## Banker Offers $1M To Solve Beal Conjecture 216

Posted
by
timothy

from the it's-no-fermat's-but-hey dept.

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.""*
## Fermat? (Score:2, 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.

## Re:Fermat? (Score:5, Informative)

## Re: (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.

## Re: (Score:3)

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.

## Brute force? (Score:3)

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

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## Re:Brute force? (Score:5, Interesting)

Maybe if I present it in the form of a cryptography scheme for terrorist communications...

## Re: (Score:3)

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 m

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## Re:Fermat? (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 [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.

## Re:Fermat? (Score:5, Funny)

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The price is more interesting.

Brute forcing to find a contradiction is no mental challenge.

Crafting a nice formal proof is.

## Re:Fermat? (Score:5, Funny)

Huh. For a second there my brain thought: "FTL? Faster Than Light (travel)? WTF!"

## Re:Fermat? (Score:5, Funny)

Huh. For a second there my brain thought: "FTL? Faster Than Light (travel)? WTF!"

WTF? World Taekwondo Federation?

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Huh. For a second there my brain thought: "FTL? Faster Than Light (travel)? WTF!"

WTF?

What The F***?FTFY, FTW!

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## Re:Fermat? (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.

## Or it could come full circle... (Score:2)

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.

## Re:Or it could come full circle... (Score:5, Interesting)

## Re: (Score:2)

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.

## Re: (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.

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.

## Re:Fermat? (Score:5, 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.

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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.

## Re: (Score:2)

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 )

## Re: (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.

## Re: (Score:2)

Not really, this conjecture if true implies FLT fairly easily.

## Have solution. Alas, subject line = too small (Score:5, Funny)

Sorry, but I promise you that the solution was very elegant.

Bert

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Is it me or are mathematical practical jokes even less funny than general mathematical jokes?

## Re:Have solution. Alas, subject line = too small (Score:5, Funny)

I suspect you are correct, but a proof remains elusive.

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That's not funny.

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## Couldn't you just make up any old equation... (Score:3, Interesting)

... along with some postulated constraints and ask people to prove them? Whats so special about this one - does it have some mathematical relevance?

## Re: (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."

## Re: (Score:2)

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?

## Re: (Score:3)

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 todisprovetrue conjectures.## Re: (Score:3)

## Re:Couldn't you just make up any old equation... (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

get100%.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? :)

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You can't always prove something that is true.

See Gödel's incompleteness theorems.

## Re:Couldn't you just make up any old equation... (Score:4, 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.

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Lest we forget that the hop farm used cow manure as fertilizer :)

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You don't sleep, and you watch the beer at all times.

## Re:Couldn't you just make up any old equation... (Score:4, 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.

## Re: (Score:2)

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

engineeris 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."## Re: (Score:2)

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.

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"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.

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"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.

## Re: (Score:3)

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.

## Re: (Score:2)

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 pro

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## Beal relates to 'abc' conjecture (Score:2)

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.

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## Re:Couldn't you just make up any old equation... (Score:5, 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.

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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).

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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.

## Re: (Score:2)

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 Be

## Re: (Score:2)

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.

## Re: (Score:2)

D'oh, good catch. But then z=2 and the conjecture specifies that all the exponents be greater than 2.

## Re: (Score:2)

Somebody got his knickers in a twist this morning.

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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.

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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.

## Re: (Score:2)

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.

## Not just "a banker" (Score:5, Informative)

Not just "a banker".. .. but the very person who proposed the conjecture.

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## Give a man a gun (Score:5, 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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Only in Texas...

## First read that as "Beale Cipher" (Score:2)

I hadn't heard of the Beal Conjecture, so the first thought that ran through my brain was the Beale Cipher [wikipedia.org], 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 (Score:5, Funny)

Oh great, yet another bank that wants a bealout.

## What's in it for him? (Score:3, 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.

## Re:What's in it for him? (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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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.

## Re: (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 i

## Related to topology? (Score:3)

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 (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.

## Beal (Score:3, Informative)

## Hmmm, Interesting ... (Score:2)

This site made it clear to me than anywhere else .. since it had code! Yesss, preciousssssss, code!

http://www.norvig.com/beal.html [norvig.com]

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 power

## The banker in question IS Beal (Score:2)

## Yeah - cat on my tongue (Score:2)

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.

## Go back to sleep Pythagoras (Score:2)

## FLT constrains this slightly. (Score:2)

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 factorFermat'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:

## To prove FLT with this (Score:2)

So if you can prove the Beal conjecture, you also get a proof of FLT by infinite descent. This would make Ferm

## Why this doesn't prove FLT alone (Score:2)

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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 factorAnd one is not considered a prime.

## Re: (Score:2)

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

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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.

## Re: A=B=C=1 (Score:2)

Cocky bastard. It'd take at least 10.