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Re: stronger than BPSW?

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  • paulunderwooduk
    ... I can go one step further: The trace of the square root of a [-1,0;0,-1], where it exists, is 0 (mod n) for prime n. Jon Grantham, in his paper Frobenius
    Message 1 of 7 , May 2, 2011
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      --- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@...> wrote:
      >
      > Hi,
      >
      > Pari/GP implements Baillie-PSW with a strong 2-SPRP test and a strong -- where the jacobi symbol is -1 -- test on x^2-P*x-1 with minimal P.
      >
      > In my paper "quadratic composite tests", available in the files section of this group, I give another 3-selfridge test based on a strong a-SPRP test and a strong lucas test on x^2-a*x+1 with minimal a.
      >
      > In a way my test is stronger in that writing n+1 as 2^s*d, I can say
      >
      > V(a,1,<(n+1)/(2^r)-1,<(n+1)/(2^r)>) is either <a,2> for r=s; or is <-a,-2> for some r: 0<r<=s
      >
      > In matrix terms I am taking the successive square roots of
      >
      > [a,-1;1,0]^(n+1) == [1,0;0,1] (for a prime)
      >
      > Neat, eh?
      >

      I can go one step further: The trace of the square root of a [-1,0;0,-1], where it exists, is 0 (mod n) for prime n.

      Jon Grantham, in his paper Frobenius Pseudoprimes, calls this sort of test "Extra strong", which is a bit confusing. I am thinking of strength in terms of the jacobi symbol being -1 and the selfridge-lucas sense. Should I call it "Extra strong strong"?

      Paul
    • whygee@f-cpu.org
      On Mon, 02 May 2011 13:24:56 -0000, paulunderwooduk ... I imagine that it s the moment when you require a number-based measurement of strength, the kind with
      Message 2 of 7 , May 2, 2011
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        On Mon, 02 May 2011 13:24:56 -0000, "paulunderwooduk"
        <paulunderwood@...> wrote:
        > Jon Grantham, in his paper Frobenius Pseudoprimes, calls this sort of
        > test "Extra strong", which is a bit confusing. I am thinking of
        > strength in terms of the jacobi symbol being -1 and the
        > selfridge-lucas sense. Should I call it "Extra strong strong"?

        I imagine that it's the moment when you require a number-based
        measurement
        of strength, the kind with a stable base unit, like the Richter scale
        :-)

        > Paul
        YG
      • paulunderwooduk
        ... This extra step is trivial: The trace the square root, where it exists, of a -I is always zero regardless of primality. One more thing I have noticed for
        Message 3 of 7 , May 3, 2011
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          --- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@...> wrote:
          >
          >
          >
          > --- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@> wrote:
          > >
          > > Hi,
          > >
          > > Pari/GP implements Baillie-PSW with a strong 2-SPRP test and a strong -- where the jacobi symbol is -1 -- test on x^2-P*x-1 with minimal P.
          > >
          > > In my paper "quadratic composite tests", available in the files section of this group, I give another 3-selfridge test based on a strong a-SPRP test and a strong lucas test on x^2-a*x+1 with minimal a.
          > >
          > > In a way my test is stronger in that writing n+1 as 2^s*d, I can say
          > >
          > > V(a,1,<(n+1)/(2^r)-1,<(n+1)/(2^r)>) is either <a,2> for r=s; or is <-a,-2> for some r: 0<r<=s
          > >
          > > In matrix terms I am taking the successive square roots of
          > >
          > > [a,-1;1,0]^(n+1) == [1,0;0,1] (for a prime)
          > >
          > > Neat, eh?
          > >
          >
          > I can go one step further: The trace of the square root of a [-1,0;0,-1], where it exists, is 0 (mod n) for prime n.
          >

          This extra step is trivial: The trace the square root, where it exists, of a -I is always zero regardless of primality.

