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RE: [PrimeNumbers] Re: primes of the form x^3 - y^2

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  • cino hilliard
    Hi David, The elliptic functions in Pari are forboding at least to me. I contacted the Pari group and got a workaround for eliptic curves from Karim. He gave
    Message 1 of 22 , Jun 18, 2009
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      Hi David,

      The elliptic functions in Pari are forboding at least to me.

      I contacted the Pari group and got a workaround for eliptic curves from
      Karim. He gave the following to find solutions for x^3-y^2 = p.



      diffcubes(n,p)=local(x,y);setintersect(vector(n,x,x^3-p),vector(n,y,y^2))
      which we found out will work in the next version of Pari. :-)



      So Karim wrote:

      > For "older" versions than that, use
      >
      > setintersect(Set(vector(n,x,x^3-p)), Set(vector(n,y,y^2)))
      >
      > ( slower but not *much* slower... )


      Ok, I did it my way with this and asked a question at the end.
      The timing aint bad. Magma and Sage would probably be faster offline.

      We need a way of knowing upper bounds a priori though.


      > diffcubes(n,p) =
      > {
      > local(j,x,y,c);
      > a=eval(setintersect(Set(vector(n,x,x^3-p)), Set(vector(n,y,y^2))));
      > c=length(a);
      > a=vecsort(a);
      > for(j=1,c,
      > y=round(sqrt(a[j])); \\ this could be iffy
      > x=round((a[j]+p)^(1/3));
      > print(j": "x"^3 - "y"^2 = "p); \\ Too fancy? Change it.
      > );
      > c;
      > }
      >
      >
      >(14:51:50) gp > diffcubes(300000,431)
      >1: 8^3 - 9^2 = 431
      >2: 11^3 - 30^2 = 431
      >3: 20^3 - 87^2 = 431
      >4: 30^3 - 163^2 = 431
      >5: 36^3 - 215^2 = 431
      >6: 138^3 - 1621^2 = 431
      >7: 150^3 - 1837^2 = 431
      >8: 575^3 - 13788^2 = 431
      >9: 3903^3 - 243836^2 = 431
      >(15:02:35) gp > ##
      > *** last result computed in 2,250 ms.
      >(15:02:39) gp >
      >
      >This finds all instances because I new a prori 243836 was the big kahuna.
      >I guess it will still be trial and error?
      >
      >Oh well, we are at least 2 quanta over what I had.
      >
      > Thank you >>>
      >Cino


      Thanks for all your work,

      Cino



      To: primenumbers@yahoogroups.com
      From: d.broadhurst@...
      Date: Thu, 18 Jun 2009 16:26:36 +0000
      Subject: [PrimeNumbers] Re: primes of the form x^3 - y^2







      Off list, Tony Noe kindly provided this link:
      http://www.research.att.com/~njas/sequences/A081120

      From the table attached thereto, one may conclude that
      Cino's sequence, "Primes not of the form x^3 - y^2", begins

      3, 5, 17, 29, 31, 37, 41, 43, 59, 73, 97, 101, 103, 113, 131,
      137, 149, 157, 163, 173, 179, 181, 197, 211, 227, 229, 241,
      257, 263, 269, 281, 283, 311, 313, 317, 331, 337, 347, 349,
      353, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 443,
      449, 457, 461, 467, 479, 491, 509, 521, 523, 541, 563, 569,
      571, 577, 601, 607, 613, 617, 619, 641, 643, 653, 659, 661,
      677, 691, 701, 709, 733, 739, 743, 751, 757, 761, 773, 787,
      809, 811, 821, 823, 827, 829, 839, 853, 857, 863, 877, 881,
      883, 907, 911, 929, 937, 941, 947, 953, 967, 977, 983, 997 ...

      This paper appears to have open access:
      http://tinyurl.com/mtvsn5
      See Section 5.1 for Cino's record holder, p = 28279,
      with 21 integral points.

      David










      [Non-text portions of this message have been removed]
    • David Broadhurst
      ... How does it compare with the brute force of issquare for the largest integer point on x^3 - y^2 = 22189 I wonder? The result is already published, but I
      Message 2 of 22 , Jun 18, 2009
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        --- In primenumbers@yahoogroups.com,
        cino hilliard <hillcino368@...> wrote:

        > I contacted the Pari group and got a workaround
        > for eliptic curves from Karim.

