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Re: mod quartic composite tests

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  • djbroadhurst
    ... Exercise 5: Show that the test loses 3 selfridges for 2*x^2 = 5 mod n, degenerating to a 1-selfridge Euler test, with base -15/16, plus a 2-selfridge Lucas
    Message 1 of 26 , Jan 18, 2013
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      --- In primenumbers@yahoogroups.com, "djbroadhurst" wrote:

      > Exercise 4: Show that the test loses 2 selfridges for 2*x^2 = 5 mod n.

      Exercise 5: Show that the test loses 3 selfridges for 2*x^2 = 5 mod n,
      degenerating to a 1-selfridge Euler test, with base -15/16, plus a
      2-selfridge Lucas test with P = 2/5 and Q = 1, and thus costs the same as BPSW.

      Comment: As in the case of BPSW, the gremlins cannot defraud this case.

      David
    • paulunderwooduk
      I failed to solve any of David s exercises... but I have done some shallow verification, n
      Message 2 of 26 , Jan 19, 2013
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        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 3 of 26 , Jan 19, 2013
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          --- 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 4 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 5 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 6 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 7 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 8 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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