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Indirect primality t. (contd)

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  • dkandadai
    This is only to illustrate the procedure. I have considered only a) the cyclotomic polynomial x^2+1 b) values of x ranging from 1 t0 100 and c) the relevant
    Message 1 of 1 , Aug 4, 2008
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      This is only to illustrate the procedure. I have
      considered only a) the cyclotomic

      polynomial x^2+1 b) values of x ranging from 1 t0 100 and c)
      the relevant failure functions



      Run the following programmes in pari {p(x) = 1 + 2*k}, {p
      (x) = 2 + 5*k},

      {p(x) = 3 + 10*k}, {p(x) = 4 + 17*k} etc. For (x = 1,
      100,print(p(x))).

      Eliminate all the values of x generated by the above
      programs. Now consider

      only the even values of x out of the remaining. The result:


      x x^2 + 1 x x^2 + 1
      Remark

      14 197 16 257
      All the numbers in

      20 401 24 577
      col. x^2 + 1 are prime



      26 677 36 1297

      40 1601 56 3137

      66 4357 74 5477

      84 7057 90 8101

      94 8837

      Conclusion: Any large prime suspect can be indirectly
      tested for primality

      provided you use sophisticated programming and
      test only whether

      the relevant value of x is generated by the
      program. If not x^2 + 1

      is prime. Logic: whenever x^2 + 1 is composite
      one of the factors is less

      than x and hence covered by atleast one of the
      failure functions.
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