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Re: [PrimeNumbers] Re: Conjecture

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  • James Merickel
    Very well.  Do that.  My reading has been that all my terms for exponents are correct and in order through the 44th terms.  It may only be the case for
    Message 1 of 30 , Sep 19, 2012
      Very well.  Do that.  My reading has been that all my terms for exponents are correct and in order through the 44th terms.  It may only be the case for (at least) the 41st term--you may wish to consider 42nd through 44th tentatively correct or likely wrong.  I'm in no position to check those (next several days).  Just add them in a note or hold off until they are checked.  The problem: It appears that Robert is saying that he is checking in sequence through a maximum exponent value.  The first 40 were found 'correctly', and it shouldn't matter for at least the 41st.
      --- On Wed, 9/19/12, Maximilian Hasler <maximilian.hasler@...> wrote:


      From: Maximilian Hasler <maximilian.hasler@...>
      Subject: Re: [PrimeNumbers] Re: Conjecture
      To: "James Merickel" <moralforce120@...>
      Date: Wednesday, September 19, 2012, 12:00 PM


      I can put your name as co-author in the three sequences already published since Monday
       (see FORMULA).


      Maximilian




      On Wed, Sep 19, 2012 at 2:52 PM, James Merickel <moralforce120@...> wrote:






      Max:
      I'm changing the titles to include your A-number.  It is a sequence I had been planning to do myself.  So, all three should link across to two others (to start with, but there's that skew-coincidence also that is going to be expanded upon as well, and probably other stuff).

      JimMe

      --- On Mon, 9/17/12, maximilian_hasler <maximilian.hasler@...> wrote:



      From: maximilian_hasler <maximilian.hasler@...>
      Subject: [PrimeNumbers] Re: Conjecture
      To: primenumbers@yahoogroups.com
      Date: Monday, September 17, 2012, 3:34 PM



       





      --- In primenumbers@yahoogroups.com, Robert Gerbicz <robert.gerbicz@...> wrote:
      >
      > Three more terms, and no more for e2,e3<40000 (if we write n=2^e2*3^e3)
      > 2^26887*3^13438
      > 2^35399*3^26448
      > 2^18350*3^38015

      At least this definitively rules out the Conjecture in its initial form.
      (To leave it open, change "40" to "a finite number".)
      Anyway, as would say a Russian colleague of mine, 40 is not a number(*).
      More surprisingly, it seems that 42 isn't the answer, either.

      (* Although I vaguely recall some people stating that 10^40 was a "fundamental" ratio occuring in several instances when looking at sizes and/or masses in the known universe... no link to dwarf Cepheid type stars though.)

      For the records, I tentatively submitted the draft http://oeis.org/draft/A216854 to store these results.
      (Not sure other editors of OEIS will not delete rather than "approve" & publish this...)

      Regards,
      Maximilian





      [Non-text portions of this message have been removed]
    • ronhallam@lineone.net
      The other day I was browsing through Riesel and opened it at Appendix 3, Legendre’s symbol, x^2 º a mod p; a thought came into my mind from where who
      Message 2 of 30 , Jul 6, 2014
        The other day I was browsing through Riesel and opened it at Appendix
        3, Legendre’s symbol,
        x^2 º a mod p; a thought came into my mind from where who knows!!
        It was just silly, that it was equivalent to N = pq.



        Conjecture

        An odd p < s (s the integer square root) will be a factor of N a
        composite odd number if and only if



        s^2 = - (N - s^2) mod p .





        I can prove the main part but not that p can be either a prime or a
        composite.

        Proof

        N = pq

        s = flr(sqr( N))

        Redefine p and q as follows:-



        p = s - t ( t is an integer - 0 < t < s )

        q = s + t + k ( k is an integer and will always be of the form of
        either (0 mod 4) or (2 mod 4) depending on N



        Substituting the redefined p and q , back into N = pq



        N = ( s - t)( s + t + k)

        N = s^2 + ks - kt - t^2 (the st values cancel out)

        Solve for t^2



        t^2 = k(s - t) - (N - s^2)

        Taking the modulus of both sides using (s - t) gives



        t^2 = 0 - ( N - s^2) (mod ( s - t))



        t^2 mod(s - t) = s^2 mod ( s- t) (any difference is a multiple of ( s
        - t))



        This gives



        s^2 = - ( N - s^2) mod p.

        QED



        Example

        I will use a number from Riesel (page 147 of my edition)

        N = 13199

        s = 114

        p = 67



        (114^2) mod 67 = 65

        (N - s^2) = 203 => 203 mod 67 = 2

        This is 67 - 2 = 65



        I have looked at some of the RSA numbers that have been solved and they
        also confirm the above.



        A number that shows that p can be composite is 1617

        N = 1617

        s = 40

        p = 33

        (40^2) mod 33 = 16

        N - s^2 = 1617 - 1600 = 17

        p - 17 = 33 - 17 = 16



        If the residues to a number N, are found using the N and then (N-s^2)
        they give 2 different sets; example using 12007001



        Residues for N =? (-1 2 3 5 7 11 13 23 29 37 43 71 73 89 97)



        Residues for (N - s^2) => (-1 2 5 23 31 43 53 59 61 67 71 89 97)



        Intersection of the 2 sets gives (-1 2 5 23 43 71 89 97)



        I am not sure that this will be of any real benefit, but who knows.





        Ron
      • djbroadhurst
        ... This is trivially equivalent to N = 0 mod p, whatever s might be. So the conjecture is that N = 0 mod p if and only if p divides N. This is true, but
        Message 3 of 30 , Jul 7, 2014
          ---In primenumbers@yahoogroups.com, <ronhallam@...> wrote :

          > s^2 = - (N - s^2) mod p .

          This is trivially equivalent to N = 0 mod p, whatever s might be.
          So the "conjecture" is that N = 0 mod p if and only if p
          divides N. This is true, but hardly noteworthy.

          David 






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