Below is n,k,Q,factorization of Q

11, 4080, 8355841, (13)(41)(61)(257)

11, 4094, 8384513, (277)(30269)

13, 12816, 104988673, (73)(673)(2137)

Adam

--- In

primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@...>

wrote:

>

> --- In primenumbers@yahoogroups.com, "Adam" <a_math_guy@> wrote:

> >

> > Below I print out prime n, (not required to be prime) k, non-

prime

> > value Q, factorization of Q, given that 2^((Q-1)/4) == 1 mod Q.

> >

>

> sorry Adam, I mistated my idea... 2^n+1 <= k <= 2*(2^n+1)

> now try...

>

> > 7, 65, 8321, (53) (157)

> > 7, 100, 12801, (3) (17) (251)

> > 11, 170, 348161, (11) (31) (1021)

> > 11, 1575, 3225601, (71) (181) (251)

> > 13, 1020, 8355841, (13) (41) (61) (257)

> > 13, 4917, 40280065, (5) (7) (67) (89) (193)

> > 17, 801, 104988673, (73) (673) (2137)

> > 17, 32768, 4294967297, (641) (6700417)

> > 19, 8192, 4294967297, (641) (6700417)

> > 19, 101628, 53282340865, (5) (13) (17) (193) (433) (577)

> > 19, 262145, 137439477761, (593) (231769777)

> > 23, 512, 4294967297, (641) (6700417)

> > 29, 8, 4294967297, (641) (6700417)

> >

> > Adam

> >

> > --- In primenumbers@yahoogroups.com, "leavemsg1" <leavemsg1@>

> > wrote:

> > >

> > > not trying to sound like a text book, but...

> > >

> > > let Q= k*2^n +1, where 'n'/is prime/ and k<= 2^n +1. I can't

> find

> > a

> > > counter-example... using... if 2^((Q-1)/4) == 1(mod Q), then Q

is

> > prime.

> > >

> > > I know that using 2 ^ limits the number of psuedo-primes in a

> search

> > > Paul Underwood found a C-E when n = 15... but 15 isn't prime.

> > >

> >

>