Banker Offers $1M To Solve Beal Conjecture

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Banker Offers $1M To Solve Beal Conjecture

Posted by timothy on Thursday June 6, 2013 @04:51AM 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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Fermat? by SleazyRidr · 2013-06-06 04:54 · Score: 1, Informative

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.

--
Is 1563649 a prime number?
1. Re:Fermat? by Samantha+Wright · 2013-06-06 04:57 · Score: 4, Informative
  
  Fermat's last theorem requires x = y = z, and argues that there is no answer. This is a generalization.
  
  --
  Bio questions? Ask me to start a Q&A journal. Computer analogies available for most topics!
2. Re:Fermat? by Kjella · 2013-06-06 05:00 · Score: 4, Informative
  
  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.
  
  --
  Live today, because you never know what tomorrow brings
3. Re:Fermat? by doublebackslash · 2013-06-06 05:04 · Score: 1
  
  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 )
  
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  md5sum /boot/vmlinuz
  d41d8cd98f00b204e9800998ecf8427e /boot/vmlinuz
4. Re:Fermat? by Anonymous Coward · 2013-06-06 05:05 · Score: 2, Informative
  
  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.
5. Re:Fermat? by Garridan · 2013-06-06 05:18 · Score: 3, Informative
  
  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.
6. Re:Fermat? by Joce640k · 2013-06-06 05:31 · Score: 2
  
  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...
7. Re:Fermat? by zrbyte · 2013-06-06 05:34 · Score: 4, Funny
  
  Huh. For a second there my brain thought: "FTL? Faster Than Light (travel)? WTF!"
8. Re:Fermat? by suutar · 2013-06-06 05:35 · Score: 1
  
  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?
9. Re:Fermat? by laejoh · 2013-06-06 05:44 · Score: 1
  
  By a small margin, there is no answer!
10. Re:Fermat? by Anonymous Coward · 2013-06-06 05:49 · 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.
11. Re:Fermat? by Steve_Ussler · 2013-06-06 05:59 · Score: 1
  
  the two are very different...
12. Re:Fermat? by formfeed · 2013-06-06 06:15 · Score: 5, Funny
  
  Huh. For a second there my brain thought: "FTL? Faster Than Light (travel)? WTF!"
  WTF? World Taekwondo Federation?
13. Re:Fermat? by OakDragon · 2013-06-06 06:28 · Score: 2
  
  Huh. For a second there my brain thought: "FTL? Faster Than Light (travel)? WTF!"
  WTF? What The F***?
  FTFY, FTW!
  
  --
  Dark Reflection
14. Re:Fermat? by Garridan · 2013-06-06 07:03 · Score: 1
  
  Sorry - I meant harder to prove. It's certainly harder to prove a non-theorem.
15. Re:Fermat? by ottothecow · 2013-06-06 07:15 · Score: 1
  
  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. If your goal was to disprove, I would guess that this would be a better use of the money than offering a prize (unless you think you can get more computer cycles out of a group of people competing for a prize than you could get just by buying them).
  So far, nobody has found an exception which means that it might be provable. To find a proof, you need actual people thinking about the question, hence the promise of a pile of cash.
  
  --
  Bottles.
16. Re:Fermat? by draconx · 2013-06-06 08:22 · 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.
17. Re:Fermat? by angel'o'sphere · 2013-06-06 09:13 · Score: 2
  
  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.
18. Re:Fermat? by gr8_phk · 2013-06-06 09:22 · Score: 1
  
  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.
19. Re:Fermat? by gr8_phk · 2013-06-06 09:23 · Score: 1
  
  the two are very different...
  Not really, this conjecture if true implies FLT fairly easily.
20. Re:Fermat? by Anonymous Coward · 2013-06-06 09:30 · Score: 1
  
  No.
  3^3 + 6^3 = 3^5
  1^3 + 2^3 != 1^5
  You can't divide out the factor like that.
21. Re:Fermat? by TC+Wilcox · 2013-06-06 09:42 · Score: 2
  
  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.
22. Re:Fermat? by Anonymous Coward · 2013-06-06 10:22 · Score: 1
  
  Let a,b,c be a minimal counter example to FLT a^n + b^n = c^n. If Beal's conjecture is true, then there is a prime p such that p divides a,b,c. (a/p)^n + (b/p)^n = (c/p)^n is then a smaller counter example of FLT, and so a,b,c is not minimal and so there cannot be a minimal counter example and FLT is true.
23. Re:Fermat? by Arancaytar · 2013-06-06 11:30 · Score: 1
  
