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16532Re: [PrimeNumbers] Digest Number 1605

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  • John W. Nicholson
    Apr 30, 2005
      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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