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Re: sufficent proof for primes ?

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  • bhelmes_1
    Dear David, ... That´s right, but does not consider the ring with an adjoined square root. If p is an odd prime, then the group of natural numbers with an
    Message 1 of 50 , Aug 9, 2010
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      Dear David,

      > 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.

      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.

      This is the difference to the natural numbers and i will try to
      explain it to you.

      If jacobi (a, p)=-1 and p a prime and A=sqrt (a), then
      (1+A)^p=(1+A^p) mod p (Is this step clear ? The binominal coefficients in the middle are always 0)

      A^p=A^(p-1+1)=A^(p-1) * A (a^((p-1)/2)= -1 mod p)
      = -1 * A
      = -A

      Therefore (1+A)^p=1-A mod p
      and with the same proof (1-A)^p= 1+A mod p

      If you combine these two equalations then you get
      (1+A)^(p^2)=1+A mod p , divison by 1+A
      (1+A)^(p^2 -1) = 1 mod p this is equal to
      (1+A)^((p+1)*(p-1)) = 1 mod p

      (1+A)^p = 1 - A , multiplication with (1+A)
      (1+A)^(p+1) = 1-A^2 = 1 - a mod p

      (1- a) is a natural number, therefore the order of the cyclic group of a natural number with adjoined part is p+1,
      in contrast to the cyclic group of natural numbers,
      which is sure p-1.

      Therefore i reveal a cyclic group with order d | p+1 in the adjoined numbers and show a contradiction.

      If you have any question, i will try to answer them.

      Nice 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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