Prev: An exact simplification challenge - 101 (tan, arctan)
Next: CREATION OF NUMBERS FROM SQR ROOT FORMULA
From: Roman B. Binder on 4 Aug 2010 08:15
Hi,
Could You imagine how to break up the TAB ideal
parameters of FLT ?
When beginning with n=3 also with equation:
X^3 + Y^3 = Z^3.................................(1)
for X;Y;Z as natural numbers and of gcd=1;
where it could be seen easy: Z>X and Z>Y and then
after TAB input: X=T+B; Y=T+A; Z=T+A+B;
eq.(1) will be: T^3 = 3AB(2T+A+B)...............(2)
where T;A;B should be also some natural numbers
and Z-X=A; Z-Y=B; X+Y=2T+A+B; etc.
and once (Z-X);(Z^2+ZX+X^2);(Z-Y);(Z^2+ZY+Y^2);
(X+Y);(X^2-XY+Y^2) should be of gcd=1
so there could be introduced such a;b;t natural
numbers where a;b;t;3 are of gcd=1 and with the
help of some natural u=>2(*) that:
A = 3^(3u-1) a^3; B = b^3; 2T+A+B = t^3;.......(I)
or: A = a^3; B = b^3; 2T+A+B = 3^(3u-1) t^3;...(II)
Such developments are so called as Abel formulae
and where known much earlier as some Abel's
publication...
and then eq.(2) will be extracted to:
t^3 = 2*3^u abt +3^(3u-1) a^3 + b^3 ..........(2.I)
or to 3^(3u-1) t^3 = 2*3^u abt +a^3 +b^3 .....(2.II)
Now such appropriate a;b;t;u numbers could exist
until we'll be able to find eq.(2.1) or (2.2) false.
It is here not so easy, does we not overlooked
something more simple ?
let see X^3 - B^3 + Y^3 = Z^3 - B^3 .............(3)
once Z^3 -B^3 =(Z-B)(Z^2+ZB+B^2)
and where Z-B = T+A+B-B = T+A = Y so RHS/Y
then X^3 -B^3 should be divided by Y
(T+B)^3 - B^3 = T^3 +3T^2 B + 3TB^2 +B^3 -B^3 =
= T(T^2 +3TB +3B^2).............................(4)
now using parameters from (I):
eq.(4) = 3^u abt[(3^u abt)^2 +3^(u+1) abt b^3 +3b^6 =
= 3^(u+1) ab^3 t[3^(2u-1) a^2 t^2 + 3^u ab^2 t +b^4]
and should be divided by Y = 3^u abt + 3^(3u-1) a^3 =
3^u a(bt + 3^(2u-1) a^2) = 3^u a *y .............(5)
We can see as some part of eq.(4) also 3^u a will
de divided and so on extract but it will remain:
3b^3 t[3^(2u-1) a^2 t^2 + 3^u ab^2 t + b^4]
once y is some factor of Y and Y is of gcd=1 to Z and X
so also to b and t as factors of X and Z are gcd=1 to y.
Therefore only 3^(2u-1) a^2 t^2 + 3^u ab^2 t =b^4....(6)
could provide some factor = y = bt + 3^(2u-1) a^2....(7)
but once b;a;3 are of gcd=1 polynomial (6) can not be
substituted once using polynomial (7):
P(6) # P(7) * P
where P should be some appropriate polynomial with
natural coefficients.................................(8)
However when it will be taken:
X^3 - B^3 = T^3 +3T^2 B + 3TB^2 +B^3 - B^3
so T^3 +3T^2B +3TB^2 = (T+A)*C
only after input T^3 = 3AB(2T+A+B)
once C = 3B(A+B) + 3BT
What finally show to us, that eq.(2) also
T^3 = 3AB(2T+A+B) supposed with natural numbers
is not true once especially analyzed with
(I) division of parameters a;b;t;u to results (8)
and similar there will be that same polynomial (6)
once for (II) division of parameters a;b;t;u
y = (3^u bt + a^2).............................(9)
You can see as in this certain point FLT for n=3
holds. Q.E.D
With some similar developments of TAB for
optional big and prime n value and for similar point
once it is taken B value and B=b^n we'll come
to similar result.
( I'll introduce it in the next post)

