[sir fermats last theorem] - A New Kind of Science: The NKS Forum

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sir fermats last theorem

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Posted by: rajesh bhowmick

a^(n)-b^(n)=c^(2)-b^(2)
like, 7^(3)-4^(3)=48^(2)-45^(2)
x^(n)=z^(2)-y^(2)
like, 3^(3)=14^(2)-13^(2)
from this information we can prove in a very elementary and simple way that it is impossilble to write
a^(n)-b^(n)=x^(n)
or
a^(n)=x^(n)+b^(n)
for n>2.


Posted by: Todd Rowland

While this is not exactly the right forum for Fermat's last theorem, it would be of NKS interest to find a way to decompose the statement of Fermat's theorem into simple primitives. Possibly many primitives, but as long as they are simple.

In this case, it is not so easy. One cannot get a contradiction from

c^2-b^2=z^2-y^2

since one never uses the hypothesis that the number in question is a nth-power or a difference of nth-powers, let alone both. Any number bigger than 2 can be written as the difference of two squares.


Posted by: rajesh bhowmick

It is a bit hard but it is much simpler then sir Andrew wiles proof of sir Fermats Last Theorem.
I will start from here that
when n>1 then in the equation
a^(n) - b^(n) = x^(n)
all x^(n) can be expressed as difference of two squares but not all a^(n)-b^(n) is not expressable in difference of two squares like 3^(3) - 1^(3).
so only those l.h.s which are expressable into difference of two squares (as the r.h.s is always expressable into difference of two squares) are remained to be proved for sir Fermats Last Theorem.
regards.


Posted by: rajesh bhowmick

a^n - b^n=(a-b)[(a-b)^(n-1)+nk]
here "k" will always be natural when n is 4 or any other prime.
a^n-b^n=(c+a-b)^2 - c^2 = (2c+a-b)(a-b)
here "c" is always natural when a^n-b^n is expressable into difference of two squares.
if a^n-b^n=x^n then
A^n - B^n = 1 where
A=a/x & B=b/x
the above two equations also gets applied for this also so,
(A-B)[(A-B)^(n-1)+nK] = (A-B)(2C+A-B)
or,[(A-B)^(n-1)+nK-2C]/(A-B)=1
there is only one solution to this equation & that is when n=2 & K=C.


Posted by: nordjohn

cool, guies! )))
---
johntvery@yahoo.com




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