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 1 hour 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









          $endgroup$





















            1












            $begingroup$

            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
              43 mins 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 1 hour ago









                John OmielanJohn Omielan

                2,046210




                2,046210























                    1












                    $begingroup$

                    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
                      43 mins ago
















                    1












                    $begingroup$

                    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
                      43 mins ago














                    1












                    1








                    1





                    $begingroup$

                    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 43 mins ago

























                    answered 1 hour ago









                    tatantatan

                    5,69062758




                    5,69062758












                    • $begingroup$
                      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
                      43 mins 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
                      43 mins 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
                    43 mins ago




                    $begingroup$
                    @JohnOmielan Edited.
                    $endgroup$
                    – tatan
                    43 mins 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

                    What other Star Trek series did the main TNG cast show up in?

                    Image
                    .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0; } 4 I'm referring to Picard, Riker, Data, Worf, Geordi, Dr. Crusher, Wesley Crusher, Chief O'Brien and family, Deanna Troi, Guinan. I know Picard and Worf show up on DS9, though I haven't seen it myself. I'm curious what happened to the characters after that last episode of TNG? star-trek star-trek-tng share | improve this question edited 10 hours ago Ham Sandwich 6,051 2 25 75 ...
                    Read more

                    Berlina muro

                    Image
                    La berlina muro ĉe Bethaniendamm, 1986 La berlina muro aŭ simple la muro [1] , nomita ankaŭ "kontraŭfaŝisma remparo" en propaganda lingvo de la iama Germana Demokratia Respubliko, estis parto de la interngermana limo, kiu dividis de la 13-a de aŭgusto 1961 ĝis la 9-a de novembro 1989 okcidentan Berlinon de la orienta parto de la urbo kaj de la ĉirkaŭa tereno de la GDR. Ĝi estis unu el la plej famaj simboloj de la malvarma milito kaj de la disdivido de Germanio. Pli ol simpla muro, la berlina muro estis fakte tuta defendokonstruaĵo kun du muroj, ena malplena strio, rondirvojo, gardturoj kaj alarmsistemoj. Dum la provo transiri la severe garditan limon al okcidenta Berlino multaj homoj estis mortigitaj. La preciza nombro de viktimoj estas pridisputita kaj ne certa; la nombroj anoncataj varias inter 86 kaj 238 mortintoj. La malplifortiĝo de Sovetunio kaj la politiko de liberigo gvidita de Miĥail Gorbaĉov ebligis al la orientgermanoj faligi la 9-an de novembro 1989 l...
                    Read more

                    Berlina aerponto

                    Image
                    Flugkoridoroj dum la blokado de Berlino La Berlina aerponto estas la provizigo de okcidenta Berlino per aviadiloj fare de la aliancanoj inter la 23-a de junio 1948 kaj la 12-a de majo 1949 dum la blokado de Berlino fare de la sovetia armeo. Enhavo 1 Historio 1.1 La aerkoridoroj 1.2 Unuaj flugoj 1.3 Plibonigo de la aerponto 1.4 Fino de la aerponto 1.5 Rozinbombiloj 1.6 Bilanco 2 Flughavenoj 2.1 Berlino 2.2 Okcidenta Germanio 3 Aviadiloj de la berlina aerponto 4 15-a datreveno de la aerponto 5 Bibliografio 6 Referencoj 7 Eksteraj ligiloj Historio | General Lucius D. Clay, militguberniestro de la usona okupadzono Jam antaŭ la berlina aerponto estiĝis la malgranda aerponto . La estro de la sovetia militadministracio en Germanio, Vasili Daniloviĉ Sokolovski, ordonis la 1-an de aprilo 1948, kiel reago al la konferenco de Londono inter britoj, francoj kaj reprezentantoj de Benelukso pri aligo de Germa...
                    Read more