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Pythagorean Triplets

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  • eharsh82
    How do I solve this equation. Find a and b for a given prime p. What properties must p have for a solution to this equation to exist. a^2 + b^2 = 0 (mod p) and
    Message 1 of 19 , Feb 2, 2004
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      How do I solve this equation. Find a and b for a given prime p.
      What properties must p have for a solution to this equation to exist.

      a^2 + b^2 = 0 (mod p)

      and what if b= a+1

      Let me know!

      Thanks,
      Harsh Aggarwal
    • Carl Devore
      ... Not much interesting about this. Since a^2 = -b^2 (mod p), we have that solutions exist iff -1 is a quadratic residue mod p. If it is, let i denote a
      Message 2 of 19 , Feb 2, 2004
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        On Tue, 3 Feb 2004, eharsh82 wrote:
        > How do I solve this equation. Find a and b for a given prime p.
        > What properties must p have for a solution to this equation to exist.
        > a^2 + b^2 = 0 (mod p)

        Not much interesting about this. Since a^2 = -b^2 (mod p), we have that
        solutions exist iff -1 is a quadratic residue mod p. If it is, let i
        denote a square root of -1 (which can be found by the algorithm we
        discussed here a few weeks ago). Then a = +/- i*b for any b are
        solutions.

        > and what if b= a+1

        Then it becomes the quadratic equation 2*a^2+2*a+1 = 0, which can be
        solved by the algorithm discussed a few weeks ago.
      • mikeoakes2@aol.com
        ... This problem is not very well posed. Solutions to that equation exist for /any/ prime p, as all it says is that p divides the lhs! I think you want the
        Message 3 of 19 , Feb 3, 2004
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          In a message dated 03/02/04 03:59:03 GMT Standard Time, harsh@... writes:


          > How do I solve this equation. Find a and b for a given prime p.
          > What properties must p have for a solution to this equation to exist.
          >
          > a^2 + b^2 = 0 (mod p)
          >

          This problem is not very well posed.
          Solutions to that equation exist for /any/ prime p, as all it says is that p
          divides the lhs!

          I think you want the restrictions that a, b are in [0,p-1], no?

          In that case, either
          (a) p = 2, or
          (b) p = 1 mod 4.

          Case (a) gives as the unique solution a=b=1.

          Case (b) has many solutions.
          Choose a = any integer in [1,p-1];
          then b = a*e mod p,
          where e is one of the 2 solutions to the congruence
          e^2 = -1 mod p.

          I think there were recent posts to this group about how to take square roots
          mod p.

          -Mike Oakes


          [Non-text portions of this message have been removed]
        • eharsh82
          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? How about the
          Message 4 of 19 , Feb 3, 2004
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            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?
            How about the distribution of a prime p where a^2+b^2=p^2 ?
            What if b=a+1, then do infinite pytahgorean primes exist?

            I think something similar to how Euler prooved there are infinite
            primes may work here but I am not sure.

            Let me know!

            Thanks,
            Harsh Aggarwal




            --- In primenumbers@yahoogroups.com, "eharsh82" <harsh@u...> wrote:
            > How do I solve this equation. Find a and b for a given prime p.
            > What properties must p have for a solution to this equation to
            exist.
            >
            > a^2 + b^2 = 0 (mod p)
            >
            > and what if b= a+1
            >
            > Let me know!
            >
            > Thanks,
            > Harsh Aggarwal
          • Andrew Swallow
            ... Or you could just call them Gaussian primes, which is more or less what they are. You should be able to find necessary information in any introductory
            Message 5 of 19 , Feb 3, 2004
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              --- 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?
              > How about the distribution of a prime p where a^2+b^2=p^2 ?
              > What if b=a+1, then do infinite pytahgorean primes exist?
              >
              > I think something similar to how Euler prooved there are infinite
              > primes may work here but I am not sure.

              Or you could just call them Gaussian primes, which is more or less
              what they are. You should be able to find necessary information in any
              introductory number theory book. Or on whatever websites there are,
              probably.

              As for b=a+1, well that changes it to just a single dimensional
              problem, and is probably more difficult to answer.
            • eharsh82
              If both a and b are nonzero then z=a+bi, is a Gaussian prime iff a^2+b^2 is an ordinary prime. So there is actually no name for primes of the form a^2+b^2
              Message 6 of 19 , Feb 3, 2004
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                If both a and b are nonzero then z=a+bi, is a Gaussian prime iff
                a^2+b^2 is an ordinary prime.

                So there is actually no name for primes of the form a^2+b^2

                Harsh Aggarwal


                --- In primenumbers@yahoogroups.com, "Andrew Swallow"
                <umistphd2003@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?
                > > How about the distribution of a prime p where a^2+b^2=p^2 ?
                > > What if b=a+1, then do infinite pytahgorean primes exist?
                > >
                > > I think something similar to how Euler prooved there are infinite
                > > primes may work here but I am not sure.
                >
                > Or you could just call them Gaussian primes, which is more or less
                > what they are. You should be able to find necessary information in
                any
                > introductory number theory book. Or on whatever websites there are,
                > probably.
                >
                > As for b=a+1, well that changes it to just a single dimensional
                > problem, and is probably more difficult to answer.
              • elevensmooth
                ... 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
                Message 7 of 19 , Feb 3, 2004
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                  --- 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
                • 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 8 of 19 , Feb 3, 2004
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                    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 9 of 19 , Feb 3, 2004
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                      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 10 of 19 , Feb 3, 2004
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                        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 11 of 19 , Feb 3, 2004
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                          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 12 of 19 , Feb 4, 2004
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                            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 13 of 19 , Feb 4, 2004
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                              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 14 of 19 , Feb 9, 2004
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                                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 15 of 19 , Feb 9, 2004
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                                  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 16 of 19 , Feb 25, 2004
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                                    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 17 of 19 , Feb 26, 2004
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                                      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


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                                    • 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 18 of 19 , Feb 26, 2004
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                                        >
                                        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 19 of 19 , Feb 26, 2004
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                                          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


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