Re: FERMAT LAST THEOREM:ANOTHER PROOF

From: Raymond Burhoe (NOSPAMrcb_at_powerpuff.com)
Date: 01/13/05 

Date: Thu, 13 Jan 2005 16:10:53 +0000 (UTC)

On 12 Jan 2005, george ghiata wrote:
>
>This is a proof of Fermat Last Theorem by Gheorghe Ghiata
>EQF: X^+Y^n=Z^n (X,Y,Z Are integers and n>=3) is impossible
>PROOF:
> X^n-X+Y^n-Y=Z^n-Z+Z-X-Y
> X+Y-Z=B; Little theorem makes B divisible n
>Case1:X*Y*Z not divisible by n
> X=B+Q
> Y=B+P
> Z=B+Q+P
> EQ1 :(B+Q)^n+(B+P)^n=(B+Q+P)^n
> From EQ1 we get:
> X+Y=2*B+Q+P=U^n=W
> Z=U*V
> Q=q^n and P=p^n
>Therefore 2*B=U^n-q^n-p^n
>Little Theorem makes (U-q-p) divisible by n:
>Now we write:
> X^n-Q^n+Y^n-P^n-Z^n +W^n=(-Q^n+Q)+(-P^n+P)+(W^n-W)+
> +(W-Q-P)
> Therefore (W-Q-P)=2*B is divisible by n^2
>Therefore B is divisible by n^2
>Now EQ1 is divided by (2*B+Q+P) and get
> EQ2:
>[B^(n-1)+(a1)*B^(n-2)+(a2)*B^(n-3).........+c*B]+[Q^n+P^n]/(Q+P)=V^n
> From EQ2 we get that (V-1) is divisible by n and that
> {V^n - [Q^n+P^N]/(Q+P)} is divisible by n^2
>If we divide EQF by (X+Y) we get :
> EQ3: X^(n-1)-Y*X^(n-2)+[Y^2}*X^(n-3)..........+Y^(n-1)=V^n
>After we multiply EQ3 With Z we write It in two ways:
>X+Y=Z+B=W
> 1.
>Z*{W^(n-1)-n*[W^(n-2)]*Y+(1/2)*n*(n-1)*[W^(n-3)]*Y^2........+n*Y^(n-1)=Z*V^n
>
> 2. The same as 1. exept we substitute Y with X
> Now we Add -up (1.+2.) and get:
> EQ4: B*G +(Z-1)*n*[Y^(n-1)+X^(n-1)] +Q^n+P^n+X^n+Y^n=2*Z*V^n

Let k be any nontrivial prime factor of W.
 
It follows from simple math that:

W = 0 (mod k)
(X + Y) = 0 (mod k)
Z = 0 (mod k)
(X + Y - Z) = 0 (mod k)
B = 0 (mod k)
(Z - Y) + (Z - X) = 0 (mod k)
Q + P = 0 (mod k)
X^(n-1) - Y^(n-1) = 0 (mod k)

Applying that to EQ4,

2n = 0 (mod k)

If you forget all about FLT, and select values for X, Y, and Z, where (X + Y) and Z share some nontrivial factor, then you seem to have proved that that nontrivial factor cannot be coprime to 2n. If I state that n = 7, then you'd have proved that (9 + 24) and 121 cannot share a factor of 11.

> Now we substitute {X^n+Y^n] with Z^n and get to the left side of
>EQ4
>the value 2*Z*V^n.Since B is divisible by n^2 we get that (Z-1) is
>divisible by n.

What's to prevent me from multiplying X, Y, and Z by -1? if (-Z - 1) is also divisible by n, then n = 1 or 2.

>But Z=U*V and (V-1) is divisible by n.
>Therefore (U-1) is divisible by n.
>Repeating the proof twice we get That (q-1) and (p-1) are divisible by
>n too.
>But We know that (u-q-p) must be divisible by n too.
>That proves Fermat last theorem for Case1.
>Case 2; Z is divisible by n
>In this case ((q+p) is divisible by n.
>As in Case 1. we proof that (q-1) and (p-1) are divisible by n.
>Therefore this proves Case2 of Fermat Last theorem
>Therfore Fermat Last theorem Is true.
>Created by Gheorghe Ghiata
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