        How does it compare with the brute force of
        "issquare" for the largest integer point on
        x^3 - y^2 = 22189
        I wonder?

        The result is already published, but I thought
        that it might be an interesting test case.
        Here is an "issquare" timing:

        c=0;gettime;
        {for(x=1,10^9,if(issquare(x^3-22189),
        print(x);c=c+1;if(c==2,break)))}
        print(ceil(gettime/10^3)" seconds")

        29585
        140292677
        98 seconds

        David
      • sta staf
        ... _________________________________________________________________ Découvrez Windows Live Spaces et créez votre site Web perso en quelques clics !
        Message 3 of 22 , Jun 19, 2009
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          ---
          >
          >
          > the collatz function to it as in f(n) = 3n+1/2 if n is odd
          >
          > and f(n) = n/2 n is even.
          >
          > Consider the number 33
          >
          >
          >
          > 33, 100, 50, 25, 76, 38, 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1
          >
          >
          >
          > The odd numbers are 33, 25 ;19,29 ,11 ,17 ,13 ,5 and so on.
          >
          >
          >
          > As u can see 6 numbre prime : 19 ;29; 11 ;17 ; 13; 5 .
          >
          >
          >
          > I believe that the collatz function can be used as a prime generating function though i am not very sure in this.
          >
          >
          >
          > i have found n = 111012973909 is the largest integer with all odd are prime
          >
          >
          >
          > 111012973909, 333038921728, 166519460864, 83259730432, 41629865216, 20814932608, 10407466304, 5203733152, 2601866576, 1300933288, 650466644, 325233322, 162616661, 487849984, 243924992, 121962496, 60981248, 30490624, 15245312, 7622656, 3811328, 1905664, 952832, 476416, 238208, 119104, 59552, 29776, 14888, 7444, 3722, 1861, 5584, 2792, 1396, 698, 349, 1048, 524, 262, 131, 394, 197, 592, 296, 148, 74, 37, 112, 56, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1
          >
          >
          >
          > all odd are prime : 111012973909 ; 162616661 ; 1861 ; 349 ; 131 ; 197 ; 37 ; 7 ;11 ; 17 ; 13 ; ; 5
          >
          >
          >
          > rachid
          >



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        • David Broadhurst
          ... The word largest makes no sense here. The 271-digit prime (5*2^897-1)/3 manifestly contains no composite odd integer in its Collatz sequence and it would
          Message 4 of 22 , Jun 19, 2009
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            --- In primenumbers@yahoogroups.com,
            sta staf <sta_staf@...> wrote:

            > i have found n = 111012973909 is the largest integer
            > with all odd are prime

            The word "largest" makes no sense here.
            The 271-digit prime (5*2^897-1)/3 manifestly contains no
            composite odd integer in its Collatz sequence and it would be
            easy to find larger primes by this, or a similar, construction.

            David
          • sta staf
            yes but also the 271-figit prime (5*2^897-1)/3 makes no sense here you think (5*2^897-1)/3) = prime numbre !!!!!!!!!!! give me n integer 111012973909
            Message 5 of 22 , Jun 19, 2009
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              yes but also the 271-figit prime (5*2^897-1)/3 makes no sense here



              you think (5*2^897-1)/3) = prime numbre !!!!!!!!!!!



              give me n integer 111012973909 < n < 2^50 with all odd integer are prime .







              becuse we canot test the sequnce :(5*2^897-1)/3





              thanks

              rachid










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            • David Broadhurst
              ... to which I calmly reply, that I do so think. Here is my proof, using APR-CL in Pari-GP: N=(5*2^897-1)/3; if(isprime(N),print(OK)) OK It appears that
              Message 6 of 22 , Jun 19, 2009
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                --- In primenumbers@yahoogroups.com,
                sta staf <sta_staf@...> wrote:

                > the 271-figit prime (5*2^897-1)/3 makes no sense here
                > you think (5*2^897-1)/3) = prime numbre !!!!!!!!!!!

                to which I calmly reply, that I do so think.
                Here is my proof, using APR-CL in Pari-GP:

                N=(5*2^897-1)/3;
                if(isprime(N),print(OK))
                OK

                It appears that coherence is in inverse proportion
                to the number of exclamation marks used by the poster.

