## Well, I'll be damned. This might be worth looking at.

Being incredibly dogged and impossibly stupid I've kept working at trying to find a solution to FLT using what I've been told by the AMS is my unique approach. I'm probably just deluding myself again but hey, why not throw it out there anyway.

Rehash:

With the def of the problem being x^p + y^p = z^p, p odd prime, x,y,z relatively prime no solution exists;

using x + dx = my; y + d = mx; z + d = mz; m=(z-y)/(z-x); d=z(x-y)/(z-x);

dx = (x-y)(x+y-z)/(z-x)

I can write

(x + dx)^p + (y + d)^p = (z+d)^p;

Before, I expanded the above and subtracted off x^p + y^p = z^p but recently I realized that was a mistake. Using p=3 as an example I worked on

(x+dx)^3 = [y(z-y)/(z-x)]^3 = x^3 + 3z(z^2 - y^2)(x-y)/(z-x) + 3z^2 (z-y)(x-y)^2/(z-x)^2

multiplying out the denominators gives

y^3 (z-y)^3 = x^3 (z-x)^3 + 3z(z^2 - y^2)(x-y)(z-x)^2 + 3z^2 (z-y)(x-y)^2
(z-x)

and it is obvious that

y^3(z-y)^3 - x^3(z-x)^3 must be divisible by z

That is the BIG conclusion of this post and if I'm wrong then someone has to show me here because the generalization is that

y^p(z-y)^p -x^p(z-x)^p must be divisible by z

Now to get a real thrill, try the above using p=2 noting that (x+y-z)^2 = 2(z-x)(z-y)
{well it gave me a thrill, yes it is divisible for p=2 but not for p odd prime]

Looking forward to hearing any comments althought I'll probably wake up in the morning and see something obviously stupid about the above and curse myself for a while and then go back to life as usual but that's neither here nor there and maybe I should lay off the Mountain Dew so that I can sleep at 1:48 when it is really 2:48 except for that silly daylight savings time thing oh well time to send this off.