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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.""
Sorry, but I promise you that the solution was very elegant.
Bert
... along with some postulated constraints and ask people to prove them? Whats so special about this one - does it have some mathematical relevance?
Fermat's last theorem requires x = y = z, and argues that there is no answer. This is a generalization.
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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.
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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.
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Give a man a gun, he can rob a bank
Give a banker an algorithm, he can rob the world.
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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.
Oh great, yet another bank that wants a bealout.
Table-ized A.I.
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.
Lost at C:>. Found at C.
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.
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.
No sig today...
Huh. For a second there my brain thought: "FTL? Faster Than Light (travel)? WTF!"
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.
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.
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).
Wha-- why would that not be mentioned in the summary?!
Huh. For a second there my brain thought: "FTL? Faster Than Light (travel)? WTF!"
WTF? World Taekwondo Federation?
Huh. For a second there my brain thought: "FTL? Faster Than Light (travel)? WTF!"
WTF? What The F***?
FTFY, FTW!
Dark Reflection
I'm no mathematician but I could write a program to brute-force this puppy. Worth trying? :-P
"When information is power, privacy is freedom" - Jah-Wren Ryel
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.
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.
for i in `facebook friends "=bday" 2>/dev/null | cut -d " " -f 3-`; do facebook wallpost $i "Happy birthday!"; done
The price is more interesting.
Brute forcing to find a contradiction is no mental challenge.
Crafting a nice formal proof is.
Cost free eBook I read (by iBook/Kobo/Amazon/ObookO/Gutenberg etc.): "The Green Odyssey" by Philip Jose Farmer.
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.
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 method cannot prove it because there's an infinite number of cases to check.
Ah, the integers are nothing. You should have seen the real interval I had on the hook last week. I tried bring it onboard using a sieve I borrowed from Eratosthenes, but the sieve was not nearly large enough. I'm telling you, the boat nearly keeled over!