                David (with no exclamation marks)
              • sta staf
                but your seqaunce :(5*2^897-1/3) this sequance of coltaze is : all Even numbers . the only odd is the ferst digit : :(5*2^897-1/3) rachid
                Message 7 of 22 , Jun 19, 2009
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                  but your seqaunce :(5*2^897-1/3)



                  this sequance of coltaze is : all Even numbers .



                  the only odd is the ferst digit : :(5*2^897-1/3)



                  rachid










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                • cino hilliard
                  Hi, ... the workaround, diffcubes2(n,p) = { local(x,y,c,a); a=eval(setintersect(Set(vector(n,y,y^2)),Set(vector(n,x,x^3-p)))); a; } is useless for large n.
                  Message 8 of 22 , Jun 19, 2009
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                    Hi,


                    >>--- In primenumbers@yahoogroups.com,
                    >>cino hilliard <hillcino368@...> wrote:
                    >> I contacted the Pari group and got a workaround
                    >> for eliptic curves from Karim.

                    the workaround,

                    diffcubes2(n,p) =
                    {
                    local(x,y,c,a);
                    a=eval(setintersect(Set(vector(n,y,y^2)),Set(vector(n,x,x^3-p))));
                    a;
                    }
                    is useless for large n. Maybe someone can tweek.


                    (10:25:03) gp > diffcubes2(10000000,22189)
                    *** Set: the PARI stack overflows !
                    current stack size: 800000000 (762.939 Mbytes)
                    [hint] you can increase GP stack with allocatemem()




                    >How does it compare with the brute force of
                    >"issquare" for the largest integer point on
                    >x^3 - y^2 = 22189
                    >I wonder?


                    >c=0;gettime;
                    >{for(x=1,10^9,if(issquare(x^3-22189),
                    >print(x);c=c+1;if(c==2,break)))}
                    >print(ceil(gettime/10^3)" seconds")

                    >29585
                    >140292677
                    >98 seconds

                    elipissq(n) =
                    {
                    c=0;
                    for(x=1,10^9,if(issquare(x^3-n),
                    print(x);c=c+1;if(c==2,break)));
                    }

                    Pari 2.3.4
                    dell 8200 2.53 ghz xp pro

                    42 minutes. say what?

                    Pari 2.4.2
                    dell 8200 2.53 ghz xp pro

                    225 sec Go figure. They must have done some stuff on issquare in v2.4.2.


                    Pari 2.4.2

                    dell 6x620 3.2 ghz vista ult 64 bit op sys
                    112 sec close to David's bench.



                    Sage online calculator

                    http://www.sagenb.com/home/hillcino368/0/



                    sage: time EllipticCurve([0,0,0,0,-22189]).integral_points()


                    [(29585 : 5088706 : 1), (140292677 : 1661699554612 : 1)]
                    Time: CPU 0.22 s, Wall: 0.87 s
                    A tough and user friendly program.

                    I wonder why Pari has not implemented the Mordell algorithm. Maybe 2.4.3?
                    Perhaps someone here can write a script or show a c program with gmp that
                    does this. A gcc/gmp would be great.




                    Dénouement

                    A child playing in a room with toys could solve the riddle of the difference between
                    two squares by arranging toy blocks and counting. Later he could puzzle with numbers
                    by squaring, subtracting, noticing patterns and conjecturing. For any positive numbers
                    a,b, a^2 - b^2 = (a-b)*(a+b). Later, he finds out how to symbolically multiply
                    (a-b)*(a+b) to get a^2-b^2 proving his prior conjectures correct.


                    So a^2 - b^2 =(a-b)*(a+b) could be the third most fundamental process in arithmetic
                    the average being second and counting being first. It is that easy of a riddle.



