Find all integers n (positive, negative, or zero) so that n^3 – 1 is divisible by n + 1












1












$begingroup$


I've tried to rearrange n^3 - 1 (turning it into (n-1)(n^2+n+1)) and play around with it, but I haven't been able to figure it out. Within the text this problem was in, the chapter taught you divisibility and the Euclidean Algorithm so I'm assuming the answer will utilize those tools.






proof-writing divisibility







share|cite







New contributor




Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$

















    1












    $begingroup$


    I've tried to rearrange n^3 - 1 (turning it into (n-1)(n^2+n+1)) and play around with it, but I haven't been able to figure it out. Within the text this problem was in, the chapter taught you divisibility and the Euclidean Algorithm so I'm assuming the answer will utilize those tools.






    proof-writing divisibility







    share|cite







    New contributor




    Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      1












      1








      1





      $begingroup$


      I've tried to rearrange n^3 - 1 (turning it into (n-1)(n^2+n+1)) and play around with it, but I haven't been able to figure it out. Within the text this problem was in, the chapter taught you divisibility and the Euclidean Algorithm so I'm assuming the answer will utilize those tools.






      proof-writing divisibility







      share|cite







      New contributor




      Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      I've tried to rearrange n^3 - 1 (turning it into (n-1)(n^2+n+1)) and play around with it, but I haven't been able to figure it out. Within the text this problem was in, the chapter taught you divisibility and the Euclidean Algorithm so I'm assuming the answer will utilize those tools.






      proof-writing divisibility




      proof-writing divisibility






      share|cite







      New contributor




      Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite







      New contributor




      Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite




      share|cite






      New contributor




      Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked 2 hours ago









      JayJay

      61




      61




      New contributor




      Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          If $n + 1$ is a factor, then note that $n equiv -1 pmod{n + 1}$. As such, $n^3 - 1 equiv left(-1right)^3 - 1 = -2 pmod{n + 1}$. This can only be the case if $n + 1 = pm 2$ or $n + 1 = pm 1$, i.e., $n = 0, 1, -2, -3$.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            John Omielan's answer is a standard mathematical approach. However if you are not comfortable with congruences you may try this:



            $$n^3-1=(n+1)(n^2-n+1)-2$$



            Dividing by $n+1$ gives
            $$frac{n^3-1}{n+1}=(n^2-n+1)-frac{2}{n+1}$$



            Since the left side & the first term on the right are integers, then $frac{2}{n+1}$ must also be an integer. So, $frac{2}{n+1}$ must be an integer. This is possible for $n=1,-3,-2,0$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              The first sentence in your last paragraph doesn't make much sense anymore after your latest change. I believe you want to say something like that since the left side & the first term on the right are integers, then $frac{2}{n + 1}$ must also be an integer.
              $endgroup$
              – John Omielan
              1 hour ago










            • $begingroup$
              @JohnOmielan Edited.
              $endgroup$
              – tatan
              1 hour ago











            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
            });


            }
            });






            Jay is a new contributor. Be nice, and check out our Code of Conduct.










            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3098147%2ffind-all-integers-n-positive-negative-or-zero-so-that-n3-1-is-divisible-b%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









            3












            $begingroup$

            If $n + 1$ is a factor, then note that $n equiv -1 pmod{n + 1}$. As such, $n^3 - 1 equiv left(-1right)^3 - 1 = -2 pmod{n + 1}$. This can only be the case if $n + 1 = pm 2$ or $n + 1 = pm 1$, i.e., $n = 0, 1, -2, -3$.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              If $n + 1$ is a factor, then note that $n equiv -1 pmod{n + 1}$. As such, $n^3 - 1 equiv left(-1right)^3 - 1 = -2 pmod{n + 1}$. This can only be the case if $n + 1 = pm 2$ or $n + 1 = pm 1$, i.e., $n = 0, 1, -2, -3$.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                If $n + 1$ is a factor, then note that $n equiv -1 pmod{n + 1}$. As such, $n^3 - 1 equiv left(-1right)^3 - 1 = -2 pmod{n + 1}$. This can only be the case if $n + 1 = pm 2$ or $n + 1 = pm 1$, i.e., $n = 0, 1, -2, -3$.






                share|cite|improve this answer









                $endgroup$



                If $n + 1$ is a factor, then note that $n equiv -1 pmod{n + 1}$. As such, $n^3 - 1 equiv left(-1right)^3 - 1 = -2 pmod{n + 1}$. This can only be the case if $n + 1 = pm 2$ or $n + 1 = pm 1$, i.e., $n = 0, 1, -2, -3$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 hours ago