  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.
24. Re:Fermat? by Garridan · 2013-06-06 11:52 · Score: 1
  
  Unless 'hit wikipedia' is your immediate intuition, I agree; that isn't immediately intuitive.
25. Re:Fermat? by Nivag064 · 2013-06-06 13:38 · Score: 1
  
  What you say is true WHEN x = y = z and x is a positive integer, but what if x, y, z are not all the same and greater than 2???
26. Re:Fermat? by arobatino · 2013-06-06 14:56 · Score: 1
  
  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.
27. Re:Fermat? by Garridan · 2013-06-06 17:05 · Score: 1
  
  You're the second person to point this out. Duh. I said "that's a clue", not "that's a clear and direct implication". Most often, more general statements are harder to prove.
28. Re:Fermat? by CodeBuster · 2013-06-06 19:03 · Score: 2
  
  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.
29. Re:Fermat? by sFurbo · 2013-06-06 20:04 · Score: 4, Funny
  
  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!
30. Re:Fermat? by Steve_Ussler · 2013-06-07 00:20 · Score: 1
  
  ok...like i know.
31. Re: Fermat? by countach · 2013-06-09 01:40 · Score: 1
  
  Well, exactly. You throw computations resources to find out if the conjecture is interesting. If you try brute forcing an exception and fail, the you've found something interesting.
Have solution. Alas, subject line = too small by kanweg · 2013-06-06 04:54 · Score: 5, Funny

Sorry, but I promise you that the solution was very elegant.
Bert
1. Re:Have solution. Alas, subject line = too small by Anonymous Coward · 2013-06-06 05:09 · Score: 1
  
  Now that would be worth something... Fermat's lost elegant proof to his own theorem.
  Of course, most mathematicians think he was either wrong, or was making a practical joke.
2. Re:Have solution. Alas, subject line = too small by Opportunist · 2013-06-06 05:24 · Score: 1
  
  Is it me or are mathematical practical jokes even less funny than general mathematical jokes?
  
  --
  We used to have a Bill of Rights. Now, with the rights gone, all we have left is the bill.
3. Re:Have solution. Alas, subject line = too small by vuke69 · 2013-06-06 05:29 · Score: 4, Funny
  
  I suspect you are correct, but a proof remains elusive.
  
  --
  Time is an illusion. Lunchtime doubly so. ~ Douglas Adams
4. Re:Have solution. Alas, subject line = too small by K.+S.+Kyosuke · 2013-06-06 05:36 · Score: 1
  
  That's very easy to prove; you simply apply the order-preserving homomorphism Mathematicalize(X -> X) to the partially ordered set of jokes. Given that practical jokes are often less funny than general jokes, given its order-preserving properties, it follows that practical mathematical jokes are often less funny than general mathematical jokes.
  
  --
  Ezekiel 23:20
5. Re:Have solution. Alas, subject line = too small by Minwee · 2013-06-06 05:43 · Score: 1
  
  Sorry, but the solution to Beal's Conjecture is in another castle.
6. Re:Have solution. Alas, subject line = too small by newcastlejon · 2013-06-06 05:44 · Score: 1
  
  I too have a solution. Unfortunately slashcode would die of shame if I ever tried to post it.
  
  --
  If God forks the Universe every time you roll a die, he'd better have a damned good memory.
7. Re:Have solution. Alas, subject line = too small by almitydave · 2013-06-06 05:57 · Score: 1
  
  That's not funny.
  
  --
  my, your, his/her/its, our, your, their
  I'm, you're, he's/she's/it's, we're, you're, they're
8. Re:Have solution. Alas, subject line = too small by SpaceLifeForm · 2013-06-06 07:48 · Score: 1
  
  Fermat was completely correct. This is a superset problem.
  
  --
  You are being MICROattacked, from various angles, in a SOFT manner.
9. Re:Have solution. Alas, subject line = too small by semi-extrinsic · 2013-06-06 08:39 · Score: 1
  
  Actually, it is.
  
