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







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    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|improve this question









    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|improve this question









      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|improve this question









      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|improve this question









      New contributor




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      share|cite|improve this question




      share|cite|improve this question








      edited 3 mins ago









      J. W. Tanner

      956113




      956113






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      asked 2 hours ago









      JayJay

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      61




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






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          2 Answers
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          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
              2 hours ago










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











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            2 Answers
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            active

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            2 Answers
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            active

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            active

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            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
                      2 hours 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
                      2 hours 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
                      2 hours 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
                      2 hours 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
                    2 hours 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
                    2 hours ago












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




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










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