# Solve This Math Problem, Win a Million Bucks

Think you've got what it takes?

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?

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.]

SharniMiller

I have a question I could be totally wrong and if so can you please help me find were I am wrong I thought I would have a go

2^3 x 5^3 = 10^3

8 x 125 = 1000

" La théorie, c'est quand on sait tout et que rien ne fonctionne.

La pratique, c'est quand tout fonctionne et que personne ne sait pourquoi"

Albert Einstein
La greffe de la moelle osseuse Albert Einstein"Si vous ne pouvez expliquer un concept à un enfant de six ans,
c'est que vous ne le comprenez pas complètement"

Albert Einstein

Le principe de la greffe est de réimplanter chez le malade (appelé receveur) des cellules souches hématopoïétiques
d’un sujet sain (appelé donneur) capables de remplacer les cellules malades après leur destruction par chimiothérapie
(complèté éventuellement par une radiothérapie), afin de fournir une moelle osseuse saine capable de fabriquer des cellules sanguines saines.
La greffe de moelle ajoute en outre une action anti leucémique par elle-même,
du fait des propriétés de défense immunitaire des cellules injectées (greffe allogénique).

source :

Conclusion donc c'est du plagia plagia = copie collé

et l'algorithme du plagia est le Karp Rabin

Alors vous allez me dire ça on le sait mais on n'a pas réussi

c'est parce que vous le faites en mode séquentiel non d'une façon parallèle et aussi non probabiliste et n'utilise pas MonteCarlo à Las Vegas.

C'est pourquoi j'ai partagé le travail de CAPA 95 la démonstration date de l'année 95 date de mon baccalauréat de mathématiques.

et je vais l'a partagé encore une fois. Vous vous dites pourqoui j'ai une tête de turc !!! ça je vous l'explequerai dans un autre post.
Je demande pardon à tous les membres du groupe du travail CAPA parce que j'ai scanné leur travail sans leur autorisation.

UN grand hommage à eux

Ils ont dit dans leur livre

parallélisme, réseaux et répartition

Parallélisme et applications irrégulières

coordinateurs Gérard Authié, Jean-Marie Garcia, Afonso Ferreira,
Jean-Louis Roch, Gilles Villard, Jean Roaman, Cathjerine Roucairol,
Bernard Virot

Edition HERMES

"Quelles classes d'applications peuvents bénificier du parallélisme ? Existe-t-il pour certains domaines industriels des logiciels parallèles performants ? Quelles sont les orientations actuelles et futures de l'algorithmique parallèle? Comment partager, distribuer ou faire migrer la charge du travail sur un grand nombre de processeurs pour obtenir de bonnes performances de calcul en parallèle ? Quelle est la difficulté posée par les applications travaillant sur des structures de données irrégulières ? comment la surmonter ? "

J'ai vu que leurs algorithme de filtrage de chaîne de caractère Karp Rabin parallèle probabiliste de Monte Carlo à Las Vegas peut servir
dans le domaine du génétique et pourra être le modèle mathématique qui expliquera le principe ou le comment de la greffe de moelle osseuse et ainsi
on pourra faire de même pour guérir les maladies comme le cancer étant données qu'on connait les pattern ADN qui sont responsables de la tumeur.

Encore Un hommage au groupe CAPA et je renouvelle mes excuses

Wadï ben Houcine ben AlHabib Mami

didipostman

Good dayI am Wadï Mami a computer science engineerI know it is a little bit odd but I found a mathematical proof for the marrow bone transplant principe in an old publication (1995) it is in french and I am not able to translate ityou will find that proof in my blog
http://didipostman.blogspot.com/2014/09/encore-merci-capa-more-explanation.html
as we know the DNA patterns for diabete we can use Karp Rabin probabilistic algorithm in parallel in order to search these spoiled sequence and replace them by inoffensive sequence we can do the same for other illnessesBest regards and thank you for reading my post
--
Minds, like parachutes, function best when open.,,,
(o o)
/ -------------------oOO--(_)--OOo---------------------\
| \-----------------------------------------------------/
|
||\--------------------------/ \------------------------/

First of all happy new year 2015

when googling I found this link

"This prize, orchestrated by sponsor Marcus Hutter, seeks to find new ways to compress data. The task is to create a compressed archive of a given 100MB file smaller than ever before as a way to create new compression algorithms.

