Re: Fermat's Last theorem short proof

In article
<8917073.1177024391960.JavaMail.jakarta@xxxxxxxxxxxxxx
orum.org>,
bassam king karzeddin <bassam@xxxxxxxxxx> wrote:

bassam king karzeddin wrote:
Dear All

As a generalization to one of my posts in this
thread


Given, two distinct, coprime non zero integers

(x & y),

Theorem- (new or old, I don¹t care), precisely
I
don't know

If, (n & m) are two positive integers, where

m = gcd ((x+y), n),

then this implies the following theorem:

Gcd ((x+y), (x^n+y^n)/(x+y)) = Rad (m),

Where Rad (m) equals the product of all the
prime
factors of (m), that is to say
Rad (m) is square free number that divides
(x^n+y^n),

Oh? Perhaps you need another condition, since

x = 15 and y = 49 are coprime and if we pick n =
8
then

m = gcd(x+y, n) = gcd(64, 8) = 8

but 15^8 + 49^8 = (16617746730113)(2) which isn't
even
divisible by x+y.


Regards,

Rick

Yes Rick,
and thank you very much for the note

In fact, and for the purpose of FLT, you may assume
either (n) is odd
positive integer

OR (x & y), are both odd-distinct-coprime-
integers,

If x = 3 and y = 1 then (x^2 + y^2) / (x + y) is not
an integer.

--
Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for
email)

But, Still Rad(m) divides (x^2+y^2), doesn't it?

Regards
B.Karzeddin
.

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