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Re: "wriggly" pseudoprime puzzle

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  • paulunderwooduk
    ... Perhaps. I am using the sequential sledge hammer technique, testing all n and a . The 6-selfridge double-lucas testing now stands at n
    Message 1 of 50 , Sep 6, 2010
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      --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
      >
      >
      >
      > --- In primenumbers@yahoogroups.com,
      > "djbroadhurst" <d.broadhurst@> wrote:
      >
      > > use a semiprime p = q*r
      > ...
      > > I chose (r-1)/(q^2-1) = 2/3
      >
      > Cf. Paul Underwood's semiprimes:
      > http://primes.utm.edu/bios/page.php?id=181
      > > 105809903
      > > 2499327041
      > found when investigating 5-selfridge double-Lucas + Fermat tests.
      > Here we have smaller ratios, namely
      >
      > ratio(p)=local(f=factor(p)[,1]);(f[2]-1)/(f[1]^2-1);
      > print([ratio(105809903),ratio(2499327041)]);
      > [1/10, 1/36]
      >
      > in http://tech.groups.yahoo.com/group/primenumbers/message/21673
      >
      > This observation led me to study a comparable ratio, namely
      >
      > print([ratio(43334121400711)]);
      > [1/33]
      >
      > in http://tech.groups.yahoo.com/group/primenumbers/message/21788
      > and then an even simpler ratio, namely
      >
      > print([ratio(555826983297468634137017311219)]);
      > [2/3]
      >
      > in http://tech.groups.yahoo.com/group/primenumbers/message/21792
      >
      > Perhaps Paul might find this observation of use,
      > for finding more double-Lucas perjuries?
      >

      Perhaps. I am using the sequential sledge hammer technique, testing all "n" and "a". The 6-selfridge double-lucas testing now stands at n<1.5*10^7 for "a+-2" and n<2.425*10^7 for "a+-1". The latter requires the test gcd(a*210)==1. At a glance it seems that n/gcd is always 6*N-1. I will get around to verifying this for my searched "n", as I have logged the the cases where gcd(a,n) is required.

      It would lighten my CPU load if I just went for such semiprimes as noted by David. My technique will take years to reach n=2^32. I have already ruled out Carmichael numbers below this limit,

      Paul
    • djbroadhurst
      ... [4] is meaningless, as it stands. You should write a double mod: 4. (1+x)^p = 1-x mod(x^2-a,p) ... There is no reason whatsoever to believe that [1] to [4]
      Message 50 of 50 , Sep 29, 2011
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        --- In primenumbers@yahoogroups.com,
        "bhelmes_1" <bhelmes@...> wrote:

        > 1. let a jacobi (a, p)=-1
        > 2. let a^(p-1)/2 = -1 mod p
        > 3. a^6 =/= 1 mod p
        > 4. (1+sqrt (a))^p = 1-sqrt (a)

        [4] is meaningless, as it stands.
        You should write a double mod:

        4. (1+x)^p = 1-x mod(x^2-a,p)

        > 1. Is it possible that there are other exceptions

        There is no reason whatsoever to believe that
        [1] to [4] establish that p is prime. Morevoer,
        some folk believe that, for every epsilon > 0,
        the number of pseudoprimes less than x may
        exceed x^(1-epsilon), for /sufficiently/ large x.

        > 2....
        > there is a cyclic order ...

        > 3....
        > there is a cyclic order ...

        The group of units (Z/nZ)* is /not/ cyclic
        if n has at least two distinct odd prime fators.

        David
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