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Beal conjecture

Synonym: 
Beal's conjecture
Type of Math Object: 
Conjecture
Major Section: 
Reference

Mathematics Subject Classification

11D41 no label found

Comments

Has anybody reviewed James Constant's proof?

http://www.coolissues.com/mathematics/Beal/beal.htm

I'd rather not read anything that begins with a claim that FLT can be proved using the binomial theorem. No such proof exists as no such attempted proofs have been shown to be correct.

First, there is a typo in the link to James Constant's proof. The correct link is:

http://fermat.coolissues.com/fermat.htm

I guess that this proof is as wrong as Pogorsky's proof which I have myself presented here not so long ago. It was also based on the binomial theorem.
However I think that this one is worth a look at least. It seems to be clear and concise, and the flaw, if it exists, must be easy to find.

Daniel

There is, indeed, a flaw in Constant's proof from http://www.coolissues.com/mathematics/Beal/beal.htm. In his "Disproof of Beal's Conjecture," after he applies the binomial theorem, he gets three results. His error comes in the second one.

He is right in that (1+y^b/x^a)^(1/m) converges to an irrational number, but he errs in saying that "z in equation (4) is an irrational number for any x^(a/m)," because there exist some x^(a/m), such that multiplying the two numbers will yield a rational number. The assumption is that multiplying an irrational number by any number will stay irrational, but there are pairs of irrational numbers that, when multiplyed, are rational (e.g. pi * 1/pi = 1).

Take the simple example of Beal's conjecture, 3^3+6^3=3^5. Inputting these variables into his binomial expansion form, we get 3=6^(3/5)*(1+3^3/6^3)^(1/5). The parentheses term,(1+27/216)^(1/5), is irrational, but multiplying by 6^(3/5), another irrational number, results in a solution of 3.

It's a simple flaw, but it means that the solution isn't as close as it may seem.

I have just modified this entry in the encyclopedia by being a little more specific as far as the definition of a "failure" goes.However, I would like to emphasise that the definition of a "failure" will depend on the problem in hand.For example if we are tackling RH the definition of a failure would be a non-root.For the purpose of illustration I am just giving one of the relevant failure functions.

Let s be a non-root( a failure).Then

zeta(s + k*zeta(s)) is a failure function.Here k belongs to N.

A.K.Devaraj

> I have just modified this entry in the encyclopedia by
> being a little more specific as far as the definition of a
> "failure" goes.However, I would like to emphasise that the
> definition of a "failure" will depend on the problem in
> hand.For example if we are tackling RH the definition of a
> failure would be a non-root.For the purpose of illustration
> I am just giving one of the relevant failure functions.
>
> Let s be a non-root( a failure).Then
>
> zeta(s + k*zeta(s)) is a failure function.Here k
> belongs to N.
>
> A.K.Devaraj

I would like to clarify that s + k*zeta(s) is the failure function and zeta(s + k*zeta(s)) generates an infinite set of failures ( non-roots).

Ok that shows Constant's disproof of Beal's Conjecture is false. What about Constant's proof of FERMAT'S LAST THEOREM? No on has address this yet.

No, he did not. At least his "proof" is short and clear enough, so that his mistakes are easy to find.
He also declares that the question why his "proof" holds for exponents greater than 2, but not for 2, is "beyond this paper"...

Unfortunately I don't understand Lagrange's form enough to know why it is wrong. I was hoping someone could explain "the why".
Thanks.

Just before his equation (6), James Constant claims, without demonstration, that "it is easy to show that the parenthesis term in equation (5) is an irrational number". If this was true, that would indeed proove Fermat's theorem. But "it is easy to find" counter-examples for exponent 2 (Pythatorean triples); James Constant has no explanation why exponent 2 is different: it is beyond the scope of his paper...
When we deal with Fermat's theorem, where the most briliant minds of mankind have failed for three centuries, we expect some detailed demonstrations, and not just hand waving with "it is easy to show...".

Daniel

Hi.

I know this is an old thread, but this guy is a joke, right ?
You just have to read the first lines of his "proof" to figure this out.

I hope he's got a real job. Maths is not for him. It's not even maths.

Listen, James ! Make a video of yourself and all your so called 'proofs'. I'm sure you'll get a lot of views in youtube's funny videos category. You could make some money.
And stop spreading your 'science' on internet. Some kids might think it's for real.
Your blindness scares me :)

The examples found by the computer are cases of the Beal‘s Conjecture . A, B and C have a common factor in every example that is written here . Here there is no disproof of the Beal‘s Conjecture . Therefor A , B and C are relatively prime , but at a superficial analisys. The Beal‘s Conjecture hold and there is no disproof of this conjecture here . I don‘t get the point of the very good mathematician , professor James Constant . A comment posted by Spanu Dumitru Viorel , from Bucharest , Romania . Email : spanuviorel@yahoo.com OK , the Beal‘s Conjecture said that A , B and C are not relatively prime . In the examples found by the computer , you must divide any equation by a common factor and you will obtain an equation where A‘ , B‘ and C‘ are not relatively prime . Let‘s take an example : 3^3 + 6^3 = 3^5 This equation has a common factor that is 3^3 . If you divide the equation written by 3^3 you will obtain

1 + 2^3 = 3^2 id est 1 +8 = 9

If you will divide any examples presented found by the computer with the common factor of that equation , you will obtain that A‘ , B‘ and C‘ are not relatively prime .

The conclusion is : The Beal‘s Conjecture Holds .

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