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24743Re: single frobenius and double fermat

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  • paulunderwooduk
    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
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