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Re: [PrimeNumbers] Re: C5 is prime!

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  • 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 1 of 9 , Sep 9, 2012
      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 2 of 9 , Sep 10, 2012
        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 3 of 9 , 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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