Re: FLT AND ITS GENERALIZATION, BEAL'S CONJECTURE

From: S. Enterprize Company (smart1234_at_aol.com)
Date: 12/29/04 

Date: 29 Dec 2004 13:02:25 GMT

>>>Nick Ancuta-Nazari wrote:
>>>> I kindly ask those interested in this subject to comment on my
>>>> approach "FLT AND ITS GENERALIZATION".
>>>> Thank you very much.
>>>> Regards,
>>>> Nick Ancuta-Nazari
>>>> nanazari@prodigy.net
>>>>
>>>> The TeX file is at
>>>>
>>>http://www.meadowdance.org/Wordsworth/Deliverables/FLT&BealConjecture.tex
>>>>
>>>> The PDF file is at
>>>>
>>>http://www.meadowdance.org/Wordsworth/Deliverables/FLT&BealConjecture.pdf
>>>
>>>The links about this conjecture, flt and its generalization no longer
>>>work, but info about Beal Conjecture is available at
>>>
>>>http://www.ams.org/new-in-math/mathnews/beal.html
>>>
>>>> As a banker in Dallas, Texas, Andrew Beal has an obvious
>>>> interest in numbers. But he has another interest that is not so
>>>> obvious: He is interested in the mathematical theory of numbers.
>>>> An amateur mathematics enthusiast, Beal came upon a question in
>>>> number theory that even the experts can't answer. The question turns
>>>> out to be at the frontier of research in the field, with connections
>>>> to other deep mysteries in mathematics. To spur mathematicians to
>>>> solve the problem, Beal has offered a prize of $5,000 for its
>>>> solution.
>>>> The prize will increase by $5,000 every year up to the amount of
>>>> $50,000.
>>>> Will the Beal Prize Problem become the next Fermat's Last
>>>> Theorem? Indeed, it is a generalization of that famous old problem,
>>>> which Pierre de Fermat proposed over 300 years ago. Like the
>>>> Fermat problem, the Beal Conjecture is easily stated:
>>>> If A^x + B^y = C^z,
>>>> then A, B, and C have a common factor. (Here all the letters
>>>> represent whole numbers, with x, y, and z bigger than 2. Two
>>>> numbers have a "common factor" if there is a number that divides
>>>> both of them evenly. For example, 12 and 63 have a common factor
>>>> of 3.)
>>>> Another resemblance between the Beal Conjecture and Fermat's Last
>>>> Theorem is that both had prizes established for their solutions. In
>>>> 1996, after Andrew Wiles made international headlines by presenting
>>>> the number theory arsenal that finally brought down Fermat's Last
>>>> Theorem, he collected the Wolfskehl Prize. Established in 1908 with
>>>> funds from the will of a German physician and amateur
>>>> mathematician, Paul Wolfskehl, the Wolfskehl Prize enormously
>>>> increased the fame of Fermat's Last Theorem by drawing thousands
>>>> of entries from all over the globe.
>>>> The article, "A Generalization of Fermat's Last Theorem: The
>>>> Beal Conjecture and Prize Problem," by Professor Daniel Mauldin,
>>>> appears in the December 1997 issue of the Notices of the AMS. This
>>>> article provides further details about Beal's question and its role
>>>> in modern number theory. See also the web site
>>>> http://www.math.unt.edu/~mauldin/beal.html.
>>>and the latest information at
>>>http://www.math.unt.edu/~mauldin/beal.html
>>>> THE BEAL CONJECTURE AND PRIZE
>>>> BEAL'S CONJECTURE: 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.
>>>> THE BEAL PRIZE. The conjecture and prize was announced in the
>>>> December 1997 issue of the Notices of the American Mathematical
>>>> Society. Since that time Andy Beal has increased the amount of the
>>>> prize for his conjecture.
>>>> The prize is now this: $100,000 for either a
>>>> proof or a counterexample of his conjecture. The prize money is being
>>>> held by the American Mathematical Society until it is awarded. In the
>>>> meantime the interest is being used to fund some AMS activities and
>>>> the annual Erdos Memorial Lecture.
>>>> CONDITIONS FOR WINNING THE PRIZE. The prize will be
>>>> awarded by the prize committee appointed by the American
>>>> Mathematical Society. The present committee members are Charles
>>>> Fefferman, Ron Graham, and Dan Mauldin. The requirements for the
>>>> award are that in the judgment of the committee,
>>>> the solution has been
>>>> recognized by the mathematics community. This includes that either a
>>>> proof has been given and the result has appeared in a reputable
>>>> refereed journal or a counterexample has been given and verified.
>>>> PRELIMINARY RESULTS. If you have believe you have solved the
>>>> problem, please submit the solution to a reputable refereed journal.
>>>> If you have questions, they can be mailed to:
>>>> The Beal Conjecture and Prize
>>>> c/o Professor R. Daniel Mauldin
>>>> Department of Mathematics
>>>> Box 311430
>>>> University of North Texas
>>>> Denton, Texas 76203
>>>
>>>> Questions and queries can also be FAXED to 940-565-4805 or sent by
>>>> e-mail to
>>>> mauldin@unt.edu
>>>> LINKS TO ARTICLES ABOUT THE CONJECTURE AND PRIZE
>>>
>>>> The Beal Conjecture
>>>> Notices American Mathematical Society, December 1997
>>>> Manchester Guardian January 8, 1998
>>>> A computer study has been carried out by Peter Norvig who is Chief of
>>>> the Computational Sciences Division at the NASA Ames Research
>>>> Center. The program and results may be found at
>>>> Beal's Conjecture: A Search for Counterexamples
>>>
>>
>>
>> I tried and I think the close as you can get to equality of both sides of
>>the equation is:
>>
>>7^3 ~= 6^3 + 5^3
>>
>>343 ~= 341
>>
>> Almost any other number combination has a larger difference.
>>
>> Would this qualify as a counter-example proof or proof?
>>
>
> You know I think that I can disprove, Fermat's Last Theorem.
>
>The basic form is established,
>
>a^n + b^n = c^n
>n >2
>a,b,c are integers
>
>Let m = 7^3 = 343 <-- the left hand side of equation
>Let n = 6^3 + 5^3 = 341 < -- the right hand side of equation
>
> 7^3 = 6^3 + 5^3
> 434 = 341
>
>(m + n)/2 = 342
>
>The average difference on each side of the equation is equal to 342.
>
> Using Surreal Non-standard Analysis,
>
> { 341 | 343 } = 342
>
> Equality IS found at the cut location.

  Ok you might say, but who says you can take the average difference here???
Ok,
Let's look at this this way then.

Surreal Non-Standard Analysis

{341| } = 342
{ | 343} = 342

   Fermat's Last Theorem is disproven at the Surreal Cut, by Smart1234. There
is a state of equality there. Both sides of the equation meet at 342.

In other words,

lim ( 343 --> k <--- 341)
k->surreal number

  Both numbers approach each other in a state of equality at k. k was found to
be 342. The equation is in a state of equality at k.

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