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

Re: mod quartic composite tests

Expand Messages
  • paulunderwooduk
    I failed to solve any of David s exercises... but I have done some shallow verification, n
    Message 1 of 26 , Jan 19, 2013
      I failed to solve any of David's exercises... but I have done some shallow verification, n<2.5*10^5, for yet another test:

      {tst(n,x)=kronecker(x^2-4,n)==-1&&
      gcd(x+1,n)==1&&
      Mod(Mod(1,n)*(L+x+1),(L^2-x*L+1)*(L^2+x*L+1))^(n)==-L^3+(x^2-2)*L+x+1;}

      It looks as though when "gcd(x+1,n)==1" is needed then n==5 (mod 6), and any of gcd(x-1,n), gcd(x,n), gcd(x^2-2,n) and gcd(x^3-3,n) greater than 1 is also a prime equal to 5 (mod 6) when "gcd(x+1,n)==1" is required,

      Paul -- subject to gcd wriggles
    • paulunderwooduk
      ... Further, it seems that if gcd(x+1,n) is needed then it is equal to 1 (mod 6) Paul
      Message 2 of 26 , Jan 19, 2013
        --- In primenumbers@yahoogroups.com, "paulunderwooduk" wrote:
        >
        >
        > I failed to solve any of David's exercises... but I have done some shallow verification, n<2.5*10^5, for yet another test:
        >
        > {tst(n,x)=kronecker(x^2-4,n)==-1&&
        > gcd(x+1,n)==1&&
        > Mod(Mod(1,n)*(L+x+1),(L^2-x*L+1)*(L^2+x*L+1))^(n)==-L^3+(x^2-2)*L+x+1;}
        >
        > It looks as though when "gcd(x+1,n)==1" is needed then n==5 (mod 6), and any of gcd(x-1,n), gcd(x,n), gcd(x^2-2,n) and gcd(x^3-3,n) greater than 1 is also a prime equal to 5 (mod 6) when "gcd(x+1,n)==1" is required,
        >

        Further, it seems that if "gcd(x+1,n)" is needed then it is equal to 1 (mod 6)

        Paul
      • paulunderwooduk
        ... These ancillary statements are mostly false, except that maybe when gcd(x+1,n) needs to be checked then gcd(x-1,n), gcd(x,n), gcd(x^2-2,n) and
        Message 3 of 26 , Jan 19, 2013
          --- In primenumbers@yahoogroups.com, "paulunderwooduk" wrote:
          >
          >
          >
          > --- In primenumbers@yahoogroups.com, "paulunderwooduk" wrote:
          > >
          > >
          > > I failed to solve any of David's exercises... but I have done some shallow verification, n<2.5*10^5, for yet another test:
          > >
          > > {tst(n,x)=kronecker(x^2-4,n)==-1&&
          > > gcd(x+1,n)==1&&
          > > Mod(Mod(1,n)*(L+x+1),(L^2-x*L+1)*(L^2+x*L+1))^(n)==-L^3+(x^2-2)*L+x+1;}
          > >
          > > It looks as though when "gcd(x+1,n)==1" is needed then n==5 (mod 6), and any of gcd(x-1,n), gcd(x,n), gcd(x^2-2,n) and gcd(x^3-3,n) greater than 1 is also a prime equal to 5 (mod 6) when "gcd(x+1,n)==1" is required,
          > >
          >
          > Further, it seems that if "gcd(x+1,n)" is needed then it is equal to 1 (mod 6)
          >

          These ancillary statements are mostly false, except that maybe when "gcd(x+1,n)" needs to be checked then gcd(x-1,n), gcd(x,n), gcd(x^2-2,n) and gcd(x^3-3,n) are either 1 or prime,

          Paul
        • paulunderwooduk
          ... Please accept my apology for my previous statements about this composite test. I am actually running tests for: (mod n, L^2-x*L+1) and (mod n, L^2+x*L+1);
          Message 4 of 26 , Jan 19, 2013
            --- In primenumbers@yahoogroups.com, "paulunderwooduk" wrote:


            > > >
            > > > {tst(n,x)=kronecker(x^2-4,n)==-1&&
            > > > gcd(x+1,n)==1&&
            > > > Mod(Mod(1,n)*(L+x+1),(L^2-x*L+1)*(L^2+x*L+1))^(n)==-L^3+(x^2-2)*L+x+1;}
            > > >

            Please accept my apology for my previous statements about this composite test. I am actually running tests for:
            (mod n, L^2-x*L+1) and (mod n, L^2+x*L+1); not the product of these. If the main test is passed then gcds with n of x-1, x, x+1, x^2-2 and x^2-3 are logged.

