Re: FLT an incurable, unending addiction

> In article
> <25136591.1133549815356.JavaMail.jakarta@xxxxxxxxxxxxx
> forum.org> "Roman B. Binder"
> <rbinder@xxxxxxxxxxxxxxxx> writes:
> ...
> I have to state that you do *not* clearly state what
> you are doing.
> For the most part I do not understand at all what you
> are meaning.
> It appears to be a language problem.
If You like to follow my math. idea so language
be less important: You like some command in german ?
I am much worse in dutch but if You comment math.
even with dutch I hope to understand:
>
> > 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
> s could
> > be extracted of any common factor and substituted
> d with
> > only natural values so let x;y<z and x;y;z of
> f gcd=1
> > now x^n + y^n = z^n could be eventually
> y substituted
> > with optional input: x=T+B; y=T+A; z=T+A+B
> > where T;A;B natural numbers
>
> You could just as well state that x <= y (or x < y as
> for x = y
> there obviously is no solution). And so we can say y
> = x + a
> and z = x + b, which makes everything shorter.
>
> > what gives to us in brief:
> > T^n = nAB(2T+A+B)*Ext
>
> Now, this does not follow, I think. At least, it is
> not clear
> how you get here.
As I used to appoint z=T+A+B; x=T+B; y=T+A
so then (T+A)^n + (T+B)^n = (T+A+B)^n
will transform to:
T^n = nAB Ex
Ex = Sum(to i+j=n-1)
(from i;j=1) [*]A^(i-1) B^(j-1) T^(n-i-j)...(1)
where [*] = (n-1)!/i!j!(n-i-j)!
what could be obtain from Newton formula for powers
once also essentially: x+y = 2T+A+B = t^n
z = T+A+B = ts
so T=t[t^(n-1)-s]
and so on T^n = t^n [t^(n-1)-s]^n....................(2)
or once x+y = 2T+A+B = n^(nu-1) t^n
z = T+A+B = n^u ts
so T=n^u t{n^[(n-1)u-1] t^(n-1) -s}
and so on T^n = n^(nu) t^n {....}^n..................(3)
therefore once we take such t and T=n^u abtp
so according to cardinal form (1)where we have
still factors A;B; so also should be 2T+A+B
(and it does but general formula for resisting form
will be not so easy so lets call it shortly Ext.:)
for n=3 T^3 = 3AB(2T+A+B)
n=5 T^5 = 5AB(2T+A+B)[2T^2+2(A+B)T+A^2+AB+B^2]
T^5 = 5AB(2T+A+B) Ext
then according to divisibility of x;y;or z or not
A=a^n or A= n^(nu-1) a^n;
B=b^n or B= n^(nu-1) b^n;
2T+A+B=t^n or 2T+A+B=n^(nu-1) t^n;
Ext=p^n or Ext=n^(nu-1) p^n;
Where once taking proper and optional values
of a;b;t;u;p the only questionable will resist:
2T+A+B= t^n or 2T+A+B= n^(nu-1) t^n
for executable cases of A;B; once
T=n^u abtp always ...
is it clear till now ?

Enjoy it,
With Compliments
Ro-bin
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