- Sep 10, 2012I 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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