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Re: single frobenius and double fermat

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  • paulunderwooduk
    ... Maybe I can make this 3.125 selfridge for small x by removing the first multiplier 4 and using x=1,3 and 6 where possible Paul
    Message 1 of 24 , Dec 11, 2012
      --- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@...> wrote:
      >
      >
      >
      > --- In primenumbers@yahoogroups.com, "djbroadhurst" <d.broadhurst@> wrote:
      >
      > > fooled
      >
      > Thanks David. However, I am going to be even more devious with this 3.5 selfridge test for small x:
      >
      > {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x,n)==1&&
      > Mod(4*(x-2),n)^((n-1)/2)==kronecker(x-2,n)&&
      > Mod(4*(x+2),n)^((n-1)/2)==kronecker(x+2,n)&&
      > Mod(Mod(1,n)*2*(L^2-4),L^2-x*L+1)^(n+1)==4*(25-4*x^2);}
      >

      Maybe I can make this 3.125 selfridge for small x by removing the first multiplier 4 and using x=1,3 and 6 where possible

      Paul
    • djbroadhurst
      ... Not really. Sticking in powers of 2 does not make forgery any harder. {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x,n)==1&&
      Message 2 of 24 , Dec 12, 2012
        --- In primenumbers@yahoogroups.com,
        "paulunderwooduk" <paulunderwood@...> wrote:

        > I am going to be even more devious

        Not really. Sticking in powers of 2 does not
        make forgery any harder.

        {tst(n,x)=kronecker(x^2-4,n)==-1&&gcd(x,n)==1&&
        Mod(4*(x-2),n)^((n-1)/2)==kronecker(x-2,n)&&
        Mod(4*(x+2),n)^((n-1)/2)==kronecker(x+2,n)&&
        Mod(Mod(1,n)*2*(L^2-4),L^2-x*L+1)^(n+1)==4*(25-4*x^2);}

        {if(tst(115886944289,3692152318),print(fooled));}

        fooled

        David
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