It still contains no valid proof for all the reasons we all have mentioned.

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- 01-01-2017, 04:03 PM #1

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## Fermat's Theorem, Revistited

**To the reader: Please read my Ignore List before continuing this thread**

--------------------------------------------------------

Since grapes (snip NF)has closed my original thread "Fermat's Theorem, Revisited (against my wishes, so I can no longer add a reference to the pdf below to the original post to save readers' time, where the full analysis is available without the flak), I wish to announce that the discussion is continued with an in depth analysis in the thread entitled "The Relativistic Unit Circle", and is available in PDF form at:

The Relativistic Unit Circle

with reference to the fundamentals of physics, math, philosophy, and many other applications.

This document continues to be actively updated with additional sections on topics relevant to its title. (See signature below)

**Proof of Fermat's Theorem**

If Fermat's expression were true (), , , and would be a Pythagorean Triple. They are not a Pythagorean Triple by the Binomial Theorem, where rem(a,b,n) > 0.

**Therefore,****, for a,b,c,n positive integers, n>2**

**QED**

Triangle.pngLast edited by BuleriaChk; 02-24-2017 at 03:00 PM. Reason: Cage Match

_______________________________________

"Flamenco Chuck" Keyser

The Relativistic Unit Circle**03/28/2017 07:40 AM PST**

Proof of Fermat's Last Theorem Updates**03/19/2017 8:23 PM PST**

**Ignore List -The Peanut Gallery.**

- 01-03-2017, 06:33 AM #2

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## Re: Fermat's Theorem, Revistited

It still contains no valid proof for all the reasons we all have mentioned.

- 01-03-2017, 07:26 AM #3

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## Re: Fermat's Theorem, Revistited

By "we" you mean yourself and the other two(snip NF)... and your "reasons" only mean you don't understand the foundation of the proof to begin with...

As far as "no valid proof", time will tell.... I'm still waiting to hear from someone knowledgeable and intelligent....Last edited by Neverfly; 01-05-2017 at 07:59 AM.

_______________________________________

"Flamenco Chuck" Keyser

The Relativistic Unit Circle**03/28/2017 07:40 AM PST**

Proof of Fermat's Last Theorem Updates**03/19/2017 8:23 PM PST**

**Ignore List -The Peanut Gallery.**

- 01-04-2017, 07:05 AM #4

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## Re: Fermat's Theorem, Revistited

With "We" I mean those here more mathematicly knowledgable than you and with "reason" I mean that your ideas are mathematicly and logically flawed.

You will keep waiting, why do you not send it to Cambridge mathematicians? They will put it in their files of crankery. Did you know that? They keep a case of all the crankery that gets sent to them?

Why not try this bloke? You might get a whole article there on your crankery

But know Malaria, in mathematics, I am far more knowledgable and intelligent than you will ever be.

For fuck sake I know how to use notation properlyLast edited by emperorzelos; 01-04-2017 at 07:07 AM.

- 01-04-2017, 08:04 AM #5

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_______________________________________

"Flamenco Chuck" Keyser

The Relativistic Unit Circle**03/28/2017 07:40 AM PST**

Proof of Fermat's Last Theorem Updates**03/19/2017 8:23 PM PST**

**Ignore List -The Peanut Gallery.**

- 01-04-2017, 12:45 PM #6
## Re: Fermat's Theorem, Revistited

FermatPoof01.jpg

There it is, in its entirety.

You assume and go on to prove that cannot equal

That is not a proof of Fermat's Last Theorem. You've only proven it for a very narrow set of values, those values where c=a+b

In order to be accepted as a general proof you have to prove it for all values of (c,a,b)

QED

- 01-04-2017, 01:18 PM #7

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## Re: Fermat's Theorem, Revistited

My proof is based on the concept that there is only one unique set of integers on a single number line, represented by (a). For generality two relate two different integers, two number lines are needed (a,b) with the relationship given ultimately by the relativistic unit circle.

This means that the only integers that are countable if 1 is chosen as initial state are those for which in a single dimension.

for the final state (1) invariant.

If the initial state = final state, then ,

if the final state is greater than the initial state, then

c= a+b means that the symbols "c" and "a+b" refer to the same unique integer on a single number line , means that c = d, so that (no b required).

