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Factorization of 8 consecutive 600-digit integers

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  • David Broadhurst
    Let x=(1320*(10^22+1932187))^3 y=(x*(5*x+9)/2-31)^2 N=(y-23^2)*(y-24^2)/55440-6 then N+k is factorized for k in [0,7]. Proof:
    Message 1 of 4 , Sep 2, 2007
      Let

      x=(1320*(10^22+1932187))^3
      y=(x*(5*x+9)/2-31)^2
      N=(y-23^2)*(y-24^2)/55440-6

      then N+k is factorized for k in [0,7].

      Proof:
      http://physics.open.ac.uk/~dbroadhu/cert/ifac8nfs.zip

      Software used:
      GGNFS (with SNFS difficulty = 151.4)
      GMP-ECM
      Msieve
      OpenPFGW
      Pari-GP
      Primo

      Neighbouring composites:
      c569=(N-1)/(2^2*25711801*125018686680668134578227)
      c597=(N+8)/(5*281)
      subjected to 400 ECM curves at B1=250000

      David Broadhurst
      2 September 2007
    • Jens Kruse Andersen
      ... Congratulations! http://hjem.get2net.dk/jka/math/consecutive_factorizations.htm is updated once again. ? x=u^3; ? y=(x*(5*x+9)/2-31)^2; ?
      Message 2 of 4 , Sep 2, 2007
        David Broadhurst wrote:
        > x=(1320*(10^22+1932187))^3
        > y=(x*(5*x+9)/2-31)^2
        > N=(y-23^2)*(y-24^2)/55440-6
        >
        > then N+k is factorized for k in [0,7].

        Congratulations!
        http://hjem.get2net.dk/jka/math/consecutive_factorizations.htm
        is updated once again.

        ? x=u^3;
        ? y=(x*(5*x+9)/2-31)^2;
        ? N=(y-23^2)*(y-24^2)/55440-6;
        ?
        ? factor(N+2)
        %108 =
        [u^3 + 4 1]
        [u^3 + 5 1]
        [u^3 + 6 1]
        [5*u^3 - 21 1]
        [5*u^3 - 16 1]
        [5*u^3 - 11 1]
        [5*u^6 + 9*u^3 + 2 1]

        ? factor(N+3)
        %109 =
        [u 3]
        [u^3 - 2 1]
        [5*u^3 + 9 1]
        [5*u^3 + 19 1]
        [5*u^6 + 9*u^3 - 124 1]
        [5*u^6 + 9*u^3 - 86 1]

        ? factor(N+6)
        %110 =
        [u - 1 1]
        [u^2 + u + 1 1]
        [5*u^3 + 14 1]
        [5*u^6 + 9*u^3 - 110 1]
        [5*u^6 + 9*u^3 - 108 1]
        [5*u^6 + 9*u^3 - 16 1]

        Nice.

        --
        Jens Kruse Andersen
      • David Broadhurst
        ... It is possible to do slightly better and reduce the number of irreducible sextics to 5/12, instead of the 6/12 that I used. The answer is (Euler s?) 41: ?
        Message 3 of 4 , Sep 4, 2007
          --- In primeform@yahoogroups.com, "Jens Kruse Andersen"
          <jens.k.a@...> wrote:

          > ? x=u^3;
          > ? y=(x*(5*x+9)/2-31)^2;
          > ? N=(y-23^2)*(y-24^2)/55440-6;
          ...
          > [5*u^6 + 9*u^3 - 16 1]
          >
          > Nice.

          It is possible to do slightly better and reduce
          the number of irreducible sextics to 5/12,
          instead of the 6/12 that I used.

          The answer is (Euler's?) 41:

          ? y=(x^3/41)^2
          ? N=(y-41)*(y-41^2)*(y-31^2)*(y-51^2)/141926400000;
          ? factor(N*(N+1)*(N+4))
          %3 =
          [x^2 - 41 1]

          [x^3 - 2091 1]

          [x^3 - 1681 1]

          [x^3 - 1271 1]

          [x^3 - 1189 1]

          [x^3 - 779 1]

          [x^3 - 369 1]

          [x^3 + 369 1]

          [x^3 + 779 1]

          [x^3 + 1189 1]

          [x^3 + 1271 1]

          [x^3 + 1681 1]

          [x^3 + 2091 1]

          [x^4 + 41*x^2 + 1681 1]

          [x^6 - 4305041 1]

          [x^6 - 3834361 1]

          [x^6 - 3767121 1]

          [x^6 - 3027481 1]

          [x^6 - 674081 1]

          but then the coefficients are larger in SNFS.

          David
        • David Broadhurst
          The Prouhet-Tarry-Escott chain for the records with k=8,9,10 is very simple. We simply use products of the 3 complex Gaussian primes with smallest odd norm: ?
          Message 4 of 4 , Sep 8, 2007
            The Prouhet-Tarry-Escott chain for the records with k=8,9,10
            is very simple. We simply use products of the 3 complex
            Gaussian primes with smallest odd norm:

            ? a=(2+I)*(3+2*I)*(4+I)
            %1 = 9 + 32*I
            ? b=(2+I)*(3+2*I)*(4-I)
            %2 = 23 + 24*I
            ? c=(2+I)*(3-2*I)*(4-I)
            %3 = 31 - 12*I
            ? 4*imag(c^2)^2==imag(b^2)^2+3*imag(a^2)^2
            %4 = 1

            Unfortunately, d=(2+I)*(3-2*I)*(4+I)=33+4*I is of no use, here.

            David (trying to relate everything to Hardy and Wright)
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