# Revised Simple Proof of the Beal's Conjecture

1. Aug 13, 2010

### MrAwojobi

SIMPLE PROOF OF BEAL’S CONJECTURE
(THE $100 000 PRIZE ANSWER) Beal’s Conjecture Beal’s conjecture states that if A^x + B^y = C^z 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. Examples ..............Common Prime Factor 2^3 + 2^3 = 2^4 => 2 2^9 + 8^3 = 4^5 => 2 3^3 + 6^3 = 3^5 => 3 3^9 + 54^3 = 3^11 => 3 27^4 + 162^3 = 9^7 => 3 7^6 + 7^7 = 98^3 => 7 33^5 + 66^5 = 33^6 => 11 34^5 + 51^4 = 85^4 => 17 19^4 + 38^3 = 57^3 => 19 Primitive Pythagorean Triples A primitive Pythagorean triple is one in which the integer lengths of the right angled triangle do not have a common prime factor. Examples are…. ( 3 , 4 , 5 ) ( 5, 12, 13) ( 7, 24, 25) ( 8, 15, 17) ( 9, 40, 41) (11, 60, 61) (12, 35, 37) (13, 84, 85) (16, 63, 65) (20, 21, 29) (28, 45, 53) (33, 56, 65) (36, 77, 85) (39, 80, 89) (48, 55, 73) (65, 72, 97) Hence the reason why x, y and z in Beal’s conjecture equation have to be greater than 2. Simple Proof It should be clear that each term in the equation A^x + B^y = C^z can be broken down into the product of its prime factors after numbers have been substituted into it. The equation can be rewritten as the 1st product + the 2nd product = the 3rd product . This holds, for x, y and z greater than 2, if and only if the left hand side of the equation can be factorized, i.e. rewritten in the form P(Q + R) where P,Q and R are positive integers. This will therefore guarantee that A, B and C share a common prime factor. The reason why this statement doesn’t always hold for the Pythagorean equation A^2 + B^2 = C^2 is due to the fact that A^2 + B^2, i.e. the left hand side of the equation, can’t be written in the form P(Q+R) where P,Q and R are positive integers. An example of an expression that could be written in the form P(Q+R), even without actual numbers substituted into the algebraic expression, is A^3 + B^3 = (A+B)(A^2-AB+B^2). One could wonder what argument stops A^3 + B^3 = (A+B)(A^2-AB+B^2) = C^z when A and B have no common prime factors. The simple argument is that even though (A+B)(A^2-AB+B^2) and C^z are products, one is a positive integer raised to a positive power i.e. C^z and the other is something else other than that. The only way that (A+B)(A^2-AB+B^2) stands a chance of being equal to C^z is if and only if A and B share a common prime factor which will give (A+B)(A^2-AB+B^2) a chance to become a positive integer raised to a positive integer power. One could also raise the argument as to why equations of the form 10^5 + 41^3 = 411^2 or 25^2 + 6^3= 29^2 do not conform with the Beal equation. The answer lies in the fact that these equations are simply derived from the right hand side of the Pythagorean triples equations 105^2 + 88^2 = 137^2 and 21^2 + 20^2 = 29^2 respectively which cannot be factorized. Note that 411^2 = (3x137)^2. 2. Aug 13, 2010 ### jgens I'm not familiar with Beal's conjecture, but first things first, you should probably prove this statement since it's the key to your 'proof'; if you can't prove this statement, then your 'proof' fails. Moreover, even if you prove that Ax + By can be written in the form P(Q+R) for integers P, Q and R (which is easy), this is different than proving that P divides Ax and By. Last edited: Aug 13, 2010 3. Aug 14, 2010 ### MrAwojobi jgens you are telling me to prove general statements which I do not need to. All I am saying is that for those types of examples I have listed in the proof, i.e. after numbers have been substituted into the equation, then the claims I make must hold. 4. Aug 14, 2010 ### jgens First, the claims you make do NOT necessarily hold; your claim is essentially a restatement of Beal's conjecture. Second, you clearly aren't familiar with the way mathematics works. Even if your general statement is obvious to you, it's not to me or a number of other people on this forum. Without a proof of that statement, the mathematical community will never accept your 'proof' of Beal's conjecture. Finally, if you don't prove the statement in the next few posts, I'm going to guess that this thread will be locked just like your last one. 5. Aug 14, 2010 ### MrAwojobi jgens maybe the best way I should have explained this was to say that for(A+B)(A^2-AB+B^2)to be a number raised to a power of 4 or greater (Fermat's last theorem doesn't permit a power of 3)then it must factorise further i.e. A and B must have at least a common factor.If they don't then (A+B)(A^2-AB+B^2) cannot factorise. The main reason that powers of 2 have purposely been removed from the statement of the conjecture is because a^2+b^2 cannot always factorise when numbers are substituted into it. When powers of 2 are involved, there may be common prime factors in some examples and none in some others, cases in point,2^5 + 88^2 = 6^5 and 13^2 + 7^3 = 2^9 respectively. 6. Aug 14, 2010 ### jgens That isn't a proof. 7. Aug 14, 2010 ### MrAwojobi jgens just as you say that 'this isn't a proof', I say to you that your response to my proof is a non-response because you haven't elaborated on why you make this statement. I guess you just can't believe that a$100 000 prize problem ended up being so simple. I am one of a very few mathematicians who see the simplicity of a mathematics problem instead of sticking to the same old complicated ways of solving mathematics problems.

