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22710Re: stronger than BPSW?

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  • paulunderwooduk
    May 3, 2011
      --- 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
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