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Re: {Spam?} [PrimeNumbers] 2^(2^m) + 1 mod prime

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  • Paul Leyland
    ... .. ... This result is (a) well-known and (b) easy to prove. Let F_i = 2^(2^i)+1 where i = 1. Then F_{i+1} = 2^(2^{i+1}) + 1 = (2^(2^{i+1}) - 1) + 2 =
    Message 1 of 2 , Jul 29, 2007
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      On Sun, 2007-07-29 at 13:16, Kermit Rose wrote:
      > 2^(2^m) + 1 mod prime
      ..
      > In general no number in this series divides any other number in the
      > series

      This result is (a) well-known and (b) easy to prove.

      Let F_i = 2^(2^i)+1 where i >= 1.

      Then F_{i+1} = 2^(2^{i+1}) + 1
      = (2^(2^{i+1}) - 1) + 2
      = (2^(2^(i) + 1) * (2^(2^(i) - 1) + 2
      = F_i * (F_i - 2) + 2

      So F_{i+1} == 2 mod F_i

      The general result follows by induction.

      Incidentally, this provides another nice proof of the infinity of the
      primes. No F_n is divisible by any other F_n and n may take on any
      integer value.


      Paul



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