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Re: sufficent proof for primes ?

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  • bhelmes_1
    ... First, there was a small error in the program, i wrote for 4. p-1 instead of p+1, i corrected it. You are right, that my program does not detect any
    Message 1 of 50 , Aug 9, 2010
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      > Dear David,
      > >
      > > > > What is your "strong witness" A for the probable primality of
      > > > > p = 170141183460469231731687303715884105727 ?

      First, there was a small error in the program, i wrote for 4.
      p-1 instead of p+1, i corrected it.

      You are right, that my program does not detect any witness for this prime, because p+1=2^n.

      I have to exclude this special case.

      Nevertheless for this special kind of primes, there is a sufficent
      prime algorithm as you know.

      Greetings from the primes
      Bernhard
    • 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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