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Re: Pythagorean Triplets

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  • eharsh82
    Elevensmooth, http://www.math.uchicago.edu/~kobotis/media/163/kim.pdf prooves there are infinite primes of the from a^2+b^2 So what about when b=a+1 Any
    Message 1 of 19 , Feb 3, 2004
      Elevensmooth,

      http://www.math.uchicago.edu/~kobotis/media/163/kim.pdf

      prooves there are infinite primes of the from a^2+b^2

      So what about when b=a+1

      Any thoughts on this

      Harsh Aggarwal

      --- In primenumbers@yahoogroups.com, "elevensmooth"
      <elevensmooth@y...> wrote:
      > --- In primenumbers@yahoogroups.com, "eharsh82" <harsh@u...> wrote:
      > > I would like to call primes of the form a^2+b^2=prime,
      pythagorean
      > > primes.
      > >
      > > Can anyone proove that there are infinite pythagorean primes?
      >
      > "Every prime of the form 4n+1 is the sum of two squares. Euler
      first
      > communicated the following elegant proof of this fact to Goldbach in
      > 1749, two years after his original proof which was rathar vague on
      > this point ..."
      >
      > Fermat's Last Theorem by Edwards, Springer Verlag, 1977.
      > --
      > ElevenSmooth: Distributed Factoring of 2^3326400-1
      > http://ElevenSmooth.com
    • Carl Devore
      ... Then it becomes a univariate quadratic polynomial with interger coefficients. No univariate polynomial of degree greater than has ever been proved to
      Message 2 of 19 , Feb 3, 2004
        On Wed, 4 Feb 2004, eharsh82 wrote:
        > So what about when b=a+1

        Then it becomes a univariate quadratic polynomial with interger
        coefficients. No univariate polynomial of degree greater than has ever
        been proved to produce an infinite number of primes when integers are
        substituted for the variable.
      • Carl Devore
        ... Should say ...degree greater than one has ever...
        Message 3 of 19 , Feb 3, 2004
          On Tue, 3 Feb 2004, Carl Devore wrote:
          > Then it becomes a univariate quadratic polynomial with interger
          > coefficients. No univariate polynomial of degree greater than has ever
          > been proved to produce an infinite number of primes when integers are
          > substituted for the variable.

          Should say "...degree greater than one has ever..."
        • eharsh82
          I am not sure if this series has finite number of primes or not. I think it has infinite primes. I have found several primes in the 10000 digit category. I
          Message 4 of 19 , Feb 3, 2004
            I am not sure if this series has finite number of primes or not.
            I think it has infinite primes.

            I have found several primes in the 10000 digit category.
            I used PFGW on (10^10000+$a)^2+ (10^10000+$a+1)^2

            I'm also working on the series:
            (2^$a)^2+(2^$a+1)^2
            and
            (2^$a)^2+(2^$a-1)^2

            It din't seem to have a problem with finding primes.

            What do you think? let me know!

            Harsh Aggarwal



            -- In primenumbers@yahoogroups.com, Carl Devore <devore@m...> wrote:
            > On Tue, 3 Feb 2004, Carl Devore wrote:
            > > Then it becomes a univariate quadratic polynomial with interger
            > > coefficients. No univariate polynomial of degree greater than
            has ever
            > > been proved to produce an infinite number of primes when integers
            are
            > > substituted for the variable.
            >
            > Should say "...degree greater than one has ever..."
          • 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 5 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 6 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 7 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 8 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 9 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 10 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 11 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 12 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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