  --
  for i in `facebook friends "=bday" 2>/dev/null | cut -d " " -f 3-`; do facebook wallpost $i "Happy birthday!"; done
Couldn't you just make up any old equation... by Viol8 · 2013-06-06 04:55 · Score: 2, Interesting

... along with some postulated constraints and ask people to prove them? Whats so special about this one - does it have some mathematical relevance?
1. Re:Couldn't you just make up any old equation... by Anonymous Coward · 2013-06-06 05:06 · Score: 3, Interesting
  
  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."
2. Re:Couldn't you just make up any old equation... by MasseKid · 2013-06-06 05:06 · Score: 2
  
  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.
3. Re:Couldn't you just make up any old equation... by girlintraining · 2013-06-06 05:11 · 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
4. Re:Couldn't you just make up any old equation... by JoshuaZ · 2013-06-06 05:21 · Score: 1
  
  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.
5. Re:Couldn't you just make up any old equation... by Maximum+Prophet · 2013-06-06 05:22 · Score: 1
  
  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?
  
  --
  All ideas^H^H^H^H^Hprocesses in this post are Patent Pending. (as well as the process of patenting all postings)
6. Re:Couldn't you just make up any old equation... by sconeu · 2013-06-06 05:28 · Score: 2
  
  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.
  
  --
  General Relativity: Space-time tells matter where to go; Matter tells space-time what shape to be.
7. Re:Couldn't you just make up any old equation... by Anonymous Coward · 2013-06-06 05:37 · Score: 1
  
  "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?"
  This isn't really true. If a computer checks a propositions to the trillions of trillions of trillions and it doesn't turn out to be false, you're not really any closer to proving the proposition true. You can't do an exhaustive search when your domain goes to infinity.
8. Re:Couldn't you just make up any old equation... by LihTox · 2013-06-06 05:39 · Score: 4, Insightful
  
  Only if x=y=z. For instance, somebody above suggested 3^3 + 6^3 = 3^5 (27+216=243). If we factor out the common 3, we get 3^2 + 2*(6^2) = 3^4 (9+72=81), which no longer has the right form because 72 is not a power of any number.
  If x=y=z, and if A^x+B^x=C^x where A,B,C had the same greatest common factor n, then you could divide all three numbers by n^x and get a new formula (A/n)^x+(B/n)^y=(C/n)^z where A/n, B/n, and C/n had no common factor, and if Beal's conjecture is true then these numbers cannot exist if x>2. Therefore A^x+B^x=C^x has no nontrivial solution for x>2, which is Fermat's last theorem.
9. Re:Couldn't you just make up any old equation... by emilper · 2013-06-06 05:42 · Score: 1
  
  you have to prove that there is no solution execept when a,b and c have a common factor
  much simpler, though maybe not easier, to try to dissaprove it ... is this a good job for OpenCL ?
10. Re:Couldn't you just make up any old equation... by aliquis · 2013-06-06 05:47 · Score: 1
  
  Relevant enough for you, or do I need to resort to a beer analogy? :)
  One million = lots of beer = solve it quickly! ... friday tomorrow!
11. Re:Couldn't you just make up any old equation... by fredrated · 2013-06-06 05:50 · Score: 1
  
  Somebody got his knickers in a twist this morning.
12. Re:Couldn't you just make up any old equation... by femtobyte · 2013-06-06 05:52 · Score: 3, Funny
  
  Beer analogy, as requested: suppose you want to be 100% certain that no one is pissing a little in your beer before you drink it. You can be 99.999% sure by ordering high-quality beer in an upscale establishment and watching the bartender fill your glass --- but you've still got that nagging fear that someone in the back room, or even the brewery, may have whizzed in the keg when no one was looking. So, instead of relying on empirical likelihoods, you go and brew your own beer, from start to finish under your watchful eye, to get that 100%-guaranteed-piss-free pint. And, ultimately, humankind's fundamental knowledge and craft of beer brewing is advanced through the initial efforts of home-brew enthusiasts.
13. Re:Couldn't you just make up any old equation... by coinreturn · 2013-06-06 05:52 · Score: 1
  
  Ah, nothing like tackling decades old math "problems"...for absofuckinglutely no reason whatsoever.
  I believe there are a million reasons to solve this old problem.
14. Re:Couldn't you just make up any old equation... by The+MAZZTer · 2013-06-06 05:57 · Score: 1
  
  Relevant xkcd
15. Re:Couldn't you just make up any old equation... by interval1066 · 2013-06-06 06:26 · Score: 1
  
  Only in so long as we don't need to provide a shopping anology for any of your questions.
  
