- Sep 9, 2012so 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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