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## Re: double parameter double frobenius

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• ... 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
• ... 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
• ... 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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