  --
  Python: 'And then suddenly you have a language which says "we're all stuck with whatever the whiniest coder wants".'
16. Re:Couldn't you just make up any old equation... by Anonymous Coward · 2013-06-06 06:39 · Score: 1
  
  And reading comprehension is like something you don't have.
17. Re:Couldn't you just make up any old equation... by Dr.+Tom · 2013-06-06 06:41 · Score: 1
  
  Boy, and here I was just pissing my life away on this chicken and egg theory.
  That's been solved. The egg came first. Evolution is true: two not-quite chickens had sex, and the mutant offspring was a modern chicken, who of course started life as an egg.
18. Re:Couldn't you just make up any old equation... by siwelwerd · 2013-06-06 06:45 · Score: 1
  
  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
19. Re:Couldn't you just make up any old equation... by gnapster · 2013-06-06 06:45 · Score: 1
  
  No, he showed that there are always statements that are undecidable. Most statements that interest mathematicians are decidable, and 100% so.
20. Re:Couldn't you just make up any old equation... by gtbritishskull · 2013-06-06 06:51 · Score: 1
  
  You could actually factor out 3^3, which would give you 1 + 2^3 = 3^2 (1 + 8 = 9), so it would be in the right form. Not sure what point that makes. Just wanted to let you know you were wrong.
21. Re:Couldn't you just make up any old equation... by Dr.+Tom · 2013-06-06 06:52 · Score: 2
  
  Gödel supported Gauss. He proved that yes, you can come up with many statements that can't be proven. However, if a statement like Beal's conjecture can't be proven, you should be able to prove *that*. Which would also qualify for the prize money, I'd say.
  The point the previous poster was making is that math is different from science, because you can prove things in math, but in science you can only disprove things.
22. Re:Couldn't you just make up any old equation... by turbidostato · 2013-06-06 06:55 · Score: 1
  
  "Most statements that interest mathematicians are decidable, and 100% so."
  And for those truly undecidable, you take them for an axiom and done with it.
23. Re:Couldn't you just make up any old equation... by cellocgw · 2013-06-06 07:11 · Score: 1
  
  Hah. No it hasn't. Is a "chicken egg" one which produces a chicken, or one which is produced BY a chicken? On such definitions does your "solution" hang by a thread... or sticky gob of eggwhite, maybe.
  
  --
  https://app.box.com/WitthoftResume Code: https://github.com/cellocgw
24. Re:Couldn't you just make up any old equation... by tibit · 2013-06-06 07:18 · Score: 1
  
  Lest we forget that the hop farm used cow manure as fertilizer :)
  
  --
  A successful API design takes a mixture of software design and pedagogy.
25. Re:Couldn't you just make up any old equation... by SleazyRidr · 2013-06-06 07:25 · Score: 1
  
  You don't sleep, and you watch the beer at all times.
  
  --
  Is 1563649 a prime number?
26. Re:Couldn't you just make up any old equation... by femtobyte · 2013-06-06 07:30 · Score: 3, Funny
  
  Every analogy breaks down somewhere (or else it wouldn't be an analogy, but the thing analogized itself). If one wishes absolute mathematical certainty for their piss-free pint, then one must be satisfied with a mathematical pint. As thus: consider a Platonic ideal beer, symbolically represented by the word "beer." Now, imagine quaffing the beer. Theoretically, this should be satisfying and delicious. If you don't consider this exercise superior to drinking an actual beer, you may just not be cut out for pure mathematics --- consider becoming a physicist instead.
27. Re:Couldn't you just make up any old equation... by Anonymous Coward · 2013-06-06 07:37 · Score: 1
  
  An undecidable problem can not generally be proven to be undecidable. Consider Fermat's Last Theorem. Let's say that this theorem is undecidable for the sake of argument. If someone was able to prove that it was undecidable then, in effect, the truth of the theorem would have been proven. This is because if the theorem was false, then a counterexample could be found (and the problem would then not be undecidable). So, for this specific case, had the theorem been undecidable it would not have been possible to prove that.
28. Re:Couldn't you just make up any old equation... by twistedcubic · 2013-06-06 07:40 · Score: 1
  
  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.
29. Re:Couldn't you just make up any old equation... by gl4ss · 2013-06-06 07:43 · Score: 1
  
  So, instead of relying on empirical likelihoods, you go and brew your own beer, from start to finish under your watchful eye, to get that 100%-guaranteed-piss-free pint.
  The primary fermentation period is a few weeks, how do you ensure that the bartender doesn't pick your lock and piss into your fermentation vessel during this time?
  by taking drugs,doh. but then you star to wonder if someone invisible has teleported into the brew and is swimming around pissing all over.
  
