Solve This Math Problem, Win a Million Bucks

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Justin Lewis / Getty Images

Want to make a quick million? All you have to do is figure out a little math problem that goes like this: Ax + By = Cz. Simple algebra, right?

(MORE: After Inhaling Hot Sauce Fumes, Three People Are Hospitalized)

Oh how deceptively innocuous a few elementary variables can seem. You’re actually looking at something inspired by one of the great mysteries of mathematics, known as Fermat’s Last Theorem and named after the 17th century French lawyer and mathematician Pierre de Fermat. Fermat came up with his own theorem back in 1637, scribbling it in the margins of his copy of the Greek text Arithmetica by Diophantus and surmising that — put your math caps on and buckle up — if n were an integer greater than 2, then the equation Xn + Yn = Zn has no positive integral solutions. The note was discovered after Fermat’s death, and it took over 350 years and untold failed attempts by others for someone to prove the theorem. In 1995, British mathematician Andrew Wiles, who’d been fascinated with the theorem since he was a child, finally got the job done, having puzzled over it in secret for roughly six years.

That’s where Texas billionaire D. Andrew Beal comes in. In 1993, he posited a closely related number theory problem hence dubbed Beal’s Conjecture (that first A-B-C equation above), where the only solution is possible when A, B and C have a common numerical factor and the exponents x, y and z are greater than 2. Beal’s been trying to prove his theorem ever since, reports ABC News, offering cash rewards in steadily increasing amounts — $5,000 in 1997, $100,000 in 2000 — to anyone with the knack to get the job done.

The prize total in 2013: $1 million, which is either a sign of Beal’s magnanimity or his skepticism that it’s actually possible. (Since Beal is worth a reported $8 billion, there’s little need to worry about whether he’ll pay the winner.)

It’s apparently not just about the money for Beal, either: In a statement, he said “I’d like to inspire young people to pursue math and science. Increasing the prize is a good way to draw attention to mathematics generally … I hope many more young people will find themselves drawn into the wonderful world of mathematics.”

[Update: The challenge is sponsored by the American Mathematical Society; the rules, including where to submit proposed solutions by email or snail mail, are available here.]

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137 comments
NadoorNadoor
NadoorNadoor

I believe that the A B equals the C so it would make sense that the X Y equals the Z so Ax+By=Cz the first two letters of alphabet would equal the third and the two second last letters of the alphabet will equal the last.

MATTHEWWEAKLEY
MATTHEWWEAKLEY

So this is saying A to the X plus B to the y equals C to the z.  A, B, C, x, y, and z are all positive integers and and the latter three are all greater than 2.  Am I missing something here?  There's an endless number of answers.  Here's one, and it's a simple one:  Say A, B, and C are all 2.  Then x = 3, y = 3 and z = 4.  That's one example.  Now the reason why A, B, and C must have a common prime factor is because A to the x and B to the y must ADD into C to the z.

Math
Math

Just finished my book How To Solve the Beal and Other Mathematical Conjectures.  It's free on Kindle for the next 3 days. It can be solved via the the Pythagorean theorem because it is the Pythagorean theorem.   Say n=4 so A^4 +B^4=C^4 or a^2A^2+b^2B^2=c^2C^2.  a,b,and c become dimensionless factors and are applied toward enlarging  the AREAS of A^2, B^2, and C^2 respectively. This is the Fermat conjecture.  Now the Beal is practically the same thing, works the same way except the exponents x and y have to be expressed in terms of z the exponent for the hypotenuse, C..

JesusFreakPanda
JesusFreakPanda

The answer is 1. Reason: anything with this high of stacks wouldn't have a very complicated answer. 

Rilurope
Rilurope

The answer is:

A= a^(2/x)

B= b^(2/y)

C= c^(2/z)

making the equation:

[a^(2/x)^x] + [b^(2/y)^y] = [c^(2/z)^z]

;)

dmoneyx
dmoneyx

4(x)+4(y)=8(z). 4(3)+4(3)=8(3). 12+12=24.

LeticiaCastro
LeticiaCastro

Has the proof of the pythagorem theorem ever been discovred? That should help explain this. Good luck for those that like mathematical proofs.

LeticiaCastro
LeticiaCastro

Pythagorym theorem will explain it. Good luck

IAMTIME
IAMTIME

this problem is easy seeing as a positive integer is above 0 and they all share a common prime factor ,a prime number is the number itself and 1 and so the most common factor is 1.I can see the million now .

nikki007hhh
nikki007hhh

the answer is D   EMAIL ME FOR MY ADDRESS RO SEND THE CHECK

THIS CONTEST BETTER BE REAL OR WE WILL MEET IN COURT :-) JOKING

NIKK007HHH@AOL.COM

pablo19590
pablo19590

It´s too easy to show that the Beal conjeture is wrong. Thanks.

theoremfinder
theoremfinder

You don't even have the theorem correct in the first sentence. Come on Matt.  Ax+ By = Cz is not the same as  Xn + Yn = Zn. Totally different.

