## Re: number of selfridges?

Expand Messages
• ... No, that does not work. It should be: Mod(Mod(1,n)*(x*l-3),l^2-x*l+1)^(n+1)+2*x^2-9==0 Mod(Mod(1,n)*(y*l-3),l^2-y*l+1)^(n+1)+2*y^2-9==0 Paul
Message 1 of 46 , Dec 4, 2011
--- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@...> wrote:
>
>
>
> --- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@> wrote:
> > > I have another test:
> > >
> > > 1d. for *any* distinct x and y:
> > > gcd(n,30)==1
> > > 0<x<=(n-1)/2
> > > 0<y<=(n-1)/2
> > > kronecker(x^2-4,n)==-1
> > > kronecker(y^2-4,n)==-1
> > > Mod(Mod(1,n)*(l^2-2),l^2-x*l+1)^(n+1)+2*x^2-9==0
> > > Mod(Mod(1,n)*(l^2-2),l^2-y*l+1)^(n+1)+2*y^2-9==0
> > > gcd(x^2-y^2,n)==1
> > >
> >
> > wriggle to 1d.:
> > gcd(x*y,n)==1
> >
>
> I have stream-lined the definition:
> gcd(n,30)==1
> gcd(x*y,n)==1
> gcd(x^2-y^2,n)==1
> kronecker(x^2-4,n)==-1
> kronecker(y^2-4,n)==-1
> Mod(Mod(1,n)*(l^2-2),l^2-x*l+1)^(n+1)+2*x^2-9==0
> Mod(Mod(1,n)*(l^2-2),l^2-y*l+1)^(n+1)+2*y^2-9==0
>
> Also fiddling with the Mod-Mod definition, I think, they can be expressed as:
>
> Mod(Mod(1,n)*(x*l-3),l^2-2)^(n+1)+2*x^2-9==0
> Mod(Mod(1,n)*(y*l-3),l^2-2)^(n+1)+2*y^2-9==0
>

No, that does not work. It should be:
Mod(Mod(1,n)*(x*l-3),l^2-x*l+1)^(n+1)+2*x^2-9==0
Mod(Mod(1,n)*(y*l-3),l^2-y*l+1)^(n+1)+2*y^2-9==0

Paul
• ... Combining fails with the composite counterexample n=256999 and x=32768, However, I have tested the 1+1+1+2 conjecture up to n
Message 46 of 46 , Apr 14, 2012
--- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@...> wrote:
>
>
>
> --- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@> wrote:
> >
> > Hi,
> >
> > I have added a Fermat test to make a 1+1+1+2 selfridge test:
> >
> > For N>5, with gcd(6,N)==1, find an integer x:
> > gcd(x^3-x,N)==1
> > kronecker(x^2-4,N)==-1
> >
> > and check:
> > (x+2)^((N-1)/2)==kronecker(x+2,N) (mod N) (Euler)
> > (x-2)^((N-1)/2)==kronecker(x-2,N) (mod N) (Euler)
> > x^(N-1)==1 (mod N) (Fermat)
> > L^(N+1) == 1 (mod N, L^2-x*L+1) (Lucas)
> >
>
> Note: I should say gcd(30,N)==1 because gcd(x^3-x,N)==1 and kronecker(x^2-4,n)==-1.
>
>
> Now consider combining the 2 Euler tests with the Lucas test:
>
> (L*D)^((n+1)/2)==D (mod N, L^2-x*L+1) (D=x^2-4.)
>
> with the restriction kronecker(x+2,N)==-1.
>
> These together with the Fermat test makes for a 1+2-selfridge test.
>
> Can you find a counterexample?
>
> So far the near-refutation from Pinch's carmichael list is:
> N,x,gcd(x^2-1)
> ------------------
> 1909001 884658 1909001
>
> Paul
>

Combining fails with the composite counterexample n=256999 and x=32768, However, I have tested the 1+1+1+2 conjecture up to n<10^7,

Paul -- restoring symmetry
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