Re: FLT an incurable, unending addiction

Dear Mr. *** ,

Following Your remarks:

> I do not exactly understand what you wish, but when a
> is an arbitrary
> integer, not equal to 1, a.x^n + y^n = z^n *can* have
> solutions in the
> integers. And indeed for some a there are solutions.
>
> When you have a proof that x^n + y^n = z^n has no
> solution, make
> sure that your proof does not also show that a.x^n +
> y^n = z^n
> has no solution for arbitrary integer n. If such is
> also shown
> by the proof, the proof is wrong. (There are many
> solutions with
> n = 5 and a = 68101.)
> --
> *** t. winter, cwi, kruislaan 413, 1098 sj
> amsterdam, nederland, +31205924131
> home: bovenover 215, 1025 jn amsterdam, nederland;
> http://www.cwi.nl/~***/

I am sorry for not to point exactly my matter:
I hope to work with the correct and approximately
simple proof for FLT, also for questionable eq.:
X^n + Y^n = Z^n once n prime >=3
where X;Y;Z integers; then such set of integers could
be extracted of any common factor and substituted with
only natural values so let x;y<z and x;y;z of gcd=1
now x^n + y^n = z^n could be eventually substituted
with optional input: x=T+B; y=T+A; z=T+A+B
where T;A;B natural numbers
what gives to us in brief:
T^n = nAB(2T+A+B)*Ext
then possible solutions could be arranged with
following natural parameters a;b;t;p;u
once T = n^u abtp allways
once x is not divisible by n so B=b^n
once x is divisible by n so B=n^(nu-1) b^n
once y could be set not divisible by n so A=a^n
once z is not divisible by n so 2T+A+B = t^n
once z is divisible by n so 2T+A+B = n^(nu-1) t^n
once one of x;y;z is divisible by n so Ext = p^n
once no one of x;y;z ia divisible by n
( 1-st fall of FLT0 so Ext = n^(nu-1)
where a;b;t;p;n of gcd=1 once u>=2
(for u=1 could be executed Eisenstein criterion )

now beginning with n=3 when Ext=1; p=1
we'll have following eq-s:
3.1) for x/3: t^3 = 2 3^u abt + a^3 + 3^(3u-1) b^3
3.2) for z/3: 3^(3u-1) t^3 = 2 3^u abt + a^3 + b^3
for prime n>3 once Ext is substituted with
its specific natural number p so questionable
will be only 2T+A+B
n.1) for x;y;z not divided by n:
t^n = 2 n^u abtp + a^n + b^n
n.2) for x/n:
t^n = 2 n^u abtp + a^n + n^(nu-1) b^n
n.3) for z/n:
n^(nu-1) t^n = 2 n^u abtp + a^n + b^n
Now in my attempt using for example eq.(n.1)
lets rewrite it as following:
b^(n-2) b^2 + 2 n^u atpb + a^n - t^n = 0
what could be checked with 2-nd degree determinant
D(n.1)= 4[(n^u atp)^2 -b^(n-2)(a^n -t^n)]
where once it is also:
t^n - a^n = b[b^(n-1)+ 2 n^u atp] = bR
so we'll have:
(n^u atp)^2 + R[b^(n-1)/2]^2 = d^2
once should be D = 4d^2 for to complete at least
one formula with b natural number:
b = (- 2 n^u atp +/-2d)/2b^(n-2)
***
Now it will be more clear, why I used to recall
possible natural solutions of ax^2 + y^2 = z^2
and why I dismiss pythagorean parameters for
the case of a=a1^2...
My final results of this proof procedure
where already been written in some of my previous
topics: "World's two proofs of FLT"
"general solutions for eq. ax^2 + y^2 = z^2 "
during the last month in sci.math .
Some years ago I used to point to property
of natural roots of eq. of n-degree once
taken to some of degenerated to 2-nd degree
shapes. I don't know if such properties were
discussed in some math. article...
Anyhow some surprise or just common flew ?

Thank You for Your attention
With Highly Appropriation of Your opinion

Ro-bin
(Roman B. Binder 91035 Jerusalem POBox 3962)
P.S. Excuse me if anywhere my written ad hock
matter contains some fault.
.