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Grok Goldbach, Grab Gold

Posted by timothy on from the proof-is-worth-a-million-dollars dept.

Caseman writes, "Are you a closet mathematician who wants to come out? British publisher Tony Faber is offering a cool million bucks to the first would-be math head to prove the infamous Goldbach conjecture. Yeah, the one about every even number being the sum of two primes. Impossible, you say? Remember Fermat's Last Theorem? It stood unproven for 350 years until a recent effort yielded a proof. Read The London Times article about the challenge. "

5 of 249 comments (clear)

  1. It doesn't make sense to offer prizes for proofs. by Paul+Crowley · · Score: 4

    Lots of work has been done that could be useful to the GC, just the same was as done for FLT. The winnings will go not to the person who does the most work towards it, but whoever supplies the piece of the puzzle that happens to be last.

    Mathematics is a great cooperative venture. It wouldn't be easy to identify one mathematician who did the largest share of the work even if they tried to do it that way, and if they did it would probably be Gauss or Poincare or some other dead heavyweight genius.

    This might encourage work on the GC, but it also might discourage publication of such work, because the mathematicians haven't quite finished the proof.
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  2. I know the proof. by Masem · · Score: 5
    I have to proof to the Goldbach conjections.

    Unfortunately, the slashdot comment submission box is too small to post the whole proof. I shall leave the proof to the reader.

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  3. Not to discourage anyone but.. by mav[LAG] · · Score: 5
    ..saying that a "recent effort yielded a proof to Fermat's Last Theorem" is a bit of an understatement. Andrew Wiles' proof of FLT runs into the hundreds of pages and makes use of vast tracts of number theory, some dating back hundreds of years. It can probably only be understood in full by a handful of top mathematicians. It took him something like seven years of concentrated work with no guarantee of success and included one horrible non-proof which he was able to fix to produce the Real Thing some months later.

    Yes, it is possible that the Goldbach conjecture may be proved or disproved by a lay person but it's far more likely to be solved by someone with a sound grasp of lots of number theory who is already studying the properties of numbers for a living.

    BTW I have a truly marvellous proof of the Goldbach conjecture but there is no margin on the submit form to write it in :)

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  4. Probability ... by sela · · Score: 5

    Disclaimer: long boring math stuff with a conclusion that may be interesting for some at the end ...

    An interesting question is: assuming prime numbers are "randomly" distributed, what is the probability that an even number won't be the sum of two odd numbers?

    If the probability is rather high, yet we did not find any such even number, it can be an indication there may be some "meat" in goldbach theorem. On the other hand, if such probability is exremely low, than the verification of goldbach's theorem up to 4e+15 means nothing ...

    Well, according to Gauss' estimate for prime number density around sufficiently large n, the density is 1/log(n). To find pi(n), which is the number of prime numbers up to n, we need to integrate this, but pi(n)=n/log(n), so we can make life simpler and over-estimate the probability by taking the smaller number.

    Now, given the number of prime numbers up to n, pi(n), if we choose x to be a prime number smaller than n, than the probability its pair: (n-x) is a prime number is: 2*pi(n)/n. (I multiplied by 2 since itcannot be possibly an even number, so no need to consider them).
    The probability that it is not a prime number is therefore (1-2*pi(n)/n).

    Since there are pi(n) different prime numbers to choose from, the probability that an even number is not the sum of two prime numbers is:
    (1-2*pi(n)/n)^pi(n).

    So what is the probability for numbers around 1e+15? If we use the estimate for pi(n), we get:
    pi(n)=2e+13
    (1-2*pi(n)/n) = 0.959

    Taking the last number by the power of pi(n), the result is that the probability is around:
    10^-(4e11)

    That is - an extremely small probability!!!

    Conclusion:
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    If Goldbach's theorem is false, the probability that an even number won't be the sum of two prime numbers is so small that we may need to check around 10^(10^11) numbers to find a counter example. This number is so big no computer could possibly handle it now or any time soon ...

    In other words: Don't bother to find a counter-example.

    Sela

  5. Didn't anyone take a college math course? by Junks+Jerzey · · Score: 5

    It's stunning how many people want to approach this as a programming problem; i.e. a fast way of checking lots of numbers. We're talking about a *proof*. Guessing and checking is missing the point.