Take a cube, add one, when is this a prime?

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up vote
2
down vote

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Take
$$
p = n^3 + 1
$$



I just realised, checking some numbers, that this only occurs for $n=1$. I was wondering if there is some general proof for this?






algebra-precalculus elementary-number-theory polynomials proof-writing prime-numbers







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  • 8




    $n^3+1=(n+1)(n^2-n+1)$.
    – Lord Shark the Unknown
    3 hours ago














up vote
2
down vote

favorite












Take
$$
p = n^3 + 1
$$



I just realised, checking some numbers, that this only occurs for $n=1$. I was wondering if there is some general proof for this?






algebra-precalculus elementary-number-theory polynomials proof-writing prime-numbers







share|cite|improve this question



















  • 8




    $n^3+1=(n+1)(n^2-n+1)$.
    – Lord Shark the Unknown
    3 hours ago












up vote
2
down vote

favorite









up vote
2
down vote

favorite











Take
$$
p = n^3 + 1
$$



I just realised, checking some numbers, that this only occurs for $n=1$. I was wondering if there is some general proof for this?






algebra-precalculus elementary-number-theory polynomials proof-writing prime-numbers







share|cite|improve this question















Take
$$
p = n^3 + 1
$$



I just realised, checking some numbers, that this only occurs for $n=1$. I was wondering if there is some general proof for this?






algebra-precalculus elementary-number-theory polynomials proof-writing prime-numbers




algebra-precalculus elementary-number-theory polynomials proof-writing prime-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 11 mins ago









greedoid

28.3k93776




28.3k93776










asked 3 hours ago









John Miller

1136




1136







  • 8




    $n^3+1=(n+1)(n^2-n+1)$.
    – Lord Shark the Unknown
    3 hours ago












  • 8




    $n^3+1=(n+1)(n^2-n+1)$.
    – Lord Shark the Unknown
    3 hours ago







8




8




$n^3+1=(n+1)(n^2-n+1)$.
– Lord Shark the Unknown
3 hours ago




$n^3+1=(n+1)(n^2-n+1)$.
– Lord Shark the Unknown
3 hours ago










2 Answers
2






active

oldest

votes

















up vote
6
down vote



accepted










You can factorize $n^3+1$, so: $$ p= (n+1)(n^2-n+1)$$ so $n+1 = 1$ (so $n=0$) or $n^2-n+1=1$ (so $n(n-1)=0$...)






share|cite|improve this answer






















  • As an extension to this, $n^2 - 1$ is also never a prime unless $n=2$. Is there a general proof for $p = n^y + z$, more specifically between $y$ and $z$.
    – John Miller
    3 hours ago






  • 1




    Heck of a "hint".
    – fleablood
    3 hours ago

















up vote
2
down vote













You can make your understanding stronger by the fact for any odd number $k > 1$, the number $p = n^k + 1$ is never a prime because it is divisible by $n+1$. The cubes or $k = 3$ is a special case of this.






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    2 Answers
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    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    6
    down vote



    accepted










    You can factorize $n^3+1$, so: $$ p= (n+1)(n^2-n+1)$$ so $n+1 = 1$ (so $n=0$) or $n^2-n+1=1$ (so $n(n-1)=0$...)






    share|cite|improve this answer






















    • As an extension to this, $n^2 - 1$ is also never a prime unless $n=2$. Is there a general proof for $p = n^y + z$, more specifically between $y$ and $z$.
      – John Miller
      3 hours ago






    • 1




      Heck of a "hint".
      – fleablood
      3 hours ago














    up vote
    6
    down vote



    accepted










    You can factorize $n^3+1$, so: $$ p= (n+1)(n^2-n+1)$$ so $n+1 = 1$ (so $n=0$) or $n^2-n+1=1$ (so $n(n-1)=0$...)






    share|cite|improve this answer






















    • As an extension to this, $n^2 - 1$ is also never a prime unless $n=2$. Is there a general proof for $p = n^y + z$, more specifically between $y$ and $z$.
      – John Miller
      3 hours ago






    • 1




      Heck of a "hint".
      – fleablood
      3 hours ago












    up vote
    6
    down vote



    accepted







    up vote
    6
    down vote



    accepted






    You can factorize $n^3+1$, so: $$ p= (n+1)(n^2-n+1)$$ so $n+1 = 1$ (so $n=0$) or $n^2-n+1=1$ (so $n(n-1)=0$...)






    share|cite|improve this answer














    You can factorize $n^3+1$, so: $$ p= (n+1)(n^2-n+1)$$ so $n+1 = 1$ (so $n=0$) or $n^2-n+1=1$ (so $n(n-1)=0$...)







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 1 hour ago

























    answered 3 hours ago









    greedoid

    28.3k93776




    28.3k93776











    • As an extension to this, $n^2 - 1$ is also never a prime unless $n=2$. Is there a general proof for $p = n^y + z$, more specifically between $y$ and $z$.
      – John Miller
      3 hours ago






    • 1




      Heck of a "hint".
      – fleablood
      3 hours ago
















    • As an extension to this, $n^2 - 1$ is also never a prime unless $n=2$. Is there a general proof for $p = n^y + z$, more specifically between $y$ and $z$.
      – John Miller
      3 hours ago






    • 1




      Heck of a "hint".
      – fleablood
      3 hours ago















    As an extension to this, $n^2 - 1$ is also never a prime unless $n=2$. Is there a general proof for $p = n^y + z$, more specifically between $y$ and $z$.
    – John Miller
    3 hours ago




    As an extension to this, $n^2 - 1$ is also never a prime unless $n=2$. Is there a general proof for $p = n^y + z$, more specifically between $y$ and $z$.
    – John Miller
    3 hours ago




    1




    1




    Heck of a "hint".
    – fleablood
    3 hours ago




    Heck of a "hint".
    – fleablood
    3 hours ago










    up vote
    2
    down vote













    You can make your understanding stronger by the fact for any odd number $k > 1$, the number $p = n^k + 1$ is never a prime because it is divisible by $n+1$. The cubes or $k = 3$ is a special case of this.






    share|cite|improve this answer
























      up vote
      2
      down vote













      You can make your understanding stronger by the fact for any odd number $k > 1$, the number $p = n^k + 1$ is never a prime because it is divisible by $n+1$. The cubes or $k = 3$ is a special case of this.






      share|cite|improve this answer






















        up vote
        2
        down vote










        up vote
        2
        down vote









        You can make your understanding stronger by the fact for any odd number $k > 1$, the number $p = n^k + 1$ is never a prime because it is divisible by $n+1$. The cubes or $k = 3$ is a special case of this.






        share|cite|improve this answer












        You can make your understanding stronger by the fact for any odd number $k > 1$, the number $p = n^k + 1$ is never a prime because it is divisible by $n+1$. The cubes or $k = 3$ is a special case of this.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 3 hours ago









        Nilotpal Kanti Sinha

        3,49621333




        3,49621333



























             

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