## 24428Re: C5 is prime!

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• Sep 10, 2012
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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.Ã I cannot read the combination and
> > this looks important.Ã You have subclass (and apparent chance exemplar!!!)
> > of Mersenne numbers that test easily as prime but contingent on proof of
> > some reasonable conjecture. (?)Ã I think I made that clear in one I sent
> > in private.Ã Maybe not.Ã 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
> > > >
> > > >
> > > >
> > > > Ã
> > > >
> > > >
> > > >
> > > > 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]
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