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Re: mod quartic composite tests

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  • paulunderwooduk
    ... Thanks for the algebra -- it was worrying me. Yes, the gremlins will not like this test: gcd(x^3-x,n) kronecker(x^2-4,n)==-1 gcd(x^2-2,n)==1
    Message 1 of 26 , Jan 9, 2013
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      --- In primenumbers@yahoogroups.com, "djbroadhurst" wrote:
      >
      >
      >
      > --- In primenumbers@yahoogroups.com,
      > Paul Underwood wrote:
      >
      > > (L+2)^n==-L^3+(x^2-2)*L+2 (mod n, (L^2-x*L+1)*(L^2+x*L+1))
      > > This test is (1)+(1)+(2)+(2)+9 selfridge for small x
      >
      > I can do it in 6 selfridges for generic x.
      > Where do your "9" selfridges come from, Paul?
      >
      > With (ed.) kronecker(x^2-4,n) = -1 and prime n, we have
      >
      > (L+a)^n = a+x-L mod(n, L^2-L*x+1) ... [1]
      > (L+a)^n = a-x-L mod(n, L^2+L*x+1) ... [2]
      >
      > and the two Frobenius tests cost 3+3=6 selfridges, generically.
      >
      > Now let f(x,L) = -L^3+(x^2-2)*L and observe
      > from simple algebra (no modularity involved) that
      >
      > f(x,L) + a = (a+x-L) - (L+x)*(L^2-x*L+1)
      > f(x,L) + a = (a-x-L) - (L-x)*(L^2+x*L+1)
      >
      > Hence [1] and [2] imply that
      >
      > (L+a)^n = f(x,L) + a mod(n, (L^2-x*L+1)*(L^2+x*L+1)) ... [3]
      >
      > and Paul has set a = 2 in [3].
      >
      > Since this is a double-Frobenius test, the gremlins
      > reserve the right to choose /both/ parameters (a,x) in [3].
      >

      Thanks for the algebra -- it was worrying me.

      Yes, the gremlins will not like this test:

      gcd(x^3-x,n)
      kronecker(x^2-4,n)==-1
      gcd(x^2-2,n)==1
      (L+x^2-2)^n==-L^3+(x^2-2)*L+(x^2-2) (mod n, (L^2-x*L+1)*(L^2+x*L+1))

      Paul
    • paulunderwooduk
      ... I have verified this test for all x for all n
      Message 2 of 26 , Jan 10, 2013
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        --- In primenumbers@yahoogroups.com, "paulunderwooduk" wrote:

        > gcd(x^3-x,n)==1
        > kronecker(x^2-4,n)==-1
        > gcd(x^2-2,n)==1
        > (L+x^2-2)^n==-L^3+(x^2-2)*L+(x^2-2) (mod n, (L^2-x*L+1)*(L^2+x*L+1))

        I have verified this test for all x for all n<10^6 co-prime to 30,

        Paul
      • paulunderwooduk
        ... ? {tst(n,x)=kronecker(x^2-4,n)==-1&& gcd(x^3-x,n)==1&& gcd(x^2-2,n)==1&& Mod(Mod(1,n)*(L+x^2-2),(L^2-x*L+1)*(L^2+x*L+1))^(n)==-L^3+(x^2-2)*L+x^2-2;} ?
        Message 3 of 26 , Jan 11, 2013
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          --- In primenumbers@yahoogroups.com, "paulunderwooduk" wrote:
          >
          >
          >
          > --- In primenumbers@yahoogroups.com, "paulunderwooduk" wrote:
          >
          > > gcd(x^3-x,n)==1
          > > kronecker(x^2-4,n)==-1
          > > gcd(x^2-2,n)==1
          > > (L+x^2-2)^n==-L^3+(x^2-2)*L+(x^2-2) (mod n, (L^2-x*L+1)*(L^2+x*L+1))
          >
          > I have verified this test for all x for all n<10^6 co-prime to 30,
          >

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

          ? {tstfile("underw297.txt");}
          2/297 counterexamples left in underw297.txt
          ? {tstfile("underw65.txt");}
          5146/12846 counterexamples left in underw65.txt

          Paul
        • paulunderwooduk
          ... Adding gcd(x^2-3,n)==1 makes sense because x^2-2==1 (mod x^2-3). So the resurrected test is: {tst(n,x)=kronecker(x^2-4,n)==-1&& gcd(x^3-x,n)==1&&
          Message 4 of 26 , Jan 11, 2013
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            --- In primenumbers@yahoogroups.com, "paulunderwooduk" wrote:

            >
            > ? {tst(n,x)=kronecker(x^2-4,n)==-1&&
            > gcd(x^3-x,n)==1&&
            > gcd(x^2-2,n)==1&&
            > Mod(Mod(1,n)*(L+x^2-2),(L^2-x*L+1)*(L^2+x*L+1))^(n)==-L^3+(x^2-2)*L+x^2-2;}
            >
            > ? {tstfile("underw297.txt");}
            > 2/297 counterexamples left in underw297.txt
            > ? {tstfile("underw65.txt");}
            > 5146/12846 counterexamples left in underw65.txt
            >

