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June 14th, 2014, 12:29 PM  #1 
Senior Member Joined: Aug 2010 Posts: 158 Thanks: 4  Fermat's last theorem proof
Fermat’s Last Theorem This theorem basically states that A^n + B^n ≠ C^n, n > 2 if A, B, C and n are all positive integers. This inequation can be rewritten as C^n – B^n ≠ A^n. Andrew Wiles, a British mathematician, produced a lengthy proof of over 100 pages in the mid 1990s. Proof Let it be initially assumed that Fermat’s inequation is true i.e. it is an equation. If A, B and C have a highest common factor other than 1, P say, P^n can be cancelled out to give a new Fermat equation. Therefore right from the onset it will be assumed that A, B and C do not have any common factors other than 1 i.e. only ‘primitive’ equations/inequations need be considered here. C^nB^n can be rewritten as [CB][C^(n1) + C^(n2) B + C^(n3)B^2 + ....CB^(n2) + B^ (n1)] Let [CB] be any of the prime factors of C^nB^n. This means that [CB] should also be a factor of [C^(n1) + C^(n2) B + C^(n3)B^2 + ....CB^(n2) + B^ (n1)] since [CB][C^(n1) + C^(n2) B + C^(n3)B^2 + ....CB^(n2) + B^ (n1)]= A^n. More importantly this also means that [CB]^(n1) is a factor of [C^(n1) + C^(n2) B + C^(n3)B^2 + ....CB^(n2) + B^ (n1)]. Algebraic long division of [C^(n1) + C^(n2) B + C^(n3)B^2 + ....CB^(n2) + B^ (n1)] by [CB] always results in an algebraic remainder of nB^(n1). This means that [CB] needs to go into nB^(n1) exactly K{[CB]^(n  2)} times for the algebraic long division process to be complete, where K is some positive integer constant and nB^(n1) = K{[CB]^(n  1)}. Therefore [CB]^(n  1) must be a factor of B^(n1) unless [CB] > B. This is a contradiction as CB cannot be a factor of B since C and B do not share any common factors other than 1. Fermat's Last Theorem is therefore proved. 
June 14th, 2014, 07:18 PM  #2  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,330 Thanks: 2457 Math Focus: Mainly analysis and algebra  $(CB)$ can't be any of the prime factors of $(C^nB^n)$. It is (at most) one of them. This is placing a restriction on $B$, $C$ and $n$. You would need to prove that such a triplet of positive integers exist. But I think I can save you the trouble. Quote:
But $(xy)$ is not a factor of $(x^2 + xy + y^2)$ unless $3y^2 = 0 \implies y=0$. This is true even if $x^3  y^3 = z^3$. A similar result goes for all $n$. If $(x  y)$ is a factor of a polynomial $p(x,y)$ then $p(x, x) = p(y, y) = 0$. For the polynomial in $B$ and $C$ in your proof, this means that $(CB)$ is a factor of $C^{n1} + C^{n2} B + C^{n3}B^2 + ....CB^{n2} + B^{n1}$ only if $nC^{n1} = 0 \implies C = 0$. This is rather fatal to your proof. Last edited by v8archie; June 14th, 2014 at 07:33 PM.  
June 15th, 2014, 03:23 AM  #3 
Senior Member Joined: Aug 2010 Posts: 158 Thanks: 4 
v8archie CB could make any prime number.

June 15th, 2014, 03:29 AM  #4 
Senior Member Joined: Aug 2010 Posts: 158 Thanks: 4 
So in this proof CB takes only prime values.

June 15th, 2014, 06:36 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,330 Thanks: 2457 Math Focus: Mainly analysis and algebra 
Well it can't be a proof of Fermat's Last Theorem then, since that states that there are no triples (A, B, C) among the positive integers for which $A^n + B^n = C^n$ for any integer $n \gt 2$. There is no mention of BC being prime.

June 15th, 2014, 07:34 AM  #6 
Senior Member Joined: Aug 2010 Posts: 158 Thanks: 4 
The approach of my proof is to show that the proof of FLT boils down to simply showing that A^n cannot be made equal to C^n  B^n by noting that A^n is made up of prime factors where each can be expressed as (C  B)^n.

June 15th, 2014, 08:26 AM  #7 
Senior Member Joined: Sep 2010 Posts: 221 Thanks: 20 
With predictable remainder nB^(n1) the assumption that CB divides C^(n1)+...+B^(n1) is invalid. But nothing prevents to assume CB=K^n; C^(n1)+...+B^(n1)=Q^n; and A=KQ

June 15th, 2014, 08:36 AM  #8 
Senior Member Joined: Aug 2010 Posts: 158 Thanks: 4 
McPogor I don't get your point. I have shown that CB will never be a factor of C^(n1)+...+B^(n1) which is the backbone of my proof.

June 15th, 2014, 08:38 AM  #9 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,330 Thanks: 2457 Math Focus: Mainly analysis and algebra  But $A^n$ does not need to have only $n$ prime factors. For example 64 is a cube, but has 6 prime factors.

June 15th, 2014, 08:58 AM  #10 
Senior Member Joined: Aug 2010 Posts: 158 Thanks: 4 
v8archie nowhere have I said or implied that A^n must have n prime factors.


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