Is it possible for a square root function,f(x), to map to a finite number of integers for all x in domain of...












2












$begingroup$


Consider the equation $$f(x) =sqrt{x^2 - x + 1}$$



Using python I checked x for $$ -100000000 leq x leq 100000000$$



and have only found two values of x, x = 0 and x = 1 that map to integers. While this range is quite large I am skeptical there is no other x that will map to an integer. How would one go about proving the choices for x that map to an integer given some square root function is finite or infinite?



Edit: $$x in mathbb{Z}$$






elementary-number-theory







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    To be clear, $x$ is an integer here? By convention, if not otherwise specified, $x,y$, are typically used for real numbers; $m,n$ are often used for integers.
    $endgroup$
    – Théophile
    9 hours ago






  • 1




    $begingroup$
    Well, if $x^2-x-3=0$ (that is, if $x=(1+sqrt{13})/2$, then $x^2-x+1= 4$, hence $f(x) = 2$. So presumably, you want $x$ to be an integer? If so, you are trying to solve the quadratic diophantine equation $x^2-x+1=y^2$.
    $endgroup$
    – Arturo Magidin
    9 hours ago










  • $begingroup$
    Justed edited, yes I meant to say $$x in mathbb{Z}$$
    $endgroup$
    – Diehardwalnut
    8 hours ago
















2












$begingroup$


Consider the equation $$f(x) =sqrt{x^2 - x + 1}$$



Using python I checked x for $$ -100000000 leq x leq 100000000$$



and have only found two values of x, x = 0 and x = 1 that map to integers. While this range is quite large I am skeptical there is no other x that will map to an integer. How would one go about proving the choices for x that map to an integer given some square root function is finite or infinite?



Edit: $$x in mathbb{Z}$$






elementary-number-theory







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    To be clear, $x$ is an integer here? By convention, if not otherwise specified, $x,y$, are typically used for real numbers; $m,n$ are often used for integers.
    $endgroup$
    – Théophile
    9 hours ago






  • 1




    $begingroup$
    Well, if $x^2-x-3=0$ (that is, if $x=(1+sqrt{13})/2$, then $x^2-x+1= 4$, hence $f(x) = 2$. So presumably, you want $x$ to be an integer? If so, you are trying to solve the quadratic diophantine equation $x^2-x+1=y^2$.
    $endgroup$
    – Arturo Magidin
    9 hours ago










  • $begingroup$
    Justed edited, yes I meant to say $$x in mathbb{Z}$$
    $endgroup$
    – Diehardwalnut
    8 hours ago














2












2








2


1



$begingroup$


Consider the equation $$f(x) =sqrt{x^2 - x + 1}$$



Using python I checked x for $$ -100000000 leq x leq 100000000$$



and have only found two values of x, x = 0 and x = 1 that map to integers. While this range is quite large I am skeptical there is no other x that will map to an integer. How would one go about proving the choices for x that map to an integer given some square root function is finite or infinite?



Edit: $$x in mathbb{Z}$$






elementary-number-theory







share|cite|improve this question











$endgroup$




Consider the equation $$f(x) =sqrt{x^2 - x + 1}$$



Using python I checked x for $$ -100000000 leq x leq 100000000$$



and have only found two values of x, x = 0 and x = 1 that map to integers. While this range is quite large I am skeptical there is no other x that will map to an integer. How would one go about proving the choices for x that map to an integer given some square root function is finite or infinite?



Edit: $$x in mathbb{Z}$$






elementary-number-theory




elementary-number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 8 hours ago







Diehardwalnut

















asked 9 hours ago









DiehardwalnutDiehardwalnut

262110




262110








  • 4




    $begingroup$
    To be clear, $x$ is an integer here? By convention, if not otherwise specified, $x,y$, are typically used for real numbers; $m,n$ are often used for integers.
    $endgroup$
    – Théophile
    9 hours ago






  • 1




    $begingroup$
    Well, if $x^2-x-3=0$ (that is, if $x=(1+sqrt{13})/2$, then $x^2-x+1= 4$, hence $f(x) = 2$. So presumably, you want $x$ to be an integer? If so, you are trying to solve the quadratic diophantine equation $x^2-x+1=y^2$.
    $endgroup$
    – Arturo Magidin
    9 hours ago










  • $begingroup$
    Justed edited, yes I meant to say $$x in mathbb{Z}$$
    $endgroup$
    – Diehardwalnut
    8 hours ago














  • 4




    $begingroup$
    To be clear, $x$ is an integer here? By convention, if not otherwise specified, $x,y$, are typically used for real numbers; $m,n$ are often used for integers.
    $endgroup$
    – Théophile
    9 hours ago






