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RE: [PrimeNumbers] Re: It is never a square

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  • cino hilliard
    Hi, I have played with consecutive primes p1,p2,p3,p4 and determined (see link) that any combinition of pi*pj + pk*pl is possibly square if 3 of them are 4k+3
    Message 1 of 5 , May 6 12:33 AM
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      Hi,
      I have played with consecutive primes p1,p2,p3,p4 and determined (see link) that any
      combinition of pi*pj + pk*pl is possibly square if

      3 of them are 4k+3 and 1 is 4k+1 or 3 are 4k+1 and 1 is 4k+3.

      Moreover for p1 > 2, this sum is even and a square sum is divisible by 4.


      http://groups.google.com/group/sumprimes/web/ps-qr-for-consecutive-primes-p-q-r-s

      Using this rule to narrow the search with the above program, I have determined that
      p1*p4+p2*p3 is not square for primes < 10^9.

      However for,

      p1*p2 + p3*p4 we find running the program:

      Squares in sum of products of consecutive primes.
      Start => 1
      Stop => 1000000
      Tue 05/05/2009
      10:51 AM
      n=18242: 203233*203249 + 203279*203293 = 82632101764
      Count = 1
      Sum = 0
      Time in sec = 16.707211

      Now
      (10:57:02) gp > factor(82632101764)
      %2 =
      [2 2]
      [143729 2]

      The 18242th prime, 203233, starts a working combination. This implies squares exist in
      the possible combinations. Eg. 203233 = 4m+1, 203249=4m+1, 203279=4m+3,
      203293=4m+1 consistent with the rule. This also implies there are possibly other squares
      for for larger prime combinations.

      Consecutive primes can be arbitrarily far apart. I do not think consecutive primes
      play a role in the truth of the assertion since if it did it would imply some underlying
      pattern in the prime numbers.

      This implies a combination exists such that p1*p4 + p2*p3 is possibly square since there is no
      known pattern in the primes.

      Consider consecutive primes p1,p2,p3,p4,p5,p6,p7

      and test p1*p2 + P5*p6

      Then output

      n=1257742: 19784717*19784731 + 19784759*19784771 = 782872229861316

      gp > factor(782872229861316)
      [2 2]
      [3 2]
      [7 2]
      [666187 2]

      is a hit
      19784717 =4m+1, 19784731,19784759,19784771 = 4m+3

      Similarly testing p1*p2 + p6*p7

      n=64155: 802751*802759 + 802793*802799 = 1288897007616
      Congruent 3,3,1,3 mod 4

      gp > factor(1288897007616)
      [2 12]
      [3 10]
      [73 2]

      and
      p1*p7 + p2*p6

      n=2321170: 38038661*38038723 + 38038669*38038709 = 2893883949908224
      Congruent 1,3,1,1 mod 4

      gp > factor(2893883949908224)
      [2 8]
      [7 2]
      [13 2]
      [36947 2]

      for p1,p2,p3,p4,p5,p6,p7,p8

      test p1*p2 + p7*p8

      Start => 1
      Stop => 1000000000
      Wed 05/06/2009
      02:25 AM
      n=77: 389*397 + 431*433 = 341056
      n=1493: 12503*12511 + 12547*12553 = 313927524
      n=169237: 2296009*2296027 + 2296081*2296097 = 10543723352100


      My guess is this process can be extended to a random selection of primes taken
      k at a time.



      Cino








      To: primenumbers@yahoogroups.com
      From: maximilian.hasler@...
      Date: Sun, 3 May 2009 21:57:09 +0000
      Subject: [PrimeNumbers] Re: It is never a square







      ...and I meant:

      "
      Proposition: If p,q,r,s are consecutive primes,
      then valuation( ps+qr, P )=1 for some prime P.

      Corollary: If p,q,r,s are consecutive primes,
      then ps+qr cannot be a square.
      "

      (sorry)
      Maximilian

      > --- In primenumbers@yahoogroups.com, Sebastian Martin Ruiz <s_m_ruiz@> wrote:
      > >
      > >
      > > Have a prime factor with exponent 1
      > > �
      > > p*s+q*r=pi^1*pj^aj*pq^aq.....
      > >
      > > --- El dom, 3/5/09, Paul Leyland <paul@> escribi�:
      > >
      > >
      > > De: Paul Leyland <paul@>
      > > Asunto: Re: [PrimeNumbers] It is never a square
      > > Para: "Sebastian Martin Ruiz" <s_m_ruiz@>
      > > Fecha: domingo, 3 mayo, 2009 8:21
      > >
      > >
      > > On Thu, 2009-04-30 at 12:07 +0000, Sebastian Martin Ruiz wrote:
      > > >
      > > >
      > > > Hello all:
      > > >�
      > > > It is true?
      > > >�
      > > > Let p<q<r<s� four consecutive prime numbers.
      > > >�
      > > > p*s+q*r has a square free prime factor.
      > > >�
      > > > Then p*s+q*r is never a square.
      > > >�
      > > > I have verified it for many values.
      > >
      > > Isn't this trivial?� Any number with a square-free prime factor is not a
      > > square.� That's what square-free means.
      > >
      > > I don't think I understand what you are trying to say.
      > >
      > >
      > > Paul
      > >
      > >










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