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Re: It is never a square

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  • Maximilian Hasler
    ...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,
    Message 1 of 5 , May 3, 2009
      ...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
      > >
      > >
    • 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 2 of 5 , May 6, 2009
        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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