  • 1




    $begingroup$
    Well, if $x^2-x-3=0$ (that is, if $x=(1+sqrt{13})/2$, then $x^2-x+1= 4$, hence $f(x) = 2$. So presumably, you want $x$ to be an integer? If so, you are trying to solve the quadratic diophantine equation $x^2-x+1=y^2$.
    $endgroup$
    – Arturo Magidin
    9 hours ago










  • $begingroup$
    Justed edited, yes I meant to say $$x in mathbb{Z}$$
    $endgroup$
    – Diehardwalnut
    8 hours ago








4




4




$begingroup$
To be clear, $x$ is an integer here? By convention, if not otherwise specified, $x,y$, are typically used for real numbers; $m,n$ are often used for integers.
$endgroup$
– Théophile
9 hours ago




$begingroup$
To be clear, $x$ is an integer here? By convention, if not otherwise specified, $x,y$, are typically used for real numbers; $m,n$ are often used for integers.
$endgroup$
– Théophile
9 hours ago




1




1




$begingroup$
Well, if $x^2-x-3=0$ (that is, if $x=(1+sqrt{13})/2$, then $x^2-x+1= 4$, hence $f(x) = 2$. So presumably, you want $x$ to be an integer? If so, you are trying to solve the quadratic diophantine equation $x^2-x+1=y^2$.
$endgroup$
– Arturo Magidin
9 hours ago




$begingroup$
Well, if $x^2-x-3=0$ (that is, if $x=(1+sqrt{13})/2$, then $x^2-x+1= 4$, hence $f(x) = 2$. So presumably, you want $x$ to be an integer? If so, you are trying to solve the quadratic diophantine equation $x^2-x+1=y^2$.
$endgroup$
– Arturo Magidin
9 hours ago












$begingroup$
Justed edited, yes I meant to say $$x in mathbb{Z}$$
$endgroup$
– Diehardwalnut
8 hours ago




$begingroup$
Justed edited, yes I meant to say $$x in mathbb{Z}$$
$endgroup$
– Diehardwalnut
8 hours ago










3 Answers
3






active

oldest

votes


















4












$begingroup$

First of all, observe that the function is defined $forall xin mathbb Z$ since $x^2+1geq2xgeq xiff x^2-x+1geq 0$.



Completing the square, we get $$x^2-x+1=(x-1)^2+color{blue}x$$



It obviously works for $x=0$. Observe now, that the nearest squares are $(x-2)^2$ and $x^2$.



Furthermore
begin{align*}(x-1)^2-(x-2)^2&=color{blue}{2x-3}tag{1}\
x^2-(x-1)^2&=color{blue}{2x-1}tag{2}
end{align*}



Can you end it now?




Hint: Observe, for instance, that $$lvert 2x-3rvert>lvert xrvert text{ unless } xin[1, 3]$$ $$lvert 2x-1rvert>lvert xrvert text{ unless } xin[frac{1}{3}, 1]$$The difference becomes then too big otherwise... Thus - and since $x$ is an integer - you just have to check the cases $xin{1, 2, 3}$.







share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    Hint: let $y = sqrt{x^2-x+1}$. Squaring both sides,
    $$y^2 = x^2-x+1,$$
    so $y^2-1=x^2-x$. That is,
    $$(y+1)(y-1) = x(x-1).$$



    So your question becomes: when can the product of two numbers with difference two (i.e., the LHS) equal the product of two numbers with difference one (i.e., the RHS)?






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Hint:



      For $x>1$, $(x-1)^2 lt x^2-x+1 lt x^2$;



      for $x<0$, $x^2<x^2-x+1<(x-1)^2$.






      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%2f3176345%2fis-it-possible-for-a-square-root-function-fx-to-map-to-a-finite-number-of-int%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        4












        $begingroup$

        First of all, observe that the function is defined $forall xin mathbb Z$ since $x^2+1geq2xgeq xiff x^2-x+1geq 0$.



        Completing the square, we get $$x^2-x+1=(x-1)^2+color{blue}x$$



        It obviously works for $x=0$. Observe now, that the nearest squares are $(x-2)^2$ and $x^2$.



        Furthermore
        begin{align*}(x-1)^2-(x-2)^2&=color{blue}{2x-3}tag{1}\
        x^2-(x-1)^2&=color{blue}{2x-1}tag{2}
        end{align*}



        Can you end it now?




        Hint: Observe, for instance, that $$lvert 2x-3rvert>lvert xrvert text{ unless } xin[1, 3]$$ $$lvert 2x-1rvert>lvert xrvert text{ unless } xin[frac{1}{3}, 1]$$The difference becomes then too big otherwise... Thus - and since $x$ is an integer - you just have to check the cases $xin{1, 2, 3}$.







        share|cite|improve this answer











        $endgroup$


















          4












          $begingroup$

          First of all, observe that the function is defined $forall xin mathbb Z$ since $x^2+1geq2xgeq xiff x^2-x+1geq 0$.



