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Re: single frobenius and double fermat

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  • paulunderwooduk
    ... n=396271;x=5042 is counterexample. I am now testing with the added wriggle: x^(n-1)==1 (mod n) Paul
    Message 1 of 24 , Dec 9, 2012
      --- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@...> wrote:
      >
      > Hi,
      >
      > here yet another composite test for a counterexample challenge,
      >
      > For n co-prime to 30 find x:
      > gcd(x^3-x,n)==1
      > kronecker(x^2-4,n)==-1
      >
      > and perform these sub-tests:
      > (x-2)^(n-1)==1 (mod n)
      > (x+2)^(n-1)==1 (mod n)
      > (x*L^3+1)^(n+1)==(x^2-1)^2 (mod n, L^2-x*L+1)
      >
      > Note: x*L^3+1==(x^3-x)*L-(x^2-1) (mod n, L^2-x*L+1)
      >
      n=396271;x=5042 is counterexample.

      I am now testing with the added wriggle:
      x^(n-1)==1 (mod n)

      Paul
    • paulunderwooduk
      ... I dropped the above wriggle. The test I am now verifying is: gcd(x^3-x,n)==1 kronecker(x^2-4,n)==-1 (x-2)^((n-1)/2)==kronecker(x-2,n) (mod n)
      Message 2 of 24 , Dec 9, 2012
        --- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@...> wrote:
        >
        >
        >
        > --- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@> wrote:
        > >
        > > Hi,
        > >
        > > here yet another composite test for a counterexample challenge,
        > >
        > > For n co-prime to 30 find x:
        > > gcd(x^3-x,n)==1
        > > kronecker(x^2-4,n)==-1
        > >
        > > and perform these sub-tests:
        > > (x-2)^(n-1)==1 (mod n)
        > > (x+2)^(n-1)==1 (mod n)
        > > (x*L^3+1)^(n+1)==(x^2-1)^2 (mod n, L^2-x*L+1)
        > >
        > > Note: x*L^3+1==(x^3-x)*L-(x^2-1) (mod n, L^2-x*L+1)
        > >
        > n=396271;x=5042 is counterexample.
        >
        > I am now testing with the added wriggle:
        > x^(n-1)==1 (mod n)
        >

        I dropped the above wriggle. The test I am now verifying is:

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

        Paul
      • paulunderwooduk
        ... I edited the sign typo. Paul
        Message 3 of 24 , Dec 9, 2012
          > The test I am now verifying is:
          >
          > gcd(x^3-x,n)==1
          > kronecker(x^2-4,n)==-1
          > (x-2)^((n-1)/2)==kronecker(x-2,n) (mod n)
          > (x+2)^((n-1)/2)==kronecker(x+2,n) (mod n)
          > (x*L^3+1)^(n+1)==(x^2-1)^2 (mod n, L^2-x*L+1)
          >

          I edited the sign typo.

          Paul
        • paulunderwooduk
          ... I think the frobenius sub-test can be simplified to: ((x^2-1)*L)^(n+1)==(x^2-1)^2 (mod n, L^2-x*L+1) Paul
          Message 4 of 24 , Dec 9, 2012
            --- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@...> wrote:
            >
            >
            > > The test I am now verifying is:
            > >
            > > gcd(x^3-x,n)==1
            > > kronecker(x^2-4,n)==-1
            > > (x-2)^((n-1)/2)==kronecker(x-2,n) (mod n)
            > > (x+2)^((n-1)/2)==kronecker(x+2,n) (mod n)
            > > (x*L^3+1)^(n+1)==(x^2-1)^2 (mod n, L^2-x*L+1)
            > >
            >

            I think the frobenius sub-test can be simplified to:

            ((x^2-1)*L)^(n+1)==(x^2-1)^2 (mod n, L^2-x*L+1)

