Loading ...
Sorry, an error occurred while loading the content.

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

Expand Messages
  • bhelmes_1
    Dear David, As remarked, no prime p = 1 mod 4 passes the test. I would like to add that only the point 5. causes the trouble. and that all primes passes the
    Message 1 of 14 , Sep 14, 2013
    • 0 Attachment
      Dear David,

      >As remarked, no prime p = 1 mod 4 passes the test.

      I would like to add that only the point 5. causes the trouble.
      and that all primes passes the points 1. - 4.

      >Here I show that the composite number
      >p = 2953*5167 = 15258151 = 7 mod 16
      >passes the test, using its minimal strong base, a = 3:

      I feel that it was not easy for you and a lot of work for you to find one counterexample.
      Besides i noticed that your selected x and y is kind of  (x+yA)^4=-1 mod p
      which is very special and shows that the proposed test is not sufficient.

      >The gremlins will entertain no further wriggle from Bernhard.

      David, i have a huge respect for your mathematical skills,
      nevertheless the description to entertain me is not what i expect from you.
      This primenumber group is the only place for me to discuss some mathematical
      ideas and to get some mathematical feedback.

      As far as i can see a proven sufficient primetest with 4 Selfridge would be a mathematical progress
      and the development of choosing the right parameters in order to build a proof is not a kind of "wriggle" but
      mathematical work.

      I appreciate deeply a fair mathematical discussion.

      David, prime numbers are something for the eternity and there is no reason to get angry about some humble
      mathematicians like me who try to learn some more details of number theory.

      I wish you a pleasant and friendly weekend.
      Bernhard



      David (their keeper)
    • djbroadhurst
      ... So remove point (5) to allow primes p = 1 mod 4. Then it s /very/ easy to find a pseudoprime: {p=3281;x=1062;y=1600; if(kronecker(3,p)==-1&&
      Message 2 of 14 , Sep 14, 2013
      • 0 Attachment
        Bernhard Helmes wrote:

        > > As remarked, no prime p = 1 mod 4 passes the test.
        > I would like to add that only the point 5. causes the trouble.

        So remove point (5) to allow primes p = 1 mod 4.
        Then it's /very/ easy to find a pseudoprime:

        {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

        David
      • 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 3 of 14 , Sep 20, 2013
        • 0 Attachment
          --- 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 4 of 14 , Sep 21, 2013
          • 0 Attachment
            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 5 of 14 , Sep 21, 2013
            • 0 Attachment
              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 6 of 14 , Oct 12, 2013
              • 0 Attachment

                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 7 of 14 , Oct 12, 2013
                • 0 Attachment

                   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

                Your message has been successfully submitted and would be delivered to recipients shortly.