RichardSRussell
RichardSRussell

For a million bucks, I'll "prove" it using by far the most popular technique ever invented by humanity: "It must be so, because God said so."


What do you mean, that doesn't count?

ImbalzanoG
ImbalzanoG

I'm unpleasant, but the assertion "La ecuación establece que si Ax + By = Cz, donde A, B, C, x, y, z son enteros positivos, siendo x, y, z mayores que 2, entonces A, B, y C deben tener un factor común primo." (FROM http://sociedad.elpais.com/sociedad/2013/06/07/actualidad/1370616631_328551.html ) is completely erroneous! Giovanni Imbalzano. SEE also: http://www.lulu.com/shop/giovanni-imbalzano/fermat-%C3%A0-la-page/paperback/product-18811968.html OR http://www.lulu.com/shop/giovanni-imbalzano/extension-of-the-fermats-numbers/paperback/product-6207910.html OR http://www.lulu.com/shop/giovanni-imbalzano/solution-of-the-fermats-last-theorem/paperback/product-4589261.html G. Imbalzano

JohnRock
JohnRock

Brilliant Hindu/ Indians discovered pretty much everything in Mathematics starting from Geometry, Algebra, Numbers 0 to 9, Trigonometry, Calculus, Arithmetic, Place value system, Decimal system and the list goes on and on.

The Fractal mindset of Brilliant Hindu Mathematicians can never be replicated by western scientists with all their models and brain theories.

Last I heard, so called Newtons unsolvable 350 yr old Math puzzle was solved by a Brilliant Hindu/Indian kid.

Mock modular forms which came from the fractal mindset of another Brilliant Hindu/Indian mathematician Srinivasa Ramanujam 100 yrs ago was proved to be correct last year/

We in the west, fail to acknowledge the source of our Knowledge and wealth since 18th century (since colonialism of India)

India was the richest country and largest economy in the world followed by China, with a GDP of 25 to 30% until the time of British invasion in the 18th century, according to a 20 yrs research done by Angus Maddison for OECD countries.

India was the number 1 manufacturer and exporter of textiles and Iron and steel and both these technologies were stolen by the British and made as twin engines of its so called Industrial revolution

Thank you




vwvan
vwvan

Checking the first 281,600 equations yields 25 solutions which by inspection all have common primes:

2^3 + 2^3 = 2^4    2^4 + 2^4 = 2^5    2^5 + 2^5 = 2^6    2^6 + 2^6 = 2^7    2^7 + 2^7 = 2^8

2^8 + 2^8 = 2^9    2^9 + 2^9 = 2^10   2^5 + 2^5 = 4^3   2^7 + 2^7 = 4^4    2^9 + 2^9 = 4^5

2^8 + 2^8 = 8^3    2^6 + 4^3 = 2^7    2^8 + 4^4 = 2^9    2^8 + 4^4 = 8^3    2^9 + 8^3 = 2^10

2^9 + 8^3 = 4^5   3^3 + 6^3 = 3^5   4^3 + 4^3 = 2^7    4^4 + 4^4 = 2^9    4^4 + 4^4 = 8^3

4^7 + 4^7 = 8^5   4^10 + 4^10 = 8^7 8^3 + 8^3 = 2^10  8^3 + 8^3 = 4^5    8^5 + 8^5 = 4^8

Further the number of candidate solutions per equation rolls off logarithmically.

So I choose to agree with Beal's conjecture, and speculate that its complexity of proof is at least Fermat, which took 358 years. That works out to about $1.40 an hour  assuming 2000 hours of effort per year.

More importantly Beal's conjecture is operationally true for the space of numbers of most common concern.

rekbhn
rekbhn

2 ^ 8 + 2 ^ 8 = 2 ^ 9


rekbhn
rekbhn

2 ^ 9 + 2 ^ 9 = 2 ^ 10

rekbhn
rekbhn

2^10 + 2 ^10 = 2 ^ 11

PramodAcharya
PramodAcharya

the problem is not correctly stated, poorly worded, written by a guy who knows nothing about mathematics

ElizabethMcBride-Lilleg
ElizabethMcBride-Lilleg

You don't SOLVE theorems...you prove (or disprove) them.  A proof requires the use of mathematical principles to demonstrate that the theorem must be true (or more accurately, that it cannot NOT be true.)  That's what you need to do to get the $1 million.  Of course, in order to disprove it, you only need to come up with one example where it fails.  I'm not sure if he'd pay you for that though.  :)

TylerWindsor
TylerWindsor

Will Hunting will be on it....after he's done seeing about a girl

TimeTunnel
TimeTunnel

As a consolation prize for you math/professor/hidden genius-types that are not going to win this $1,000,000 prize (and even more so if you love college football), you might want to Google “Shoelace Shootout”. Game theory finds a new application in real life, and the short read will be like a breath of fresh air compared to banging your head against the wall trying to prove Beal’s Conjecture. Your claim to fame will be you got a glimpse of the future, two years before it happens.

cspanb2
cspanb2

This has already been proven by Andrew Wiles

Math
Math

@MATTHEWWEAKLEY 

Yes you are missing something.  In the Beal conjecture A,B,C,x,y,z all have to be integers and all have to be different .  In the Fermat conjecture A,B,C all have to be different integers but x,y,z all must be the same integer.