  --
  world was created 5 seconds before this post as it is.
30. Re:Couldn't you just make up any old equation... by Anonymous Coward · 2013-06-06 08:04 · Score: 1
  
  Your idea doesn't work. If A is an undecidable statement, then so is (not A).
  So your system is inconsistent, being full of contradictory pairs of axioms.
31. Re:Couldn't you just make up any old equation... by loufoque · 2013-06-06 08:25 · Score: 1
  
  You can't always prove something that is true.
  See Gödel's incompleteness theorems.
32. Re:Couldn't you just make up any old equation... by LostOne · 2013-06-06 08:27 · Score: 1
  
  Just to point out that 3^3 is *not* a factor of A, B, or C in this case since A, B, and C are the bases (3, 6, and 3 in this case).
  
  --
  
  If it works in theory, try something else in practice.
33. Re:Couldn't you just make up any old equation... by Zalbik · 2013-06-06 08:57 · Score: 1
  
  You could actually factor out 3^3, which would give you 1 + 2^3 = 3^2 (1 + 8 = 9), so it would be in the right form. Not sure what point that makes. Just wanted to let you know you were wrong.
  Huh? 3^3 (27) is not a common factor of A, B and C for the formula the GP gave (3^3 + 6^3 = 3^5)
  The specific point is that just because A, B, and C have a common factor (in this case 3), it does not mean you can use the common factor to simplify the equation.
  Because of this fact, Fermat's Last Theorem follows from Beal's conjecture quite nicely.
  The GP was not wrong.
34. Re:Couldn't you just make up any old equation... by gtbritishskull · 2013-06-06 10:43 · Score: 1
  
  3^3 = 27 which is a factor of 27
  6^3 = 216 = 27 * 8 (or 3^3 * 2^3) so is a factor of 27
  3^5 = 243 = 27 * 9 (or 3^3 * 3^2) so is a factor of 27
  Either you are completely wrong or I have complete forgotten how to do math.
35. Re:Couldn't you just make up any old equation... by turbidostato · 2013-06-06 11:34 · Score: 1
  
  "If A is an undecidable statement, then so is (not A)."
  Which happens to be a very operational (modern) definition of an axiom: only one straight line goes through two points... or not.
36. Re:Couldn't you just make up any old equation... by Anonymous Coward · 2013-06-06 12:01 · Score: 1
  
  A=3, does not have a common factor of 27, B = 6, does not have a common factor of 27, C=3, does not have a common factor of 27.
  1+2^3=3^2 does not count, as the conjecture requires that the powers must be greater than 2.
37. Re:Couldn't you just make up any old equation... by LihTox · 2013-06-06 13:07 · Score: 1
  
  D'oh, good catch. But then z=2 and the conjecture specifies that all the exponents be greater than 2.
38. Re:Couldn't you just make up any old equation... by LihTox · 2013-06-06 13:12 · Score: 1
  
  gtbritshskull is correct; s/he merely said that if there were a common factor (3 in my example), then you could simplify the expression, which is true: 3^3+6^3=3^5 CAN be simplified by dividing both sides by 3^3.
  In general, you can divide both sides by [gcd(A,B,C)]^min(x,y,z) I believe.
39. Re:Couldn't you just make up any old equation... by dido · 2013-06-06 13:43 · Score: 1
  
  You don't seem to get what it means for a theorem to be undecidable. Let's take for example the Epimenides Paradox, which is a simple example: "All Cretans are liars," says Epimenides, the Cretan. Is his statement true? If it is true, then how can Epimenides, a Cretan, be telling the truth? If it is false, then Epimenides, a Cretan was telling the truth, contradicting the very statement he made! The statement is a simple example of such an undecidable statement. Showing a theorem is undecidable does not prove its truth or falsity, it is by definition neither true nor false within the confines of the axiomatic system under which it is formulated.
  