If you can get it compressed to smaller than the current record — around 16MB — you get a portion of the prize money. So far, Alexander Ratushnya is the three-time winner.

The prize? The percent you improve the compression is equal to the percent of the 50,000€ you win, with a minimum of 3%."

Well you can find a solution in this link

• Reduce file size up to 95%, reduce transmission times, and improve processing efficiencies

Of course it is not free but once buyed you can use it in your program plz visit

for a free solution

thank you

please ask them all to pay me If you're not mathematically endowed, this may be the best way to get paid to "do math."

\$5,000 — Erdős conjecture on arithmetic progressions

While Erdős passed away in 1996, Ronald Graham remains the current administrator of any outstanding Erdős problem prizewinners.

You can make \$5,000 by proving one of the remaining Erdős problems, the Erdős conjecture on arithmetic progressions

If the sum of the reciprocals of the elements of a set A (comprised of positive integers) diverges, then A contains arbitrarily long sequences of numbers with constant differences between terms.  (1)

So you've got this set A, made of positive integers. You take the reciprocal of those numbers — for a number x, the reciprocal is 1/x — and you sum them, and find that they never converge to a single number, they just keep adding to infinity.

Well, this conjecture says that when that happens, you'll notice A contains sequences of numbers with an arbitrary distance between them. (2)

Prove that works every time, let Graham know, and your check — signed by Graham if you want to cash it, signed by Erdős if you make the right call and frame it — is in the mail.

Well it is me didipostman  there is a contradiction betwwen (1) and (2) you can't prove that and I can't reach Graham as well

My millenium Prize

didipostman's Conjecture

Well U₀ = 7
Un+1 = Un + (n+1) * 4
proove that we always obtain prime numbers﻿

My millenium Prize

didipostman's Conjecture

Well U₀ = 7
Un+1 = Un + (n+1) * 4
proove  that we always obtain prime numbers﻿

as Riemann Hypothesis can not be proven as beal's conjecture can not be proven please see below "the yesterday email "
Birch and Swinnerton-Dyer Conjecture can not be proven as

the Birch and Swinnerton-Dyer conjecture asserts that the size of the group of rational points is related to the behavior of an associated zeta function ζ(s) near the point s=1. In particular this amazing conjecture asserts that if ζ(1) is equal to 0, then there are an infinite number of rational points (solutions), and conversely, if ζ(1) is not equal to 0, then there is only a finite number of such points.
I mean see the yesterday comments
thank you﻿

I should win \$1,000,000 — Riemann Hypothesis

too as the erdos prize and the hutters prize and \$1,000,000 — Prove Beal's Conjecture see above
Good day

ζ(s) = 1 + 1/2s + 1/3s + 1/4s + ... + 1/cs  + .....

called the Riemann Zeta function. The Riemann hypothesis asserts that all interesting solutions of the equation

ζ(s) = 0

that was proven for the first 10,000,000,000 solutions

and Beal's Conjecture is related. If a, b, c, x, y, and z are all positive integers and x, y, s are greater than two,

ax + by = cs
is only possible when a, b and c have a common prime factor

and Beal's Conjecture can not be proven see

as Gödel has answred Hilbert Beal you canot verify your conjecture
Gödel's two incompleteness theorems were his crowning achievement. They have often been misunderstood in their popular presentations. We must first be clear about what the theorems do not say. They do not state that there are absolutely unprovable truths. What the first theorem asserts is that if a system of axioms can express the arithmetic of both addition and multiplication of counting numbers (1, 2, 3 and so on), and if the system is consistent (that is, if it does not imply a contradiction), then there will be a sentence G of the system that is unprovable in the system but is nevertheless true. If this new sentence G is now added to the axioms of the system, then applying Gödel's procedure to this enlarged system will generate a new sentence H, which will be true but unprovable in the system with the added axiom G. So it would be more accurate to describe Gödel's theorems as having to do with "incompletability" rather than "incompleteness." The second incompleteness theorem states that, given the same hypothesis as the first, one such true but unprovable sentence expresses the consistency of the system.

