- Sep 9, 2012For 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.Â I cannot read the combination and

> this looks important.Â 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.Â Maybe not.Â 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]

> > >

> >

>

>

>

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