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Another try...an innovative approach to FLT
Posted:
Sep 25, 1996 8:41 AM


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+yz)^n = n(zx)(zy)(x+y)Q
where I use Q for all those other terms. For n=3, Q=1 because
(x+yz)^3 = 3(zx)(zy)(x+y)
My conclusion is that
n(zy)(xy)^{n2}[(x^2  y^2)Q  z^{n1}] must be divisible by
(zx) which means that
[{x^2  y^2)Q  z^{n1}] must be divisible by (zx) 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 (zx) = 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(zy)(delta y)^{n1} from (delta x)^n
and using the substitution n(zx)(zy)(x+y)Q = (x+yz)^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