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40. Re:Couldn't you just make up any old equation... by girlintraining · 2013-06-06 15:02 · Score: 1
  
  Theoretically, this should be satisfying and delicious. If you don't consider this exercise superior to drinking an actual beer, you may just not be cut out for pure mathematics --- consider becoming a physicist instead.
  You're close. A physicist is the one who brews the beer; An engineer is the one who would drink the beer. Of course, it would only be an approximation to actual beer, in much the same way Budweiser is. And this is why a room full of mathematicians, physicists, and engineers inevitably leads to an article in the police blotter that ends with "...authorities believe liquor may have been involved."
  
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  #fuckbeta #iamslashdot #dicemustdie
41. Re:Couldn't you just make up any old equation... by Alsee · 2013-06-07 04:58 · Score: 1
  
  I've solved the physics for ascertaining that no one is pissing your beer before you drink it.
  However it only works for spherical beer in a vacuum.
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42. Re:Couldn't you just make up any old equation... by Alsee · 2013-06-07 05:31 · Score: 1
  
  if a statement like Beal's conjecture can't be proven, you should be able to prove *that*.
  Urk? How did you reach that odd conclusion?
  "X is unprovable" is itself a mathematical statement, and I see no reason to actively expect it never to fall into the category of unprovable statements.
  I would find it rather surprising if there didn't exist a recursively unprovable chain, and by that I mean a chain where all of the following were unprovable:
  X
  X1: X is unprovable
  X2: X1 is unprovable.
  X3: X2 is unprovable.
  X(n): X(n-1) is unprovable.
  If we want to get meta, it may well be possible to prove that such a chain exists. However it's easy to prove it could only be a non-constructive proof. Any mathematics successfully identifying a specific statement "X" for such a chain would itself constitute a proof of statement X1, in contradiction with the definition of a recursively unprovable chain.
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  - - You can't take something off the Internet! That's like trying to take pee out of a swimming pool.
Comment removed by account_deleted · 2013-06-06 05:07 · Score: 5, Informative

Comment removed based on user account deletion
Give a man a gun by sl4shd0rk · 2013-06-06 05:08 · Score: 4, Funny

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

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1. Re:Give a man a gun by steelfood · 2013-06-06 05:51 · Score: 1
  
  Only in Texas...
  
  --
  "If a nation expects to be ignorant and free in a state of civilization, it expects what never was and never will be."
2. Re:Give a man a gun by Alsee · 2013-06-07 05:44 · Score: 1
  
  Give a banker an algorithm, he can rob the world.
  Shore? Sure.
  Sieve of Eratosthenes? Not so much.
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  - - You can't take something off the Internet! That's like trying to take pee out of a swimming pool.
First read that as "Beale Cipher" by jfengel · 2013-06-06 05:15 · Score: 1

I hadn't heard of the Beal Conjecture, so the first thought that ran through my brain was the Beale Cipher, 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.
Too big to solve by Tablizer · 2013-06-06 05:20 · Score: 5, Funny

Oh great, yet another bank that wants a bealout.

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Table-ized A.I.
1. Re:Too big to solve by Alsee · 2013-06-07 05:56 · Score: 1
  
  And all the Chicken Littles running around screaming there will be a global economic collapse if RSA isn't Too Big To Factor.
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  - - You can't take something off the Internet! That's like trying to take pee out of a swimming pool.
What's in it for him? by gstoddart · 2013-06-06 05:23 · Score: 2, Insightful

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.

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Lost at C:>. Found at C.
1. Re:What's in it for him? by gstoddart · 2013-06-06 05:33 · Score: 2
  
  In this case, I think the motivation is more likely to just be remembered in the history books as the guy who financed the solution.
  New rule, only the people who solve the problem get credit.
  Otherwise we'll end up with the Fermat/Coca Cola theorem, or The Theory of Relativity, brought to you by Kinkos.
  
  --
  Lost at C:>. Found at C.
2. Re:What's in it for him? by Captain+Spam · 2013-06-06 05:39 · Score: 5, Funny
  
  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.
  
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  Demanding constant attention will only lead to attention.
3. Re:What's in it for him? by suutar · 2013-06-06 05:42 · Score: 1
  
  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.
4. Re:What's in it for him? by Anonymous Coward · 2013-06-06 06:02 · Score: 2, Informative
  
  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.
5. Re:What's in it for him? by scheme · 2013-06-06 06:05 · Score: 1
  
  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.
  
