beal initially offered a prize of US $5,000 in 1997, rising to $50,000 over ten years[1], but has since raised it to US $1,000,000.

      Slashdot, once upon a time, you put more relevant information in posts.

  • by kanweg (771128) on Thursday June 06, 2013 @12:54PM (#43926965)

    Sorry, but I promise you that the solution was very elegant.

    Bert

  • by Viol8 (599362) on Thursday June 06, 2013 @12:55PM (#43926975)

    ... along with some postulated constraints and ask people to prove them? Whats so special about this one - does it have some mathematical relevance?

    • Re: (Score:3, Interesting)

      by Anonymous Coward

      That's essentially what Carl Friedrich Gauss said when he was challenged to prove Fermat's Last Theorem. Something on the lines of: "I have no real interest in such endeavors since I could easily put forward a multitude of propositions which one could neither prove nor disprove."

      • That's essentially what Carl Friedrich Gauss said when he was challenged to prove Fermat's Last Theorem. Something on the lines of: "I have no real interest in such endeavors since I could easily put forward a multitude of propositions which one could neither prove nor disprove."

        Did Gauss "put forward a multitude of propositions which one could neither prove nor disprove"?

        Especially now that we have very fast computers, it seems like the false conjectures would be quickly disproven, and the true ones might take a bit longer. If we eliminate needlessly complicated conjectures, are we left with only "interesting" ones?

        • by sconeu (64226)

          it seems like the false conjectures would be quickly disproven, and the true ones might take a bit longer

          Yeah, I think it would take *quite* a bit longer to disprove true conjectures.

    • by MasseKid (1294554)
      Sure you could. In this case he's offering cash for people to solve it. If you are offering a million dollars for simple math problems, you are going to go broke very, very quickly.
    • by girlintraining (1395911) on Thursday June 06, 2013 @01:11PM (#43927171)

      Whats so special about this one - does it have some mathematical relevance?

      Yes, it's relevance is that mathematicians don't like empirical evidence that a statement is only 99.9999% accurate; They demand 100%. And in mathematics, you can get 100%.

      And just like prime numbers, fermat's last theorem, etc., an enhanced understanding of the relationships laid out by certain formulas can, and often does, lead to an enhanced understanding of the universe -- which for some strange reason, seems to have the quality of being well-described, if not completely described, by the body of knowledge known as mathematics. And by understanding the universe better, we understand ourselves, and can make our lives easier. Creating most of our modern technology requires an understanding of mathematics -- so better math means better technology.

      Relevant enough for you, or do I need to resort to a beer analogy? :)

      • Relevant xkcd [xkcd.com]
      • Only in so long as we don't need to provide a shopping anology for any of your questions.
      • by loufoque (1400831)

        You can't always prove something that is true.
        See Gödel's incompleteness theorems.

    • by JoshuaZ (1134087)
      There really isn't anything that interesting. Most of the interest in Fermat's Last Theorem extended from its history, and to a large extent mathematicians cared much more about what Wiles proved that implied FLT. In this context, this looks like an extremely difficult Diophantine equation, and unlike some such equations this one has little obvious impact on other areas.
    • by siwelwerd (869956)
      Arguably the fact that it's an open question is what makes it special. But while Beal more or less came up with the question in a vacuum, it's related to other questions posed over the years. Some more details can be found here: http://www.ams.org/notices/199711/beal.pdf [ams.org]
      • The quoted article relates the Beal conjecture to 'abc' (which is the deepest and most remarkable of them), which might have been proven by Shin Mochizuki. Nobody really knows yet because he appeared to invent a large new branch of mathematics along the way.

        This could be a way to pay Shin bunch of money.

    • It certainly has educational relevance, among other things. Anyone who thinks long and hard about an interesting (to them) mathematics problem is not wasting anyone's time. A proof, or even and attempted proof, might have relevance in fields outside of abstract mathematics.
  • Not just "a banker" (Score:5, Informative)

    by deego (587575) on Thursday June 06, 2013 @01:07PM (#43927129)

    Not just "a banker".. .. but the very person who proposed the conjecture.

    • by SammyIAm (1348279)
      Wha-- why would that not be mentioned in the summary?!
      • by gnapster (1401889)
        I have no idea; it's in the second paragraph of TFA. What's more, this isn't the first prize he's offered: he's just upping this ante to $1M.
  • by sl4shd0rk (755837) on Thursday June 06, 2013 @01:08PM (#43927135)

    Give a man a gun, he can rob a bank
    Give a banker an algorithm, he can rob the world.

  • I hadn't heard of the Beal Conjecture, so the first thought that ran through my brain was the Beale Cipher [wikipedia.org], a set of coded 19th century messages supposedly leading to millions of dollars in treasure.

