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Re: [PrimeNumbers] Re: Pythagorean Triplets

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  • Andy Swallow
    ... But that is a question you will never be able to answer, one way or another, if all you re doing is search for primes of this type using computer methods.
    Message 1 of 19 , Feb 4, 2004
      On Wed, Feb 04, 2004 at 05:31:24AM -0000, eharsh82 wrote:
      > I am not sure if this series has finite number of primes or not.
      > I think it has infinite primes.

      But that is a question you will never be able to answer, one way or
      another, if all you're doing is search for primes of this type using
      computer methods. So wouldn't it be more interesting to study the
      abstract theory? Your original question was about Gaussian primes, or
      primes congruent to 1 mod 4. That's all interesting and fairly basic
      stuff. I would have thought that more informative answers would be found
      in there.

      Apologies if I'm talking rubbish. It just seems strange that on the one
      hand you're interested in whether certain sets contain infinitely many
      primes, yet on the other hand you're studying the sets using methods
      guaranteed to not be able to answer the question, :-)

      Anyway, that's my morning rant out of the way...

      Andy
    • eharsh82
      Here is my proof for infiniteness of these primes if b=a+1 then we get 2*a^2+2*a+1 =p solving this a is an integer if there is a prime p such that 2*p-1=m^2 or
      Message 2 of 19 , Feb 4, 2004
        Here is my proof for infiniteness of these primes

        if b=a+1
        then we get 2*a^2+2*a+1 =p
        solving this
        a is an integer if there is a prime p such that 2*p-1=m^2
        or m^2+1/2 is a prime

        The distribution of such primes would follow the distribution of
        primes with the formula n^2+1

        it is conjectured that such primes are infinite.

        ----
        Taken from primepages.com

        Are there infinitely many primes of the form n2+1?
        There are infinitely many of the forms n2+m2 and n2+m2+1. A more
        general form of this conjecture is if a, b, c are relatively prime, a
        is positive, a+b and c are not both even,and b2-4ac is not a perfect
        square, then there are infinitely many primes an2+bn+c [HW79, p19].

        ---


        What do you all think?

        Also the series I talked about, what do you think about's it
        distribution.

        a=2^n or 2^n-1
        b=a+1 = 2^n+1 or 2^n

        the series reduce to ((2^n-1)^2+1)/2 and ((2^n+1)^2+1)/2

        What about the distribution of primes in such a series?

        let me know!


        Harsh Aggarwal



        --- In primenumbers@yahoogroups.com, Andy Swallow <umistphd2003@y...>
        wrote:
        > On Wed, Feb 04, 2004 at 05:31:24AM -0000, eharsh82 wrote:
        > > I am not sure if this series has finite number of primes or not.
        > > I think it has infinite primes.
        >
        > But that is a question you will never be able to answer, one way or
        > another, if all you're doing is search for primes of this type using
        > computer methods. So wouldn't it be more interesting to study the
        > abstract theory? Your original question was about Gaussian primes,
        or
        > primes congruent to 1 mod 4. That's all interesting and fairly basic
        > stuff. I would have thought that more informative answers would be
        found
        > in there.
        >
        > Apologies if I'm talking rubbish. It just seems strange that on the
        one
        > hand you're interested in whether certain sets contain infinitely
        many
        > primes, yet on the other hand you're studying the sets using methods
        > guaranteed to not be able to answer the question, :-)
        >
        > Anyway, that's my morning rant out of the way...
        >
        > Andy
      • pop_stack
        Hi, I m a newby so please be gentle. OK There is no integer solution for X^2 + Y^2 = Z^2 when X and Y are prime. (In other words, no two primes, squared and
        Message 3 of 19 , Feb 9, 2004
          Hi, I'm a newby so please be gentle.
          OK

          There is no integer solution for
          X^2 + Y^2 = Z^2 when X and Y are prime.

          (In other words, no two primes, squared and sumed can equal a perfect
          square).

          This came out of my Pythagorean triplets program
          that seems to show that either X AND/OR Z are always prime
          and Y is never prime.

