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Re: mod quartic composite tests

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  • 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 1 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 2 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 3 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 4 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 5 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 6 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 7 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 8 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 9 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 10 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 11 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 12 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 13 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 14 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 15 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 16 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 17 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 18 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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