With The Best Regards
Ro-Bin
From: Roman B. Binder on 7 Aug 2010 14:27
> Hi,
> Could You imagine how to break up the TAB ideal
> parameters of FLT ?
> When beginning with n=3 also with equation:
> X^3 + Y^3 = Z^3.................................(1)
> for X;Y;Z as natural numbers and of gcd=1;
> where it could be seen easy: Z>X and Z>Y and then
> after TAB input: X=T+B; Y=T+A; Z=T+A+B;
> eq.(1) will be: T^3 = 3AB(2T+A+B)...............(2)
> where T;A;B should be also some natural numbers
> and Z-X=A; Z-Y=B; X+Y=2T+A+B; etc.
> and once (Z-X);(Z^2+ZX+X^2);(Z-Y);(Z^2+ZY+Y^2);
> (X+Y);(X^2-XY+Y^2) should be of gcd=1
> so there could be introduced such a;b;t natural
> numbers where a;b;t;3 are of gcd=1 and with the
> help of some natural u=>2(*) that:
> A = 3^(3u-1) a^3; B = b^3; 2T+A+B = t^3;.......(I)
> or: A = a^3; B = b^3; 2T+A+B = 3^(3u-1) t^3;...(II)
> Such developments are so called as Abel formulae
> and where known much earlier as some Abel's
> publication...
> and then eq.(2) will be extracted to:
> t^3 = 2*3^u abt +3^(3u-1) a^3 + b^3 ..........(2.I)
> or to 3^(3u-1) t^3 = 2*3^u abt +a^3 +b^3 .....(2.II)
> Now such appropriate a;b;t;u numbers could exist
> until we'll be able to find eq.(2.1) or (2.2) false.
> It is here not so easy, does we not overlooked
> something more simple ?
> let see X^3 - B^3 + Y^3 = Z^3 - B^3 .............(3)
> once Z^3 -B^3 =(Z-B)(Z^2+ZB+B^2)
> and where Z-B = T+A+B-B = T+A = Y so RHS/Y
> then X^3 -B^3 should be divided by Y
> (T+B)^3 - B^3 = T^3 +3T^2 B + 3TB^2 +B^3 -B^3 =
> = T(T^2 +3TB +3B^2).............................(4)
> )
> now using parameters from (I):
> eq.(4) = 3^u abt[(3^u abt)^2 +3^(u+1) abt b^3 +3b^6
> =
> = 3^(u+1) ab^3 t[3^(2u-1) a^2 t^2 + 3^u ab^2 t +b^4]
> and should be divided by Y = 3^u abt + 3^(3u-1) a^3
> =
> 3^u a(bt + 3^(2u-1) a^2) = 3^u a *y .............(5)
> We can see as some part of eq.(4) also 3^u a will
> de divided and so on extract but it will remain:
> 3b^3 t[3^(2u-1) a^2 t^2 + 3^u ab^2 t + b^4]
> once y is some factor of Y and Y is of gcd=1 to Z and
> X
> so also to b and t as factors of X and Z are gcd=1 to
> y.
> Therefore only 3^(2u-1) a^2 t^2 + 3^u ab^2 t
> =b^4....(6)
> could provide some factor = y = bt + 3^(2u-1)
> a^2....(7)
> but once b;a;3 are of gcd=1 polynomial (6) can not
> be
> substituted once using polynomial (7):
> P(6) # P(7) * P
> where P should be some appropriate polynomial with
> natural
> coefficients.................................(8)
> However when it will be taken:
> X^3 - B^3 = T^3 +3T^2 B + 3TB^2 +B^3 - B^3
> so T^3 +3T^2B +3TB^2 = (T+A)*C
> only after input T^3 = 3AB(2T+A+B)
> once C = 3B(A+B) + 3BT
> What finally show to us, that eq.(2) also
> T^3 = 3AB(2T+A+B) supposed with natural numbers
> is not true once especially analyzed with
> (I) division of parameters a;b;t;u to results (8)
> and similar there will be that same polynomial (6)
> once for (II) division of parameters a;b;t;u
> y = (3^u bt + a^2).............................(9)
> You can see as in this certain point FLT for n=3
> holds. Q.E.D
> With some similar developments of TAB for
> optional big and prime n value and for similar point
> once it is taken B value and B=b^n we'll come
> to similar result.
> ( I'll introduce it in the next post)
>
> With The Best Regards
> Ro-Bin

Hi,
Continuing my thread for some optional and extremely
big n and prime number lets rewrite:
Lhs = X^n - B^n + Y^n = Z^n - B^n = Rhs.............(10)
once Rhs = (Z-B)*Ext = Y*Ext
so reminds only X^n - B^n to be divided by Y at Lhs
.......................till k=n
X^n - B^n = (X-B){Sum from k=1 X^(n-k) B^(k-1)}
where X-B = T = n^u abtp
then the {Sum} could be substituted with X-B=T and XB terms and the T^(n-1)=Oldst will be the oldest term once taking t as changing value and a;b;p;u; as parameters
where once n(XB)^(n-1)/2 will be the smallest term so on Absolute.term expressed with b value will be:
for X=T+B Abst = nB^(n-1) =nb^(n-1)n;
Thorough all these terms in {Sum} You can see possibility
of extraction of common nb^(n-1) and so on after extr.:
Oldst = n^[(n-1u-1] (abp)^(n-1)
Abst = b^(n-1)(n-1)
Once Y=T+A where T=n^u abtp and A=a^n or A=n^(nu-1) a^b
so Y = a[ n^u btp +a^(n-1)] = a*y
or Y = n^u a{btp +n^[(n-1)u -1] a^n} = n^u a*y
Now once X^n - B^n can be rewritten as:
T*{Sum} = nb^(n-1) T*{Sum'} = n^(u+1) ab^n tp*(Sum'}
so a or n^u a factors of Y will be completed but once
a;b;t;p;y; are of gcd=1 so {Sum'} should be divided by y
once y=n^u bpt +a^(n-1) or y=bpt + n^[(n-1)u-1] a^(n-1)
but here at least b and a terms of gcd=1 will disturb
to construct the proper division of X^n - B^n / Y
until we again input here the full form of T^n:
T^n = nAB(2T+A+B)*Ext...............................(11)
where once 2T+A+B = t^n so Ext = n^(nu-1) p^n
or once 2T+A+B = n^(nu-1) t^n so Ext = p^n
What show to us, that such ideal form (11) is not true:
one of taken ideal and natural numbers a;b;t;p;
should be irrational.
Q.E.D. (in Brief)

Best Regards
Ro-Bin

P.S. With some short of time I was able to close this
my thread only so briefly as above. Some concerns could
come here in some areas of polynomials and their real
properties: it looks so for so called primitive
polynomials, am I mistaken ?
From: Roman B. Binder on 9 Aug 2010 14:06
Hi,
When using some imagination, such result is not
only because of irreducibility of this one of
polynomials we'd like to factorize: the two of
these polynomials we like to divide have not enough
common factors: in the very end we'll find at least
one of them irreducible after extracting the common
factors or two of them will contain completely
different factors also different prime numbers
spectrum...

Best Regards
Ro-Bin
 | 
Pages: 1
Prev: An exact simplification challenge - 101 (tan, arctan)
Next: CREATION OF NUMBERS FROM SQR ROOT FORMULA