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  "When you sit with a nice girl for two hours, it seems like two minutes. When you sit on a hot stove for two minutes, it
6. Re:What's in it for him? by siwelwerd · 2013-06-06 06:39 · Score: 1
  
  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
7. Re:What's in it for him? by Impy+the+Impiuos+Imp · 2013-06-06 08:44 · Score: 1
  
  When I was a CS grad student, I had a scraggly 2" beard and a big heavy winter coat. One, coincidentally, particularly frigid day I went and did what I frequently did, go stand for hours in the bookstore looking thru many books.
  One of the young male workers, bless his soul, had seen me off and on over those hours, and said to me, "It sure is cold outside today, isn't it, sir?"
  I am not a bum. I am a grad student.
  
  --
  (-1: Post disagrees with my already-settled worldview) is not a valid mod option.
Related to topology? by mederbil · 2013-06-06 05:27 · Score: 2

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 by Anonymous Coward · 2013-06-06 05:36 · Score: 2, Insightful

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.
Common Factor? by trum4n · 2013-06-06 05:43 · Score: 1

Even if A,B,C,X,Y and Z are prime, they all have a common factor of one. Does one not count?
1. Re:Common Factor? by trum4n · 2013-06-06 06:20 · Score: 1
  
  Not trolling, legit question. I mean, if one counts, then there is no solution!
2. Re:Common Factor? by thrich81 · 2013-06-06 07:36 · Score: 1
  
  1, as the multiplicative unit in the algebraic ring of integers, has a special significance and does not work like the other integers in almost all number theory discussions. Same goes for 0 as the additive unit. These two "numbers" have special properties in most algebraic constructions (groups, fields, rings, etc.). A legitimate question you ask there, but the answer pretty much is just that 1 is "special" and doesn't "count" for this issue. Perhaps a better answer might be that because of 1's special properties in the ring of integers, including it in many theorems, formulas, etc, causes the construction under study to collapse to a trivial subset of the complexity of the actual problem and so become not useful in studying the general problem -- not sure what I just said there!
3. Re:Common Factor? by gstrickler · 2013-06-06 08:23 · Score: 1
  
  Even if A,B,C,X,Y and Z are prime, they all have a common factor of one. Does one not count?
  From the link in the sumary: 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
  And one is not considered a prime.
  
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  make imaginary.friends COUNT=100 VISIBLE=false
4. Re:Common Factor? by trum4n · 2013-06-06 08:38 · Score: 1
  
  TFS did not say "prime." Thanks for the clarification tho.
A great read on Fermat's Last Theorem by Anonymous Coward · 2013-06-06 06:03 · Score: 1

Fermat's Enigma by SImon Singh
The book covers many interesting figures (pun intended) that contributed directly or indirectly to the solution.
Beal by bbartlog · 2013-06-06 06:06 · Score: 3, Informative

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:Not just "a banker" by SammyIAm · 2013-06-06 06:13 · Score: 2

Wha-- why would that not be mentioned in the summary?!
Howard Beal's Conjecture by Anonymous Coward · 2013-06-06 06:13 · Score: 1

I tried to prove Beal's Conjecture. I got frustrated and yelled out my window, "I'm mad as hell, and I'm not going to take this anymore!"
Differing incentives by Abroun · 2013-06-06 06:22 · Score: 1

Having spent lots of time around bankers, it's no surprise to me that one would view money as incentive to solve an interesting mathematical problem. I doubt Andrew Wiles spent all that time in his attic hoping to get rich, and I wonder whether other mathematicians with the chops to solve this challenge will be influenced to try harder or change focus for something as cheap as money.
1. Re:Differing incentives by Xest · 2013-06-06 20:59 · Score: 1
  
  No but it does at least tell people looking for money that there's money in the field of maths and hence get more people into the field than otherwise would.
Hmmm, Interesting ... by Toad-san · 2013-06-06 06:29 · Score: 1

This site made it clear to me than anywhere else .. since it had code! Yesss, preciousssssss, code!
http://www.norvig.com/beal.html
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 powers and factoring, of course) for integers as big as you had hard drive storage for .. which was kind of big even then :-)
I might just have to reengineer that; it was ever so much fun. And simple too :-) Then solve (well, disprove) the conjecture, pocket the million, be suitably modest at the Nobel Prize awards ...
Toad, mafematakul Toad
The banker in question IS Beal by wonkey_monkey · 2013-06-06 06:31 · Score: 1

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.

