Rational solutions of $x^4(k^2+4)-4kx^2=z^2$, where $k>0$ is given rational.












0












$begingroup$


How to generate some parametric family of rational solutions to $x^4(k^2+4)-4kx^2=z^2$, where $k>0$ is given rational.






number-theory diophantine-equations parametrization







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    How to generate some parametric family of rational solutions to $x^4(k^2+4)-4kx^2=z^2$, where $k>0$ is given rational.






    number-theory diophantine-equations parametrization







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      How to generate some parametric family of rational solutions to $x^4(k^2+4)-4kx^2=z^2$, where $k>0$ is given rational.






      number-theory diophantine-equations parametrization







      share|cite|improve this question











      $endgroup$




      How to generate some parametric family of rational solutions to $x^4(k^2+4)-4kx^2=z^2$, where $k>0$ is given rational.






      number-theory diophantine-equations parametrization




      number-theory diophantine-equations parametrization






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 2 at 19:03







      ersh

















      asked Feb 2 at 18:57









      ershersh

      538113




      538113






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          For $k=4$, Above equation is shown below:



          $20x^4=(4x)^2+z^2$ ------(A)



          Equation $(A)$ has parametric solution:



          $x=2p(5m^2-4m+1)$



          $z=16(p^2)(25m^4-45m^3+30m^2-9m+1)$



          Where, $p=[1/(5m^2-1)]$



          For $m=3$ we have:



          $(x,z,k)=[(17/11),(1054/121),(4)]$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            @ Sam, How did you arrived at that parametric solution?
            $endgroup$
            – ersh
            Feb 3 at 14:33








          • 1




            $begingroup$
            Equation (A) is equivalent to (5x^2-4)=[(z)/(2x)]^2=(y)^2
            $endgroup$
            – Sam
            Feb 3 at 23:10






          • 1




            $begingroup$
            Since (x,z)=(2,16) is numerical solution of equation (A), it implies (y) =(4). Hence parametrize (5x^2-4)=y^2 at (x,y) =(2,4). After arriving at parameter's of (x,y) we use the equality (z)=( 2xy) to find (z)
            $endgroup$
            – Sam
            Feb 3 at 23:17










          • $begingroup$
            Good Thank you!!
            $endgroup$
            – ersh
            Feb 3 at 23:26



















          1












          $begingroup$

          The equation can be written
          begin{equation*}
          x^2((k^2+4)x^2-4k)=z^2
          end{equation*}
          and, so we need
          begin{equation*}
          (k^2+4)x^2-4k=y^2
          end{equation*}



          If $x=1$, then $y=pm (k-2)$, so the line $y=k-2+m(x-1)$ will meet the quadric at the further point
          begin{equation*}
          x=frac{-(k^2-2km+m^2+4m+4)}{k^2-m^2+4}
          end{equation*}



          For example, if $k=1/2$, putting $m=4$ gives
          begin{equation*}
          z^2=frac{3^2*5^2*43^2*103^2}{2^2*47^4}
          end{equation*}



          Allan Macleod






          share|cite|improve this answer











          $endgroup$














            Your Answer








            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3097650%2frational-solutions-of-x4k24-4kx2-z2-where-k0-is-given-rational%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            For $k=4$, Above equation is shown below:



            $20x^4=(4x)^2+z^2$ ------(A)



            Equation $(A)$ has parametric solution:



            $x=2p(5m^2-4m+1)$



            $z=16(p^2)(25m^4-45m^3+30m^2-9m+1)$



            Where, $p=[1/(5m^2-1)]$



            For $m=3$ we have:



            $(x,z,k)=[(17/11),(1054/121),(4)]$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              @ Sam, How did you arrived at that parametric solution?
              $endgroup$
              – ersh
              Feb 3 at 14:33








            • 1




              $begingroup$
              Equation (A) is equivalent to (5x^2-4)=[(z)/(2x)]^2=(y)^2
              $endgroup$
              – Sam
              Feb 3 at 23:10






            • 1




              $begingroup$
              Since (x,z)=(2,16) is numerical solution of equation (A), it implies (y) =(4). Hence parametrize (5x^2-4)=y^2 at (x,y) =(2,4). After arriving at parameter's of (x,y) we use the equality (z)=( 2xy) to find (z)
              $endgroup$
              – Sam
              Feb 3 at 23:17










            • $begingroup$
              Good Thank you!!
              $endgroup$
              – ersh
              Feb 3 at 23:26
















            1












            $begingroup$

            For $k=4$, Above equation is shown below:



            $20x^4=(4x)^2+z^2$ ------(A)



            Equation $(A)$ has parametric solution:



            $x=2p(5m^2-4m+1)$



            $z=16(p^2)(25m^4-45m^3+30m^2-9m+1)$



            Where, $p=[1/(5m^2-1)]$



            For $m=3$ we have:



            $(x,z,k)=[(17/11),(1054/121),(4)]$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              @ Sam, How did you arrived at that parametric solution?
              $endgroup$
              – ersh
              Feb 3 at 14:33








            • 1




              $begingroup$
              Equation (A) is equivalent to (5x^2-4)=[(z)/(2x)]^2=(y)^2
              $endgroup$
              – Sam
              Feb 3 at 23:10






            • 1




              $begingroup$
              Since (x,z)=(2,16) is numerical solution of equation (A), it implies (y) =(4). Hence parametrize (5x^2-4)=y^2 at (x,y) =(2,4). After arriving at parameter's of (x,y) we use the equality (z)=( 2xy) to find (z)
              $endgroup$
              – Sam
              Feb 3 at 23:17










            • $begingroup$
              Good Thank you!!
              $endgroup$
              – ersh
              Feb 3 at 23:26














            1












            1








            1





            $begingroup$

            For $k=4$, Above equation is shown below:



            $20x^4=(4x)^2+z^2$ ------(A)



            Equation $(A)$ has parametric solution:



            $x=2p(5m^2-4m+1)$



            $z=16(p^2)(25m^4-45m^3+30m^2-9m+1)$



            Where, $p=[1/(5m^2-1)]$



            For $m=3$ we have:



            $(x,z,k)=[(17/11),(1054/121),(4)]$






            share|cite|improve this answer









            $endgroup$



            For $k=4$, Above equation is shown below:



            $20x^4=(4x)^2+z^2$ ------(A)



            Equation $(A)$ has parametric solution:



            $x=2p(5m^2-4m+1)$



            $z=16(p^2)(25m^4-45m^3+30m^2-9m+1)$



            Where, $p=[1/(5m^2-1)]$



            For $m=3$ we have:



            $(x,z,k)=[(17/11),(1054/121),(4)]$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Feb 3 at 10:11









            SamSam

            261




            261












            • $begingroup$
              @ Sam, How did you arrived at that parametric solution?
              $endgroup$
              – ersh
              Feb 3 at 14:33








            • 1




              $begingroup$
              Equation (A) is equivalent to (5x^2-4)=[(z)/(2x)]^2=(y)^2
              $endgroup$
              – Sam
              Feb 3 at 23:10






            • 1




              $begingroup$
              Since (x,z)=(2,16) is numerical solution of equation (A), it implies (y) =(4). Hence parametrize (5x^2-4)=y^2 at (x,y) =(2,4). After arriving at parameter's of (x,y) we use the equality (z)=( 2xy) to find (z)
              $endgroup$
              – Sam
              Feb 3 at 23:17










            • $begingroup$
              Good Thank you!!
              $endgroup$
              – ersh
              Feb 3 at 23:26


















            • $begingroup$
              @ Sam, How did you arrived at that parametric solution?
              $endgroup$
              – ersh
              Feb 3 at 14:33








            • 1




              $begingroup$
              Equation (A) is equivalent to (5x^2-4)=[(z)/(2x)]^2=(y)^2
              $endgroup$
              – Sam
              Feb 3 at 23:10






            • 1




              $begingroup$
              Since (x,z)=(2,16) is numerical solution of equation (A), it implies (y) =(4). Hence parametrize (5x^2-4)=y^2 at (x,y) =(2,4). After arriving at parameter's of (x,y) we use the equality (z)=( 2xy) to find (z)
              $endgroup$
              – Sam
              Feb 3 at 23:17










            • $begingroup$
              Good Thank you!!
              $endgroup$
              – ersh
              Feb 3 at 23:26
















            $begingroup$
            @ Sam, How did you arrived at that parametric solution?
            $endgroup$
            – ersh
            Feb 3 at 14:33






            $begingroup$
            @ Sam, How did you arrived at that parametric solution?
            $endgroup$
            – ersh
            Feb 3 at 14:33






            1




            1




            $begingroup$
            Equation (A) is equivalent to (5x^2-4)=[(z)/(2x)]^2=(y)^2
            $endgroup$
            – Sam
            Feb 3 at 23:10




            $begingroup$
            Equation (A) is equivalent to (5x^2-4)=[(z)/(2x)]^2=(y)^2
            $endgroup$
            – Sam
            Feb 3 at 23:10




            1




            1




            $begingroup$
            Since (x,z)=(2,16) is numerical solution of equation (A), it implies (y) =(4). Hence parametrize (5x^2-4)=y^2 at (x,y) =(2,4). After arriving at parameter's of (x,y) we use the equality (z)=( 2xy) to find (z)
            $endgroup$
            – Sam
            Feb 3 at 23:17