            Paul
          • djbroadhurst
            ... {tst(n,x)=gcd(x^3-x,n)==1&&kronecker(x^2-4,n)==-1&& Mod(x-2,n)^((n-1)/2)==kronecker(x-2,n)&& Mod(x+2,n)^((n-1)/2)==kronecker(x+2,n)&&
            Message 5 of 24 , Dec 10, 2012
              --- In primenumbers@yahoogroups.com,
              "paulunderwooduk" <paulunderwood@...> wrote:

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

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

              {v=readvec("underwg.txt");
              \\ http://physics.open.ac.uk/~dbroadhu/cert/underwg.txt.gz
              print(sum(k=1,#v,tst(v[k][1],v[k][2]))" counterexamples");}

              19959 counterexamples

              David
            • paulunderwooduk
              ... Thanks again, David. I had an mistake in one of my equations that caused erroneous checking of your files. I have another test. For small x, and choosing
              Message 6 of 24 , Dec 10, 2012
                --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
                >
                >
                >
                > --- In primenumbers@yahoogroups.com,
                > "paulunderwooduk" <paulunderwood@> wrote:
                >
                > > gcd(x^3-x,n)==1
                > > kronecker(x^2-4,n)==-1
                > > (x-2)^((n-1)/2)==kronecker(x-2,n) (mod n)
                > > (x+2)^((n-1)/2)==kronecker(x+2,n) (mod n)
                > > (x*L^3+1)^(n+1)==(x^2-1)^2 (mod n, L^2-x*L+1)
                >
                > {tst(n,x)=gcd(x^3-x,n)==1&&kronecker(x^2-4,n)==-1&&
                > Mod(x-2,n)^((n-1)/2)==kronecker(x-2,n)&&
                > Mod(x+2,n)^((n-1)/2)==kronecker(x+2,n)&&
                > Mod(Mod(1,n)*(x*L^3+1),L^2-x*L+1)^(n+1)==(x^2-1)^2;}
                >
                > {v=readvec("underwg.txt");
                > \\ http://physics.open.ac.uk/~dbroadhu/cert/underwg.txt.gz
                > print(sum(k=1,#v,tst(v[k][1],v[k][2]))" counterexamples");}
                >
                > 19959 counterexamples
                >

                Thanks again, David. I had an mistake in one of my equations that caused erroneous checking of your files.

                I have another test. For small x, and choosing where possible x==3 or x==6, the test is on average 3.25 selfridge. For n co-prime to 30 find any x:

                gcd(x^3-x,n)==1
                kronecker(x^2-4,n)==-1
                (x-2)^((n-1)/2)==kronecker(x-2,n) (mod n)
                (x+2)^((n-1)/2)==kronecker(x+2,n) (mod n)
                (x^3-x)*(L^2-4)^(n+1)==(x^3-x)^2*(25-4*x^2) (mod n, L^2-x*L+1)

                I this tested against your files:

                {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x^3-x,n)==1&&
                Mod(x-2,n)^((n-1)/2)==kronecker(x-2,n)&&
                Mod(x+2,n)^((n-1)/2)==kronecker(x+2,n)&&
                Mod(Mod(1,n)*(x^3-x)*(L^2-4),L^2-x*L+1)^(n+1)==(x^3-x)^2*(25-4*x^2);}

                {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
                ... should be: ((x^3-x)*(L^2-4))^(n+1)==(x^3-x)^2*(25-4*x^2) (mod n, L^2-x*L+1) Paul
                Message 7 of 24 , Dec 10, 2012
                  > (x^3-x)*(L^2-4)^(n+1)==(x^3-x)^2*(25-4*x^2) (mod n, L^2-x*L+1)

                  should be:

                  ((x^3-x)*(L^2-4))^(n+1)==(x^3-x)^2*(25-4*x^2) (mod n, L^2-x*L+1)

