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Re: sufficent proof for primes ?

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  • djbroadhurst
    ... I invited you to consider that simpler case in order to impress on you the importance of p being a prime. ... In this case you also begin with the crucial
    Message 1 of 50 , Aug 9, 2010
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      --- In primenumbers@yahoogroups.com,
      "bhelmes_1" <bhelmes@...> wrote:

      > > If p is an odd prime, then the group (Z/Zp)*
      > > of integers from 1 to p-1 is cyclic under
      > > multiplication modulo p.
      >
      > That´s right, but does not consider the
      > ring with an adjoined square root.

      I invited you to consider that simpler case
      in order to impress on you the importance
      of p being a prime.

      > If p is an odd prime, then the group of natural numbers with an
      > adjoined square root is cyclic with the order of p+1 under
      > multiplication modulo p.

      In this case you also begin with the crucial caveat:
      "if p is an odd prime".

      That means that you have no hope of proving the primality
      of a probable prime, say n, that passes your test.
      Any "cyclicity" that you use must be modulo a *prime* divisor
      of n. All the classical proofs begin with the wording:
      "Let p be a prime divisor of n" and then attempt to show that
      (for example) p > sqrt(n), as in Morrison's theorem.

      But in your "proof" of primality you considered the case
      n = f*g where you claim that you do not need f and g
      to be prime. That's simply crazy, to my mind.

      David
    • 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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