**You keep repeating the same misconception about the nature of dimensions (not to mention trigonometry, calculus, polynomials, all of which require more than one dimension. Including the equation .**

Sheesh, have you EVER graphed a function? Even in high school? Any trigonometry even?

(You might learn something if you read the rest of the proof....)

(snip NF)Last edited by Neverfly; 01-05-2017 at 08:01 AM. Reason: Cage Match

"Flamenco Chuck" Keyser

The Relativistic Unit Circle**03/28/2017 07:40 AM PST**

Proof of Fermat's Last Theorem Updates**03/19/2017 8:23 PM PST**

**Ignore List -The Peanut Gallery.**

- 01-04-2017, 01:32 PM #8
## Re: Fermat's Theorem, Revistited

Anybody can read the entire proof. It's nonsense, as I said. You cannot just assume that c=a+b

that there is only one unique set of integers on a single number line, represented by (a). For generality two relate two different integers, two number lines are needed (a,b) with the relationship given ultimately by the relativistic unit circle.

This means that the only integers that are countable if 1 is chosen as initial state are those for which in a single dimension.

for the final state (1) invariant.

If the initial state = final state, then ,

if the final state is greater than the initial state, then

c= a+b means that the symbols "c" and "a+b" refer to the same unique integer on a single number line, given a metric. Therefore, it means that for , so that c = a.

**You keep repeating the same misconception about the nature of dimensions (not to mention trigonometry, calculus, polynomials, all of which require more than one dimension. Including the equation .**

Sheesh, have you EVER graphed a function? Even in high school? Any trigonometry even?

- 01-04-2017, 01:43 PM #9

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## Re: Fermat's Theorem, Revistited

You say I have to prove "it" for all a, b, c. ?

So I have to prove what? (my proof contains**n**The proof is obvious by inspection. But the relativistic unit circle nails it, not only for integers, but for ALL positive real numbers (integers are just a special case, and the 3,4,5 right triangle exposes an interesting symmetry relevant to QFT).**o**other elements then the general variables a, b, and c, and neither does the Binomial Theorem.. and Fermat's theorem is expressed in its context.

**c=a+b isn't even really necessary for the proof if the Binomial Theorem is valid**; I just introduced it to show the vector nature of the expression when a and b are independent variables and c is the dependent variable.

If a and b are NOT independent variables, then some other variable must be involved, but is not expressed in either Fermat's theorem or the Binomial Theorem.

Time will tell; I'm finally making contacts at the university. (snip NF)

Others are reading this, and you are on record... Good luck with that....

I'm sure you don't understand the "rest of the proof", since you don't understand trigonometry.

(snip NF)

and you're still a despicable moderator who keeps trying to silence me.... by deluging my threads with repetitive spam, if not closing them outright.Last edited by Neverfly; 01-05-2017 at 08:03 AM. Reason: Cage Match

"Flamenco Chuck" Keyser

The Relativistic Unit Circle**03/28/2017 07:40 AM PST**

Proof of Fermat's Last Theorem Updates**03/19/2017 8:23 PM PST**

**Ignore List -The Peanut Gallery.**

- 01-04-2017, 02:11 PM #10
## Re: Fermat's Theorem, Revistited

You have to prove Fermat's Last Theorem for all possible values of (c,a,b), not just the ones where c=a+b

That's why you think the proof is "trivial"

It's not.

?

So I have to prove what? (my proof contains**n**The proof is obvious by inspection. But the relativistic unit circle nails it, not only for integers, but for ALL positive real numbers (integers are just a special case, and the 3,4,5 right triangle exposes an interesting symmetry relevant to QFT).**o**other elements then the general variables a, b, and c, and neither does the Binomial Theorem.. and Fermat's theorem is expressed in its context.

c=a+b isn't even really necessary for the proof if the Binomial Theorem is valid; I just introduced it to show the vector nature of the expression when a and b are independent variables and c is the dependent variable. If a and b are NOT independent variables, then some other variable must be involved, but is not expressed in either Fermat's theorem or the Binomial Theorem.

Time will tell. But you will always remain a village idiot. Along with the rest of the Peanut Gallery.

Others are reading this, and you are on record... Good luck with that....

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