                    Then later, the child goes into another room with more blocks. He starts making pyramids,
                    cubes, and squares. Suddenly, with a flash of insight, he about cube - square = number
                    or a^3 - b^2 = n? It is not that easy of a riddle.



                    I doubt my Mathematica I bought in 80's and Maple in the 90's does this as Sage.

                    Maybe the latest versions do.



                    [Rant]

                    It is disheartning that students can get these for $125 and seniors only get 1/2 off
                    of retail ($2000) for Mathematica.

                    [End Rant]



                    Cheers and Roebuck,

                    Cino





                    [Non-text portions of this message have been removed]
                  • David Broadhurst
                    ... As indeed I had expected :-) Sometimes brute force is best, as in my prior hack: c=0;gettime; {for(x=1,10^9,if(issquare(x^3-22189),
                    Message 9 of 22 , Jun 19, 2009
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                      --- In primenumbers@yahoogroups.com,
                      cino hilliard <hillcino368@...> wrote:

                      > the workaround,
                      > diffcubes2(n,p) =
                      > {
                      > local(x,y,c,a);
                      > a=eval(setintersect(Set(vector(n,y,y^2)),Set(vector(n,x,x^3-p))));
                      > a;
                      > }
                      > is useless for large n.

                      As indeed I had expected :-)

                      Sometimes brute force is best, as in my prior hack:

                      c=0;gettime;
                      {for(x=1,10^9,if(issquare(x^3-22189),
                      print(x);c=c+1;if(c==2,break)))}
                      print(ceil(gettime/10^3)" seconds")

                      29585
                      140292677
                      98 seconds

                      Thanks, Cino, for your fascinating thread, which
                      probes deeply into the theory of elliptic curves.

                      Best regards

                      David
                    • Peter Kosinar
                      Hello Rachid, ... It d still satisfy the property you re expecting it to satisfy: All odd terms of the sequence are primes. However, unless I m very much
                      Message 10 of 22 , Jun 19, 2009
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                        Hello Rachid,

                        Even if this was true:
                        > this sequance of coltaze is : all Even numbers .
                        > the only odd is the ferst digit : :(5*2^897-1/3)

                        It'd still satisfy the property you're expecting it to satisfy: "All odd
                        terms of the sequence are primes."

                        However, unless I'm very much mistaken, the sequence looks like this:
                        (5*2^897-1)/3, 5*2^896, 5*2^895, ..., 5*2^2, 5*2^1, 5, 8, 4, 2, 1 and as
                        far as my memory goes, the term "5" looks like an odd prime to me.

                        However, as you wanted a small example of this kind (< 2^50), consider
                        297784399189. Any objections against it?

                        Peter

                        --
                        [Name] Peter Kosinar [Quote] 2B | ~2B = exp(i*PI) [ICQ] 134813278
                      • David Broadhurst
                        ... sta staf wrote, ... Dear Rachid: Until you learn to be more disciplined, I shall no longer reply. Please isolate your error in the above
                        Message 11 of 22 , Jun 19, 2009
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                          --- In primenumbers@yahoogroups.com,
                          sta staf <sta_staf@...> wrote,
                          untidily and incorrectly:

                          > but your seqaunce :(5*2^897-1/3)
                          > this sequance of coltaze is : all Even numbers .
                          > the only odd is the ferst digit : :(5*2^897-1/3)
                          > rachid

                          Dear Rachid: Until you learn to be more disciplined,
                          I shall no longer reply. Please isolate your error
                          in the above statement, bearing in mind that 5
                          is an odd prime. Then consider that you asked
                          only for a Collatz sequence in which no odd
                          member is composite, with which request my
                          simple construction perfectly complies.

                          Discipline is a stern master:
                          I shall not allow you to "wriggle"
                          out of the hole you have carelessly
                          and untidily dug yourself into.

                          You may think me stern. But please recall
                          that mathematics is far sterner than I am.

                          Stay well,

                          David
                        • David Broadhurst
                          ... After reading http://www.primepuzzles.net/puzzles/puzz_476.htm Rachid will perhaps not object to my next claim, namely that (2^1322*(5*2^897-1)/3-1)/3 is a
                          Message 12 of 22 , Jun 20, 2009
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                            --- In primenumbers@yahoogroups.com,
                            "David Broadhurst" <d.broadhurst@...> wrote:

                            > The 271-digit prime (5*2^897-1)/3 manifestly contains no
                            > composite odd integer in its Collatz sequence and it would be
                            > easy to find larger primes by this, or a similar, construction.

