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24423RE: C5 is prime!

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  • leavemsg1
    Sep 7, 2012
      either if (Mp) mod p = 1, and N is a square, then (Mp)
      is prime as well, or simply iff (Mp) mod p == 1, then
      choose a different 'p'. I believe it works now.

      --- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@> wrote:
      >
      > ...
      > if p= 4*k +1, and q= 2*p +3 are both prime, then if
      > [(Mp)^p -p] mod q== N, and q mod N== +/-1, then (Mp),
      > the base... is prime.
      > (someone would have to prove this conjecture.)
      > ...
      > let 2^127 -1 = 170141183460469231731687303715884105727
      > ...
      > C5 = 2^(2*27*49*19*43*73*127*337*5419*92737*649657*
      > 77158673929+1) -1
      > ...
      > let p= 5, q= 13 such that [(C5)^5 -5] mod 13 = N; and
      > 2^54 mod 13 == 12 == (-1) (by chance!!!), and then...
      > ...
      > (-1)^(odd power)*2 -1 ==
      > (-1)^49*19*43*73*127*337*5419*92737*649657*
      > 77158673929*2^1 -1 ==
      > (-1)*2 -1

      {typo correction}

      = -2 -1 and (-3)^5 -5 = -248 and [(C5)^5 -5] mod
      > 13 == 12
      > ...
      > thus, if [(C5)^5 -5] mod 13 == 12, and 13 mod 12 == 1,
      > then C5 must be prime!
      > ...
      > Rewards,
      > ...
      > Bill Bouris
      >
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