What is the validity of my father's proof for Fermat's Last Theorem?

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The discussion revolves around the validity of a proposed proof for Fermat's Last Theorem, attributed to the author's father. Key points include skepticism about the proof's claims regarding generating Pythagorean triples and its reliance on reductio ad absurdum arguments, which some participants find unconvincing. Critics highlight that the proof contains ambiguous notations and potentially flawed reasoning, particularly in its treatment of cases for n=3 and n=4. Despite the author's confidence in the proof's originality and the mathematical models presented, many participants urge for more rigorous examination and clarity. The conversation underscores the challenges of validating new mathematical proofs against established theories.
  • #31
Would you mind helping out the lesser mortals round here like me who don't see what you're getting at? What riddle, what answer, what ellipsis?
 
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  • #32
[x+y]^2 - x^2 = 2xy + y^2

3^2 = 2*4*1 + 1^2

5^2 = 2*12*1 + 1^2

7^2 = 2*24*1 + 1^2

[x+y]^3 - x^3 = 3yx^2 + 3xy^2 + y^3


y[3x^2 +3xy + y^2]

cannot be a cube.

y[2x+y] is a square.

y[3x^2 +3xy + y^2] is not a cube

y[4x^3 + 6yx^2 + 4xy^2 + y^3] is not a 4th power
 
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  • #33
Yes. And? I think it was archimedes who proved there were infinitely many pythagorean triples because there are infinitely many odd squares (plus some coprimality). If you happen to have a good test lying around to check every single case there then you've got a proof of FLT.
 
  • #34
The proof boils down to the fact that

[binomial expansion]^n - [first term]^n

cannot be an integer nth power, for n > 2

reducto, and Wiles is
 
  • #35
So you've got a proof lying around that none of those expressions can possibly the n'th power of an integer? Cor, it's amazing no one thought of this line of attack!
 
  • #36
Why don't you start by showing why y[3x^2 +3xy + y^2] can't be a cube, and then we'll go on from there.
 
  • #37
Originally posted by selfAdjoint
Why don't you start by showing why y[3x^2 +3xy + y^2] can't be a cube, and then we'll go on from there.

For the simple case x = x and y = 1:

[x+1]^3 - x^3 = 3x^2 + 3x + 1

equals

6*x*[x+1]/2 + 1

6*1+1 = 6+1

6*2+1 = 12+1

6*3+1 = 18+1

6*6+1 = 36+1

etc...

equals

6*N*[N+1]/2 + 1
equals

6*[1+2+3+...+ N] + 1

not a cube...
 
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  • #38
Originally posted by Russell E. Rierson
[x+y]^2 - x^2 = 2xy + y^2

3^2 = 2*4*1 + 1^2

5^2 = 2*12*1 + 1^2

7^2 = 2*24*1 + 1^2

[x+y]^3 - x^3 = 3yx^2 + 3xy^2 + y^3


y[3x^2 +3xy + y^2]

cannot be a cube.

y[2x+y] is a square.

y[3x^2 +3xy + y^2] is not a cube

y[4x^3 + 6yx^2 + 4xy^2 + y^3] is not a 4th power








y[3x^2 + 3xy + y^2]

x = x

y = n

1[3x^2 + 3x + 1]

2[3x^2 + 6x + 4]

3[3x^2 + 9x + 9]

4[3x^2 + 12x + 16]

5[3x^2 + 15x + 25]

6[3x^2 + 18x + 36]

etc...

etc...

etc...

Now let x = 1, with y = n

1[3+3+1] = 2^3 - 1^3

2[3+6+4] = 3^3 - 1^3

3[3+9+9] = 4^3 - 1^3

4[3+12+16] = 5^3 - 1^3

5[3+15+25] = 6^3 - 1^3

6[3+18+36] = 7^3 - 1^3

etc...

etc...

etc...
 
  • #39
And that is never a cube for what reason? And the other infinite set of cases for n=3 are also false because?

I don't remember a proof that 1 more than 6 times the sum of the first N numbers is a cube, so fill in the blanks for me.
 
  • #40
Originally posted by matt grime
And that is never a cube for what reason? And the other infinite set of cases for n=3 are also false because?

I don't remember a proof that 1 more than 6 times the sum of the first N numbers is a cube, so fill in the blanks for me.

It means that the "Universal Set" is finitely representable...

Another simplistic example:

x = 1, y = n

y[3 + 3y + y^2]

1*[3 + 3 + 1] = 2^3 - 1^3

=

1*[2^2 + 3]

2*[3 + 6 + 4] = 3^3 - 1^3

=

2*[3^2 + 4]

3*[3 + 9 + 9] = 4^3 - 1^3

=

3*[4^2 + 5]

etc...

etc...

etc...