                            After reading http://www.primepuzzles.net/puzzles/puzz_476.htm
                            Rachid will perhaps not object to my next claim, namely that
                            (2^1322*(5*2^897-1)/3-1)/3 is a 668-digit prime whose Collatz
                            sequence contains no composite odd integer.

                            As both Jens and I have noted, left extensibility is trivial,
                            given enough computing power. Hence my comment to Rachid:

                            > The word "largest" makes no sense here.

                            David
                          • Adriano Palma
                            ** For Your Eyes Only ** ** High Priority ** the agent never got back to me, in spite of three different demands I assume they sold it I looked at another one
                            Message 13 of 22 , Jun 20, 2009
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                              ** For Your Eyes Only **
                              ** High Priority **


                              the agent never got back to me, in spite of three different demands
                              I assume they sold it
                              I looked at another one so I may decide to move after all.
                              I trust & hope your vacation be goo

                              |||||||||||||||||||||||||||||||||||||||||||||||||||||
                              ξε ν’, γγέλλειν Λακεδαιμονίοις ἀ ὅτι τ δε
                              κείμεθα, το ς κείνων ῥήμασι πειθόμενοι.
                              /begin/read__>sig.file: postal address
                              palma
                              University of KwaZulu-Natal Philosophy
                              3rd floor of Memorial Tower Building
                              Howard College Campus
                              Durban 4041
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                              Tel off: [+27] 031 2601591 (sec: Mrs. Yolanda Hordyk) [+27]
                              031-2602292
                              Fax [+27] 031-2603031
                              mobile 07 62 36 23 91 calling from overseas +[27] 76 2362391
                              EMAIL: palma@...
                              EMAIL: palma@...
                              MY OFFICE # IS 290@Mtb
                              *only when in Europe*: inst. J. Nicod
                              29 rue d'Ulm
                              f-75005 paris france
                              email me for details if needed at palma@...
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                              received this e-mail in error please destroy the original and notify
                              the sender.d
                              >>> "David Broadhurst" <d.broadhurst@...> 6/20/2009 3:39 PM >>>



                              --- In primenumbers@yahoogroups.com (
                              mailto:primenumbers%40yahoogroups.com ),
                              "David Broadhurst" <d.broadhurst@...> wrote:

                              > The 271-digit prime (5*2^897-1)/3 manifestly contains no
                              > composite odd integer in its Collatz sequence and it would be
                              > easy to find larger primes by this, or a similar, construction.

                              After reading http://www.primepuzzles.net/puzzles/puzz_476.htm (
                              http://www.primepuzzles.net/puzzles/puzz_476.htm )
                              Rachid will perhaps not object to my next claim, namely that
                              (2^1322*(5*2^897-1)/3-1)/3 is a 668-digit prime whose Collatz
                              sequence contains no composite odd integer.

                              As both Jens and I have noted, left extensibility is trivial,
                              given enough computing power. Hence my comment to Rachid:

                              > The word "largest" makes no sense here.

                              David



                              Please find our Email Disclaimer here: http://www.ukzn.ac.za/disclaimer/


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                            • sta staf
                              you have the right david there is always another prime can often be extended to the left like your numbre :(2^1322*(5*2^897-1)/3-1)/3) is a Construction
                              Message 14 of 22 , Jun 20, 2009
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                                you have the right david there is always another prime can often be extended to the left
                                like your numbre :(2^1322*(5*2^897-1)/3-1)/3)

                                is a Construction Sequence . very easy to Making the collatz sequence contains no composite odd integer.


                                yes ( The word "largest" makes no sense here)



                                rachid










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                              • David Broadhurst
                                ... En Angleterre, nous avons l habitude de tirer à gauche. Amitiés David
                                Message 15 of 22 , Jun 20, 2009
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                                  --- In primenumbers@yahoogroups.com,
                                  sta staf <sta_staf@...> wrote:

                                  > you have the right david there is always another prime
                                  > can often be extended to the left
                                  > like your numbre : (2^1322*(5*2^897-1)/3-1)/3

                                  En Angleterre, nous avons l'habitude de tirer à gauche.

