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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^n-B^n can be rewritten as [C-B][C^(n-1) + C^(n-2) B + C^(n-3)B^2 + ....CB^(n-2) + B^ (n-1)] Let [C-B] be any of the prime factors of C^n-B^n. This means that [C-B] should also be a factor of [C^(n-1) + C^(n-2) B + C^(n-3)B^2 + ....CB^(n-2) + B^ (n-1)] since [C-B][C^(n-1) + C^(n-2) B + C^(n-3)B^2 + ....CB^(n-2) + B^ (n-1)]= A^n. More importantly this also means that [C-B]^(n-1) is a factor of [C^(n-1) + C^(n-2) B + C^(n-3)B^2 + ....CB^(n-2) + B^ (n-1)]. Algebraic long division of [C^(n-1) + C^(n-2) B + C^(n-3)B^2 + ....CB^(n-2) + B^ (n-1)] by [C-B] always results in an algebraic remainder of nB^(n-1). This means that [C-B] needs to go into nB^(n-1) exactly K{[C-B]^(n - 2)} times for the algebraic long division process to be complete, where K is some positive integer constant and nB^(n-1) = K{[C-B]^(n - 1)}. Therefore [C-B]^(n - 1) must be a factor of B^(n-1) unless [C-B] > B. This is a contradiction as C-B 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
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Quote:
 Originally Posted by MrAwojobi Let [C-B] be any of the prime factors of C^n-B^n.
$(C-B)$ can't be any of the prime factors of $(C^n-B^n)$. 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:
 Originally Posted by MrAwojobi [C-B] should also be a factor of [C^(n-1) + C^(n-2) B + C^(n-3)B^2 + ....CB^(n-2) + B^ (n-1)] since [C-B][C^(n-1) + C^(n-2) B + C^(n-3)B^2 + ....CB^(n-2) + B^ (n-1)]= A^n
$$x^3 - y^3 = (x-y)(x^2 + xy + y^2)$$
But $(x-y)$ 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 $(C-B)$ is a factor of $C^{n-1} + C^{n-2} B + C^{n-3}B^2 + ....CB^{n-2} + B^{n-1}$ only if $nC^{n-1} = 0 \implies C = 0$.

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 C-B 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 C-B 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 B-C 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^(n-1) the assumption that C-B divides C^(n-1)+...+B^(n-1) is invalid. But nothing prevents to assume C-B=K^n; C^(n-1)+...+B^(n-1)=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 C-B will never be a factor of C^(n-1)+...+B^(n-1) which is the backbone of my proof.
June 15th, 2014, 08:38 AM   #9
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Quote:
 Originally Posted by MrAwojobi 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.
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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