          I am not sophisticated enough to know whether the above is trivial.

          But, I'm excited to find a group for prime numbers.

          BTW, I have Visual Basic and or Excel demonstrations of
          the statements above. No proofs, of course.

          Thanks for any input.

          pop_stack

          leppart@...
        • Jud McCranie
          ... This holds if x and y are both odd (not just prime) because x^2+y^2 must be congruent to 2 mod 4 and z^2 must be congruent to 0 mod 4. It holds when x=2
          Message 4 of 19 , Feb 9, 2004
            At 11:30 PM 2/9/2004, pop_stack wrote:
            >Hi, I'm a newby so please be gentle.
            >OK
            >
            >There is no integer solution for
            >X^2 + Y^2 = Z^2 when X and Y are prime.

            This holds if x and y are both odd (not just prime) because x^2+y^2 must be
            congruent to 2 mod 4 and z^2 must be congruent to 0 mod 4. It holds when
            x=2 (more generally when x == 2 mod 4) and y is odd for similar reasons.
          • eharsh82
            I realized that the series (2^n+1)^2 + (2^n)^2 and (2^n-1)^2 + (2^n) ^2 are nothing but Aurifeuillian Factors. They have some special properties that I have
            Message 5 of 19 , Feb 25, 2004
              I realized that the series (2^n+1)^2 + (2^n)^2 and (2^n-1)^2 + (2^n)
              ^2 are nothing but Aurifeuillian Factors.

              They have some special properties that I have discovered.

              1) Basically these numbers are 2^p+2^((p+1)/2)+1 and 2^p-2^((p+1)/2)+1

              2^(2p)+1=(2^p+2^((p+1)/2)+1)*(2^p-2^((p+1)/2)+1)
              2^(2p)+1= M2p * L2p (Notation used in literature)

              2) p must be prime so that either L or M can be a base 2-PRP.

              3) Either L or M is divisible by 5. When p%8=1 or 5 M is divisible by
              5 and when p%8=3 or 7 L is divisible by 5. So only one candidate
              remains for each prime.

              4) So the factors of these numbers are of the form 4*p*k+1.

              5) I think their distribution is really similar to mersenne primes,
              as most of their properties. I have searched these numbers up to
              p=35000 and am continuing to search higher. I have found them to
              produce an equal number of primes as mersenne numbers. I think these
              primes are the Gaussian equivalents of mersenne primes.

              6) I think a top 20 list of these numbers can be started on the
              primepages.org web page since these numbers are well known and have
              been discussed in a lot of papers.

              7) I am not sure if DWT can be used productively, with this series.
              But if anyone knows how it can be used productively, please let me
              know.

              In order to speed up the search to higher n's, I am looking for a
              sieve/ Trial factorer. Could someone with the required skill please
              write me a program to sieve? I did try myself to write one but it is
              not very fast. I am currently using that and sieving all numbers up
              to 25G before moving to PRPing. (takes about 30 sec to take a
              candidate to 25 G)

              Let me know, if any one can help.

              -- Harsh Aggarwal


              Here are the primes I have found so far.

              2^1+2^((1+1)/2)+1
              2^1-2^((1+1)/2)+1
              - Complete Set -
              2^3+2^((3+1)/2)+1
              2^3-2^((3+1)/2)+1
              - Complete Set -
              2^5+2^((5+1)/2)+1
              2^7-2^((7+1)/2)+1
              2^11+2^((11+1)/2)+1
              2^19+2^((19+1)/2)+1
              2^29+2^((29+1)/2)+1
              2^47-2^((47+1)/2)+1
              2^73-2^((73+1)/2)+1
              2^79-2^((79+1)/2)+1
              2^113-2^((113+1)/2)+1
              2^151-2^((151+1)/2)+1
              2^157+2^((157+1)/2)+1
              2^163+2^((163+1)/2)+1
              2^167-2^((167+1)/2)+1
              2^239-2^((239+1)/2)+1
              2^241-2^((241+1)/2)+1
              2^283+2^((283+1)/2)+1
              2^353-2^((353+1)/2)+1
              2^367-2^((367+1)/2)+1
              2^379+2^((379+1)/2)+1
              2^457-2^((457+1)/2)+1
              2^997+2^((997+1)/2)+1
              2^1367-2^((1367+1)/2)+1
              2^3041-2^((3041+1)/2)+1
              2^10141+2^((10141+1)/2)+1
              2^14699+2^((14699+1)/2)+1
              2^27529-2^((27529+1)/2)+1