                  Paul
                • djbroadhurst
                  ... {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x^3-x,n)==1&& Mod(x-2,n)^((n-1)/2)==kronecker(x-2,n)&& Mod(x+2,n)^((n-1)/2)==kronecker(x+2,n)&&
                  Message 8 of 24 , Dec 10, 2012
                    --- In primenumbers@yahoogroups.com,
                    "paulunderwooduk" <paulunderwood@...> wrote:

                    > gcd(x^3-x,n)==1
                    > kronecker(x^2-4,n)==-1
                    > (x-2)^((n-1)/2)==kronecker(x-2,n) (mod n)
                    > (x+2)^((n-1)/2)==kronecker(x+2,n) (mod n)
                    > (x^3-x)*(L^2-4)^(n+1)==(x^3-x)^2*(25-4*x^2) (mod n, L^2-x*L+1)

                    {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x^3-x,n)==1&&
                    Mod(x-2,n)^((n-1)/2)==kronecker(x-2,n)&&
                    Mod(x+2,n)^((n-1)/2)==kronecker(x+2,n)&&
                    Mod(Mod(1,n)*(x^3-x)*(L^2-4),L^2-x*L+1)^(n+1)==
                    (x^3-x)^2*(25-4*x^2);}

                    {if(tst(115886944289,3692152318),print("fooled"));}

                    fooled

                    David
                  • paulunderwooduk
                    ... Thanks. Here is another composite test: {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x^3-x,n)==1&&kronecker(x^2-1,n)==-1&&
                    Message 9 of 24 , Dec 10, 2012
                      --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
                      >
                      >
                      >
                      > --- In primenumbers@yahoogroups.com,
                      > "paulunderwooduk" <paulunderwood@> wrote:
                      >
                      > > gcd(x^3-x,n)==1
                      > > kronecker(x^2-4,n)==-1
                      > > (x-2)^((n-1)/2)==kronecker(x-2,n) (mod n)
                      > > (x+2)^((n-1)/2)==kronecker(x+2,n) (mod n)
                      > > (x^3-x)*(L^2-4)^(n+1)==(x^3-x)^2*(25-4*x^2) (mod n, L^2-x*L+1)
                      >
                      > {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x^3-x,n)==1&&
                      > Mod(x-2,n)^((n-1)/2)==kronecker(x-2,n)&&
                      > Mod(x+2,n)^((n-1)/2)==kronecker(x+2,n)&&
                      > Mod(Mod(1,n)*(x^3-x)*(L^2-4),L^2-x*L+1)^(n+1)==
                      > (x^3-x)^2*(25-4*x^2);}
                      >
                      > {if(tst(115886944289,3692152318),print("fooled"));}
                      >
                      > fooled
                      >

                      Thanks. Here is another composite test:

                      {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x^3-x,n)==1&&kronecker(x^2-1,n)==-1&&
                      Mod(x-2,n)^((n-1)/2)==kronecker(x-2,n)&&
                      Mod(x+2,n)^((n-1)/2)==kronecker(x+2,n)&&
                      Mod(x-1,n)^((n-1)/2)==kronecker(x-1,n)&&
                      Mod(x+1,n)^((n-1)/2)==kronecker(x+1,n)&&
                      Mod(Mod(1,n)*x*(L^2-4),L^2-x*L+1)^(n+1)==x^2*(25-4*x^2);}

                      It's on average about 5 selfridge for carefully chosen x,

                      Paul
                    • djbroadhurst
                      ... Generically, that is 7 selfridges: 4 Euler tests and one Frobenius. However, it s reasonably easy to fool:
                      Message 10 of 24 , Dec 10, 2012
                        --- In primenumbers@yahoogroups.com,
                        "paulunderwooduk" <paulunderwood@...> wrote:

                        > Here is another composite test
                        > It's on average about 5 selfridge

                        Generically, that is 7 selfridges: 4 Euler tests
                        and one Frobenius. However, it's reasonably easy to fool:

                        {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x^3-x,n)==1&&
                        kronecker(x^2-1,n)==-1&&
                        Mod(x-2,n)^((n-1)/2)==kronecker(x-2,n)&&
                        Mod(x+2,n)^((n-1)/2)==kronecker(x+2,n)&&
                        Mod(x-1,n)^((n-1)/2)==kronecker(x-1,n)&&
                        Mod(x+1,n)^((n-1)/2)==kronecker(x+1,n)&&
                        Mod(Mod(1,n)*x*(L^2-4),L^2-x*L+1)^(n+1)==x^2*(25-4*x^2);}

                        {if(tst(312432294658994604401,2805168083964928859),
                        print(fooled));}

                        fooled

                        David


                        > It's on average about 5 selfridge for carefully chosen x,
                        >
                        > Paul
                        >
                      • paulunderwooduk
                        ... Thanks once more. Now a last ditch test: {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x^3-x,n)==1&& kronecker(x^2-1,n)==-1&&
                        Message 11 of 24 , Dec 10, 2012
                          --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
                          >
                          > --- In primenumbers@yahoogroups.com,
                          > "paulunderwooduk" <paulunderwood@> wrote:
                          >
                          > > Here is another composite test
                          > > It's on average about 5 selfridge
                          >
                          > Generically, that is 7 selfridges: 4 Euler tests
                          > and one Frobenius. However, it's reasonably easy to fool:
                          >
                          > {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x^3-x,n)==1&&
                          > kronecker(x^2-1,n)==-1&&
                          > Mod(x-2,n)^((n-1)/2)==kronecker(x-2,n)&&
                          > Mod(x+2,n)^((n-1)/2)==kronecker(x+2,n)&&
                          > Mod(x-1,n)^((n-1)/2)==kronecker(x-1,n)&&
                          > Mod(x+1,n)^((n-1)/2)==kronecker(x+1,n)&&
                          > Mod(Mod(1,n)*x*(L^2-4),L^2-x*L+1)^(n+1)==x^2*(25-4*x^2);}
                          >
                          > {if(tst(312432294658994604401,2805168083964928859),
                          > print(fooled));}
                          >
                          > fooled
                          >

                          Thanks once more. Now a last ditch test:

                          {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x^3-x,n)==1&&
                          kronecker(x^2-1,n)==-1&&
                          Mod(x-2,n)^((n-1)/2)==kronecker(x-2,n)&&
                          Mod(x+2,n)^((n-1)/2)==kronecker(x+2,n)&&
                          Mod(x-1,n)^((n-1)/2)==kronecker(x-1,n)&&
                          Mod(x+1,n)^((n-1)/2)==kronecker(x+1,n)&&
                          Mod(x,n)^(n-1)==1&&
                          Mod(Mod(1,n)*(L^2-4),L^2-x*L+1)^(n+1)==25-4*x^2;}

                          Paul
                        • djbroadhurst
                          ... That s 5 + 3 = 8 selfrideges, but still easy to fool: {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x^3-x,n)==1&& kronecker(x^2-1,n)==-1&&
                          Message 12 of 24 , Dec 10, 2012
                            --- In primenumbers@yahoogroups.com,
                            "paulunderwooduk" <paulunderwood@...> wrote:

                            > Now a last ditch test

                            That's 5 + 3 = 8 selfrideges, but still easy to fool:

                            {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x^3-x,n)==1&&
                            kronecker(x^2-1,n)==-1&&
                            Mod(x-2,n)^((n-1)/2)==kronecker(x-2,n)&&
                            Mod(x+2,n)^((n-1)/2)==kronecker(x+2,n)&&
                            Mod(x-1,n)^((n-1)/2)==kronecker(x-1,n)&&
                            Mod(x+1,n)^((n-1)/2)==kronecker(x+1,n)&&
                            Mod(x,n)^(n-1)==1&&
                            Mod(Mod(1,n)*(L^2-4),L^2-x*L+1)^(n+1)==25-4*x^2;}