Rilurope
Rilurope

@dmoneyx  

if the equation were: Ax+By=Cz, then the solution would be:

a^n*a^(-n+2) + b^n*b^(-n+2) = c^n*c^(-n+2)

KayneHiggin
KayneHiggin

@LeticiaCastro Long, long ago. It's easily demonstrable.


Beal's theorem says that the exponents must be different integers. So Pythagorean theorem doesn't work.

lonewolf333
lonewolf333

@RichardSRussell Trolling the religious much?  Mocking people with beliefs different than your own shows how stupid you are.  I'd call you ignorant, but I don't want to lie.

analyze.holes
analyze.holes

@JohnRock I was born in India. And I have proved the Beal conjecture. 

It is true : India has some problems. But who does not have a problem? On individual basis, group basis, community basis, country basis, religion basis or whatever basis. Who is perfect?

The issue here is the Beal conjecture. And it is correct.

EmpyreanOrder
EmpyreanOrder

@JohnRock Sure, discovery is great, but ADVANCEMENT is even better. India "was this" and "was that." What is it today? The headlines I recall from the New York Times is that India is still an oppressive regime that limits free speech, and has no rape laws. Until government & social structure improves, I don't see India contributing as much as its Western rivals in advancements in math/science/technology, solely due to lack of funding and current state of technology. In some parts of the country, one would have to scour through junkyards, looking for rubber tubes and glass bulbs just to have certain surgeries performed on them (they have a about a year or two to search, since that's how long the waiting list for the surgery would be). 

lonewolf333
lonewolf333

@JohnRock How about some facts from the dark side on your pro-Indian rant?  They still have one of the most oppressive caste systems in the world, and many of them look down on marrying outside Indian blood.  India isn't the Mecca of civilization you make it out to be.

analyze.holes
analyze.holes

@vwvan Did you check all 281,600 equations? I just finished my book "The Beal Conjecture - Detailed Analysis, Complete Proof, Concrete Examples" . The first edition is available for purchase. I prove that it works in all cases. Soon my proof will be accepted and there will be no more the Beal conjecture. It will be known as the Beal theorem. I hope Mr. Beal will be happy about it. 

For more info, I can be reached at 416-725-0909.

analyze.holes
analyze.holes

@rekbhn All these three solutions fall in one class. They all have 2 as a common factor. My book provides the proof. 

analyze.holes
analyze.holes

@PramodAcharya  No. I disagree with you.

The problem is correctly stated. I have proved the conjecture in my book. "The Beal Conjecture - Detailed Analysis, Complete Proof, Concrete Examples" . For more info, I can be reached at 416-725-0909.


cjh2nd
cjh2nd

@PramodAcharya 

sounds like it might have been written by my high school calculus teacher then hahh

analyze.holes
analyze.holes

@ElizabethMcBride-Lilleg  I have proved that it is correct. There cannot be a counterexample. It will always work.  The details are in my book titled, "The Beal Conjecture - Detailed Analysis, Complete Proof, Concrete Examples" .

You are right about mathematical principles. Indeed,. I have used mathematical principles to prove that it will work in all cases.

SuperDave
SuperDave

How do you like them apples!

analyze.holes
analyze.holes

@TimeTunnel Who knows they may not wait for two years. 

I believe my proof is convincing enough that it is the best way to prove it. Hence, they may not wait for any better proof or a dispute to that proof.

analyze.holes
analyze.holes

@cspanb2 Andrew Wiles proved the Fermat's Last Theorem. Isn't it?


I have proved the Beal conjecture. 

cspanb2
cspanb2

I see the article mentioned that. 1782^12 + 1841^12 = 1922^12

tygrr94
tygrr94

@Math Thanks for that clarification b/c I wasn't sure about that either.

cjh2nd
cjh2nd

@lonewolf333

"Mocking people with beliefs different than your own shows how stupid you are."

hi pot, my name is kettle. hypocritical much?



TheMuffinman
TheMuffinman

Lol u wish. If u had solved it y r u wasting ur time writing in this comments. Go spend ur money

KaushikDas
KaushikDas

@cspanb2 I am not sure this is true: 1782^12 is even number, 1841^12 is odd number, sum of odd and even should be odd

analyze.holes
analyze.holes

@KaushikDas @cspanb2 You are right. When you add an even number to an odd number, the result will be an odd number. 

And that is one class of solutions I have examined thoroughly, in my recent book.

cspanb2
cspanb2

@KaushikDas @cspanb2 

 No its a Fermat near miss. On a calculator (9-digit) it will appear to be correct (seemingly disproving the theorem), but is not after the 9th digit.