              ----------------------------------------------------------------------




              --- In primenumbers@yahoogroups.com, "eharsh82" <harsh@u...> wrote:
              > Here is my proof for infiniteness of these primes
              >
              > if b=a+1
              > then we get 2*a^2+2*a+1 =p
              > solving this
              > a is an integer if there is a prime p such that 2*p-1=m^2
              > or m^2+1/2 is a prime
              >
              > The distribution of such primes would follow the distribution of
              > primes with the formula n^2+1
              >
              > it is conjectured that such primes are infinite.
              >
              > ----
              > Taken from primepages.com
              >
              > Are there infinitely many primes of the form n2+1?
              > There are infinitely many of the forms n2+m2 and n2+m2+1. A more
              > general form of this conjecture is if a, b, c are relatively prime,
              a
              > is positive, a+b and c are not both even,and b2-4ac is not a
              perfect
              > square, then there are infinitely many primes an2+bn+c [HW79, p19].
              >
              > ---
              >
              >
              > What do you all think?
              >
              > Also the series I talked about, what do you think about's it
              > distribution.
              >
              > a=2^n or 2^n-1
              > b=a+1 = 2^n+1 or 2^n
              >
              > the series reduce to ((2^n-1)^2+1)/2 and ((2^n+1)^2+1)/2
              >
              > What about the distribution of primes in such a series?
              >
              > let me know!
              >
              >
              > Harsh Aggarwal
              >
              >
              >
              > --- In primenumbers@yahoogroups.com, Andy Swallow
              <umistphd2003@y...>
              > wrote:
              > > On Wed, Feb 04, 2004 at 05:31:24AM -0000, eharsh82 wrote:
              > > > I am not sure if this series has finite number of primes or not.
              > > > I think it has infinite primes.
              > >
              > > But that is a question you will never be able to answer, one way
              or
              > > another, if all you're doing is search for primes of this type
              using
              > > computer methods. So wouldn't it be more interesting to study the
              > > abstract theory? Your original question was about Gaussian
              primes,
              > or
              > > primes congruent to 1 mod 4. That's all interesting and fairly
              basic
              > > stuff. I would have thought that more informative answers would
              be
              > found
              > > in there.
              > >
              > > Apologies if I'm talking rubbish. It just seems strange that on
              the
              > one
              > > hand you're interested in whether certain sets contain infinitely
              > many
              > > primes, yet on the other hand you're studying the sets using
              methods
              > > guaranteed to not be able to answer the question, :-)
              > >
              > > Anyway, that's my morning rant out of the way...
              > >
              > > Andy
            • mikeoakes2@aol.com
              ... You are confirming known results. I introduced these Gaussian Mersennes in a post to the Mersenne mailing list in 2000:
              Message 6 of 19 , Feb 26, 2004
                In a message dated 26/02/04 01:58:11 GMT Standard Time, harsh@... writes:


