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Primes in the field of |a+b*sqrt(3)*I|=1

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  • bhelmes_1
    A peaceful day to all group members and especially for Kermit and David, i would like to start a discussion about primes and some special theory concerning the
    Message 1 of 1 , Feb 24, 2014
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      A peaceful day to all group members and especially for Kermit and David,

      i would like to start a discussion about primes and some special theory
      concerning the imaginary quadratic field a+b*sqrt(3)*I mod  p
      with the reduction to the norm = 1 mod p of these elements

      I hope that this field could be useful for developing a sufficient prime test.
      It would be nice if some more mathematicians could participate in the discussion.

      As far as i see there exists always a trivial third root for all odd numbers:

      Let A=sqrt (3)

       1. |(p-1)/2+[(p-1)/2]AI|=1 mod p is always true

          Proof:

           |(p-1)/2+[(p-1)/2]AI|                                 mod p
          =[(p-1)/2+[(p-1)/2]AI]*[(p-1)/2-[(p-1)/2]AI]   mod p | with A²=3
          =[(p-1)/2]²+3*[(p-1)/2]²                             mod p | adding
          =4[(p-1)/2]²                                             mod p | squaring
          =4[(p²-2p+1)/4]                                       mod p | cancel down
          =  (p²-2p+1)                                            mod p | with p² = 0 mod p and 2p = 0 mod p
          = 1

          2. {(p-1)/2+[(p-1)/2]AI}³=1 mod p is always true

          Proof:

          {(p-1)/2+[(p-1)/2]AI}²  mod p
          =[(p-1)/2]²  -[(p-1)/2]²A²   +2[(p-1)/2]²     AI mod p | with A²=3
          =[(p-1)/2]²  -3[(p-1)/2]²     +2[(p-1)/2]²     AI mod p | subtraction by the second term
          =-2[(p-1)/2]²                 +2[(p-1)/2]²       AI mod p | squaring
          =-2(p-1)(p-1)/4               +2[(p-1)(p-1)/4]AI mod p | cancel down
          = -(p-1)(p-1)/2               + [(p-1)(p-1)/2]  AI mod p | p-1=-1 mod p
          =       (p-1)/2               - [     (p-1)/2]      AI mod p

          Proof:

          =  { (p-1)/2 + [(p-1)/2]AI }³                   mod p |  with  { (p-1)/2 + [(p-1)/2]AI }²= { (p-1)/2 - [(p-1)/2]AI }  mod p
          =  { (p-1)/2 - [(p-1)/2]AI } * { (p-1)/2 + [(p-1)/2]*AI } mod p
          =  | (p-1)/2 + [(p-1)/2]AI |                     mod p |  with proof 1.
          =         1

          3. Choosing an element of this group

          If
          a = (c^2 - 3 d^2) / (c^2 + 3 d^2) mod p
          and
          b = (2 c d) / (c^2 + 3 d^2) mod p,
          then
          a^2 + 3 b^2 = 1 mod p.

          Proof:

          a^2 + 3 b^2

          = [c^2 - 3 d^2)   / (c^2 + 3 d^2)]^2 + 3 [ (2 c d)   / (c^2 + 3 d^2)]^2  | squaring
          = (c^2 - 3 d^2)^2 / (c^2 + 3 d^2)^2  + 3 (4 c^2 d^2) / (c^2 + 3 d^2)^2  | ???
          = (1 / (c^2 + 3 d^2)^2 ) ( (c^2 - 3 d^2)^2 + 12 c^2 d^2 )
          = (1 / (c^2 + 3 d^2)^2 ) (c^4 - 6 c^2 d^2 + 9 d^4 + 12 c^2 d^2)
          = (1 / (c^2 + 3 d^2)^2 ) (c^4 + 6 c^2 d^2 + 9 d^4)
          = (1 / (c^2 + 3 d^2)^2 ) (c^2 + 3 d^2)^2
          = 1
          (Proof friendly transmitted by Kermit Rose)

      If someone likes more an illustrate picture of the different number in this field,
      there is a web application in order to see the different kind of fields
      http://devalco.de 4. k) with adjoined √3 in the complex field with norm=1 and a metric

      Some questions for those you had enough patience to follow the text:
      1. Is it possible to distinguish the elements in quadratic and non quadratic residues
      2. For primes p = 1 mod 6 there seems to be always p-1 elements and
          for primes  p = 5 mod 6 there seems to be always p+1 elements
          Kermit, can you transmit your proof for the group elements ?
      3. Is it sensful to develop a kind of rabin-miller test in order to check the 2. and 3. root of 1

      The best greetings from the primes
      Bernhard


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