                John OmielanJohn Omielan

                2,046210




                2,046210























                    2












                    $begingroup$

                    John Omielan's answer is a standard mathematical approach. However if you are not comfortable with congruences you may try this:



                    $$n^3-1=(n+1)(n^2-n+1)-2$$



                    Dividing by $n+1$ gives
                    $$frac{n^3-1}{n+1}=(n^2-n+1)-frac{2}{n+1}$$



                    Since the left side & the first term on the right are integers, then $frac{2}{n+1}$ must also be an integer. So, $frac{2}{n+1}$ must be an integer. This is possible for $n=1,-3,-2,0$






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      The first sentence in your last paragraph doesn't make much sense anymore after your latest change. I believe you want to say something like that since the left side & the first term on the right are integers, then $frac{2}{n + 1}$ must also be an integer.
                      $endgroup$
                      – John Omielan
                      1 hour ago










                    • $begingroup$
                      @JohnOmielan Edited.
                      $endgroup$
                      – tatan
                      1 hour ago
















                    2












                    $begingroup$

                    John Omielan's answer is a standard mathematical approach. However if you are not comfortable with congruences you may try this:



                    $$n^3-1=(n+1)(n^2-n+1)-2$$



                    Dividing by $n+1$ gives
                    $$frac{n^3-1}{n+1}=(n^2-n+1)-frac{2}{n+1}$$



                    Since the left side & the first term on the right are integers, then $frac{2}{n+1}$ must also be an integer. So, $frac{2}{n+1}$ must be an integer. This is possible for $n=1,-3,-2,0$






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      The first sentence in your last paragraph doesn't make much sense anymore after your latest change. I believe you want to say something like that since the left side & the first term on the right are integers, then $frac{2}{n + 1}$ must also be an integer.
                      $endgroup$
                      – John Omielan
                      1 hour ago










                    • $begingroup$
                      @JohnOmielan Edited.
                      $endgroup$
                      – tatan
                      1 hour ago














                    2












                    2








                    2





                    $begingroup$

                    John Omielan's answer is a standard mathematical approach. However if you are not comfortable with congruences you may try this:



                    $$n^3-1=(n+1)(n^2-n+1)-2$$



                    Dividing by $n+1$ gives
                    $$frac{n^3-1}{n+1}=(n^2-n+1)-frac{2}{n+1}$$



                    Since the left side & the first term on the right are integers, then $frac{2}{n+1}$ must also be an integer. So, $frac{2}{n+1}$ must be an integer. This is possible for $n=1,-3,-2,0$






                    share|cite|improve this answer











                    $endgroup$



                    John Omielan's answer is a standard mathematical approach. However if you are not comfortable with congruences you may try this:



                    $$n^3-1=(n+1)(n^2-n+1)-2$$



                    Dividing by $n+1$ gives
                    $$frac{n^3-1}{n+1}=(n^2-n+1)-frac{2}{n+1}$$



                    Since the left side & the first term on the right are integers, then $frac{2}{n+1}$ must also be an integer. So, $frac{2}{n+1}$ must be an integer. This is possible for $n=1,-3,-2,0$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 1 hour ago

























                    answered 2 hours ago









                    tatantatan

                    5,70062758




                    5,70062758












                    • $begingroup$
                      The first sentence in your last paragraph doesn't make much sense anymore after your latest change. I believe you want to say something like that since the left side & the first term on the right are integers, then $frac{2}{n + 1}$ must also be an integer.
                      $endgroup$
                      – John Omielan
                      1 hour ago










                    • $begingroup$
                      @JohnOmielan Edited.
                      $endgroup$
                      – tatan
                      1 hour ago


















                    • $begingroup$
                      The first sentence in your last paragraph doesn't make much sense anymore after your latest change. I believe you want to say something like that since the left side & the first term on the right are integers, then $frac{2}{n + 1}$ must also be an integer.
                      $endgroup$
                      – John Omielan
                      1 hour ago










                    • $begingroup$
                      @JohnOmielan Edited.
                      $endgroup$
                      – tatan
                      1 hour ago
















                    $begingroup$
                    The first sentence in your last paragraph doesn't make much sense anymore after your latest change. I believe you want to say something like that since the left side & the first term on the right are integers, then $frac{2}{n + 1}$ must also be an integer.
                    $endgroup$
                    – John Omielan
                    1 hour ago




