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Math The Almighty Buck

Banker Offers $1M To Solve Beal Conjecture 216

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 ) on Thursday June 06, 2013 @12:54PM (#43926955)

    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: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.

  • Re:Fermat? (Score:2, Informative)

    by Anonymous Coward on Thursday June 06, 2013 @01:05PM (#43927109)

    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.

  • Comment removed (Score:5, Informative)

    by account_deleted ( 4530225 ) on Thursday June 06, 2013 @01:07PM (#43927129)
    Comment removed based on user account deletion
  • 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? :)

  • Re:Fermat? (Score:3, Informative)

    by Garridan ( 597129 ) on Thursday June 06, 2013 @01:18PM (#43927241)

    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.

  • 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.

  • by Anonymous Coward on Thursday June 06, 2013 @02:02PM (#43927727)

    Given that the person who came up with conjecture and the banker offering the $1M are the same Andy Beal, this particular case isn't an example of some rich fucker trying to steal credit from an actual mathematician. Of course, "only the people who solve the problem get credit" has never historically been the rule --- most conjectures still stay named after the person who first publicly conjectured them, even if it took someone else decades later to actually prove the statement. For example, we still call it "Fermat's Last Theorem," not the "Wiles-Ribet Theorem." Much the same occurs in physics: things usually get named after the theorist who proposes they might be possible, instead of the experimentalists who prove so.

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

    by draconx ( 1643235 ) on Thursday June 06, 2013 @04:22PM (#43929429) Homepage

    Obviously nobody has found an exception to disprove it yet. The dude wouldn't be offering a pile of money if he were just looking to disprove it...he would just funnel the money into some supercomputer time to step through an absurd amount of integers until he comes up with an exception.

    The set of integers to test is big. Really big. You just won't believe how vastly, hugely, mindbogglingly big it is. I mean, you may think that Graham's number [wikipedia.org] is a lot, but that's just peanuts compared to the integers.

    If a conjecture could be disproven by simply throwing computational resources at the problem, chances are that it's not particularly interesting. Many open problems in number theory have known lower bounds well above anything that could possibly be tested by a computer. For example, there is no odd perfect number [wikipedia.org] less than 10**1500.

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