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Re: number of selfridges?

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  • paulunderwooduk
    ... When calculating the left to right binary ladder, for each bit a square is performed and if the bit is 1 then a multiplication by the base is done. The
    Message 1 of 46 , Dec 5, 2011
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      --- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@...> wrote:

      > > 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)*(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
      >

      When calculating the left to right binary ladder, for each bit a square is performed and if the bit is "1" then a multiplication by the base is done. The results also need to be reduced modulo "n".

      For squaring of an intermediate value a*l+b and noting l^2=x*l-1:
      (a*l+b)^2==a^2*l^2+2*a*b*l+b^2
      ==a^2*(x*l-1)+2*a*b*l+b^2
      ==a*(a*x+2*b)*l-(a^2-b^2)
      ==a*(a*x-2*b)*l-(a-b)*(a+b)
      Since "x" is small there are:
      2 small multiplications
      3 small additions
      3 small modular reductions
      2 multiplications
      2 modular reductions

      If there is a bit set multiplication by the base is performed:
      (a*l+b)(x*l-3)==a*x*l^2-3*a*l+b*x*l-3*b
      ==a*x*(x*l-1)-(3*a-b*x)*l-3*b
      ==(a*x^2-3*a+b*x)*l-a*x-3*b
      The operations are:
      5 small multiplications
      3 small additions
      2 small modular reductions

      So for a random bit:
      4.5 small multiplications
      4.5 small additions
      3.5 small modular reductions
      2 multiplications
      2 modular reductions

      By choosing just a minimal "x>0", this (single) test (on "x") is faster than Baillie-PSW, which requires at least 3 full multiplications and modular reductions for each bit,

      I have just checked the algorithm to 10^8...

      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 10:19 AM
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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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