Hilbert insisted that it is possible to prove the consistency of all of classical mathematics (and, in particular, of both the counting numbers and the real numbers) by using only the truths of the elementary arithmetic of the counting numbers. The attempt to do so became known as "Hilbert's program." Gödel's incompleteness theorems showed that Hilbert was wrong; the arithmetic of the counting numbers could not prove even its own consistency, much less that of the real numbers!

For Gödel's proof to work, however, it is essential that arithmetic have both addition and multiplication available. In fact, if we consider the counting numbers with just addition (and not multiplication) defined on them, then we can find a set of axioms that prove all true first-order sentences in the language of this system. So Hilbert was right—for this weaker system.

http://www.americanscientist.org/bookshelf/pub/the-incomplete-g-del

Well Gödel has answred Hilbert Beal you canot verify your conjecture

Gödel's two incompleteness theorems were his crowning achievement. They have often been misunderstood in their popular presentations. We must first be clear about what the theorems do not say. They do not state that there are absolutely unprovable truths. What the first theorem asserts is that if a system of axioms can express the arithmetic of both addition and multiplication of counting numbers (1, 2, 3 and so on), and if the system is consistent (that is, if it does not imply a contradiction), then there will be a sentence G of the system that is unprovable in the system but is nevertheless true. If this new sentence G is now added to the axioms of the system, then applying Gödel's procedure to this enlarged system will generate a new sentence H, which will be true but unprovable in the system with the added axiom G. So it would be more accurate to describe Gödel's theorems as having to do with "incompletability" rather than "incompleteness." The second incompleteness theorem states that, given the same hypothesis as the first, one such true but unprovable sentence expresses the consistency of the system.

Hilbert insisted that it is possible to prove the consistency of all of classical mathematics (and, in particular, of both the counting numbers and the real numbers) by using only the truths of the elementary arithmetic of the counting numbers. The attempt to do so became known as "Hilbert's program." Gödel's incompleteness theorems showed that Hilbert was wrong; the arithmetic of the counting numbers could not prove even its own consistency, much less that of the real numbers!

For Gödel's proof to work, however, it is essential that arithmetic have both addition and multiplication available. In fact, if we consider the counting numbers with just addition (and not multiplication) defined on them, then we can find a set of axioms that prove all true first-order sentences in the language of this system. So Hilbert was right—for this weaker system.

http://www.americanscientist.org/bookshelf/pub/the-incomplete-g-del

nyvanod

the answer is 1 from nyvanod

nyvanod

how do i send the answer to you

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

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

@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.

tygrr94

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

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

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

Rilurope

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

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

Rilurope

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)

LeticiaCastro

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

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.

LeticiaCastro

Pythagorym theorem will explain it. Good luck

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

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

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

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

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

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

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

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

2 ^ 8 + 2 ^ 8 = 2 ^ 9

rekbhn

2 ^ 9 + 2 ^ 9 = 2 ^ 10

rekbhn

2^10 + 2 ^10 = 2 ^ 11

PramodAcharya

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

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

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

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.

aeseef

Max, you wanna give it a shot?

cspanb2

This has already been proven by Andrew Wiles

SamSam

27^4 + 162^3 = 9^7

MikeAlan

Side note on M. Beal. He is quite the avid poker player and decided he wanted to play the best in the world. He figured if he could make the stakes high enough, the professional players would be out of their comfort zone. At first it worked but then Phil Ivey cleaned his clock to the tune of about 20M dollars. Beal has since quit the game. lol

jenni.julep

It's a little misleading - the link states that solutions must be published in a respected journal and then to notify the committee.  We can't just solve it and send them the solution.  That kinda sucks.