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24434Re: C5 is prime!

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  • leavemsg1
    Sep 10, 2012
      the 4 equations for a Mersenne number (Mr) where r is prime.

      (either one or the other is true)
      p=4k+1, q=2p+3 (both prime) [(Mr)^p-p] mod q == -1, or
      p=4k+3, q=2p+3 (again) [(Mr)^p-p] mod q == -1 (can't be +1)
      (but, both of the following must be true)
      p=4k+1, q=2p+1 (again) [(Mr)^p-p] mod q == p
      p=4k+3, q=2p+1 (again) [(Mr)^p-p] mod q == p+2

      look!!!
      > let 2^127 -1 = 170141183460469231731687303715884105727
      > ...
      > C5 = 2^(2*27*49*19*43*73*127*337*5419*92737*649657*77158673929+1) -1
      > ...
      > let p= 3, q= 7 such that [(C5)^3 -3] mod 7 = N; and 2^27 mod 7
      > == 1 (again, by chance!!!) and then...
      > ...
      > (1)^(left over exponent)*2 -1 ==
      > (1)^2*49*19*43*73*127*337*5419*92737*649657*77158673929*2^1 -1 ==
      > (1)*2 -1 = 2 -1 and (1)^3 -3 = 1 -3 = -2 and [(C5)^3 -3] mod 7 == 5
      > ...
      > thus, if [(C5)^3 -3] mod 7 == 5, or p +2 = 3 +2 = 5, then C5 must
      > be prime!
      > ...
      C5 is definitely prime, if my study is correct.

      > --- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@> wrote:
      >
      > so many times, there are typos when using e-mail. in math, Mp
      > is often used to describe a Mersenne number with a prime expo-
      > nent. I have corrected it to (Mr). this conjecture cannot be
      > found in print. it's my own new idea which may need tweaking.
      > ...
      > (conjecture)
      > if p= 4*k +1, and q= 2*p +3 are both prime, then if [(Mr)^p -p]
      > mod q == N, & q mod N == +/-1, then (Mr), the base... is prime.
      > also, if (Mr) mod p = 1, then choose a different 'p', or if N is
      > a square, then that (Mr) is prime. finally, the exponent cannot
      > be such that q mod r == 1. there may be other small restrictions.
      > (someone would have to prove this conjecture.)
      > ...
      > Now, I think that it is stated IN FULL. It's a brand new idea.
      > ...
      > 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 = -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!
      > ...

      Bill Bouris
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