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. (?) 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

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

>

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