September 6th, 2015, 04:20 AM  #1 
Senior Member Joined: Oct 2011 Posts: 122 Thanks: 1  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+BC=0 and $\displaystyle A^n+B^n<C^n$. Notations are done in a number system with a prime base n: $\displaystyle A_{(t)} $ – tth 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]} $ – tdigits 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 $\displaystyle [(a_{[t+1]})^{n1}]_{(t+1)}=[x]=nx$. So, let us assume that for a prime number n>2, relatively prime A, B, C, and A'[or B']≠0 1°) $\displaystyle A^n=(CB)P$ [$\displaystyle =aP=C^nB^n$, where $\displaystyle P=p^n$ and /for convenience/ a=CB] where, as known, 1a°) P'=p'=1 (a consequence of Fermat's little theorem), 1b°) $\displaystyle [U=] A+BC=un^k$, 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$\displaystyle ^n=(CB)P$, where P$\displaystyle =Pg^{n1}$, 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 (CB)$\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 ^{n1}+$Q$\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)^{n1}]_{(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$\displaystyle _{[k+1]}=$(CB)$\displaystyle _{[k+1]}=1$ and 5°) [A(CB)]$\displaystyle _{[k+1]}=$[A+BC]$\displaystyle _{[k+1]}=$U$\displaystyle _{[k+1]}=0$, which contradicts to 1b°. Thus FLT proved. Victor SOROKINE 
September 6th, 2015, 04:24 AM  #2  
Senior Member Joined: Sep 2015 From: 4th Dimension Posts: 146 Thanks: 13 Math Focus: Everything (a little bit) 
Quite scary Quote:  
September 7th, 2015, 05:28 AM  #3 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 1,928 Thanks: 628 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
Could a moderator please move this to the number theory subforum? This is not high school algebra.

September 7th, 2015, 05:45 AM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,557 Thanks: 2148 Math Focus: Mainly analysis and algebra  
September 7th, 2015, 12:44 PM  #5 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,555 Thanks: 597 Math Focus: Wibbly wobbly timeywimey stuff. 
@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 
September 7th, 2015, 05:01 PM  #6 
Senior Member Joined: Dec 2013 Posts: 1,050 Thanks: 31 
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! 
September 7th, 2015, 06:04 PM  #7  
Senior Member Joined: Apr 2015 From: Barto PA Posts: 154 Thanks: 18  Quote:
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.  
September 7th, 2015, 06:58 PM  #8  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,555 Thanks: 597 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Dan  
September 8th, 2015, 09:54 PM  #9 
Senior Member Joined: Oct 2011 Posts: 122 Thanks: 1  Flt 
September 8th, 2015, 10:04 PM  #10 
Senior Member Joined: Oct 2011 Posts: 122 Thanks: 1  Flt 

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