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systemd is Roko's Basilisk.
Re:Not just "a banker" by gnapster · 2013-06-06 06:51 · Score: 1

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.
Re:AC Sees Banker's $1M, Raises $2M by gnapster · 2013-06-06 07:08 · Score: 1

Will Beale fold, or see your bid? This is gripping!
Brute force? by GameboyRMH · 2013-06-06 07:13 · Score: 2

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

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"When information is power, privacy is freedom" - Jah-Wren Ryel
1. Re:Brute force? by Iniamyen · 2013-06-06 08:14 · Score: 3, Funny
  
  I dunno. Do you have access to infinite compute cycles?
2. Re:Brute force? by GameboyRMH · 2013-06-06 08:31 · Score: 4, Interesting
  
  Maybe if I present it in the form of a cryptography scheme for terrorist communications...
  
  --
  "When information is power, privacy is freedom" - Jah-Wren Ryel
3. Re:Brute force? by RockDoctor · 2013-06-08 23:12 · Score: 1
  
  Maybe if I present it in the form of a cryptography scheme for terrorist communications...
  ... that would really grab the attention of a banker. If anyone likes a good, solid, unbreakable encryption scheme more than a terrorist, it's a banker. For much the same reasons. Which makes one wonder.
  
  --
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Or it could come full circle... by ZeroPly · 2013-06-06 07:43 · Score: 1

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.

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1. Re:Or it could come full circle... by semi-extrinsic · 2013-06-06 08:22 · Score: 4, Interesting
  
  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.
  
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Re:A=B=C=1 by aardwolf64 · 2013-06-06 07:45 · Score: 1

That is A solution to the equation. To win the money, you need to prove conclusively that the conjecture is either true or false for ALL numbers that fall within the range.
Yeah - cat on my tongue by bytesex · 2013-06-06 07:47 · Score: 1

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.

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Go back to sleep Pythagoras by SpaceLifeForm · 2013-06-06 08:13 · Score: 1

Exponents must be greater than 2.

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FLT constrains this slightly. by gstrickler · 2013-06-06 08:48 · Score: 1
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)
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make imaginary.friends COUNT=100 VISIBLE=false
To prove FLT with this by gr8_phk · 2013-06-06 09:31 · Score: 1

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 Fermat proud.
Why this doesn't prove FLT alone by gr8_phk · 2013-06-06 09:35 · Score: 1

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.
Re: A=B=C=1 by oobayly · 2013-06-06 09:45 · Score: 1

Cocky bastard. It'd take at least 10.
Like the journey to Ithaca by hooiberg · 2013-06-06 09:49 · Score: 1

It is not really the destination, but the journey, which matters. During the proof, new mechanisms will be constructed, which may (or may not) have applications elsewhere. New approaches to problems may be tried. While the proofs of many of many of the great conjectures have no direct relevance to the quality of life, the new mathematical insight will slowly permeate into other sciences.
Beal relates to 'abc' conjecture by mbkennel · 2013-06-06 10:24 · Score: 1

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.
My New Secret: ALIEN Theorem about:Beal Conjecture by Kosova · 2013-06-16 20:31 · Score: 1

Dear Gentlemen: Dr.Andrew Beal and High math Staff of AMS, Now I would like to Show all of You My Secret Theorem for your Beal Conjucture... First I will used My Secret ALIEN Theorem and from this we will Take this: Agron (Theorem) for the Beal Conjucture: My ALIEN Proof: We will take : A^x + B^y = C^z Now we take: A=x, x=n, B=y, y=n, C=z, z=n why we take: n because A, B, C it's Greater then 2 so now we have: x^n + y^n =z^n For: n=2 we have this: x^2 + y^2=z^2 For : x=m^2 - n^2 , y=2mn and z=m^2 + n^2 , when m= positive integer then => [(m^2)-(n^2)]^2 + (2mn)^2 = [(m^2)+(n^2)]^2 , => m^4+n^4-2(m^2)(n^2)+4(m^2)(n^2) =m^4+n^4+2(m^2)(n^2) , => 2(m^2)(n^2)=2(m^2)(n^2). Author: Agron Hoxha, (Ferizaj) Kosova a.51@collector.org me1@null.net Original Date: 10.June.2013 (Copy Rights),