Re: Binomial Theorem for X^n + Y^n

From: Russell E. Rierson (analog57_at_yahoo.com)
Date: 08/23/04 

Date: 23 Aug 2004 03:11:21 -0700

deepkdeb@yahoo.com (Deep K. Deb) wrote in message news:<4a3bc6b5.0408220822.ba1d7a7@posting.google.com>...
> analog57@yahoo.com (Russell E. Rierson) wrote in message news:<c410ffa5.0408212104.51a16dfb@posting.google.com>...

> Very interesting observations. It can further be simplified.
> As an example, write x^5 + y^5 = Q5/16
> Then x^5 + y^5 = Q5/2^(5-1)
> So in general, x^m + y^m = Qm/2^(m-1)
> Now, the challenge is to write Qm in more compact form so that it can
> be recognized that Qm/(2^m-1) cannot be an m-th power of an integer
> without applying FLT.

If A = B

[(A+A)/2]^3 + [(A-A)/2]^3 = [(A+A)/2]^3 + 0 = A^3

[(A+A)/2]^m + 0 = A^m

If B = n*A

[(A + n*A)/2]^m + [(A - n*A)/2]^m = [(A*(n+1))/2]^m + [(A*(1-n))/2]^m

...

x+y = A

x-y = B

(A+B)/2 = x

(A-B)/2 = y

Let B = A + K

A+B = 2A+K

A-B = -K

[(2A+K)/2]^m + [(-K)/2]^m
 

x^m + y^m = (A + K/2)^m + (-K/2)^m