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Re: double parameter double frobenius

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  • paulunderwooduk
    ... I meant: (L-x^2)^(n+1)==1+x^2-x^3 (mod n, L^2-x*L+1)
    Message 1 of 4 , Mar 17 8:35 PM
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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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