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Factorization of 8 consecutive 850-digit integers

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  • djbroadhurst
    Application of a multigrade equality found by Tarry: factorization of 8 consecutive 850-digit integers Let x = 429*(2^50 + 9096237); y = (16*x^6 - 72*x^4 +
    Message 1 of 4 , Dec 22, 2009
      Application of a multigrade equality found by Tarry:
      factorization of 8 consecutive 850-digit integers

      Let x = 429*(2^50 + 9096237);
      y = (16*x^6 - 72*x^4 + 81*x^2 - 25)^2;
      N = 72*(y - 2^2)*(y - 2^8)*(y - 5^4)*(y - 21^2)/14! - 3;
      then
      p849 = N/3;
      p812 = (N + 1)/(2*9277*9862129*3520556383*110169881898372133);
      p850a = (N + 2);
      p850b = (N + 5)/2;
      p845 = (N + 6)/(3*19*1009);
      p844 = (N + 7)/(2^2*208577);
      are prime and the factorizations of (N + 3) and (N + 4) in
      http://physics.open.ac.uk/~dbroadhu/cert/ifacgast.zip
      contain no prime with more than 108 digits.

      Comments: Algebraic factorization of (N + 3)*(N + 4) comes from
      a sextic substitution in a multigrade equality discovered in
      1913 by Gaston Tarry, amateur extraordinaire:
      http://www-history.mcs.st-and.ac.uk/Biographies/Tarry.html
      OpenPFGW, GMP-ECM, Pari-GP and Primo achieved the rest.
      Remarkably, p849, p850a and p850b have BLS proofs, with
      helpers of less than 16 digits from the factorization of
      N + 3 = 3*(p849 + 1) = p850a + 1 = 2*(p850b - 1).
      Neither of the composites
      c837 = (N - 1)/(2^2*95819*27609121);
      c824 = (N + 8)/(5*311137*114685080442874011127);
      is likely to have a prime divisor with less than 25 digits.

      David Broadhurst, 22 December, 2009
    • Jens Kruse Andersen
      ... Congratulations! The record page is updated again. -- Jens Kruse Andersen
      Message 2 of 4 , Dec 22, 2009
        David wrote:
        > Application of a multigrade equality found by Tarry:
        > factorization of 8 consecutive 850-digit integers

        Congratulations!
        The record page is updated again.

        --
        Jens Kruse Andersen
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