Solve $an^2+bn+c=k^2$ over the integers.












2












$begingroup$


I need to solve $5n^2+14n+1=k^2$ over integers $n$ and $k$. I was wondering if there is any general theory for solving diophantine equations of the form $an^2+bn+c=k^2$. For my specific case, I already found the first $21$ solutions.



$$(2,7);(5,14);(21,50);(42,97);(152,343);(296,665);(1050,2351);(2037,4558);(7205,16114);(13970,31241);(49392,110447);(95760,214129);(338546,757015);(656357,1467662);(2320437,5188658);(4498746,10059505);(15904520,35563591);(30834872,68948873);(109011210,243756479);(211345365,472582606);(747173957,1670731762)$$



After this I get overflow problems. I noticed that the sequence grows exponentially, so I wondered what the base is. The ratio of terms seems to oscelate between $1.939$ and $3.535$. I then noticed that $3.535approxfrac52sqrt{2}$. Unfortunately I could not find a closed expression that is approximately $1.939$.



So is there a general way of solving such equations? And is there an obvious reason why the sequence grows exponentially?






diophantine-equations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    math.stackexchange.com/questions/2424587/…
    $endgroup$
    – individ
    Jan 5 at 5:09
















2












$begingroup$


I need to solve $5n^2+14n+1=k^2$ over integers $n$ and $k$. I was wondering if there is any general theory for solving diophantine equations of the form $an^2+bn+c=k^2$. For my specific case, I already found the first $21$ solutions.



$$(2,7);(5,14);(21,50);(42,97);(152,343);(296,665);(1050,2351);(2037,4558);(7205,16114);(13970,31241);(49392,110447);(95760,214129);(338546,757015);(656357,1467662);(2320437,5188658);(4498746,10059505);(15904520,35563591);(30834872,68948873);(109011210,243756479);(211345365,472582606);(747173957,1670731762)$$



After this I get overflow problems. I noticed that the sequence grows exponentially, so I wondered what the base is. The ratio of terms seems to oscelate between $1.939$ and $3.535$. I then noticed that $3.535approxfrac52sqrt{2}$. Unfortunately I could not find a closed expression that is approximately $1.939$.



So is there a general way of solving such equations? And is there an obvious reason why the sequence grows exponentially?






diophantine-equations







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    math.stackexchange.com/questions/2424587/…
    $endgroup$
    – individ
    Jan 5 at 5:09














2












2








2





$begingroup$


I need to solve $5n^2+14n+1=k^2$ over integers $n$ and $k$. I was wondering if there is any general theory for solving diophantine equations of the form $an^2+bn+c=k^2$. For my specific case, I already found the first $21$ solutions.



$$(2,7);(5,14);(21,50);(42,97);(152,343);(296,665);(1050,2351);(2037,4558);(7205,16114);(13970,31241);(49392,110447);(95760,214129);(338546,757015);(656357,1467662);(2320437,5188658);(4498746,10059505);(15904520,35563591);(30834872,68948873);(109011210,243756479);(211345365,472582606);(747173957,1670731762)$$



After this I get overflow problems. I noticed that the sequence grows exponentially, so I wondered what the base is. The ratio of terms seems to oscelate between $1.939$ and $3.535$. I then noticed that $3.535approxfrac52sqrt{2}$. Unfortunately I could not find a closed expression that is approximately $1.939$.



So is there a general way of solving such equations? And is there an obvious reason why the sequence grows exponentially?






diophantine-equations







share|cite|improve this question









$endgroup$




I need to solve $5n^2+14n+1=k^2$ over integers $n$ and $k$. I was wondering if there is any general theory for solving diophantine equations of the form $an^2+bn+c=k^2$. For my specific case, I already found the first $21$ solutions.



$$(2,7);(5,14);(21,50);(42,97);(152,343);(296,665);(1050,2351);(2037,4558);(7205,16114);(13970,31241);(49392,110447);(95760,214129);(338546,757015);(656357,1467662);(2320437,5188658);(4498746,10059505);(15904520,35563591);(30834872,68948873);(109011210,243756479);(211345365,472582606);(747173957,1670731762)$$



After this I get overflow problems. I noticed that the sequence grows exponentially, so I wondered what the base is. The ratio of terms seems to oscelate between $1.939$ and $3.535$. I then noticed that $3.535approxfrac52sqrt{2}$. Unfortunately I could not find a closed expression that is approximately $1.939$.



So is there a general way of solving such equations? And is there an obvious reason why the sequence grows exponentially?






diophantine-equations




diophantine-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 4 at 20:17









SmileyCraftSmileyCraft

3,561518




3,561518








  • 1




    $begingroup$
    math.stackexchange.com/questions/2424587/…
    $endgroup$
    – individ
    Jan 5 at 5:09














  • 1




    $begingroup$
    math.stackexchange.com/questions/2424587/…
    $endgroup$
    – individ
    Jan 5 at 5:09








1




1




$begingroup$
math.stackexchange.com/questions/2424587/…
$endgroup$
– individ
Jan 5 at 5:09




$begingroup$
math.stackexchange.com/questions/2424587/…
$endgroup$
– individ
Jan 5 at 5:09










1 Answer
1






active

oldest

votes


















4












$begingroup$

Complete the square, and you get
$$ 5 left(n + frac{7}{5}right)^2 - frac{44}{5} = k^2 $$
or $$ (5n+7)^2 - 44 = 5 k^2 $$
With $x = 5n+7$, this is a Pell-type equation
$$ x^2 - 44 = 5 k^2 $$
where you only want solutions where $x equiv 2 mod 5$.






share|cite|improve this answer









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    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%2f3062058%2fsolve-an2bnc-k2-over-the-integers%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    Complete the square, and you get
    $$ 5 left(n + frac{7}{5}right)^2 - frac{44}{5} = k^2 $$
    or $$ (5n+7)^2 - 44 = 5 k^2 $$
    With $x = 5n+7$, this is a Pell-type equation
    $$ x^2 - 44 = 5 k^2 $$
    where you only want solutions where $x equiv 2 mod 5$.






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      Complete the square, and you get
      $$ 5 left(n + frac{7}{5}right)^2 - frac{44}{5} = k^2 $$
      or $$ (5n+7)^2 - 44 = 5 k^2 $$
      With $x = 5n+7$, this is a Pell-type equation
      $$ x^2 - 44 = 5 k^2 $$
      where you only want solutions where $x equiv 2 mod 5$.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        Complete the square, and you get
        $$ 5 left(n + frac{7}{5}right)^2 - frac{44}{5} = k^2 $$
        or $$ (5n+7)^2 - 44 = 5 k^2 $$
        With $x = 5n+7$, this is a Pell-type equation
        $$ x^2 - 44 = 5 k^2 $$
        where you only want solutions where $x equiv 2 mod 5$.






        share|cite|improve this answer









        $endgroup$



        Complete the square, and you get
        $$ 5 left(n + frac{7}{5}right)^2 - frac{44}{5} = k^2 $$
        or $$ (5n+7)^2 - 44 = 5 k^2 $$
        With $x = 5n+7$, this is a Pell-type equation
        $$ x^2 - 44 = 5 k^2 $$
        where you only want solutions where $x equiv 2 mod 5$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 20:32









        Robert IsraelRobert Israel

        326k23215469




        326k23215469






























            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%2f3062058%2fsolve-an2bnc-k2-over-the-integers%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

            Ellipse (mathématiques)

            Image
            Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
            Read more

            Quarter-circle Tiles

            Image
            up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
            Read more

            Mont Emei

            Image
            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
            Read more