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double parameter double frobenius

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  • paulunderwooduk
    Hi, perhaps the gremlins will like this puzzling composite test. For non-square n coprime to 30 find x and a: kronecker(x^2-4,n)==-1 gcd(x^3-x,n)==1
    Message 1 of 4 , Mar 14, 2013
      Hi,

      perhaps the gremlins will like this puzzling composite test. For non-square n coprime to 30 find x and a:
      kronecker(x^2-4,n)==-1
      gcd(x^3-x,n)==1
      gcd((x+a)^3-(x+a),n)==1

      and do the sub-test:
      (L+(x+a)^2)^n==-l^3+(x^2-2)*L+(x+a)^2 (mod n, (L^2-x*L+1)*(L^2+x*L+1))

      Or in pari-speak:
      {tst(n,x,a)=kronecker(x^2-4,n)==-1&&
      gcd(x^3-x,n)==1&&
      gcd((a+x)^3-(x+a),n)==1&&
      Mod(Mod(1,n)*(L+(x+a)^2),(L^2-x*L+1)*(L^2+x*L+1))^n==-L^3+(x^2-2)*L+(x+a)^2;}

      Paul -- in triple loop hell
    • paulunderwooduk
      ... Dispelled with n=17261;x=676;a=65 Paul
      Message 2 of 4 , Mar 14, 2013
        --- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@...> wrote:
        >
        > Hi,
        >
        > perhaps the gremlins will like this puzzling composite test. For non-square n coprime to 30 find x and a:
        > kronecker(x^2-4,n)==-1
        > gcd(x^3-x,n)==1
        > gcd((x+a)^3-(x+a),n)==1
        >
        > and do the sub-test:
        > (L+(x+a)^2)^n==-l^3+(x^2-2)*L+(x+a)^2 (mod n, (L^2-x*L+1)*(L^2+x*L+1))
        >
        > Or in pari-speak:
        > {tst(n,x,a)=kronecker(x^2-4,n)==-1&&
        > gcd(x^3-x,n)==1&&
        > gcd((a+x)^3-(x+a),n)==1&&
        > Mod(Mod(1,n)*(L+(x+a)^2),(L^2-x*L+1)*(L^2+x*L+1))^n==-L^3+(x^2-2)*L+(x+a)^2;}
        >

        Dispelled with n=17261;x=676;a=65

        Paul
      • paulunderwooduk
        ... I am now testing for a=0, having covered n
        Message 3 of 4 , Mar 17, 2013
          > --- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@> wrote:
          > >
          > > Hi,
          > >
          > > perhaps the gremlins will like this puzzling composite test. For non-square n coprime to 30 find x and a:
          > > kronecker(x^2-4,n)==-1
          > > gcd(x^3-x,n)==1
          > > gcd((x+a)^3-(x+a),n)==1
          > >
          > > and do the sub-test:
          > > (L+(x+a)^2)^n==-l^3+(x^2-2)*L+(x+a)^2 (mod n, (L^2-x*L+1)*(L^2+x*L+1))
          > >
          > > Or in pari-speak:
          > > {tst(n,x,a)=kronecker(x^2-4,n)==-1&&
          > > gcd(x^3-x,n)==1&&
          > > gcd((a+x)^3-(x+a),n)==1&&
          > > Mod(Mod(1,n)*(L+(x+a)^2),(L^2-x*L+1)*(L^2+x*L+1))^n==-L^3+(x^2-2)*L+(x+a)^2;}
          > >
          >
          > Dispelled with n=17261;x=676;a=65
          >

          I am now testing for a=0, having covered n<2*10^6. I have a question. I test separately:
          (L+x^2)^(n+1)==1+x^2+x^3 (mod n, L^2-x*L+1)
          (L-x^2)^(n+1)==1+x^2-x^3 (mod n, L^2+x*L+1)

          Is gcd(x,n)==1 implied by the combined test I posted before?

          Paul
        • paulunderwooduk
          ... I meant: (L-x^2)^(n+1)==1+x^2-x^3 (mod n, L^2-x*L+1)
          Message 4 of 4 , Mar 17, 2013
            --- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@...> wrote:
            >
            >
            >
            >
            > > --- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@> wrote:
            > > >
            > > > Hi,
            > > >
            > > > perhaps the gremlins will like this puzzling composite test. For non-square n coprime to 30 find x and a:
            > > > kronecker(x^2-4,n)==-1
            > > > gcd(x^3-x,n)==1
            > > > gcd((x+a)^3-(x+a),n)==1
            > > >
            > > > and do the sub-test:
            > > > (L+(x+a)^2)^n==-l^3+(x^2-2)*L+(x+a)^2 (mod n, (L^2-x*L+1)*(L^2+x*L+1))
            > > >
            > > > Or in pari-speak:
            > > > {tst(n,x,a)=kronecker(x^2-4,n)==-1&&
            > > > gcd(x^3-x,n)==1&&
            > > > gcd((a+x)^3-(x+a),n)==1&&
            > > > Mod(Mod(1,n)*(L+(x+a)^2),(L^2-x*L+1)*(L^2+x*L+1))^n==-L^3+(x^2-2)*L+(x+a)^2;}
            > > >
            > >
            > > Dispelled with n=17261;x=676;a=65
            > >
            >
            > I am now testing for a=0, having covered n<2*10^6. I have a question. I test separately:
            > (L+x^2)^(n+1)==1+x^2+x^3 (mod n, L^2-x*L+1)
            > (L-x^2)^(n+1)==1+x^2-x^3 (mod n, L^2+x*L+1)
            >

            I meant:
            (L-x^2)^(n+1)==1+x^2-x^3 (mod n, L^2-x*L+1)

            > Is gcd(x,n)==1 implied by the combined test I posted before?
            >
            > Paul
            >
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