Let x = 2, y = n

y[3x^2 + 3xy + y^2] = [x+y]^3 - x^3

y[12+ 6y + y^2]

1[12+6+1] = 3^3 - 2^3 = not a cube

2[12+12+4] = 4^3 - 2^3

3[12+18+9] = 5^3 - 2^3

4[12+24+16] = 6^3 - 2^3

etc...

etc...

etc...

1*[4^2+3] = 3^3 - 2^3

2*[5^2+3] = 4^3 - 2^3

3*[6^2+3] = 5^3 - 2^3

4*[7^2+3] = 6^3 - 2^3

etc...

etc...

etc...

x = 3, y = n

y[27 + 9y + y^2]

1*[27 + 9 + 1] = 4^3 - 3^3

2*[27 + 18 + 4] = 5^3 - 3^3

3*[27 + 27 + 9] = 6^3 - 3^3

4*[27 + 36 + 16] = 7^3 - 3^3

etc...

etc...

etc...

1*[5^2 + 12] = 4^3 - 3^3

2*[6^2 + 13] = 5^3 - 3^3

3*[7^2 + 14] = 6^3 - 3^3

etc...

etc...

etc...

x = 4, y = n

y*[48 + 12y + y^2]

1*[7^2 + 12] = 5^3 - 4^3

2*[8^2 + 12] = 6^3 - 4^3

3*[9^2 + 12] = 7^3 - 4^3

etc...

etc...

etc...

Interesting...
 
  • #41
The universal set? Representable? Do you practise at this or does it come naturally?
 
  • #42
Originally posted by matt grime
The universal set? Representable? Do you practise at this or does it come naturally?

The mathematical structure that is isomorphic to the universe, yes, a "natural" mathematics.
 
  • #43
Q@A

A^3 + B^3 = [A+B]*[A^2 - AB + B^2]


[x+y]^3 + x^3 = [x+y+x]*[(x+y)^2 -x*(x+y) + x^2]

= [2x+y]*[x^2 + xy + y^2]

x = x

y = 1

[2x+1]*[x^2 + x + 1]

interesting...

[derivative]*[antiderivative] = x^3 + [x+1]^3
 
  • #44
Interesting...

x^3 + y^3 = [x+y]*[x^2-xy+y^2] = [x+y]*[(x+y)^2-3xy]

Let x+y = A

Let xy = B

[x+y]*[(x+y)^2-3xy] = A*[A^2-3B] = [A^3 - 3AB]

[A^3 - 3AB] cannot be a cube

8 - 6B is not a cube

27 - 9B is not a cube

64 - 12B is not a cube

125 - 15B is not a cube

A^3 - 3AB

8 - 6*1 = 2 = 1^3 + 1

27 - 9*1 = 2^3 + 20

27 - 9*2 = 2^3 + 1

64 - 12*1 = 3^3 + 25

64 - 12*2 = 3^3 + 13

64 - 12*3 = 3^3 + 1

interesting...
 
  • #45
A possible clue, to the riddle of Fermat? :

x^p + y^p

=

[x+y]*[x^(p-1)+y^(p-1) - xy*[{x^(p-2)+y^(p-2)}/(x+y)]]


x^2+y^2 = [x+y]*[(x+y)-xy*[2/(x+y)]

=

(x+y)^2 - 2xy

x^3+y^3 = [x+y]*[(x^2+y^2)-xy*[(x+y)/(x+y)]

etc.

etc.

etc.
 
  • #46
5^1 = 1*0 + 5

5^2 = 2*10 + 5

5^3 = 3*40 + 5

5^p = p*a + 5

x^p = p*a + x



x^p = p*a + x

y^p = p*b + y

z^p = p*c + z



...x^p + y^p = z^p



p*a + x + p*b + y = p*c + z

p*[a + b - c] = z - [x + y]

p = [z - (x + y)]/[a + b - c]
 
  • #47
Tom Ballard work

Digiflux:

Thank you very much for sharing us your father work on FLT.
I make already a print of it and sent it to some of my friends.
It looks a very interesting and also professional mathematical work.
I like the Geometry attitude to this problem and I will do my best to study it.

It will be very surprising if the 350 years problem can be solve so shortly but you never know. This is what so nice in mathematics. maybe it is the prove from the book as Erdos was saying. Still I think that Wiles he is the one to say if it corrects prove or there is hole somewhere. I hope that no but if there is one it may be filled one day.

Best
Moshek
:smile:

https://www.physicsforums.com/showthread.php?t=17243
 
  • #48
Hi:

First question come to my mind after passing on your father nice and elegant work is: Maybe there are solution to x^n+y^n=z^2 that can come not from his geometrical attitude to the case n=2. I will continue to learned his work.

Best
Moshek
:smile:
 
  • #49
Digiflux:

Have you got already any respond from A.Willes who solved this problem after he fix his mistake , to your father geometric attitude to this problem?

