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

To:

paulmillscv@...;

primenumbers@yahoogroups.com
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.

Tom Hadley

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