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












1












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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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    $begingroup$


    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.
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      asked 1 hour ago









      JayJay

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      New contributor




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      New contributor





      Jay is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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          2 Answers
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          $begingroup$

          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
              45 mins ago











            Your Answer





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            2 Answers
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            active

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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









            $endgroup$


















              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























                    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
                      45 mins 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
                      45 mins 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 45 mins ago

























                    answered 1 hour 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
                      45 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
                      45 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
                    45 mins ago




                    $begingroup$
                    @JohnOmielan Edited.
                    $endgroup$
                    – tatan
                    45 mins ago










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