                > I realized that the series (2^n+1)^2 + (2^n)^2 and (2^n-1)^2 + (2^n)
                > ^2 are nothing but Aurifeuillian Factors.
                >
                > They have some special properties that I have discovered.
                >
                > 1) Basically these numbers are 2^p+2^((p+1)/2)+1 and 2^p-2^((p+1)/2)+1
                >
                > 2^(2p)+1=(2^p+2^((p+1)/2)+1)*(2^p-2^((p+1)/2)+1)
                > 2^(2p)+1= M2p * L2p (Notation used in literature)
                >
                > 2) p must be prime so that either L or M can be a base 2-PRP.
                >
                > 3) Either L or M is divisible by 5. When p%8=1 or 5 M is divisible by
                > 5 and when p%8=3 or 7 L is divisible by 5. So only one candidate
                > remains for each prime.
                >
                > 4) So the factors of these numbers are of the form 4*p*k+1.
                >
                > 5) I think their distribution is really similar to mersenne primes,
                > as most of their properties. I have searched these numbers up to
                > p=35000 and am continuing to search higher. I have found them to
                > produce an equal number of primes as mersenne numbers. I think these
                > primes are the Gaussian equivalents of mersenne primes.
                >
                > 6) I think a top 20 list of these numbers can be started on the
                > primepages.org web page since these numbers are well known and have
                > been discussed in a lot of papers.
                >
                > 7) I am not sure if DWT can be used productively, with this series.
                > But if anyone knows how it can be used productively, please let me
                > know.
                >
                > In order to speed up the search to higher n's, I am looking for a
                > sieve/ Trial factorer. Could someone with the required skill please
                > write me a program to sieve? I did try myself to write one but it is
                > not very fast. I am currently using that and sieving all numbers up
                > to 25G before moving to PRPing. (takes about 30 sec to take a
                > candidate to 25 G)
                >
                > Let me know, if any one can help.
                >

                You are confirming known results.

                I introduced these "Gaussian Mersennes" in a post to the Mersenne mailing
                list in 2000:
                http://www.mail-archive.com/mersenne@.../msg05162.html

                Chris Caldwell already has a "Top-20" page about these numbers:
                http://primes.utm.edu/top20/page.php?id=41

                You might also like to look at my post "Gaussian analogues of the Cullen and
                Woodall primes" of Dec 2000:
                http://listserv.nodak.edu/scripts/wa.exe?A2=ind0012&L=nmbrthry&P=R529&D=0&H=0&
                O=T&T=1

                -Mike Oakes


                [Non-text portions of this message have been removed]
              • Mark Rodenkirch
                ... http://listserv.nodak.edu/scripts/wa.exe?A2=ind0012&L=nmbrthry&P=R529&D=0&H=0&O=T&T=1 I looked at your link and it is quite interesting. I have a couple
                Message 7 of 19 , Feb 26, 2004
                  >
                  http://listserv.nodak.edu/scripts/wa.exe?A2=ind0012&L=nmbrthry&P=R529&D=0&H=0&O=T&T=1

                  I looked at your link and it is quite interesting. I have a couple of
                  comments though. You mention that G0(n) = n*(1+i)^n + 1 and is
                  related to Cullens but later state that G0(n) = n*2^(n/2) + 1 and then
                  go on to show primes of that form. Am I missing something? n*(1+i)^n
                  + 1 =/= n*2^(n/2) + 1. The same could be said of G2(n) and Woodalls.

                  You also have Ne(n) as n^2*2^n + 1, which pre-dates the Hyper-Cullen
                  search of Steven Harvey. He has noted your finds as he searches up to
                  200000.

                  --Mark
                • mikeoakes2@aol.com
                  In a message dated 26/02/04 14:20:52 GMT Standard Time, mgrogue@wi.rr.com ... It is if n = 0 mod 8, which was (one of) the values I was talking about.
                  Message 8 of 19 , Feb 26, 2004
                    In a message dated 26/02/04 14:20:52 GMT Standard Time, mgrogue@...
                    writes:


                    > You mention that G0(n) = n*(1+i)^n + 1 and is
                    > related to Cullens but later state that G0(n) = n*2^(n/2) + 1 and then
                    > go on to show primes of that form. Am I missing something? n*(1+i)^n
                    > + 1 =/= n*2^(n/2) + 1.
                    >
                    It is if n = 0 mod 8, which was (one of) the values I was talking about.
                    Remember: (1+i)^2 = 2*i.

                    -Mike Oakes


                    [Non-text portions of this message have been removed]
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