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Re: [PrimeNumbers] Interesting divisor of a GF number

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  • Yves Gallot
    ... If P divides F_n then 2^2^n = -1 (mod P) and (2^2^n)^k = (2^k)^2^n = (-1)^k (mod P) ... if (2^k)^2^n = -1 (mod P) and k is odd, you have only 1/k chance
    Message 1 of 2 , Jul 27, 2001
      > Why is it that F(n) implies GF(8,n) and GF(32,n),

      If P divides F_n then
      2^2^n = -1 (mod P)
      and (2^2^n)^k = (2^k)^2^n = (-1)^k (mod P)

      > while GF(32,n) or GF(8,n) does not imply F(n) ??

      if (2^k)^2^n = -1 (mod P) and k is odd,
      you have only 1/k chance that 2^2^n = -1 (mod P).

      Yves
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