- 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.

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:>

a

> 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

> 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.

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

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

> 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.

>

now try...

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

find

> 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

> a

search

> > 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

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

> >

> - 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:>

prime

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

> >

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

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

is

> >

>

> 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

> > 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.

> > >

> >

>