                                  Amitiés

                                  David
                                • cino hilliard
                                  Hi, I did a brute force in gcc/gmp. http://groups.google.com/group/elliptic-curves/web/gcc-gmp-brute-force-elliptic-curve Output: n start = 22189 n stop =
                                  Message 16 of 22 , Jun 23, 2009
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                                    Hi,

                                    I did a brute force in gcc/gmp.

                                    http://groups.google.com/group/elliptic-curves/web/gcc-gmp-brute-force-elliptic-curve



                                    Output:

                                    n start => 22189
                                    n stop => 22189
                                    x range => 150000000
                                    max dups => 1
                                    n prime => 1
                                    29585 22189 5088706
                                    140292677 22189 1661699554612
                                    count = 2
                                    31.9994

                                    for dell 8200 2.53 ghz



                                    24.9 sec

                                    for dell gx620 3.2 ghz vista ult



                                    Working with a developer of sage I was pointed to ratpoints.

                                    >> http://www.mathe2.uni-bayreuth.de/stoll/programs/index.html, is free
                                    >> (GPL) and only requires gmp.



                                    >> ratpoints looks promising but I can't install it. Using Msys, I keep getting
                                    >> the cannot find -lgmp.

                                    I installed gmp 4.3.1 with msys which was used for the above timings.


                                    I searched long for this. There ain't much out there on it. In fact, google -lgmp
                                    gets a no hit.



                                    Still asking, how do we determine the x range for ratpoints and how does sage
                                    do it?



                                    Reply:

                                    >ratpoints takes 0.09s to fine all integral points with x<10^8! But
                                    >there is no proof, which Sage provides. Sage takes longer as the
                                    >curve has rank 5.

                                    >I cannot help with Windows problems. gmp is a standard library (Gnu
                                    >MultiPrecision) but I do not know how to use it on Windows.

                                    >Very soon ratpoints will be available within Sage -- we are just testing it now.

                                    Maybe someone in primenumbers can load ratpoints in a windows xp pro of vista
                                    platform.



                                    Anyway, the gcc/gmp brute force is considerably faster than Pari for x range 10^8-10^9.



                                    Cino





                                    To: primenumbers@yahoogroups.com
                                    From: d.broadhurst@...
                                    Date: Fri, 19 Jun 2009 23:14:46 +0000
                                    Subject: [PrimeNumbers] Re: primes of the form x^3 - y^2







                                    --- In primenumbers@yahoogroups.com,
                                    cino hilliard <hillcino368@...> wrote:

                                    > the workaround,
                                    > diffcubes2(n,p) =
                                    > {
                                    > local(x,y,c,a);
                                    > a=eval(setintersect(Set(vector(n,y,y^2)),Set(vector(n,x,x^3-p))));
                                    > a;
                                    > }
                                    > is useless for large n.

                                    As indeed I had expected :-)

                                    Sometimes brute force is best, as in my prior hack:

                                    c=0;gettime;
                                    {for(x=1,10^9,if(issquare(x^3-22189),
                                    print(x);c=c+1;if(c==2,break)))}
                                    print(ceil(gettime/10^3)" seconds")

                                    29585
                                    140292677
                                    98 seconds





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                                    [Non-text portions of this message have been removed]
                                  • David Broadhurst
                                    ... Sage is being sagacious, here. It s good that you can access such wisdom without paying big dollars for Magma. Congrats to Cino, for persistence. David
                                    Message 17 of 22 , Jun 23, 2009
                                    • 0 Attachment
                                      --- In primenumbers@yahoogroups.com,
                                      cino hilliard <hillcino368@...> wrote:

                                      > Sage takes longer as the curve has rank 5.

                                      Sage is being sagacious, here.
                                      It's good that you can access
                                      such wisdom without paying big
                                      dollars for Magma.

                                      Congrats to Cino, for persistence.

                                      David
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