                    $begingroup$
                    The first sentence in your last paragraph doesn't make much sense anymore after your latest change. I believe you want to say something like that since the left side & the first term on the right are integers, then $frac{2}{n + 1}$ must also be an integer.
                    $endgroup$
                    – John Omielan
                    1 hour ago












                    $begingroup$
                    @JohnOmielan Edited.
                    $endgroup$
                    – tatan
                    1 hour ago




                    $begingroup$
                    @JohnOmielan Edited.
                    $endgroup$
                    – tatan
                    1 hour ago










                    Jay is a new contributor. Be nice, and check out our Code of Conduct.










                    draft saved

                    draft discarded


















                    Jay is a new contributor. Be nice, and check out our Code of Conduct.













                    Jay is a new contributor. Be nice, and check out our Code of Conduct.












                    Jay is a new contributor. Be nice, and check out our Code of Conduct.
















                    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%2f3098147%2ffind-all-integers-n-positive-negative-or-zero-so-that-n3-1-is-divisible-b%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

                    flock() on closed filehandle LOCK_FILE at /usr/bin/apt-mirror

                    Image
                    0 I recently upgraded an Ubuntu 16.04 virtual machine to 18.04. I regularly use this VM to run apt-mirror at school on their high-speed internet, then rsync these files to my rack-mount server at home for updating multiple computers at home without wasting the tiny amount of bandwidth I have there. Since updating I can't get apt-mirror to run on my VM. It keeps giving the error: flock() on closed filehandle LOCK_FILE at /usr/bin/apt-mirror line 214 immediately followed by: apt-mirror is already running, exiting at /usr/bin/apt-mirror line 217 apt-mirror is NOT running, and does not have a cron job. I've tried rebooting, and even looked for the old-style lockfile to delete, but I guess that method is no longer used. Is there some way to force unlock this so I can run apt-mirror? UPDATE: So... I f...
                    Read more

                    Mangá

                    Image
                      Nota:  Não confundir com Manga. Mangás e Animes Mangá Mangaká • Tankōbon Film comic La nouvelle manga Mangá original em inglês Anime História Animação influenciada por animes • Filler • OVA • Seiyū Públicos Kodomo • Shōjo • Shōnen Josei • Seinen Gêneros Mahō shōjo • Mecha • Harém Ecchi • Hentai ( Yaoi • Yuri Futanari • Shotacon • Lolicon Toddlercon ) Termos Bishōjo • Chibi • Fan service Gaiden • Kawaii • Kaitō • Light novel Mahō shōnen • Moe (antropomorfismo • Kemonomimi Nekomimi ) • Super deformed Omake • Tsundere • Yonkoma Listas Mangás • Animes • Ecchi • Mangaka • Estúdios • Revistas Fãs ( otaku ) Anime Music Video Convenções • Cosplay Dōjinshi • Fansub • Fandub • Fujoshi OS-tan • Scanlation Portal de Anime e mangá v d e O mangá (português brasileiro) ou manga (português europeu) [ 1 ] (em japonês: 漫画 , manga ? , lit. “história em quadrinhos”) , é a ...
                    Read more

                    Eduardo VII do Reino Unido

                    Image
                    Eduardo VII Rei do Reino Unido da Grã-Bretanha e Irlanda e dos Domínios Britânicos de Além-mar Imperador da Índia Rei do Reino Unido e Imperador da Índia Reinado 22 de janeiro de 1901 a 6 de maio de 1910 Coroação 9 de agosto de 1902 Delhi Durbar 1 de janeiro de 1903 Predecessora Vitória Sucessor Jorge V   Esposa Alexandra da Dinamarca Descendência Alberto Vitor, Duque de Clarence e Avondale Jorge V do Reino Unido Luísa, Princesa Real Vitória de Gales Maud de Gales Alexandre João de Gales Casa Saxe-Coburgo-Gota Nome completo Alberto Eduardo Nascimento 9 de novembro de 1841   Palácio de Buckingham, Londres, Reino Unido Morte 6 de maio de 1910 (68 anos)   Palácio de Buckingham, Londres, Reino Unido Enterro Capela de São Jorge, Windsor, Berkshire, Reino Unido 20 de maio de 1910 Pai Alberto de Saxe-Coburgo-Gota Mãe Vitória do Reino Unido Assinatura ...
                    Read more