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Re: no counter-example???

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  • leavemsg1
    ... sorry Adam, I mistated my idea... 2^n+1
    Message 1 of 4 , Feb 15, 2008
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      --- 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.
      > >
      >
    • Adam
      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 ... prime ...
      Message 2 of 4 , Feb 16, 2008
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        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.
        > > >
        > >
        >
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