          One more thing I have noticed for primes +-2 (mod 5) and of the form 4*N+3 is:

          [3,-1;1,0]^(n+1)/2) == [-1,0;0,-1] (mod n)
          [7,-1;1,0]^(n+1)/4) == [-1,0;0,-1] (mod n)

          Paul
        • paulunderwooduk
          ... It is an interesting fact that: trace( [3,-1;1,0]^(2*N) ) = trace( [7,-1;1,0]^N ) If, for my 6 selfridge test for {a-2;a+2}, I first find:
          Message 4 of 7 , May 26, 2011
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            --- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@...> wrote:
            >
            >
            >
            > --- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@> wrote:
            > >
            > >
            > >
            > > --- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@> wrote:
            > > >
            > > > Hi,
            > > >
            > > > Pari/GP implements Baillie-PSW with a strong 2-SPRP test and a strong -- where the jacobi symbol is -1 -- test on x^2-P*x-1 with minimal P.
            > > >
            > > > In my paper "quadratic composite tests", available in the files section of this group, I give another 3-selfridge test based on a strong a-SPRP test and a strong lucas test on x^2-a*x+1 with minimal a.
            > > >
            > > > In a way my test is stronger in that writing n+1 as 2^s*d, I can say
            > > >
            > > > V(a,1,<(n+1)/(2^r)-1,<(n+1)/(2^r)>) is either <a,2> for r=s; or is <-a,-2> for some r: 0<r<=s
            > > >
            > > > In matrix terms I am taking the successive square roots of
            > > >
            > > > [a,-1;1,0]^(n+1) == [1,0;0,1] (for a prime)
            > > >
            > > > Neat, eh?
            > > >
            > >
            > > I can go one step further: The trace of the square root of a [-1,0;0,-1], where it exists, is 0 (mod n) for prime n.
            > >
            >
            > This extra step is trivial: The trace the square root, where it exists, of a -I is always zero regardless of primality.
            >
            > One more thing I have noticed for primes +-2 (mod 5) and of the form 4*N+3 is:
            >
            > [3,-1;1,0]^(n+1)/2) == [-1,0;0,-1] (mod n)
            > [7,-1;1,0]^(n+1)/4) == [-1,0;0,-1] (mod n)
            >

            It is an interesting fact that:

            trace( [3,-1;1,0]^(2*N) ) = trace( [7,-1;1,0]^N )

            If, for my 6 selfridge test for {a-2;a+2}, I first find:

            Mod([3,-1;1,0],n)^(n+1)==I

            I can immediately deduce Mod([7,-1;1,0],n)^((n+1)/2) is also I. This deduction can be made since the traces are the same and the determinants are all 1, and remembering that gcd(30,n)==1. If:

            Mod([7,-1;1,0],n)^((n+1)/2) = [r,-s;s,-r+2] (for the trace condition) then the determinant is -r^2+2*r+s^2==1. Rearranging terms:

            r-1 == +-s

            Mod([7,-1;1,0],n)^((n+1)/2+1) = [7*r-s,-7*s+r-2;r,-s]

            The "non-trace" diagonal always sums to zero:

            -7*s+2*(r-1)==0 (mod n)

            Substituting:

            -7*s+-2*s == 0 (mod n)

            but gcd(30,n)==1. So s is 0 and consequently r is 1.

            So a 4-selfridge (PRP) test follows for numbers +-2 (mod 5)

            (i) gcd(30,n)==1
            (ii) Mod(3,n)^n==3
            (iii) Mod(7,n)^n==7
            (iv) Mod([3,-1;1,0],n)^(n+1)==1 (*)

            As David has pointed out, (*) can be more quickly calculated as:

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

            Paul
          • paulunderwooduk
            ... Further, if kronecker(-3,n)==-1, this can be reduced to 3 selfridges by dropping the Mod(7,n)^n==7 test because the composite test is based a-2==-1 and
            Message 5 of 7 , May 31, 2011
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              --- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@...> wrote:
              >
              >
              >
              > --- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@> wrote:
              > >
              > >
              > >
              > > --- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@> wrote:
              > > >
              > > >
              > > >
              > > > --- In primenumbers@yahoogroups.com, "paulunderwooduk" <paulunderwood@> wrote:
              > > > >
              > > > > Hi,
              > > > >
              > > > > Pari/GP implements Baillie-PSW with a strong 2-SPRP test and a strong -- where the jacobi symbol is -1 -- test on x^2-P*x-1 with minimal P.
              > > > >
              > > > > In my paper "quadratic composite tests", available in the files section of this group, I give another 3-selfridge test based on a strong a-SPRP test and a strong lucas test on x^2-a*x+1 with minimal a.
              > > > >
              > > > > In a way my test is stronger in that writing n+1 as 2^s*d, I can say
              > > > >
              > > > > V(a,1,<(n+1)/(2^r)-1,<(n+1)/(2^r)>) is either <a,2> for r=s; or is <-a,-2> for some r: 0<r<=s
              > > > >
              > > > > In matrix terms I am taking the successive square roots of
              > > > >
              > > > > [a,-1;1,0]^(n+1) == [1,0;0,1] (for a prime)
              > > > >
              > > > > Neat, eh?
              > > > >
              > > >
              > > > I can go one step further: The trace of the square root of a [-1,0;0,-1], where it exists, is 0 (mod n) for prime n.
              > > >
              > >
              > > This extra step is trivial: The trace the square root, where it exists, of a -I is always zero regardless of primality.
              > >
              > > One more thing I have noticed for primes +-2 (mod 5) and of the form 4*N+3 is:
              > >
              > > [3,-1;1,0]^(n+1)/2) == [-1,0;0,-1] (mod n)
              > > [7,-1;1,0]^(n+1)/4) == [-1,0;0,-1] (mod n)
              > >
              >
              > It is an interesting fact that:
              >
              > trace( [3,-1;1,0]^(2*N) ) = trace( [7,-1;1,0]^N )
              >
              > If, for my 6 selfridge test for {a-2;a+2}, I first find:
              >
              > Mod([3,-1;1,0],n)^(n+1)==I
              >
              > I can immediately deduce Mod([7,-1;1,0],n)^((n+1)/2) is also I. This deduction can be made since the traces are the same and the determinants are all 1, and remembering that gcd(30,n)==1. If:
              >
              > Mod([7,-1;1,0],n)^((n+1)/2) = [r,-s;s,-r+2] (for the trace condition) then the determinant is -r^2+2*r+s^2==1. Rearranging terms:
              >
              > r-1 == +-s
              >
              > Mod([7,-1;1,0],n)^((n+1)/2+1) = [7*r-s,-7*s+r-2;r,-s]
              >
              > The "non-trace" diagonal always sums to zero:
              >
              > -7*s+2*(r-1)==0 (mod n)
              >
              > Substituting:
              >
              > -7*s+-2*s == 0 (mod n)
              >
              > but gcd(30,n)==1. So s is 0 and consequently r is 1.
              >
              > So a 4-selfridge (PRP) test follows for numbers +-2 (mod 5)
              >
              > (i) gcd(30,n)==1
              > (ii) Mod(3,n)^n==3
              > (iii) Mod(7,n)^n==7
              > (iv) Mod([3,-1;1,0],n)^(n+1)==1 (*)
              >
              > As David has pointed out, (*) can be more quickly calculated as:
              >
              > Mod(Mod(1,n)*l,l^2-3*l+1)^(n+1)==1
              >

              Further, if kronecker(-3,n)==-1, this can be reduced to 3 selfridges by dropping the Mod(7,n)^n==7 test because the composite test is based
              a-2==-1 and a+2==3.

              All in this thread will be added to the next release of my paper. I am still testing the ideas for small n, n<10^8,

              Paul
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