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2515RE: [PrimeNumbers] FLT

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  • Hadley, Thomas H (Tom), NLCIO
    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
      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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