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Re: prime observation/question

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  • Bill Bouris
    ... maybe if q can only be odd???, then a statement that all Fermat primes greater than F4 are composite may arise??? just tinkering with the idea!
    Message 1 of 2 , Jul 8, 2007
      --- In primenumbers@yahoogroups.com, "leavemsg1"
      > wrote:
      >
      > maybe... let Z = 2^(2^(p+1))+1 ; p is prime
      >
      > Z is prime iff [Z (mod (2^p+1))] == 2^q ; for some q

      maybe if q can only be odd???, then a statement that
      all Fermat primes greater than F4 are composite may
      arise??? just tinkering with the idea! comments???

      >
      > for p = 2, 3,..., next???
      >
      > eg. p=2, 2^8+1 mod 5 == 2^1 and...
      > p=3, 2^16+1 mod 9 == 2^3 and...
      >
      > I searched up to p=389 using 'Try GMP!' interpreter
      > and the result for all the primes thus far were a
      > residue of 17... not 2^5 as expected.
      >
      > If it were to jump up to 2^5,... I think that the
      Z as stated in
      the
      > first statement would be prime.
      >
      > Are p=2,3,... the only generators for a prime Z ????
      >
      > Thanks in advance for any commentary.
      >
      I ran it up a little further to p=509 and I'm still
      not able to produce a residue other than 17 for any
      prime up to 509




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