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Re: number of selfridges?

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  • paulunderwooduk
    ... 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
    Message 1 of 46 , Dec 4, 2011
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      --- 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

      Paul
    • paulunderwooduk
      ... 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
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        --- 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.
        >
        > 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
        >

        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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