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C5 is prime!

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  • leavemsg1
    ... 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
    Message 1 of 9 , Sep 7 7:52 AM
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      ...
      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 = -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
    • leavemsg1
      clarification: also, if (Mp) mod p = 1 and N = 0, then if N = 0, then that (Mp) is prime as well. ... {typo correction} = -2 -1 and (-3)^5 -5 = -248 and
      Message 2 of 9 , Sep 7 11:29 AM
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        clarification: also, if (Mp) mod p = 1 and N = 0, then
        if N = 0, then that (Mp) is prime as well.

        --- 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
        >
      • leavemsg1
        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.
        Message 3 of 9 , Sep 7 10:20 PM
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          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
          >
        • James Merickel
          Please repeat in full corrected state or give us notice of where in print that is to be found at some time.  I cannot read the combination and this looks
          Message 4 of 9 , Sep 9 5:35 AM
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            Please repeat in full corrected state or give us notice of where in print that is to be found at some time.  I cannot read the combination and this looks important.  You have subclass (and apparent chance exemplar!!!) of Mersenne numbers that test easily as prime but contingent on proof of some reasonable conjecture. (?)  I think I made that clear in one I sent in private.  Maybe not.  If not, that's what I meant.
            JGM

            --- On Fri, 9/7/12, leavemsg1 <leavemsg1@...> wrote:


            From: leavemsg1 <leavemsg1@...>
            Subject: [PrimeNumbers] RE: C5 is prime!
            To: primenumbers@yahoogroups.com
            Date: Friday, September 7, 2012, 10:20 PM



             



            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
            >








            [Non-text portions of this message have been removed]
          • leavemsg1
            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).
            Message 5 of 9 , Sep 9 10:30 AM
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              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!
              ...

              --- In primenumbers@yahoogroups.com, James Merickel <moralforce120@...> wrote:
              >
              > Please repeat in full corrected state or give us notice of where in print that is to be found at some time.  I cannot read the combination and this looks important.  You have subclass (and apparent chance exemplar!!!) of Mersenne numbers that test easily as prime but contingent on proof of some reasonable conjecture. (?)  I think I made that clear in one I sent in private.  Maybe not.  If not, that's what I meant.
              > JGM
              >
              > --- On Fri, 9/7/12, leavemsg1 <leavemsg1@...> wrote:
              >
              >
              > From: leavemsg1 <leavemsg1@...>
              > Subject: [PrimeNumbers] RE: C5 is prime!
              > To: primenumbers@yahoogroups.com
              > Date: Friday, September 7, 2012, 10:20 PM
              >
              >
              >
              >  
              >
              >
              >
              > 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
              > >
              >
              >
              >
              >
              >
              >
              >
              >
              > [Non-text portions of this message have been removed]
              >
            • leavemsg1
              one such additional and logical restriction is (Mr) mod q =/= p.
              Message 6 of 9 , Sep 9 9:19 PM
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                one such additional and logical restriction is
                (Mr) mod q =/= p.

                --- 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!
                > ...
                >
                > --- In primenumbers@yahoogroups.com, James Merickel <moralforce120@> wrote:
                > >
                > > Please repeat in full corrected state or give us notice of where in print that is to be found at some time.  I cannot read the combination and this looks important.  You have subclass (and apparent chance exemplar!!!) of Mersenne numbers that test easily as prime but contingent on proof of some reasonable conjecture. (?)  I think I made that clear in one I sent in private.  Maybe not.  If not, that's what I meant.
                > > JGM
                > >
                > > --- On Fri, 9/7/12, leavemsg1 <leavemsg1@> wrote:
                > >
                > >
                > > From: leavemsg1 <leavemsg1@>
                > > Subject: [PrimeNumbers] RE: C5 is prime!
                > > To: primenumbers@yahoogroups.com
                > > Date: Friday, September 7, 2012, 10:20 PM
                > >
                > >
                > >
                > >  
                > >
                > >
                > >
                > > 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
                > > >
                > >
                > >
                > >
                > >
                > >
                > >
                > >
                > >
                > > [Non-text portions of this message have been removed]
                > >
                >
              • Maximilian Hasler
                For r=6, Mr is composite but p=5, q=13, N=2 yields a counter-example. What about r=37, p=23593, N=1 ? all your conditions are satisfied, but Mr is not prime.
                Message 7 of 9 , Sep 9 9:59 PM
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                  For r=6, Mr is composite but p=5, q=13, N=2 yields a counter-example.

                  What about r=37, p=23593, N=1 ?
                  all your conditions are satisfied, but Mr is not prime.

                  You wrote, "when N is a square, then Mr is prime".
                  It is easily seen that this is wrong
                  there are many small primes r,p,q such that N>1 is a square and Mr
                  composite,
                  e.g. r=11, p=13, N=4.
                  So I assume that somehow you meant
                  "when N is a square, choose another p" ?

                  Then ideed it will be difficult to find a counter-example:
                  I think you excluded all possible cases...