            Adding gcd(x^2-3,n)==1 makes sense because x^2-2==1 (mod x^2-3). So the resurrected test is:

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

            Then checking David's files at:
            http://physics.open.ac.uk/~dbroadhu/cert/

            {tstfile(file)=local(c,n,x,v=readvec(file));
            for(k=1,#v,n=v[k][1];x=v[k][2];
            if(tst(n,x)&&!isprime(n),c++));
            print(c"/"#v" counterexamples left in "file);c;}

            ? {tstfile("underbh4.txt");}
            0/33445 counterexamples left in underbh4.txt
            ? {tstfile("underbh6.txt");}
            0/308619 counterexamples left in underbh6.txt
            ? {tstfile("underw97.txt");}
            0/97 counterexamples left in underw97.txt
            ? {tstfile("underw297.txt");}
            0/297 counterexamples left in underw297.txt
            ? {tstfile("underw65.txt");}
            0/12846 counterexamples left in underw65.txt
            ? {tstfile("underw65x.txt");}
            0/10220 counterexamples left in underw65x.txt
            ? {tstfile("underwg.txt");}
            0/100000 counterexamples left in underwg.txt

            Paul
          • paulunderwooduk
            ... n=2672279 and x=89805 forms a near-counterexample, saved only by gcd(x^3-x,n)==2672279. Since gcd(x^3-x,n)==1 is required I do not have to verify numbers
            Message 5 of 26 , Jan 15, 2013
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              --- In primenumbers@yahoogroups.com, "paulunderwooduk" wrote:

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

              n=2672279 and x=89805 forms a near-counterexample, saved only by gcd(x^3-x,n)==2672279.

              Since gcd(x^3-x,n)==1 is required I do not have to verify numbers divisible by 2 or 3, and with kronecker(x^2-4,n)==-1, I do not have to verify numbers divisible by 5. Also the squares modulo 7 are 0, 1, 4, and 2, and because I am checking gcd(x^3-x,n)==1, kronecker(x^2-4,n)==-1 and gcd(x^2-2,n)==1, I can skip numbers divisible by 7. All in all, I need only verify numbers co-prime to 210,

              Paul
            • paulunderwooduk
              ... That should read gcd(x^2-3,n)==2672279. Paul
              Message 6 of 26 , Jan 15, 2013
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                --- In primenumbers@yahoogroups.com, "paulunderwooduk" wrote:

                > n=2672279 and x=89805 forms a near-counterexample, saved only by gcd(x^3-x,n)==2672279.

                That should read gcd(x^2-3,n)==2672279.

                Paul
              • djbroadhurst
                ... A wise wriggle :-) Without it, you left yourself wide open to fraud. With it, you seem to be much more secure. I believe, yet cannot show, that there are
                Message 7 of 26 , Jan 15, 2013
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                  --- In primenumbers@yahoogroups.com,
                  "paulunderwooduk" wrote:

                  > Adding gcd(x^2-3,n)==1

                  A wise "wriggle" :-)

                  Without it, you left yourself wide open to fraud.
                  With it, you seem to be much more secure.

                  I believe, yet cannot show, that there are zillions of
                  counterexamples to your single-parameter, double-Frobenius
                  test, yet the prospects of finding one, now that you
                  have bolted the stable door, seem to be minuscule.

                  Thanks, Paul, for your responsiveness.

                  David
                • djbroadhurst
                  ... http://physics.open.ac.uk/~dbroadhu/cert/underwqd.txt.gz provides 352869 such frauds: {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x^3-x,n)==1&&gcd(x^2-2,n)==1&&
                  Message 8 of 26 , Jan 17, 2013
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                    --- In primenumbers@yahoogroups.com,
                    "paulunderwooduk" wrote:

                    > n=2672279 and x=89805

                    http://physics.open.ac.uk/~dbroadhu/cert/underwqd.txt.gz
                    provides 352869 such frauds:

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

                    {tstfile(file)=local(v=readvec(file),c);
                    c=sum(k=1,#v,tst(v[k][1],v[k][2])&&!isprime(v[k][1]));
                    print(c"/"#v" counterexamples left in "file);c;}

                    tstfile("underwqd.txt");

                    352869/352869 counterexamples left in underwqd.txt

                    All are trapped by Paul's latest wriggle,
                    which requires x^2-3 to be coprime to n.

                    David
                  • paulunderwooduk
                    ... Thanks for these, David. Is it a comprehensive list for all n
                    Message 9 of 26 , Jan 17, 2013
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                      --- In primenumbers@yahoogroups.com, "djbroadhurst" wrote:

                      > http://physics.open.ac.uk/~dbroadhu/cert/underwqd.txt.gz

                      >
                      > 352869/352869 counterexamples left in underwqd.txt
                      >
                      > All are trapped by Paul's latest wriggle,
                      > which requires x^2-3 to be coprime to n.
                      >

                      Thanks for these, David. Is it a comprehensive list for all n <= 97847746461047271599?

