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Re: mod quartic composite tests

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  • 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 1 of 26 , Jan 19, 2013
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      --- 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 2 of 26 , Jan 19, 2013
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        --- 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 3 of 26 , Jan 21, 2013
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          --- 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 4 of 26 , Jan 27, 2013
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            --- 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 5 of 26 , Jan 30, 2013
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              > --- 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
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