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RE twist on FlittleT

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  • Jose Ramón Brox
    ... From: Bill Bouris [a^ (p-2)] is equiv. to [p - (p-1)/a] (mod p) Interesting? or not? ... It s trivial: a^(p-1) = = 1 a^(p-2)*a = =
    Message 1 of 1 , Jul 1, 2005
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      ----- Original Message -----
      From: "Bill Bouris" <leavemsg1@...>

      [a^ (p-2)] is equiv. to [p - (p-1)/a] (mod p)

      Interesting? or not?

      ------------------------------------------

      It's trivial:

      a^(p-1) = = 1
      a^(p-2)*a = = 1
      a^(p-2)*a = = p-p+1
      a^(p-2)*a = = p-(p-1)
      a^(p-2) = = [p-(p-1)]/a
      a^(p-2) = = p/a-(p-1)/a

      And we have that GCD(a,p)=1
      p/a = = x (mod p)
      a*x = = p = = 0 (mod p)
      x = = 0 (mod p)
      p/a = = 0 = = p (mod p).

      So:
      a^(p-2) = = p-(p-1)/a.

      QED

      Jose Brox
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