Re: Unexpected shot for FLT !

tbonepower07@xxxxxxxxx wrote:
Roman B. Binder wrote:
matt271829-news@xxxxxxxxxxx wrote:
Roman B. Binder wrote:
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.

OK: I'll dismiss not necessary words and some
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

Unless I misunderstood, you said

Ls + c^n = (x^n + y^n + c^n)/xy

but you also said

Ls = x^n + y^n

so you are saying

x^n + y^n+c^n = (x^n + y^n + c^n)/xy
1= 1/xy
If x and y are integers,
x = 1 and y = 1

So that's a flaw right there.
obviously not: for c = xy -(x+y); x=1;y=1
c = 1 - 2 = -1
1^n + 1^n -1^n = 1^n divisible by 1 =xy
Initially Lemma was indicated for x;y natural
numbers of gcd=1, like these resisting in FLT
question. I have seen chance for to retrieve more
general results, but for optional integers I can
now check some special configurations:
1) x=-y L=x^n +y^n +c^n
for c = kxy -(x+y)
L= x^n -x^n +(kxy -x+x)^n = (kxy)^n /xy
I can also return to general forms:
L = x^n +y^n +(kxy-x-y)^n configures to
x[(ky-1)Ext +x^(n-1)] = y[(kx-1)Ext +y^(n-1)]
so anyhow L/xy
May be You'll find better counter example ?

Compliments
Ro-bin
.

Relevant Pages

  • Re: Unexpected shot for FLT !
    ... I think your posts are generally very difficult ... for c = kxy -x -y ... More two my steps and FLT is ours in much more ... You are probably trolling, but in the offchance you are sincere, you must realize the proble stems from your use highly non standard of /, ...
    (sci.math)
  • Re: Unexpected shot for FLT !
    ... I think your posts are generally very difficult ... 1)-st of my steps for FLT: ... for c = kxy -x -y ... You are probably trolling, ...
    (sci.math)
  • Re: Unexpected shot for FLT !
    ... " Silence tells me everything " ... I think your posts are generally very difficult to ... for c = kxy -x -y ... More two my steps and FLT is ours in much more ...
    (sci.math)
  • Re: Unexpected shot for FLT !
    ... because of Your boss? ... " Silence tells me everything " ... I think your posts are generally very difficult to ... More two my steps and FLT is ours in much more ...
    (sci.math)
  • Re: Unexpected shot for FLT !
    ... might discourage people from trying to work ... 1)-st of my steps for FLT: ... for c = kxy -x -y ... You are probably trolling, ...
    (sci.math)
Quantcast