                  Maximilian



                  On Mon, Sep 10, 2012 at 12:19 AM, leavemsg1 <leavemsg1@...> wrote:

                  > **
                  >
                  >
                  > one such additional and logical restriction is
                  > (Mr) mod q =/= p.
                  >
                  >
                  > --- 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!
                  > > ...
                  > >
                  > > --- In primenumbers@yahoogroups.com, James Merickel <moralforce120@>
                  > wrote:
                  > > >
                  > > > Please repeat in full corrected state or give us notice of where in
                  > print that is to be found at some time. I cannot read the combination and
                  > this looks important. You have subclass (and apparent chance exemplar!!!)
                  > of Mersenne numbers that test easily as prime but contingent on proof of
                  > some reasonable conjecture. (?)Â I think I made that clear in one I sent
                  > in private. Maybe not. If not, that's what I meant.
                  > > > JGM
                  > > >
                  > > > --- On Fri, 9/7/12, leavemsg1 <leavemsg1@> wrote:
                  > > >
                  > > >
                  > > > From: leavemsg1 <leavemsg1@>
                  >
                  > > > Subject: [PrimeNumbers] RE: C5 is prime!
                  > > > To: primenumbers@yahoogroups.com
                  > > > Date: Friday, September 7, 2012, 10:20 PM
                  > > >
                  > > >
                  > > >
                  > > > Â
                  > > >
                  > > >
                  > > >
                  > > > 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
                  > > > >
                  > > >
                  > > >
                  > > >
                  > > >
                  > > >
                  > > >
                  > > >
                  > > >
                  > > > [Non-text portions of this message have been removed]
                  > > >
                  > >
                  >
                  >
                  >


                  [Non-text portions of this message have been removed]
                • leavemsg1
                  I agree. it will be difficult to formalize the conditions without knowing how to construct the Lagrangian-style proof. it wouldn t be when r is not prime, and
                  Message 8 of 9 , Sep 10 6:59 AM
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                    I agree. it will be difficult to formalize the conditions
                    without knowing how to construct the Lagrangian-style proof.
                    it wouldn't be when r is not prime, and maybe p would have
                    size restrictions. it seems that there's a theorem in there.
                    I'll have to think about it some more.

                    --- In primenumbers@yahoogroups.com, Maximilian Hasler <maximilian.hasler@...> wrote:
                    >
                    > For r=6, Mr is composite but p=5, q=13, N=2 yields a counter-example.
                    >
                    > What about r=37, p=23593, N=1 ?
                    > all your conditions are satisfied, but Mr is not prime.
                    >
                    > You wrote, "when N is a square, then Mr is prime".
                    > It is easily seen that this is wrong
                    > there are many small primes r,p,q such that N>1 is a square and Mr
                    > composite,
                    > e.g. r=11, p=13, N=4.
                    > So I assume that somehow you meant
                    > "when N is a square, choose another p" ?
                    >
                    > Then ideed it will be difficult to find a counter-example:
                    > I think you excluded all possible cases...
                    >
                    > Maximilian
                    >
                    >
                    >
                    > On Mon, Sep 10, 2012 at 12:19 AM, leavemsg1 <leavemsg1@...> wrote:
                    >
                    > > **
                    > >
                    > >
                    > > one such additional and logical restriction is
                    > > (Mr) mod q =/= p.
                    > >
                    > >
                    > > --- 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!
                    > > > ...
                    > > >
                    > > > --- In primenumbers@yahoogroups.com, James Merickel <moralforce120@>
                    > > wrote:
                    > > > >
                    > > > > Please repeat in full corrected state or give us notice of where in
                    > > print that is to be found at some time. I cannot read the combination and
                    > > this looks important. You have subclass (and apparent chance exemplar!!!)
                    > > of Mersenne numbers that test easily as prime but contingent on proof of
                    > > some reasonable conjecture. (?)Â I think I made that clear in one I sent
                    > > in private. Maybe not. If not, that's what I meant.
                    > > > > JGM
                    > > > >
                    > > > > --- On Fri, 9/7/12, leavemsg1 <leavemsg1@> wrote:
                    > > > >
                    > > > >
                    > > > > From: leavemsg1 <leavemsg1@>
                    > >
                    > > > > Subject: [PrimeNumbers] RE: C5 is prime!
                    > > > > To: primenumbers@yahoogroups.com
                    > > > > Date: Friday, September 7, 2012, 10:20 PM
                    > > > >
                    > > > >
                    > > > >
                    > > > > Â
                    > > > >
                    > > > >
                    > > > >
                    > > > > 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
                    > > > > >
                    > > > >
                    > > > >
                    > > > >
                    > > > >
                    > > > >
                    > > > >
                    > > > >
                    > > > >
                    > > > > [Non-text portions of this message have been removed]
                    > > > >
                    > > >
                    > >
                    > >
                    > >
                    >
                    >
                    > [Non-text portions of this message have been removed]
                    >
                  • leavemsg1
                    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,
                    Message 9 of 9 , Sep 10 10:45 PM
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                      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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