          Completing the square, we get $$x^2-x+1=(x-1)^2+color{blue}x$$



          It obviously works for $x=0$. Observe now, that the nearest squares are $(x-2)^2$ and $x^2$.



          Furthermore
          begin{align*}(x-1)^2-(x-2)^2&=color{blue}{2x-3}tag{1}\
          x^2-(x-1)^2&=color{blue}{2x-1}tag{2}
          end{align*}



          Can you end it now?




          Hint: Observe, for instance, that $$lvert 2x-3rvert>lvert xrvert text{ unless } xin[1, 3]$$ $$lvert 2x-1rvert>lvert xrvert text{ unless } xin[frac{1}{3}, 1]$$The difference becomes then too big otherwise... Thus - and since $x$ is an integer - you just have to check the cases $xin{1, 2, 3}$.







          share|cite|improve this answer











          $endgroup$
















            4












            4








            4





            $begingroup$

            First of all, observe that the function is defined $forall xin mathbb Z$ since $x^2+1geq2xgeq xiff x^2-x+1geq 0$.



            Completing the square, we get $$x^2-x+1=(x-1)^2+color{blue}x$$



            It obviously works for $x=0$. Observe now, that the nearest squares are $(x-2)^2$ and $x^2$.



            Furthermore
            begin{align*}(x-1)^2-(x-2)^2&=color{blue}{2x-3}tag{1}\
            x^2-(x-1)^2&=color{blue}{2x-1}tag{2}
            end{align*}



            Can you end it now?




            Hint: Observe, for instance, that $$lvert 2x-3rvert>lvert xrvert text{ unless } xin[1, 3]$$ $$lvert 2x-1rvert>lvert xrvert text{ unless } xin[frac{1}{3}, 1]$$The difference becomes then too big otherwise... Thus - and since $x$ is an integer - you just have to check the cases $xin{1, 2, 3}$.







            share|cite|improve this answer











            $endgroup$



            First of all, observe that the function is defined $forall xin mathbb Z$ since $x^2+1geq2xgeq xiff x^2-x+1geq 0$.



            Completing the square, we get $$x^2-x+1=(x-1)^2+color{blue}x$$



            It obviously works for $x=0$. Observe now, that the nearest squares are $(x-2)^2$ and $x^2$.



            Furthermore
            begin{align*}(x-1)^2-(x-2)^2&=color{blue}{2x-3}tag{1}\
            x^2-(x-1)^2&=color{blue}{2x-1}tag{2}
            end{align*}



            Can you end it now?




            Hint: Observe, for instance, that $$lvert 2x-3rvert>lvert xrvert text{ unless } xin[1, 3]$$ $$lvert 2x-1rvert>lvert xrvert text{ unless } xin[frac{1}{3}, 1]$$The difference becomes then too big otherwise... Thus - and since $x$ is an integer - you just have to check the cases $xin{1, 2, 3}$.








            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 4 hours ago









            Carmeister

            2,8592924




            2,8592924










            answered 9 hours ago









            Dr. MathvaDr. Mathva

            3,225630




            3,225630























                3












                $begingroup$

                Hint: let $y = sqrt{x^2-x+1}$. Squaring both sides,
                $$y^2 = x^2-x+1,$$
                so $y^2-1=x^2-x$. That is,
                $$(y+1)(y-1) = x(x-1).$$



                So your question becomes: when can the product of two numbers with difference two (i.e., the LHS) equal the product of two numbers with difference one (i.e., the RHS)?






                share|cite|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  Hint: let $y = sqrt{x^2-x+1}$. Squaring both sides,
                  $$y^2 = x^2-x+1,$$
                  so $y^2-1=x^2-x$. That is,
                  $$(y+1)(y-1) = x(x-1).$$



                  So your question becomes: when can the product of two numbers with difference two (i.e., the LHS) equal the product of two numbers with difference one (i.e., the RHS)?






                  share|cite|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    Hint: let $y = sqrt{x^2-x+1}$. Squaring both sides,
                    $$y^2 = x^2-x+1,$$
                    so $y^2-1=x^2-x$. That is,
                    $$(y+1)(y-1) = x(x-1).$$



                    So your question becomes: when can the product of two numbers with difference two (i.e., the LHS) equal the product of two numbers with difference one (i.e., the RHS)?






                    share|cite|improve this answer









                    $endgroup$



                    Hint: let $y = sqrt{x^2-x+1}$. Squaring both sides,
                    $$y^2 = x^2-x+1,$$
                    so $y^2-1=x^2-x$. That is,
                    $$(y+1)(y-1) = x(x-1).$$



                    So your question becomes: when can the product of two numbers with difference two (i.e., the LHS) equal the product of two numbers with difference one (i.e., the RHS)?







