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September 6th, 2015, 04:20 AM   #1
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An elementary proof of the Fermat’s Last Theorem

Fermat's Last Theorem (main case: n is prime):
For integers A, B, C and prime n>2 the equal $\displaystyle A^n+B^n=C^n$ does not exist.

The essence of the contradiction: If Fermat’s equality $\displaystyle A^n+B^n=C^n$ exists then A+B-C=0 and $\displaystyle A^n+B^n<C^n$.

Notations are done in a number system with a prime base n:
$\displaystyle A_{(t)} $ – t-th digit from the end in the number A; for convenience: $\displaystyle A_{(1)}=A'$, $\displaystyle A_{(2)}=A''$, $\displaystyle A_{(3)}=A'''$;
$\displaystyle A_{[t]} $ – t-digits ending of the number A; $\displaystyle A_{/t/}$, where A=pq…r, – the product of $\displaystyle p_{[t]}*q_{[t]}*...*r_{[t]}$.
From binomial theorem (for prime n and) its follow two simple lemmas:
0a°) if $\displaystyle A_{[t+1]}=xn^t+1$, where t>0 and A is the base of a degree $\displaystyle A^n$, then the digit $\displaystyle (A^n)_{(t+2)}=x$;
0b°) if $\displaystyle a_{[t+1]}=xn^t+1$, where digit $\displaystyle a_{(t+1)}=x>0$ и t>0, then the digit $\displaystyle [(a_{[t+1]})^{n-1}]_{(t+1)}=[-x]=n-x$.

So, let us assume that for a prime number n>2, relatively prime A, B, C, and A'[or B']≠0
1°) $\displaystyle A^n=(C-B)P$ [$\displaystyle =aP=C^n-B^n$, where $\displaystyle P=p^n$ and /for convenience/ a=C-B] where, as known,
1a°) P'=p'=1 (a consequence of Fermat's little theorem),
1b°) $\displaystyle [U=] A+B-C=un^k$, where k [>0] – the number of zeroes after the digit u' (i.e. $\displaystyle U_{[k+1]}≠0$).
1c°) g – any integer solution [which exists!] of the equation $\displaystyle (Ag)_{[k+2]}=1$.

An elementary proof of the Fermat’s Last Theorem

Let's multiply the equation 1° by the number $\displaystyle g^n$ from 1c° received the new equality 1°:
1°) A$\displaystyle ^n=(C-B)P$, where P$\displaystyle =Pg^{n-1}$, A$\displaystyle =Ag$, A$\displaystyle ^n=A^n*g^n$ and A$\displaystyle _{[k+2]}=$A$\displaystyle ^n_{[k+2]}=1$; k and n are const.
Let us show that the ending (C-B)$\displaystyle _{[k+2]} $, or a$\displaystyle _{[k+2]} $, is also equal to 1.
To do this, the number P will be represented in the following form: P=q$\displaystyle ^{n-1}+$Q$\displaystyle n^{k+2}$
[this is the KEY to the demonstration], where q and Q are integers.

2°) a'= q'=1, which is deduced from 1°b.

3°) From the identity A$\displaystyle ^n_{(2)}=[(a''n+1)(q''n+1)^{n-1}]_{(2)}= $ (cf. 0b°)=
$\displaystyle =[(a''n+1)(-q''n+1)]_{(2)} $ [=0] we find: a''=q'' and the degree of endings A$\displaystyle ^n_{/2/}=[(a''n+1)_{[2]}]^n$, from here (cf. 0a°) we find the digit [A$\displaystyle ^n]_{(3)} $:
4°) [A$\displaystyle ^n]_{(3)} $ (=0 – cf. 1°) = a'' and therefore a''=q''=0 (otherwise [A$\displaystyle ^n]_{(3)}≠0$).

And then, we makes calculations 3°-4° with all subsequent digits [until the (k+1)-th] of the numbers A, P and a, with the result equality A$\displaystyle _{[k+1]}=$P$\displaystyle _{[k+1]}=$a$\displaystyle _{[k+1]}=$(C-B)$\displaystyle _{[k+1]}=1$ and

5°) [A-(C-B)]$\displaystyle _{[k+1]}=$[A+B-C]$\displaystyle _{[k+1]}=$U$\displaystyle _{[k+1]}=0$, which contradicts to 1b°. Thus FLT proved.

Victor SOROKINE
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September 6th, 2015, 04:24 AM   #2
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Quite scary
Quote:
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September 7th, 2015, 05:28 AM   #3
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Could a moderator please move this to the number theory subforum? This is not high school algebra.
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September 7th, 2015, 05:45 AM   #4
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Originally Posted by Yash Malik View Post
Quite scary
It's almost certainly nonsense. I wouldn't waste my time reading it.
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September 7th, 2015, 12:44 PM   #5
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@victorsorokin

I can't read it myself, but if you feel you have an actual proof why are you posting it here and not trying to get published??

-Dan
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September 7th, 2015, 05:01 PM   #6
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This guy is genius and he deserves the Fields medal!
He is right! All the international mathematical community confirms his proof.
Let him send it to some mathematical journal. He will surely be published.
Do not debate with him. It will be loss of time.
Do not waste your time. Just applaud loudly!
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September 7th, 2015, 06:04 PM   #7
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Quote:
Originally Posted by mobel View Post
This guy is genius and he deserves the Fields medal!
He is right! All the international mathematical community confirms his proof.
Let him send it to some mathematical journal. He will surely be published.
Do not debate with him. It will be loss of time.
Do not waste your time. Just applaud loudly!
Somehow I detect just a slight note of facetiousness in your reply. Evidently
the OP has posted various editions of
his 'proof' on numerous sites for many
years using the last name 'Sorokine'.
His 'math' is well beyond my ability to
comprehend. Perhaps Professor Wiles
could have a look.
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September 7th, 2015, 06:58 PM   #8
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Quote:
Originally Posted by uvkajed View Post
Somehow I detect just a slight note of facetiousness in your reply. Evidently
the OP has posted various editions of
his 'proof' on numerous sites for many
years using the last name 'Sorokine'.
His 'math' is well beyond my ability to
comprehend. Perhaps Professor Wiles
could have a look.
Professor Wiles is currently working on Fermat's Second to Last Theorem.

-Dan
Thanks from agentredlum
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September 8th, 2015, 09:54 PM   #9
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Flt

Quote:
Originally Posted by topsquark View Post
@victorsorokin

I can't read it myself, but if you feel you have an actual proof why are you posting it here and not trying to get published??

-Dan
This topic is forbidden in all major forums.
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September 8th, 2015, 10:04 PM   #10
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Flt

Quote:
Originally Posted by topsquark View Post
Professor Wiles is currently working on Fermat's Second to Last Theorem.

-Dan
An elementary proof - this is a task for the crazy.
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