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Re: [PrimeNumbers] Re: Pythagorean Triplets

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  • 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 1 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 2 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 3 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 4 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 5 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 6 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 7 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 8 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 9 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 10 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 11 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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