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Re: double parameter double frobenius

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  • paulunderwooduk
    ... Dispelled with n=17261;x=676;a=65 Paul
    Message 1 of 4 , Mar 14, 2013
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      --- 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 2 of 4 , Mar 17, 2013
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        > --- 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 3 of 4 , Mar 17, 2013
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          --- 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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