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generalised 6-selfridge double fermat+lucas

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  • paulunderwooduk
    Hi, I will use capital letters to represent 2 by 2 matrices and lower case for integers. Consider: R^2-r*R+1==0 I call this trivial if r=0 (mod d) or r=+-1
    Message 1 of 5 , Sep 14, 2010
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      Hi,

      I will use capital letters to represent 2 by 2 matrices and lower case for integers.

      Consider:

      R^2-r*R+1==0

      I call this "trivial" if r=0 (mod d) or r=+-1 (mod d) for some proper divisor "d" of a given "n", because the equation is cyclic.

      Now on to the double equations:

      M^2-x*M+1==0
      N^2-y*N+1==0

      I do not want x=+-y (mod d) because they will be identical for that divisor of "n".

      The composite test for "n" is:

      First find x and y:
      gcd(x^3-x,n)==1
      gcd(y^3-y,n)==1
      gcd(x^2-y^2,n)==1
      jacobi(x^2-4,n)==-1
      jacobi(Y^2-4,n)==-1

      Secondly, check
      x^(n-1) == 1 (mod n)
      y^(n-1) == 1 (mod n)
      M^(n+1) == I (mod n)
      N^(n+1) == I (mod n)

      I have checked n<2*10^4 with gcd(30,n)==1,

      Paul
    • paulunderwooduk
      ... Back to the drawing board: n=41159; x=3547; y=3225; ... Paul
      Message 2 of 5 , Sep 14, 2010
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        --- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@...> wrote:
        >
        > Hi,
        >
        > I will use capital letters to represent 2 by 2 matrices and lower case for integers.
        >
        > Consider:
        >
        > R^2-r*R+1==0
        >
        > I call this "trivial" if r=0 (mod d) or r=+-1 (mod d) for some proper divisor "d" of a given "n", because the equation is cyclic.
        >
        > Now on to the double equations:
        >
        > M^2-x*M+1==0
        > N^2-y*N+1==0
        >
        > I do not want x=+-y (mod d) because they will be identical for that divisor of "n".
        >
        > The composite test for "n" is:
        >
        > First find x and y:
        > gcd(x^3-x,n)==1
        > gcd(y^3-y,n)==1
        > gcd(x^2-y^2,n)==1
        > jacobi(x^2-4,n)==-1
        > jacobi(Y^2-4,n)==-1
        >
        > Secondly, check
        > x^(n-1) == 1 (mod n)
        > y^(n-1) == 1 (mod n)
        > M^(n+1) == I (mod n)
        > N^(n+1) == I (mod n)
        >
        > I have checked n<2*10^4 with gcd(30,n)==1,
        >

        Back to the drawing board:
        n=41159;
        x=3547;
        y=3225;

        :-(

        Paul
      • djbroadhurst
        ... n = q*r with (r-1)/(q^2-1)=1/12 and each kronecker getting its minus sign from the smaller prime factor q. I do recommend such semiprimes for your study
        Message 3 of 5 , Sep 14, 2010
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          > Back to the drawing board:
          > n=41159;
          > x=3547;
          > y=3225;

          n = q*r with (r-1)/(q^2-1)=1/12
          and each kronecker getting its minus sign from
          the smaller prime factor q.

          I do recommend such semiprimes for your study :-)

          David
        • paulunderwooduk
          ... I found one that was not a semiprime, but the ratio rule still applies: n=41159=79*521;ratio=12 n=45629=103*443;ratio=24 n=64079=139*461;ratio=42
          Message 4 of 5 , Sep 15, 2010
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            --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@...> wrote:
            >
            > > Back to the drawing board:
            > > n=41159;
            > > x=3547;
            > > y=3225;
            >
            > n = q*r with (r-1)/(q^2-1)=1/12
            > and each kronecker getting its minus sign from
            > the smaller prime factor q.
            >
            > I do recommend such semiprimes for your study :-)
            >

            I found one that was not a semiprime, but the ratio rule still applies:
            n=41159=79*521;ratio=12
            n=45629=103*443;ratio=24
            n=64079=139*461;ratio=42
            n=96049=139*691;ratio=28
            n=197209=199*991;ratio=40
            n=228241=181*(13*97);ratio=26
            n=287051=151*1901;ratio=12
            n=330929=149*2221;ratio=10
            n=638189=619*1031;ratio=372
            n=853469=239*3571;ratio=16
            n=875071=241*3631;ratio=16

            Note: the above "n" are all +-1 (mod 5)

            Paul
          • djbroadhurst
            ... Here the kronecker is (-1)^3 = -1 and we are into Arnault territory: {pu3(n,x)=gcd(x^3-x,n)==1&&kronecker(x^2-4,n)==-1&&
            Message 5 of 5 , Sep 15, 2010
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              --- In primenumbers@yahoogroups.com,
              "paulunderwooduk" <paulunderwood@...> wrote:

              > n=228241=181*(13*97)

              Here the kronecker is (-1)^3 = -1 and we are into Arnault territory:

              {pu3(n,x)=gcd(x^3-x,n)==1&&kronecker(x^2-4,n)==-1&&
              Mod(x,n)^(n-1)==1&&Mod((x+s)/Mod(2,n),s^2+4-x^2)^(n+1)==1;}

              {pu6(n,x,y)=gcd(x^2-y^2,n)==1&&pu3(n,x)&&pu3(n,y);}

              {n=13*97*181; x=218; y=824; if(pu6(n,x,y),print("refuted"));}

              refuted

              Note that there are no "matrices" above: a double mod is enough.
              Congratulations again, Paul, on testing your ideas to destruction.

              Best regards

              David
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