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 9 hours ago









                    ThéophileThéophile

                    20.4k13047




                    20.4k13047























                        1












                        $begingroup$

                        Hint:



                        For $x>1$, $(x-1)^2 lt x^2-x+1 lt x^2$;



                        for $x<0$, $x^2<x^2-x+1<(x-1)^2$.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Hint:



                          For $x>1$, $(x-1)^2 lt x^2-x+1 lt x^2$;



                          for $x<0$, $x^2<x^2-x+1<(x-1)^2$.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Hint:



                            For $x>1$, $(x-1)^2 lt x^2-x+1 lt x^2$;



                            for $x<0$, $x^2<x^2-x+1<(x-1)^2$.






                            share|cite|improve this answer









                            $endgroup$



                            Hint:



                            For $x>1$, $(x-1)^2 lt x^2-x+1 lt x^2$;



                            for $x<0$, $x^2<x^2-x+1<(x-1)^2$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 8 hours ago









                            J. W. TannerJ. W. Tanner

                            4,4791320




                            4,4791320






























                                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%2f3176345%2fis-it-possible-for-a-square-root-function-fx-to-map-to-a-finite-number-of-int%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

                                Knooppunt Holsloot

                                Image
                                Knooppunt Holsloot Kenmerken Wegen A37 × N34 Type knooppunt klaverblad Portaal     Verkeer & Vervoer Knooppunt Holsloot is een verkeersknooppunt op de kruising van de autosnelweg A37 en de N34 in Drenthe, tussen Emmen en Coevorden. Oorspronkelijk was deze aansluiting een afritconstructie van het type Haarlemmermeeraansluiting, zoals bij Knooppunt Hooipolder. Sinds 2002 is het knooppunt omgebouwd tot een volledig klaverblad. Als noviteit werd hier voor het eerst gebruikgemaakt van dimbare verlichting in plaats van de normale vaste tijden waarop de verlichting aan en uit gaat, wat een grote kostenbesparing opleverde. [1] Dat Holsloot als volwaardig knooppunt is uitgevoerd, waarbij alle verkeersstromen en afbuigingen ononderbroken kunnen doorgaan, is uitzonderlijk. Het betreft hier namelijk een aansluiting tussen een autosnelweg en niet-snelweg. Een vergelijkbare situatie treft men nabij Hoensbroek bij het knooppunt Ten Esschen.     Aa
                                Read more

                                Altaar (religie)

                                Image
                                Deel van de serie over kerkelijk gerei waaronder paramenten en liturgisch vaatwerk gebruikt in de liturgie Liturgisch vaatwerk Vasa sacra Miskelk · Pateen Kelklepeltje Ciborie · Monstrans Pyxis · Custodiale Vasa non sacra Ampullen · Wijwatervat Olievaatje Wierookvat · Ablutievat Paramenten Amict · Albe · Baarkleed · Cingel Tuniek · Dalmatiek · Fanon Kazuifel · Manipel · Stola Gremiale · Benedictievelum Mijter Koorkledij Rochet · Superplie Koorkap · Cappa magna Kovel Kelkgerei Bursa · Kelkvelum Ciborievelum Kerklinnen Corporale · Kelkdoekje Lavabodoekje · Palla Altaardwaal Kerkinterieur Altaar · Ambo Biechtstoel · Communiebank Doksaal · Doopvont Faldistorium Godslamp · Hoogaltaar Heilig Kruisaltaar Katheder · Preekstoel Sedilia Tabernakel · Volksaltaar Liturgische boeken Altaarmissaal Benedictionale · Brevier Evangeliarium · Evangelistarium Graduale · Kyriale Lectionarium Psalter · Rituaal Sacramentarium · Volksmissaal Overige Flambou
                                Read more

                                Gregoriusmis

                                Image
                                Een afbeelding van de Gregoriusmis uit een getijdenboek gemaakt in de zuidelijke Nederlanden tussen 1500 en 1525 Gregoriusmis, Meester van het Akense Altaar, ca. 1500-1525. Collectie Museum Catharijneconvent De Gregoriusmis of de Mis van Sint Gregorius is een motief in katholieke kunst, dat het eerst verscheen in de late Middeleeuwen en dat nog steeds werd gebruikt tijdens de contrareformatie. Paus Gregorius I (ca. 540-604) is afgebeeld, de H. Mis opdragend, terwijl Christus als de Man van Smarten verschijnt op het altaar dat voor Sint Gregorius staat. Volgens de legende had St. Gregorius gebeden om een teken voor een betwijfelaar van de transsubstantiatieleer, waarop de lijdende Christus tijdens het opdragen van de Mis verscheen om hem te overtuigen en waarbij zijn bloed in de altaarkelk liep. Vaak zijn ook de passiewerktuigen afgebeeld. Zie de categorie Mass of Saint Gregory van Wikimedia Commons voor mediabestanden over dit onderwerp. This page is onl
                                Read more