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no counter-example???

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  • leavemsg1
    not trying to sound like a text book, but... let Q= k*2^n +1, where n /is prime/ and k
    Message 1 of 4 , Feb 13, 2008
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      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 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,
      Message 2 of 4 , Feb 14, 2008
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        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:
        >
        > 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.
        >
      • leavemsg1
        ... sorry Adam, I mistated my idea... 2^n+1
        Message 3 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 4 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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