Moshek
 
  • #50
My Father's Theorem

Moshek:

Thanks for your kind words regarding my father's theorem. If he were still alive, I have no doubt that he would clearly answer all of your questions. I am an artist, so I cannot.

My father sent Wiles a copy but he received NO REPLY, and Princeton is conveniently, no longer reviewing Fermat proof submissions . They don't want to lose the notoriety that they have received from The Wiles "Proof". I put "proof" in quotes because, the Wiles "proof" is so complex that only a handful of people on the planet claim to understand it and they could be wrong. It's as big as a phone book and Wiles had a committee of Princeton mathematicians helping him. I don't buy it. The Wiles "proof" SUCKS.

My father had this proof for decades in his head and finally decided to write it down.

Thanks also to everyone who has studied and commented on my father's work.
 
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  • #51
Posted by matt grime
***************
yes, it;s a conspiracy, damn you you've figured it out, we're all charlatans and your father an unrecognized genius that puts us all to shame... oh, please. how, if your father's work is so simple (and wiles's so hard) can you not figure out what it was and answer questions on it? any mathematician would be able with sufficient time and inclination learn and defend Wiles's proof. :zzz: :zzz:
***************
Reply:

Princeton has gotten tremendous publicity from Wiles and his bogus "proof". That means $ pour in. They ARE protecting their crybaby, Wiles.

Which independent mathematicians have confirmed Wiles proof? None that I know of. You are just taking Wiles word on faith. Faith and math don't mix. Yer not one of those "creation evidence" religious nuts are you?
 
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  • #52
digiflux said:
Princeton has gotten tremendous publicity from Wiles and his bogus "proof". That means $ pour in. They ARE protecting their crybaby, Wiles.

Sure, hardly anyone had even heard of Princeton before Wiles. :rolleyes:

This was the first time I saw this thread and I was honestly considering giving your fathers work a read until I saw your last two posts. Your attitude and conspiracy theory is counter productive to your goal of publicising your fathers work. If it's correct then it will stand on it's own without you crying foul about how Wiles proof is just a "proof" because you can't understand it. If you want to prove Wiles proof is wrong you'll have to find an error in his work, otherwise you're just spouting unfounded slander that really has nothing to do with whether your fathers work is correct or not and it's not motivating me to help you in any way.

By the way, even if a simple proof does turn up someday it will in no way diminish Wiles accomplishment, he will always be the first to have tackled the problem. So he (and the manipulating overlords at Princeton) really would have no reason to supress such a thing. They, and I'm sure many other universities, probably don't review submissions on Fermat anymore because of the sheer volume of attempts and the historic waste of time it's been to find errors in random submissions. I would bet that the announcement of a proof inaccessible to people not willing to do the work required to understand it has only furthur motivated elementary attempts.

Before you claim that I'm a "defender of the faith" let me say something about my belief of Wiles proof. It's been around for quite some time and has been read and understood by people much smarter than I am, and much more qualified to find errors. No errors have been found to date. Mathematicians are pretty ruthless when it comes to finding errors in proofs, so I'm pretty confident that it's correct. However if my life, or even just my own work, was dependant on the proofs correctness you can be sure that I would do everything possible to verify it on my own. As it is I'm satisfied to be "pretty confidant".
 
  • #53
If it matters, I deleted my post before that reply came in because i thought it overly aggressive. perhaps i was correct in that assumption given your borderline libellous reply but incorrect to remove it.

Back to the same old misunderstandings from the non-mathemticians (I do love being told what mathematics is by someone who knows nothing about it).

wiles's proof has undergone peer review, but of course by other mathematicians so in your opinion it cannot be correct for they cannot be independent (anyone capable of understanding it on first reading is too dependent on wiles one presumes from your assertion that no 'independent' verifiers exist).

incidentally, any mathematician would be happy to find an error in wiles's proof, and many will hve poured through it to see how he came up with an idea they missed, they would then be even happier to find a correction as happened with the original version.

i am not taking wiles's word, i am taking that of the mathematical community whom i respect and who have had the proof availiable to read for several years now and who have publisehd it in a perr reviewed jounral. this does not mean it cannot be incorrect and no mathematician would ever say otherwise, but maths isn't quite the cut and dried subject you appear to think it is. Please feel free to look through the proof and find an error in it. just because you don't understand it means others cannot. (of course he didn't prove FLT directly he proved that all semistable elliptic curves are modular, but I'm sure you knew that). it is a long and difficult proof but it certainly appears correct. Devlin wrote an interesting article on the soundness of mathematical proofs in his AMA articles once.
 
  • #54
Digiflux,

sorry, but according to the comments in the guest book (at
http://books.dreambook.com/pokerface/fermatproof.html ),
your father's proof is wrong.
 
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