Loading ...
Sorry, an error occurred while loading the content.

24428Re: C5 is prime!

Expand Messages
  • leavemsg1
    Sep 10, 2012
    • 0 Attachment
      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]
      >
    • Show all 9 messages in this topic