                            {if(tst(28928708996670209,9176237087918226),
                            print(fooled));}
                          • djbroadhurst
                            ... Moreover, we may allow 7 Euler tests before the Frobenius test and still obtain a counterexample: {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x^3-x,n)==1&&
                            Message 13 of 24 , Dec 11, 2012
                              --- In primenumbers@yahoogroups.com,
                              "djbroadhurst" <d.broadhurst@...> wrote:

                              > > Now a last ditch test
                              > That's 5 + 3 = 8 selfridges, but still easy to fool

                              Moreover, we may allow 7 Euler tests before the
                              Frobenius test and still obtain a counterexample:

                              {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x^3-x,n)==1&&
                              kronecker(x^2-1,n)==-1&&kronecker(25-4*x^2,n)==-1&&
                              Mod(x-2,n)^((n-1)/2)==kronecker(x-2,n)&&
                              Mod(x+2,n)^((n-1)/2)==kronecker(x+2,n)&&
                              Mod(x-1,n)^((n-1)/2)==kronecker(x-1,n)&&
                              Mod(x+1,n)^((n-1)/2)==kronecker(x+1,n)&&
                              Mod(5-2*x,n)^((n-1)/2)==kronecker(5-2*x,n)&&
                              Mod(5+2*x,n)^((n-1)/2)==kronecker(5+2*x,n)&&
                              Mod(x,n)^((n-1)/2)==kronecker(x,n)&&
                              Mod(Mod(1,n)*(L^2-4),L^2-x*L+1)^(n+1)==25-4*x^2;}

                              {if(tst(500813768599682335601,104655233442347018047),
                              print(fooled));}

                              fooled

                              David
                            • paulunderwooduk
                              I have another new composite test for odd n: {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x,n)==1&& Mod(x-2,n)^((n-1)/2)==kronecker(x-2,n)&&
                              Message 14 of 24 , Dec 11, 2012
                                I have another new composite test for odd n:

                                {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x,n)==1&&
                                Mod(x-2,n)^((n-1)/2)==kronecker(x-2,n)&&
                                Mod(x+2,n)^((n-1)/2)==kronecker(x+2,n)&&
                                Mod(Mod(1,n)*(x*L-4),L^2-x*L+1)^(n+1)==16-3*x^2;}

                                Since for x==1,3 or 6 it is 3 selfridge, a single test with carefully chosen small x is on average 3.125 selfridge,

                                Paul
                              • djbroadhurst
                                ... {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x,n)==1&& Mod(x-2,n)^((n-1)/2)==kronecker(x-2,n)&& Mod(x+2,n)^((n-1)/2)==kronecker(x+2,n)&&
                                Message 15 of 24 , Dec 11, 2012
                                  --- In primenumbers@yahoogroups.com,
                                  "paulunderwooduk" <paulunderwood@...> wrote:

                                  > I have another new composite test

                                  {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x,n)==1&&
                                  Mod(x-2,n)^((n-1)/2)==kronecker(x-2,n)&&
                                  Mod(x+2,n)^((n-1)/2)==kronecker(x+2,n)&&
                                  Mod(Mod(1,n)*(x*L-4),L^2-x*L+1)^(n+1)==16-3*x^2;}

                                  {quote=" Vain are the thousand creeds that move men's hearts,
                                  unutterably vain, worthless as wither'd weeds.";
                                  if(tst(10538495213,2237176843),print(quote));}

                                  Vain are the thousand creeds that move men's hearts,
                                  unutterably vain, worthless as wither'd weeds.