            Now for some speculation about the results so far:

            1) taking the mod with "the product" implies gcd(x,n)==1.

            2) for "the product", gcds of n with x-1, x^2-2 and x^2-3 are either 1 or a prime congruent to 5 (mod 6) where "gcd(x+1,n)" is logged.

            3) logged gcd(x+1,n) is not 1

            4) the logged n are all congruent to 5 (mod 6).

            Paul
          • paulunderwooduk
            ... Here is another test, on the same theme, for which I cannot also easily find a fraud: {tst(n,x)=kronecker(x^2-4,n)==-1&& gcd(x^2-1,n)==1&&
            Message 5 of 26 , Jan 21, 2013
              --- In primenumbers@yahoogroups.com, "paulunderwooduk" wrote:

              >
              > > > >
              > > > > {tst(n,x)=kronecker(x^2-4,n)==-1&&
              > > > > gcd(x+1,n)==1&&
              > > > > Mod(Mod(1,n)*(L+x+1),(L^2-x*L+1)*(L^2+x*L+1))^(n)==-L^3+(x^2-2)*L+x+1;}
              > > > >
              >

              Here is another test, on the same theme, for which I cannot also easily find a fraud:

              {tst(n,x)=kronecker(x^2-4,n)==-1&&
              gcd(x^2-1,n)==1&&
              Mod(Mod(1,n)*(L+x^2-1),(L^2-x*L+1)*(L^2+x*L+1))^(n)==-L^3+(x^2-2)*L+x^2-1;}

              Paul
            • paulunderwooduk
              ... n=2953711;x=285843 is a near-counterexample that comes from me testing over the two quadratics that form the quartic. gcd(x,n)==95281 and
              Message 6 of 26 , Jan 27, 2013
                --- In primenumbers@yahoogroups.com, "paulunderwooduk" wrote:

                > Here is another test, on the same theme, for which I cannot also easily find a fraud:
                >
                > {tst(n,x)=kronecker(x^2-4,n)==-1&&
                > gcd(x^2-1,n)==1&&
                > Mod(Mod(1,n)*(L+x^2-1),(L^2-x*L+1)*(L^2+x*L+1))^(n)==-L^3+(x^2-2)*L+x^2-1;}
                >

                n=2953711;x=285843 is a near-counterexample that comes from me testing over the two quadratics that form the quartic. gcd(x,n)==95281 and gcd(x^2-2,n)==31,

                Paul
              • paulunderwooduk
                ... n=1934765;x=1219266 has gcd(x-1,n)==265. So 2) might be 1 or a product of primes each congruent 5 (mod 6) , but as greater n get tested I guess this rule
                Message 7 of 26 , Jan 30, 2013
                  > --- In primenumbers@yahoogroups.com, "paulunderwooduk" wrote:
                  >
                  >
                  > > > >
                  > > > > {tst(n,x)=kronecker(x^2-4,n)==-1&&
                  > > > > gcd(x+1,n)==1&&
                  > > > > Mod(Mod(1,n)*(L+x+1),(L^2-x*L+1)*(L^2+x*L+1))^(n)==-L^3+(x^2-2)*L+x+1;}
                  > > > >
                  >
                  > Please accept my apology for my previous statements about this composite test. I am actually running tests for:
                  > (mod n, L^2-x*L+1) and (mod n, L^2+x*L+1); not the product of these. If the main test is passed then gcds with n of x-1, x, x+1, x^2-2 and x^2-3 are logged.
                  >
                  > Now for some speculation about the results so far:
                  >
                  > 1) taking the mod with "the product" implies gcd(x,n)==1.
                  >
                  > 2) for "the product", gcds of n with x-1, x^2-2 and x^2-3 are either 1 or a prime congruent to 5 (mod 6) where "gcd(x+1,n)" is logged.
                  >

                  n=1934765;x=1219266 has gcd(x-1,n)==265. So 2) might be "1 or a product of primes each congruent 5 (mod 6)", but as greater n get tested I guess this rule will break too...

                  > 3) logged gcd(x+1,n) is not 1
                  >
                  > 4) the logged n are all congruent to 5 (mod 6).
                  >

                  I have verified all n<1.95*10^6

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