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Topic: Another innovative approach to FLT
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James Harris

Posts: 16
Registered: 12/12/04
Another innovative approach to FLT
Posted: Sep 25, 1996 8:41 AM
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I've tried to get most of the hysterical emotion out of my approach.
I'll show what I have first and then show how I got it.

First off, this is about FLT so the usual about x^n + y^n = z^n applies.

The extra is that (x+y-z)^n = n(z-x)(z-y)(x+y)Q

where I use Q for all those other terms. For n=3, Q=1 because

(x+y-z)^3 = 3(z-x)(z-y)(x+y)

My conclusion is that

n(z-y)(x-y)^{n-2}[(x^2 - y^2)Q - z^{n-1}] must be divisible by

(z-x) which means that

[{x^2 - y^2)Q - z^{n-1}] must be divisible by (z-x) since none

of the other expressions are.

But letting y=af the moduli of z and x with respect to f are the same
as (z-x) = f^n and looking at the interior of the expression we see that
then Q must have the same moduli as z and x with respect to f. But then
the expression is only divisible by f^2 because of the (x^2 - y^2)Q.

For n=3, it's more obvious because Q=1 and you get x^2 - y^2 - z^2

which is obviously not divisible by f^n but only by f^2.

Where all of this come from is my expression for x^n + y^n = z^n where I
use x + delta x = my; y + delta y = mx; z + delta y = mz; to expand the
original expression.

I've done this before but I found that at the first level, my
denominators divided out, but then I finally got around to adding those
two terms and get the result from above.

It comes from subtracting

n(z-y)(delta y)^{n-1} from (delta x)^n

and using the substitution n(z-x)(z-y)(x+y)Q = (x+y-z)^n

and is easily verifiable.

I wonder if anyone will check after all of the other posts but that's
neither here nor there.

James Harris

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