                                  David (per proxy Emily)
                                • paulunderwooduk
                                  Here yet another new composite test. This time there is a sneaky use of 2: {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x,n)==1&&
                                  Message 16 of 24 , Dec 11, 2012
                                    Here yet another new composite test. This time there is a sneaky use of 2:

                                    {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x,n)==1&&
                                    Mod(x-2,n)^((n-1)/2)==kronecker(x-2,n)&&
                                    Mod(x+2,n)^((n-1)/2)==kronecker(x+2,n)&&
                                    Mod(Mod(1,n)*2*(L^2-4),L^2-x*L+1)^(n+1)==4*(25-4*x^2);}

                                    Paul
                                  • djbroadhurst
                                    ... But still easy to fool: {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x,n)==1&& Mod(x-2,n)^((n-1)/2)==kronecker(x-2,n)&& Mod(x+2,n)^((n-1)/2)==kronecker(x+2,n)&&
                                    Message 17 of 24 , Dec 11, 2012
                                      --- In primenumbers@yahoogroups.com,
                                      "paulunderwooduk" <paulunderwood@...> wrote:

                                      > Here yet another new composite test.
                                      > This time there is a sneaky use of 2

                                      But still easy to fool:

                                      {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x,n)==1&&
                                      Mod(x-2,n)^((n-1)/2)==kronecker(x-2,n)&&
                                      Mod(x+2,n)^((n-1)/2)==kronecker(x+2,n)&&
                                      Mod(Mod(1,n)*2*(L^2-4),L^2-x*L+1)^(n+1)==4*(25-4*x^2);}

                                      {if(tst(37423804289,10332300710),print(fooled));}

                                      fooled

                                      David
                                    • paulunderwooduk
                                      ... Thanks David. However, I am going to be even more devious with this 3.5 selfridge test for small x: {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x,n)==1&&
                                      Message 18 of 24 , Dec 11, 2012
                                        --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:

                                        > fooled

                                        Thanks David. However, I am going to be even more devious with this 3.5 selfridge test for small x:

                                        {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x,n)==1&&
                                        Mod(4*(x-2),n)^((n-1)/2)==kronecker(x-2,n)&&
                                        Mod(4*(x+2),n)^((n-1)/2)==kronecker(x+2,n)&&
                                        Mod(Mod(1,n)*2*(L^2-4),L^2-x*L+1)^(n+1)==4*(25-4*x^2);}

                                        Paul
                                      • paulunderwooduk
                                        ... Maybe I can make this 3.125 selfridge for small x by removing the first multiplier 4 and using x=1,3 and 6 where possible Paul
                                        Message 19 of 24 , Dec 11, 2012
                                          --- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@...> wrote:
                                          >
                                          >
                                          >
                                          > --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@> wrote:
                                          >
                                          > > fooled
                                          >
                                          > Thanks David. However, I am going to be even more devious with this 3.5 selfridge test for small x:
                                          >
                                          > {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x,n)==1&&
                                          > Mod(4*(x-2),n)^((n-1)/2)==kronecker(x-2,n)&&
                                          > Mod(4*(x+2),n)^((n-1)/2)==kronecker(x+2,n)&&
                                          > Mod(Mod(1,n)*2*(L^2-4),L^2-x*L+1)^(n+1)==4*(25-4*x^2);}
                                          >

                                          Maybe I can make this 3.125 selfridge for small x by removing the first multiplier 4 and using x=1,3 and 6 where possible

                                          Paul
                                        • djbroadhurst
                                          ... Not really. Sticking in powers of 2 does not make forgery any harder. {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x,n)==1&&
                                          Message 20 of 24 , Dec 12, 2012
                                            --- In primenumbers@yahoogroups.com,
                                            "paulunderwooduk" <paulunderwood@...> wrote:

                                            > I am going to be even more devious

                                            Not really. Sticking in powers of 2 does not
                                            make forgery any harder.

                                            {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x,n)==1&&
                                            Mod(4*(x-2),n)^((n-1)/2)==kronecker(x-2,n)&&
                                            Mod(4*(x+2),n)^((n-1)/2)==kronecker(x+2,n)&&
                                            Mod(Mod(1,n)*2*(L^2-4),L^2-x*L+1)^(n+1)==4*(25-4*x^2);}

                                            {if(tst(115886944289,3692152318),print(fooled));}

                                            fooled

                                            David
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