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• Sep 4, 2001
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Typo: Second line should read "even when a and b are not integers".
Sorry, Tom

-----Original Message-----
From: Hadley, Thomas H (Tom), NLCIO
Sent: Tuesday, September 04, 2001 10:29 AM
Subject: RE: [PrimeNumbers] FLT

Paul,
You claim in your corrected proof that (a+b)^p is congruent to a^p + b^p
(mod p) even when a and b are not rationals, given a definition of
modularity that says for example t = 1/2 mod p means that t = 1/2 + k.p for
some integer k, which seems to me a reasonable definition.

This is not true. A counter-example is easy:

let a = 2, b = 1/5, p = 3
(a+b)^p mod p would be (2+1/5)^3
= (11/5)^3
= 1331/125
= 9 + 206/125
= 3*3 + 206/125
= 206/125 mod 3

However,

a^p + b^p would be 2^3 + (1/5)^3
= 8 + 1/125
= 3*2 + 2 + 1/125
= 3*2 + 251/125
= 251/125 mod 3

These are not equal. Your "Binomial prime theorem" is false. Sorry, but
it's back to the tee for you.

I agree with Phil that you should probably be working with Usenet's sci.math
group, not this list. I'm not going to spend any more time reviewing your
stuff. I'm an amateur. Even if I _don't_ find a flaw, that would mean
nothing.