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Re: generalised 6-selfridge double fermat+lucas

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  • 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 1 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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