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Re: generalised 6-selfridge double fermat+lucas

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  • 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 1 of 5 , Sep 14, 2010
      > 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 2 of 5 , Sep 15, 2010
        --- 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 3 of 5 , Sep 15, 2010
          --- 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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