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Factorization of 9 consecutive 552-digit numbers

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  • djbroadhurst
    Let x = (1320*(10^20 + 13065906))^3 y = (x*(5*x + 9)/2 - 31)^2 N = (y - 23^2)*(y - 24^2)/55440 - 7 then p544 = N/(2^2*85580203) p552a = (N + 1)/3 p546 = (N
    Message 1 of 178 , Nov 11, 2009
      Let

      x = (1320*(10^20 + 13065906))^3
      y = (x*(5*x + 9)/2 - 31)^2
      N = (y - 23^2)*(y - 24^2)/55440 - 7

      then

      p544 = N/(2^2*85580203)
      p552a = (N + 1)/3
      p546 = (N + 2)/(2*7^2*16127)
      p540 = (N + 5)/(223*9887*252181)
      p552b = (N + 6)/2
      p533 = (N + 8)/(2^2*9479*90863*15503173813)

      are prime. Moreover N + 3, N + 4 and N + 7 are completely
      factorized, with no prime factor having more than 128 digits.
      These 9 consecutive factorizations at 552 digits are in
      http://physics.open.ac.uk/~dbroadhu/cert/ifacgg1.zip

      Notes: I used OpenPFGW, GMP-ECM, Msieve, ggnfs, Pari-GP.
      All primality proofs were performed by Pari-GP.
      The primality proof for p552b is easy, since
      we know the factorization of p552b - 1 = (N + 4)/2.
      The composite numbers
      c542 = (N - 1)/(5*53*151*259499)
      c524 = (N + 9)/(5*7*31*37*288232349*593280102498871)
      are unlikely to have any prime divisor less than 10^30.
      This particular constellation was studied because
      p539 = (N + 10)/(2*3*60647*24846803)
      is prime. Thus a complete factorization of N + 9
      would have yielded 11 consecutive factorizations.

      David Broadhurst, Armistice Day, 2009
    • djbroadhurst
      ... Yes. Moreover the proof is rather easy. Let C be a Carmicahel number and let p be its largest prime factor. Then p-1|C-1. and hence p-1|(C/p)-1. Thus p
      Message 178 of 178 , Feb 25, 2010
        --- In primeform@yahoogroups.com,
        Devaraj Kandadai <dkandadai@...> asked:

        > Are all Carmichael numbers " quadratically smooth"?

        Yes. Moreover the proof is rather easy.

        Let C be a Carmicahel number and let p
        be its largest prime factor. Then p-1|C-1.
        and hence p-1|(C/p)-1. Thus p < C/p,
        and C is quadratically smooth.

        David
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