8. Aug 14, 2010

### CRGreathouse

Ah. But you have claimed that yours is a proof, while jgens has not claimed his post as more than a comment. You need to defend your statement; you can't expect others to explain to you where you've made mistakes.

9. Aug 15, 2010

### MrAwojobi

CRGreathouse

not you again!!!! You seem to be everywhere (I bet you're thinking to yourself 'see who's talking'). I would prefer that your responses are based on proper mathematical criticisms.
I thought that this is what maths forums are all about when I discovered them about a month ago or so. I think I might have been wrong. Also in your other forum you keep mentioning that I should pay this, that and the other (including yourself) money to rewrite my proofs or disproofs. Are maths forums money making ventures i.e. helping people with their research work and getting paid for it? Is this why most of you do not use your real names? Is this why when some of you recognise a real proof you just knock it down even when you know it is correct? I do wonder.

I will now try and get my proof published in a recognised mathematics journal. Failing this, I will write a book about my proofs so that it is accessible to a wider audience who I'm sure will agree with them. I think mathematics is being controlled my a mathematics establishment that wouldn't recognise a proof simply because it is too simple.

10. Aug 15, 2010

### jgens

If you're referring to your 'proof' of Beal's Conjecture, I don't think that anyone who has commented here (or in your other thread) recognizes it as a proof. Sorry.

Good luck with this endeavor. The mathematical community will likely be more critical than the members here at PF.

As someone who hasn't really begun a study of higher mathematics and probably belongs to this "wider audience", I would wager that anyone who knows what a correct proof looks like won't agree with your 'proof' of Beal's Conjecture.

I think that you happen to be a crackpot who doesn't have any idea what a correct proof actually looks like. But hey, we all think things.

Last edited: Aug 15, 2010
11. Aug 15, 2010

### CRGreathouse

Yes, forums are generally money-making endeavors. No, people don't (generally?) make money by helping in research in that manner -- I'm simply suggesting that since you aren't paying people, you can't expect them to work the way you'd prefer.

I can't speak for the others. I so use my real name. Do you?

I'm sure no one here, or on mymathforum, thinks that you have a valid proof.

I'll leave this link here, just in case:
http://en.wikipedia.org/wiki/Vanity_press

I'm sure your proof of this is every bit as valid as your proof of the T-Z conjecture.

12. Aug 16, 2010

### zetafunction

i think everyone SHOULD encourage this people trying to solve math puzzles...

it is a pity that society ONLY wants people solving math problems but do not take respect :) by people that have made 'advances' into a mathematical problem, advances that sometimes are ignored by math community

keep on working mr. Awojoby, the best think is to keep on trying and to ask questions to see where the mistakes are , and correct them

13. Aug 16, 2010

### jgens

Out of curiosity, what advances into the problem do you think that the OP has made? His key to the proof (concerning the 'factorization' of Ax+By) amounts to nothing more than a restatement of Beal's Conjecture; and thus, I think that it's safe to say that new ground has NOT been broken here.

14. Aug 16, 2010

### JCVD

15. Aug 16, 2010

### CRGreathouse

Indeed. I posted some other examples (including a version of the one you link to) in the other thread. Or, at least, the only other version of this thread I'm following -- I'm sure there are more if you search Google.

16. Sep 2, 2010

### MrAwojobi

Hi y'all

I believe this latest write up of the proof should make more sense. Tell me what you think by reviewing the attachment. Many thanks.

17. Sep 2, 2010

### CRGreathouse

I don't see an attachment.