## 24423RE: C5 is prime!

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• Sep 7, 2012
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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