            $begingroup$
            Since (x,z)=(2,16) is numerical solution of equation (A), it implies (y) =(4). Hence parametrize (5x^2-4)=y^2 at (x,y) =(2,4). After arriving at parameter's of (x,y) we use the equality (z)=( 2xy) to find (z)
            $endgroup$
            – Sam
            Feb 3 at 23:17












            $begingroup$
            Good Thank you!!
            $endgroup$
            – ersh
            Feb 3 at 23:26




            $begingroup$
            Good Thank you!!
            $endgroup$
            – ersh
            Feb 3 at 23:26











            1












            $begingroup$

            The equation can be written
            begin{equation*}
            x^2((k^2+4)x^2-4k)=z^2
            end{equation*}
            and, so we need
            begin{equation*}
            (k^2+4)x^2-4k=y^2
            end{equation*}



            If $x=1$, then $y=pm (k-2)$, so the line $y=k-2+m(x-1)$ will meet the quadric at the further point
            begin{equation*}
            x=frac{-(k^2-2km+m^2+4m+4)}{k^2-m^2+4}
            end{equation*}



            For example, if $k=1/2$, putting $m=4$ gives
            begin{equation*}
            z^2=frac{3^2*5^2*43^2*103^2}{2^2*47^4}
            end{equation*}



            Allan Macleod






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              The equation can be written
              begin{equation*}
              x^2((k^2+4)x^2-4k)=z^2
              end{equation*}
              and, so we need
              begin{equation*}
              (k^2+4)x^2-4k=y^2
              end{equation*}



              If $x=1$, then $y=pm (k-2)$, so the line $y=k-2+m(x-1)$ will meet the quadric at the further point
              begin{equation*}
              x=frac{-(k^2-2km+m^2+4m+4)}{k^2-m^2+4}
              end{equation*}



              For example, if $k=1/2$, putting $m=4$ gives
              begin{equation*}
              z^2=frac{3^2*5^2*43^2*103^2}{2^2*47^4}
              end{equation*}



              Allan Macleod






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                The equation can be written
                begin{equation*}
                x^2((k^2+4)x^2-4k)=z^2
                end{equation*}
                and, so we need
                begin{equation*}
                (k^2+4)x^2-4k=y^2
                end{equation*}



                If $x=1$, then $y=pm (k-2)$, so the line $y=k-2+m(x-1)$ will meet the quadric at the further point
                begin{equation*}
                x=frac{-(k^2-2km+m^2+4m+4)}{k^2-m^2+4}
                end{equation*}



                For example, if $k=1/2$, putting $m=4$ gives
                begin{equation*}
                z^2=frac{3^2*5^2*43^2*103^2}{2^2*47^4}
                end{equation*}



                Allan Macleod






                share|cite|improve this answer











                $endgroup$



                The equation can be written
                begin{equation*}
                x^2((k^2+4)x^2-4k)=z^2
                end{equation*}
                and, so we need
                begin{equation*}
                (k^2+4)x^2-4k=y^2
                end{equation*}



                If $x=1$, then $y=pm (k-2)$, so the line $y=k-2+m(x-1)$ will meet the quadric at the further point
                begin{equation*}
                x=frac{-(k^2-2km+m^2+4m+4)}{k^2-m^2+4}
                end{equation*}



                For example, if $k=1/2$, putting $m=4$ gives
                begin{equation*}
                z^2=frac{3^2*5^2*43^2*103^2}{2^2*47^4}
                end{equation*}



                Allan Macleod







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Feb 3 at 15:51

























                answered Feb 3 at 10:00









                Allan MacLeodAllan MacLeod

                1,24975




                1,24975






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3097650%2frational-solutions-of-x4k24-4kx2-z2-where-k0-is-given-rational%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    MongoDB - Not Authorized To Execute Command

                    Image
                    .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
                    Read more

                    How to fix TextFormField cause rebuild widget in Flutter

                    Image
                    .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I've got the same issue at #9280. I tried all solutions from that topic with/without GlobalKey for Scaffold and Form but the problem wasn't fixed. I tried to build a authentication form with login and signup. But the keyboard doesn't show when I tap the TextFormField at the first time. Then I try to input some letter from physical keyboard, the widget is rebuilt. Like this: login screen Nothing to be log when I run flutter run --verbose, so I logged manually. Every tapping on field, the widget called dispose -> init -> build. [√] Flutter (Channel beta, v1.0.0, on Microsoft Windows [Version 10.0.17763.195], ...
                    Read more

                    in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

                    Image
                    1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
                    Read more