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Re: Yet another factoring puzzle

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  • djbroadhurst
    ... Indeed, it was I who added http://factordb.com/index.php?id=1100000000464478896 with factors p404 and p414. Jean-Louis gets full marks for exploiting
    Message 1 of 23 , Sep 16 5:08 PM
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      Jean-Louis Charton wrote:

      > For sure one can find this solution quickly using openpfgw.
      > But the smart guy now knowing that you (or others) have filled
      > in factordb with many numbers of this form just has to type...

      Indeed, it was I who added
      http://factordb.com/index.php?id=1100000000464478896
      with factors p404 and p414.

      Jean-Louis gets full marks for exploiting factordb,
      instead of doing what I intended, along the lines of

      $ more abc

      ABC2 25^$a-4*5^$a-1&25^$a+4*5^$a-1
      a: from 301 to 304

      $ pfgw -f -d -e20000000 abc

      25^301-4*5^301-1 has factors: 2^2*7229
      25^302-4*5^302-1 has factors: 2^2*3173791
      25^303-4*5^303-1 has factors: 2^2*1931*934579
      25^304-4*5^304-1 has factors: 2^2*11*29*1289*1759*9511*27851
      (25^304-4*5^304-1)/(2^2*11*29*1289*1759*9511*27851) is 3-PRP!
      (0.0057s+0.2797s)
      25^304+4*5^304-1 has factors: 2^2*1439*17390951
      (25^304+4*5^304-1)/(2^2*1439*17390951) is 3-PRP!
      (0.0058s+0.3055s)

      But now, J-L, are you able to explain my comment:

      > What is remarkable about this exercise is that it can be
      > solved so quickly. Heuristically, that was not to be expected.

      Can you quantify my suprise?

      AmitiƩs

      David
    • j_chrtn
      ... Well, I would say that for n = 600, both algebraic factors are more than 415 digits and trial factoring them with pfgw up to 20000000 with -d option you
      Message 2 of 23 , Sep 18 4:39 PM
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        --- In primenumbers@yahoogroups.com, <david.broadhurst@...> wrote:

        >But now, J-L, are you able to explain my comment:
        >
        >> What is remarkable about this exercise is that it can be
        >> solved so quickly. Heuristically, that was not to be expected.
        >
        >Can you quantify my suprise?
        >

        Well, I would say that for n >= 600, both algebraic factors are more than 415 digits and trial factoring them with pfgw up to 20000000 with -d option you may remove a 20000000-smooth composite factor of say 10 or 15 digits. This leave 2 factors of more than 400 digits. The "probability" for one such factor to be prime is roughly 1/l(10^400), that is about 0.001. So for both factors to be prime the probability is about 0.000001.

        Am I right ?

        Amicalement,

        J-L
      • djbroadhurst
        ... We seek complete factorization of both of the cofactors 25^k-4*5^k-1 and 25^k+4*5^k-1 for some k 300, where each has more than 400 decimal digits. We had
        Message 3 of 23 , Sep 19 12:26 PM
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          Jean-Louis Carton wrote:

          > So for both factors to be prime the probability is about 0.000001.

          We seek complete factorization of both of the cofactors
          25^k-4*5^k-1 and 25^k+4*5^k-1
          for some k > 300, where each has more than
          400 decimal digits. We had better avoid
          the case k = 0 mod 3, where each cofactor
          has an algebraic factorization, which is
          here a distinct disadvantage.

          Suppose that we sieve out primes to depth d
          and hope for what is left to yield a
          a pair of PRPs as here, with k = 304:

          (25^304-4*5^304-1)/(2^2*11*29*1289*1759*9511*27851) is 3-PRP!
          (25^304+4*5^304-1)/(2^2*1439*17390951) is 3-PRP!

          The probability for success for a single value of k
          coprime to 3 is of order

          (exp(Euler)*log(p)/log(25))^2/k^2

          Setting p = 2*10^7 and summing over /all/ k > 300,
          the probability that Exercise 6 has /no/ solution is

          exp(-2/3*(exp(Euler)*log(2*10^7)/log(25))^2/301) =~ 83%

          In fact it has a solution almost immediately, at k = 304.

          David
        • Phil Carmody
          ... But is that more or less remarkable than the expectation of any one of Phil Taylor s darts landing in the region around where it actually landed? You only
          Message 4 of 23 , Sep 25 3:08 PM
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            On Mon, 9/16/13, djbroadhurst wrote:
            --- "djbroadhurst" <d.broadhurst@...> wrote:
            > > Exercise 6: Find the complete factorization of F(n) for at
            > > least one even integer n > 600.
            > As far as I can tell, no-one (apart from the setter)
            > yet solved Exercise 6, which can be done in less
            > than 2 minutes, using OpenPFGW. What is remarkable
            > about this exercise is that it can be solved so
            > quickly. Heuristically, that was not to be expected.

            But is that more or less remarkable than the expectation of
            any one of Phil Taylor's darts landing in the region around
            where it actually landed? You only chose that target after
            the arrow had landed, I'm sure.

            How many mathematical diversions have you looked at
            via the medium of numerical computation? How many of
            them would you expect to be remarkably easier than
            expected to solve? Probably a non-zero answer. Don't
            be surprised that one particular example was one.

            Knowing what he's trying to say, even if he's not getting
            it across clearly,
            Phil
            --
            () ASCII ribbon campaign () Hopeless ribbon campaign
            /\ against HTML mail /\ against gratuitous bloodshed

            [stolen with permission from Daniel B. Cristofani]
          • djbroadhurst
            ... It happened thus: 1) I determined to factorize F(n)=((n^2-9)/4)^2-5 for n
            Message 5 of 23 , Sep 27 2:52 PM
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               ---In primenumbers@yahoogroups.com, <thefatphil@...> wrote:

               

              > You only chose that target after
              > the arrow had landed, I'm sure.

               

              It happened thus:

               

              1) I determined to factorize F(n)=((n^2-9)/4)^2-5 for
              n <= 300, completely. As later shown in "factordb", I succeeded.

               

              2) Meanwhile I ran OpenPFGW on n in [301,600], hoping for a
              quick outlier and found none.

               

              3) I estimated the probability of an easily discoverable
              completely factorization for n>600 and found it to be small.

               

              4) Recalling how I had once been caught out before by
              a "probably no more" heuristic, I set a lone process running on
              n in [601, 10000]  so as not to be caught out again by Jens.

               

              5) When I later looked  and pfgw.log, it had found a hit at
              n=608.

               

              So yes, Phil, you are quite correct that the puzzle was set
              after this finding. However the heuristic that I gave was
              made prior to my discovery, else I would not have said that
              I was surprised.

               

              The point that you are making (I think) is that I do such 
              expsriments often and only notice when the result is unexpected.
              I don't tell folk about all the boring times when a negative
              heuristic is borne out by a null result. That is the selection

              effect.

               

              David (guilty of not boring folk with what is routine)

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