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Re: Proposition for a better p-1 and p+1 test

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  • djbroadhurst
    ... The gremlins now have more than 23,000 pseudoprimes that fool Bernhard s test. But before revealing them, they await a response to the very simple case,
    Message 1 of 14 , Sep 20, 2013
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      --- In primenumbers@yahoogroups.com,
      "djbroadhurst" <david.broadhurst@...> wrote:

      > {p=3281;x=1062;y=1600;
      > if(kronecker(3,p)==-1&&
      > Mod(3,p)^((p-1)/2)==-1&&
      > x^2-3*y^2==Mod(1,p)&&
      > Mod(Mod(1,p)*(x+y*A),A^2-3)^p==x-y*A&&
      > !isprime(p),print(" fooled"));}
      >
      > fooled

      The gremlins now have more than 23,000 pseudoprimes
      that fool Bernhard's test. But before revealing them,
      they await a response to the very simple case, above.

      David
    • bhelmes_1
      Dear David, the amount of counterexamples sounds impressively. I suppose that there are a lot of p=1 mod 4. I think that i will publish soon a test which
      Message 2 of 14 , Sep 21, 2013
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        Dear David,

        the amount of counterexamples sounds impressively.
        I suppose that there are a lot of p=1 mod 4.
        I think that i will publish soon a test which separates between p=1 mod 4 and p=3 mod 4,
        but i would like to check first some list by myself in order to save some time for you and to get
        some more informations.

        How does it feel to be a "keeper" of such a lot of interesting gremlins :-)

        A new and better proposition:

        A. Let p an element of N with p=3 mod 4,
           m the greatest power of 2 with 2^m*r=p+1
           and 2^m < r
         
        1. choose a prime a with jacobi (a, p)=-1
        2. Verify a^(p-1)/2=-1 mod p
        3. Depending on the a choose the pell solution of x, y with
           x^2-ay^2=1 mod p
           http://mathworld.wolfram.com/PellEquation.html
        4. Verify that (x+yA)^(2^m)<>1 mod p  with A=sqrt (a)
           because gcd (p-1, p+1) = 2
        5. Calculate and verify (x+yA)^p=x-yA mod p
        6. Calculate (x+yA)^r=x(r)+y(r)A mod p and
           verify that  gcd (y(r), p)=1

           (The point 6. derives of a factorisation test, is cheap and powerful
            because gcd (y(n), p) > 1 results gcd (y(m), p) > 1 with m=n*t)

        if 1.-6. is ok. then p is a prime.

        David, you will realize that i abstain form the choose of the smallest a
        and that i add the point 4. for good reasons

        I wish you a peaceful and pleasant weekend
        Bernhard
      • djbroadhurst
        ... The gremlins refuse to consider a test that is not true for primes. Consider, for example, the prime p = 1037*2^8-1 = 265471 and set a = 3, x = 2, y = 1.
        Message 3 of 14 , Sep 21, 2013
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          Bernhard Helmes wrote:

          > A new and better proposition

          The gremlins refuse to consider a
          test that is not true for primes.

          Consider, for example, the prime
          p = 1037*2^8-1 = 265471
          and set a = 3, x = 2, y = 1.

          Then this prime fails your latest wriggle:

          > (x+yA)^(2^m)<>1 mod p with A=sqrt(a)

          print(Mod(Mod(1,265471)*(2+A),A^2-3)^(2^8));
          Mod(Mod(1, 265471), A^2 - 3)

          Please do not post probable primality "tests"
          that are failed by primes.

          David (trying to control the grumpy gremlins:-)
        • bhelmes_1
          A peaceful day for all group members, dear David, I changed only the point 3. Is this changement a sharper condition or is this mathematically seen the same ?
          Message 4 of 14 , Oct 12, 2013
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            A peaceful day for all group members,

            dear David,


            I changed only the point 3.

            Is this changement a sharper condition or is this mathematically seen the same ?


            I. For p=3 mod 4


            1. choose the smallest prime a with jacobi (a, p)=-1
            2. Verify a^(p-1)/2=-1 mod p
                [This is the p-1 test which proves that a is a non quadric residuum]
            3. Depending on the a choose the pell solution of x, y with
                x^2-ay^2=-1 mod p (the minus is the change :-)
            4. Calculate and verify (x+yA)^p=x-yA mod p with A=sqrt (a)
            5. m the greatest power of 2 with 2^m*r=p+1 and 2^m < r

                Calculate (x+yA)^r=x(r)+y(r)A mod p and
               verify that  gcd (y(r), p)=1
               [This is a strong p+1 test with a solution of the according pell equation]
             
            if 1. - 5. is true then p should be a prime


            I think that this "wriggle" will include that there exist a cycle group with order t > 2 and t | r

            There is no need to hurry up for the gremlins,
            the prime numbers will exist a little longer. ;-)

            Friendly greetings from the primes
            Bernhard

          • djbroadhurst
            ... For a=3 mod 4 there is no such thing. Please do not post tests that refer to non-existent entities. David
            Message 5 of 14 , Oct 12, 2013
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               Bernhard Helmes wrote:

               

              > the pell solution of x, y with x^2-ay^2=-1

               

              For a=3 mod 4 there is no such thing.

               

              Please do not post "tests" that refer
              to non-existent entities.

               

              David

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