- Hi,

I have devised a couple of composite tests in pari-gp and now I am wondering how many selfridge units they are. Please someone here enlighten me:

1. for kronecker(5,n)==-1:

Mod(Mod(1,n)*(l^2-2),l^2-3*l+1)^(n+1)+9==0

2. for kronecker(5,n)==1 and n!=(r+1)*(2*r+1):

Mod(Mod(1,n)^(l^2-2),l^2-3*l+1)*((n+1)/2)+3*l-3==0

(gcd(n,30)==1)

Paul - --- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@...> wrote:
>

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,

>

>

> --- 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.

>

> Re: http://tech.groups.yahoo.com/group/primenumbers/message/24090?l=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

>

Paul -- restoring symmetry