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Re: mod quartic composite tests

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  • paulunderwooduk
    ... n=287051 and x=3988 is a counterexample. ... This test is (1)+(1)+(2)+(2)+9 selfridge for small x, where (1) is a fermat test and (2) is a frobenius test
    Message 1 of 26 , Jan 9, 2013
      --- In primenumbers@yahoogroups.com, "paulunderwooduk" wrote:
      >
      > Hi,
      >
      > I have devised a new composite test for odd n with x:
      > gcd(x^3-x,n)==1 (mod n)
      > kronecker(x^2-4,n)==-1
      >
      > and the sub-test:
      > (L+1)^n==-L^3+(x^2-2)*L+1 (mod n, (L^2-x*L+1)*(L^2+x*L+1))
      >
      > Can you find a counterexample?

      n=287051 and x=3988 is a counterexample.

      > If this is too easy for you, please try:
      >
      > (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, where (1) is a fermat test and (2) is a frobenius test over L^2-x*L+1 or L^2+x*L+1.

      I already have:
      "Non-square N>1 is prime if and only if for any integer x such that
      KoneckerSymbol(x^2-4,N)==-1
      then both (L+-2)^(N+1)==5+-2*x (mod N, L^2-x*L+1) (Both L+-2 needed.)
      Verified for odd N< 10^7."

      Paul
    • djbroadhurst
      ... I can do it in 6 selfridges for generic x. Where do your 9 selfridges come from, Paul? With kronecker(x^4-2,n) = -1 and prime n, we have (L+a)^n = a+x-L
      Message 2 of 26 , Jan 9, 2013
        --- 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 kronecker(x^4-2,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].

        David
      • djbroadhurst
        ... but meant to write x^2-4, not x^4-2
        Message 3 of 26 , Jan 9, 2013
          --- In primenumbers@yahoogroups.com,
          "djbroadhurst" wrote:
          > With kronecker(x^4-2,n) = -1
          but meant to write x^2-4, not x^4-2
        • 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 4 of 26 , Jan 9, 2013
            --- 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 5 of 26 , Jan 10, 2013
              --- 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 6 of 26 , Jan 11, 2013
                --- 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 7 of 26 , Jan 11, 2013
                  --- 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 8 of 26 , Jan 15, 2013
                    --- 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 9 of 26 , Jan 15, 2013
                      --- 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 10 of 26 , Jan 15, 2013
                        --- 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 11 of 26 , Jan 17, 2013
                          --- 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 12 of 26 , Jan 17, 2013
                            --- 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 13 of 26 , Jan 17, 2013
                              --- 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 14 of 26 , Jan 18, 2013
                                > 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 15 of 26 , Jan 18, 2013
                                  --- 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 16 of 26 , Jan 18, 2013
                                    --- 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 17 of 26 , Jan 18, 2013
                                      --- 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 18 of 26 , Jan 18, 2013
                                        --- 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 19 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 20 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 21 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 22 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 23 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 24 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 25 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
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