    Alas, they appear to be a hoax, and the Beal Conjecture is probably more solvable.

  • by Tablizer (95088) on Thursday June 06, 2013 @01:20PM (#43927271) Homepage Journal

    Oh great, yet another bank that wants a bealout.

  • by gstoddart (321705) on Thursday June 06, 2013 @01:23PM (#43927303) Homepage

    So, being quite cynical about such things, in what way would a proof of this conjecture allow him to make more money?

    Philanthropy and advancing science are good, but my first thoughts is that if someone can prove this he stands to make massive amounts of money.

    • by Captain Spam (66120) on Thursday June 06, 2013 @01:39PM (#43927485) Homepage

      So, being quite cynical about such things, in what way would a proof of this conjecture allow him to make more money?

      Philanthropy and advancing science are good, but my first thoughts is that if someone can prove this he stands to make massive amounts of money.

      You know the old jokes about rich people paying bums on the street to fight for their own amusement? Well, extend that to mathematicians.

    • by suutar (1860506)
      maybe he's a closet cryptographer and if this is proven his new cipher scheme will be proven unbreakable and he can sell software for zillions.
    • by scheme (19778)
      Well, the banker is the Beal in the name of the conjecture so he's directly involved since he stumbled across this while looking at solutions to a more general version of the equation in fermat's last theorem. He's rich, has extra cash, and probably is curious to know either way about his conjecture.
    • by siwelwerd (869956)
      He apparently does mathematics in his spare time and has spent a good amount of time working on this question (that's why it's named after him). The prize has been offered for several years actually, the news is that it has increased to $1M. The origin of the prize, along with some more details, here: http://www.ams.org/notices/199711/beal.pdf [ams.org]
  • by mederbil (1756400) on Thursday June 06, 2013 @01:27PM (#43927335)

    Given the proof of FLT required proving an isomorphism between topology and number theory (among other things), I wonder if these problems are so similar that this will again be the case. It's interesting how the number system is related to geometry.

  • or maybe (Score:2, Insightful)

    by Anonymous Coward

    Why don't they just put up a simple problem and pay out $1 each for up to 1 million students who can solve it.

  • Beal (Score:3, Informative)

    by bbartlog (1853116) on Thursday June 06, 2013 @02:06PM (#43927781)
    This is the same Beal who founded Beal Aerospace. Also the same Beal who challenged the world's best professional poker players to the highest one-on-one Texas Hold'Em games ever played ($100,000/$200,000 IIRC). Also, this isn't an equation to be 'solved'. It's a conjecture to be proved (or disproved).
  • This site made it clear to me than anywhere else .. since it had code! Yesss, preciousssssss, code!

    http://www.norvig.com/beal.html [norvig.com]

    Python, not perl, but that's okay, close enough. It even had a working algorithm, love it!

    Except even a simple toad can see that the numbers are going to get a wee bit big (even if they are all integers). Sure wish I hadn't lost the source to Toad's Infinite Math [tm], hacked back in the 80's. It was in Turbo Pascal, but let you do all the common math things (to include power

  • TFS has left out the rather pertinent fact that the "Texas banker" is the same Beal who came up with the conjecture in the first place.
  • x = y = z = 2, and A^2 + B^2 = C^2 is true for A=3, B=4, C=5. Common factor between 3,4 and 5 (other than 1)? None.

  • BEAL'S CONJECTURE: If Ax + By = Cz, where A, B, C, x, y and z are positive integers and x, y and z are all greater than 2, then A, B and C must have a common prime factor

    Fermat's Last Theorem is a subset problem, where x=y=z, Since FLT has been proven (that no such solution exists), then we can say that:

    • - x, y, and z must all be greater than 2 (as specified in the original conjecture)
    • - and x, y, and z cannot all be equal (because then FLT would be wrong)
  • If we assume the Beal Conjecture is true it implies FLT is true. FLT would require A^n+B^n = C^n with n>2. Beal says that in such a case A,B, and C must have a common factor (F). We could then divide through by F^n and get a smaller triple a,b,c that also satisfy the equation. Under Beal this could be repeated until the only common factor is 1 and then we'd have 1^n+1^n=1^n which is impossible.

    So if you can prove the Beal conjecture, you also get a proof of FLT by infinite descent. This would make Ferm
    • In the above, the assumption of FLT is that the exponents are the same. This allows the division of the equation by F^n which reduces the size of the numbers involved. The Beal Conjecture allows the exponents to be different so this division is not always possible and this same method doesn't prove Beal. Just wanted to clarify that point.

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