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Factorization of 11 consecutive 515-digit numbers

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  • djbroadhurst
    Let x = 3*10^28 + 45140566 y = (2*x^3 - 10*x - 3)^2 N = y*(2*y - 11)^2/9 - 10 then for k = 0 to 10 the factorization of N + k is in
    Message 1 of 178 , Nov 29, 2009
      Let
      x = 3*10^28 + 45140566
      y = (2*x^3 - 10*x - 3)^2
      N = y*(2*y - 11)^2/9 - 10
      then for k = 0 to 10 the factorization of N + k is in
      http://physics.open.ac.uk/~dbroadhu/cert/ifac515g.zip
      It is likely that neither of
      c450 = (N - 1)/(2*11*2377*1861437640330348699939070011
      *483058876581817566371212578148357)
      c455 = (N + 11)/(2*5*18289*5453317*4322671468897
      *105475130567783166422952459096726461)
      has a prime divisor less than 10^35.

      David Broadhurst, 29 November, 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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