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

>