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Re: [PrimeNumbers] Digest Number 1605

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  • John W. Nicholson
    Hi all, I wanted to keep these messages together and I add another at the bottom. I will let you think about how they relate. I no doubt (2nd, no! 3rd, no!)
    Message 1 of 2 , Apr 30, 2005
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      Hi all,

      I wanted to keep these messages together and I add another at the bottom. I
      will let you think about how they relate. I no doubt (2nd, no! 3rd, no!) "n-th"
      a "Big congratulations on an impressive record" also. ;-)



      --- primenumbers@yahoogroups.com wrote:

      >
      > There are 3 messages in this issue.
      >
      > Topics in this digest:
      >
      > 1. Re: new simult. prime record with 2058 digits
      > From: "Jens Kruse Andersen" <jens.k.a@...>
      > 2. Re: new simult. prime record with 2058 digits
      > From: Gary Chaffey <garychaffey2@...>
      > 3. Re: x^y - y^x
      > From: "Mark Underwood" <mark.underwood@...>
      >
      >
      > ________________________________________________________________________
      > ________________________________________________________________________
      >
      > Message: 1
      > Date: Fri, 29 Apr 2005 16:56:56 +0200
      > From: "Jens Kruse Andersen" <jens.k.a@...>
      > Subject: Re: new simult. prime record with 2058 digits
      >
      > Norman Luhn wrote:
      >
      > > The name of the big 4-quadruplet is :
      > >
      > > 4104082046.4800#+5651 {+0,+2,+6,+8}
      >
      > Big congratulations on an impressive record.
      >
      > k=4 is the most varied simultaneous record in recent years.
      > Since the latest quadruplet, it has been CC 1st kind, CC 2nd kind, BiTwin,
      > CPAP,
      > CC 2nd kind again:
      > http://hjem.get2net.dk/jka/math/simultprime.htm#history
      >
      > --
      > Jens Kruse Andersen
      >
      >
      >
      > ________________________________________________________________________
      > ________________________________________________________________________
      >
      > Message: 2
      > Date: Fri, 29 Apr 2005 16:41:13 +0100 (BST)
      > From: Gary Chaffey <garychaffey2@...>
      > Subject: Re: new simult. prime record with 2058 digits
      >
      > I would like to re-itterate Jens congratulations...As a searcher of various
      > forms of simultaneous primes I know how hard it must have been to find this.
      > Gary
      >
      > Jens Kruse Andersen <jens.k.a@...> wrote:
      > Norman Luhn wrote:
      >
      > > The name of the big 4-quadruplet is :
      > >
      > > 4104082046.4800#+5651 {+0,+2,+6,+8}
      >
      > Big congratulations on an impressive record.
      >
      > k=4 is the most varied simultaneous record in recent years.
      > Since the latest quadruplet, it has been CC 1st kind, CC 2nd kind, BiTwin,
      > CPAP,
      > CC 2nd kind again:
      > http://hjem.get2net.dk/jka/math/simultprime.htm#history
      >
      > --
      > Jens Kruse Andersen
      >
      >
      >
      > Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
      > The Prime Pages : http://www.primepages.org/
      >
      >
      >
      >
      >
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      > ________________________________________________________________________
      > ________________________________________________________________________
      >
      > Message: 3
      > Date: Fri, 29 Apr 2005 20:12:12 -0000
      > From: "Mark Underwood" <mark.underwood@...>
      > Subject: Re: x^y - y^x
      >
      > --- In primenumbers@yahoogroups.com, D�cio Luiz Gazzoni Filho
      > <decio@d...> wrote:
      > > Of note, divisibility of x^y - y^x by p means that x^y == y^x mod
      > p. Surely
      > > this equation must have interesting properties?
      > >
      >
      > Sure. Of course, if x and y share a factor of p then x^y - y^x does.
      > That's why we make sure x and y are relatively prime.
      >
      > But did we know that x-1 and y-1 must also be relatively prime? It is
      > easy to show that if x-1 and y-1 share a factor of p, then x^y - y^x
      > does!
      >
      > Futhermore, the even number -1 must also be relatively prime to the
      > other number + 1. It is easy to show that if the even number - 1
      > shares a factor p with the other (odd) number + 1, then x^y - y^n has
      > the same factor p as well!
      >
      >
      > In a related fashion with Paul's equation of x^y + y^x, not only must
      > x and y be relatively prime, but x+1 and y+1 must be relatively
      > prime. If x+1 and y+1 share a common factor of p, then so does x^y +
      > y^x !
      >
      > Furthermore, the even number + 1 must be relatively prime to the
      > other number -1. If they share a common factor of p, then so does x^y
      > + y^x !
      >
      >
      > When we count up all the qualifying x,y pairs up to 500 for x^y - y^x
      > and for x^y + y^x (with Paul's equation, x and y's one less than a
      > prime are disqualified), we find that about 2.2 times more pairs
      > qualify for your expression than for Paul's. This is starting to
      > better approach the prime count difference for yours and Paul's
      > expressions.
      >
      >
      > Mark
      >


