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Just kidding! Here's the real proof of FLT.
Posted:
Sep 7, 1996 3:40 PM


This is a strange one but it works, so what can I say?
Prove x^n + y^n = z^n false when x,y,z relatively prime integers and n a Natural prime greater than 2.
use x  delta x = my, y  delta y = mx, z + delta y = mz
this requires that m = (z+y)/(z+x), delta x = (yx)(x+y+z)/(z+x) delta y = z(yx)/(z+x)
substituting into the first equation gives
(x  delta x)^n + (y  delta y)^n = (z + delta y)^n
Multiplying out and subtracting off x^n + y^n = z^n , you have all terms multiplied by n except
(delta x)^n + 2(delta y)^n
(I added (delta x)^n to both sides)
But that proves that (x+y+z)^n + 2z^n must be divisible by n
so by F. Little T. x+y+z + 2z must be divisible by n, and therefore
z [x+yz+x+y+z+2z=2(x+y+z)]must be divisible by n. (yes it's ugly but it works)
Note that z+x and (yx) can't be divisible by n, which is obvious.
So, just in case you think that I've just managed to prove that z is divisible by n take the following
x  delta x = mz, y  delta y = my, z = delta y = mx
do the same thing and you prove that y must be divisible by n.
Well, that does it and I just have to say that the above is ugly. I can understand why a certain someone didn't publish it before. Personally, I'm feeling a bit nauseous.
Signing out.
James S.
do the rest of the above and you prove that y must be divisible by n
Work out everything the same as above and you get that y must be divisible by n.



