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Banker Offers $1M To Solve Beal Conjecture

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.""

4 of 216 comments (clear)

  1. Comment removed by account_deleted · · Score: 5, Informative

    Comment removed based on user account deletion

  2. Re:Couldn't you just make up any old equation... by girlintraining · · Score: 5, Informative

    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? :)

    --
    #fuckbeta #iamslashdot #dicemustdie
  3. Re:Fermat? by Anonymous Coward · · Score: 5, Informative

    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.

  4. Re:Fermat? by draconx · · Score: 5, Informative

    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 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 less than 10**1500.