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Factorization of 9 consecutive 641-digit numbers

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  • djbroadhurst
    Let x = (1320*(5*10^23 + 7574922))^3 y = (x*(5*x + 9)/2 - 31)^2 N = (y - 3^4)*(y - 2^10)/55440 then p628 = (N + 2)/(7621*10607*83911) p641 = (N + 3)/2 p637 =
    Message 1 of 178 , Nov 23, 2009
      Let

      x = (1320*(5*10^23 + 7574922))^3
      y = (x*(5*x + 9)/2 - 31)^2
      N = (y - 3^4)*(y - 2^10)/55440

      then

      p628 = (N + 2)/(7621*10607*83911)
      p641 = (N + 3)/2
      p637 = (N + 5)/(2^2*17*281)
      p631 = (N + 6)/(5*7*71*3027473)
      p498 = (N + 7)/(2*3*773*1699*10103853977418233687
      *41681046033582287059951
      *36621959455077044549010329
      *4211861143455373066167093525382847
      *86528790939635630219352122503699213)
      p615 = (N + 8)/(5960137*45290243150024180117)

      are prime. Moreover N, N + 1 and N + 4 are completely
      factorized, with no prime factor having more than 160 digits.
      These 9 consecutive factorizations at 641 digits
      have been mailed to Jens Kruse Andersen.

      Notes: I used OpenPFGW, GMP-ECM, Msieve, ggnfs, Pari-GP.
      All primality proofs were performed by Pari-GP.
      The primality proof for p641 is easy, since
      we know the factorization of p641 - 1 = (N + 1)/2.
      The composite numbers
      c610 = (N - 1)/(2*7*88309036301*7938470054221217579)
      c578 = (N + 9)/(2^3*1138534864709243017
      *2353150702179440759951
      *28486552286765416493593)
      are unlikely to have any prime divisor less than 10^30.

      David Broadhurst, 23 November, 2009

      PS: I expect this record to be short lived:
      Joe Crump has a 804-digit target awaiting SNFS.
    • 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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