      Replace x with a^2+b^2, and y with c^2+d^2 then you have this

      http://mathworld.wolfram.com/SquareNumber.html

      Equation (19).


      See equations (20) and (21) for error with s={-3, -1, 1, 3} as the
      "r_0=3^2-s^3 bases". r_0 is the orgin error radius r away from O the true
      orgin.

      Use equation (7) for an simple proof of Wiles-FLT.
      Use equation (33) for proof of a Twin between each prime squared. Or in other
      words a prime between n^2 and (n+1)^2.

      One might note on the page the 5x^2+/-4 = y^2 iff too. These are the equations
      required for twins along with this congruence y +/- 5 == 0 mod (y+6)(y+8).
      which is the same as (y+1)^2 == 0 mod (y)(y+2). From a twin one can find "zeros
      of admissibility" for other primes.

      Another note is that a 3,4,5 and/or 5,12,13 triangles can be used as a logic
      sandwich. Remember how close a prime is from a multiple of these numbers. This
      is similar to a Dubber Chain of Pythagorean triangles.

      The twin Goldbach failure is between 2^6-1 = 7*3^2 (the Primitive Prime factor
      exception by Ribenboim's book) and 4900=70^2=2^2*5^*7^2 (Cannonball Problem)is
      a "
      This proves Goldbach's and Dickson's with the same induction that Erdos used
      with Choquet Theory (with different starting and ending points as fitting for
      each k-tuple). http://mathworld.wolfram.com/ChoquetTheory.html and
      http://mathforum.org/library/drmath/view/51505.html

      Thanks to Mark for the the "failure" which is required for Wiles-FLT success
      (otherwise there would not be "enough" composites).

      >>
      >> In addition, I've been looking at the number of prime twins in
      >> successively higher, non intersecting intervals from
      >>
      >> n to n^(1 + 2 / n^(1/e))
      >>
      >> and the number of prime triplets in intervals from
      >>
      >> n to n^(1 + 3 / n^(1/e))
      >>
      >> They have also appeared to only increase, but I have not tested very
      >> much.
      >
      >
      >Of course, I have just found a counterexample in the prime triplets.
      >A slight modification is in order. :o
      >
      >Mark


      John W. Nicholson

      Sorry if this does not make sense, let the questions start.

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    • Dick
      O.K., I ll bite ... ... if all of the prime powers in the factorizations of both x and y are such that none of the p^a==-1 mod 4, then the substitutions
      Message 2 of 2 , May 1, 2005
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        O.K., I'll bite

        --- In primenumbers@yahoogroups.com, "John W. Nicholson"

        <snip>

        > Replace x with a^2+b^2, and y with c^2+d^2 then you have this
        >
        > http://mathworld.wolfram.com/SquareNumber.html
        >
        > Equation (19).

        if all of the prime powers in the factorizations of both x and y are
        such that none of the p^a==-1 mod 4, then the substitutions you
        suggest can be made, otherwise they cannot (in integers anyway)

        So the first question (after giving up on the WTF!? question) is, how
        are the x & y that cannot be integrally expressed as the sum of two
        squares to be disposed.

        The second question is how do we leap to Wiles-FLT when our x & y have
        differing exponents with severe constraints (base equals exponent of
        the other and vice versa)?


        > John W. Nicholson
        >
        > Sorry if this does not make sense, let the questions start.

        If you sense it doesn't make sense (you are quite correct in my case
        by the way), you should be able to anticipate the most likely
        questions, thus you could start answering and clarifying those without
        need of further input.

        Are you saying Mark's findings ultimately represent proof of FLT,
        Goldbach, twin prime conjecture and primes between squares?

        Curiously yours,

        Dick Boland
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