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Topic: Just kidding! Here's the real proof of FLT.
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James Harris

Posts: 16
Registered: 12/12/04
Just kidding! Here's the real proof of FLT.
Posted: Sep 7, 1996 3:40 PM
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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 = -(y-x)(x+y+z)/(z+x)

delta y = z(y-x)/(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+y-z+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 (y-x) 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.







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