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Re: mod quartic composite tests

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  • djbroadhurst
    ... Exercise 4: Show that the test loses 2 selfridges for 2*x^2 = 5 mod n. David
    Message 1 of 26 , Jan 18, 2013
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      --- In primenumbers@yahoogroups.com, "djbroadhurst" wrote:
      > Exercise 3: Show that the test loses 1 selfridge for 2*x^2 = 5 mod n.

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

      David
    • 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 2 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 3 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 4 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 5 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 6 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 7 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 8 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 9 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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