Re: Unexpected shot for FLT !
- From: "Roman B. Binder" <rbinder@xxxxxxxxxxxxxxxx>
- Date: Thu, 29 Jun 2006 21:20:29 EDT
matt271829-news@xxxxxxxxxxx wrote:
Roman B. Binder wrote:OK: I'll dismiss not necessary words and some
Hi,
You are afraid to say something
because of Your boss ?
" Silence tells me everything "
Ro-bin
I think your posts are generally very difficult to
understand. This
might discourage people from trying to work through
them.
very well known conditions:
1)-st of my steps for FLT:
X^n + Y^n = Z^n ??? for prime n>=3; X;Y;Z integers;
then there are such natural x;y;z of gcd=1 that:
Ls = x^n + y^n = z^n = Rs
Lets look for such natural c that:
Ls + c^n = (x^n + y^n + c^n)/xy
c = kxy -(x+y)
once Ls + c^n = (x+c)Ext + y^n = (y+c)Ext + x^n
Ls+c^n =(x+kxy-x-y)Ext +y^n =(y+kxy-x-y)Ext +x^n
Ls+c^n =y(kx -1)Ext + y^n = x(ky -1)Ext + x^n
Ls+c^n = y[(kx-1)Ext + y^(n-1)] =
= x[(ky-1)Ext + x^(n-1)}
Ls+c^n = (x^n + y^n + c^n)/ xy
for c = kxy -x -y
( generally true also for c;k integers
and for all n positive odd numbers
so for sure for n>=3 and prime numbers
and for k natural numbers )
is it clear till now ???
This is first step introduced also as
Lemma: let X^n + Y^n + C^n = L for odd n >0
then for C = KXY -(X+Y) and X;Y;K integers L/XY
More two my steps and FLT is ours in much more
easy way: if it is difficult for You to understand
so may be it is not so elementary way:
2) In the second step we'll try to find such
simple z that z^n + our c^n will be also divisible
by xy
3) In the third step we'll try to generalize all
possible cases of factorization of z^n + c^n
for to achieve (z^n + c^n)/xy
Again,is it clear till now ?
Can I continue my step 2 ?
With Compliments
Ro-bin
.
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