                      Paul
                    • djbroadhurst
                      ... By no means. However the updated file http://physics.open.ac.uk/~dbroadhu/cert/underwqd.txt.gz now has more:
                      Message 10 of 26 , Jan 17, 2013
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                        --- In primenumbers@yahoogroups.com,
                        "paulunderwooduk" wrote:

                        > Is it a comprehensive list for all n <= 97847746461047271599?

                        By no means. However the updated file
                        http://physics.open.ac.uk/~dbroadhu/cert/underwqd.txt.gz
                        now has more:

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

                        {tstfile(file)=local(v=readvec(file),c);
                        c=sum(k=1,#v,tst(v[k][1],v[k][2])&&!isprime(v[k][1]));
                        print(c"/"#v" counterexamples left in "file);c;}

                        tstfile("underwqd.txt");

                        422355/422355 counterexamples left in underwqd.txt

                        Challenge: Find a composite 10^10-smooth positive
                        integer, n, such that:
                        1) there exist an integer x that passes tst(n,x),
                        2) n is not in underwqd.txt.

                        Comment: I do not know of any such integer. Nor do I know
                        how to search for one. So I guess that means my gremlins
                        are comprehensively exhausted, though the question is not.

                        David
                      • djbroadhurst
                        ... Exercise: Show that Paul s test Mod(Mod(1,n)*(L+x^2-2),(L^2-x*L+1)*(L^2+x*L+1))^n==-L^3+(x^2-2)*L+x^2-2 requires n to be a Fermat pseudoprime in base
                        Message 11 of 26 , Jan 18, 2013
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                          > All are trapped by Paul's latest wriggle,
                          > which requires x^2-3 to be coprime to n.

                          Exercise: Show that Paul's test
                          Mod(Mod(1,n)*(L+x^2-2),(L^2-x*L+1)*(L^2+x*L+1))^n==-L^3+(x^2-2)*L+x^2-2
                          requires n to be a Fermat pseudoprime in base 1+(x^2-3)*(x^2-2)^3
                          and thus loses (at least) one selfridge of potency for x^2 = 3 mod n.

                          David
                        • paulunderwooduk
                          ... ? M=[0,(x^2-2),0,-1;1,0,0,0;0,1,0,0;0,0,1,0];matdet(M+x^2-2)==1+(x^2-3)*(x^2-2)^3 1 Thanks for the insight. This can be split into 2 Fermat tests: ?
                          Message 12 of 26 , Jan 18, 2013
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                            --- In primenumbers@yahoogroups.com, "djbroadhurst" wrote:
                            >
                            >
                            >
                            > > All are trapped by Paul's latest wriggle,
                            > > which requires x^2-3 to be coprime to n.
                            >
                            > Exercise: Show that Paul's test
                            > Mod(Mod(1,n)*(L+x^2-2),(L^2-x*L+1)*(L^2+x*L+1))^n==-L^3+(x^2-2)*L+x^2-2
                            > requires n to be a Fermat pseudoprime in base 1+(x^2-3)*(x^2-2)^3
                            > and thus loses (at least) one selfridge of potency for x^2 = 3 mod n.
                            >

                            ? M=[0,(x^2-2),0,-1;1,0,0,0;0,1,0,0;0,0,1,0];matdet(M+x^2-2)==1+(x^2-3)*(x^2-2)^3
                            1

                            Thanks for the insight.

                            This can be split into 2 Fermat tests:

                            ? M=[x,-1;1,0];matdet(M+x^2-2)
                            x^4 + x^3 - 4*x^2 - 2*x + 5
                            ? M=[-x,-1;1,0];matdet(M+x^2-2)
                            x^4 - x^3 - 4*x^2 + 2*x + 5

                            which are equal to:

                            ? M=[x,-1;1,0];matdet(M+x^2-2)==(x^2-2)*(x+2)*(x-1)+1
                            1
                            ? M=[-x,-1;1,0];matdet(M+x^2-2)==(x^2-2)*(x-2)*(x+1)+1
                            1

                            Paul
                          • djbroadhurst
                            ... Exercise 2: Show that the test loses 3 selfrides for x^2 = 3 mod n. Comment 2: Hence the happy gremlins, in this case. Exercise 3: Show that the test loses
                            Message 13 of 26 , Jan 18, 2013
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                              --- In primenumbers@yahoogroups.com, "paulunderwooduk" wrote:

                              > > loses (at least) one selfridge of potency for x^2 = 3 mod n.
                              > Thanks for the insight.

                              Exercise 2: Show that the test loses 3 selfrides for x^2 = 3 mod n.
                              Comment 2: Hence the happy gremlins, in this case.

                              Exercise 3: Show that the test loses 1 selfride for 2*x^2 = 5 mod n.
                              Comment 3: The gremlins were not able to fool it in this case.

                              David
                            • djbroadhurst
                              ... Exercise 4: Show that the test loses 2 selfridges for 2*x^2 = 5 mod n. David
                              Message 14 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 15 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 16 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 17 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 18 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 19 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 20 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 21 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 22 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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