Prove $n$ having to be an exponent of 2 for $b^n + 1 =$ a prime number












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I have been having problems finding a solution for this problem and honestly have no ideas left how to solve this, please help.



Assume that $b^n + 1= $ a prime number for some integers $b,n$ where $b>1$ and $n>1$. Prove that $n$ must take the form $n=2^k$ for some positive integers $k$



What i have been looking at are the generalized Fermat numbers due to $b^n + 1$ taking the form ${b^2}^k + 1$ for $n=2^k$ and Fermat's little theorem but seemingly without any progress.






number-theory prime-numbers fermat-numbers







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  • $begingroup$
    See this question, three days ago.
    $endgroup$
    – Dietrich Burde
    Jan 30 at 18:01
















2












$begingroup$


I have been having problems finding a solution for this problem and honestly have no ideas left how to solve this, please help.



Assume that $b^n + 1= $ a prime number for some integers $b,n$ where $b>1$ and $n>1$. Prove that $n$ must take the form $n=2^k$ for some positive integers $k$



What i have been looking at are the generalized Fermat numbers due to $b^n + 1$ taking the form ${b^2}^k + 1$ for $n=2^k$ and Fermat's little theorem but seemingly without any progress.






number-theory prime-numbers fermat-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    See this question, three days ago.
    $endgroup$
    – Dietrich Burde
    Jan 30 at 18:01














2












2








2





$begingroup$


I have been having problems finding a solution for this problem and honestly have no ideas left how to solve this, please help.



Assume that $b^n + 1= $ a prime number for some integers $b,n$ where $b>1$ and $n>1$. Prove that $n$ must take the form $n=2^k$ for some positive integers $k$



What i have been looking at are the generalized Fermat numbers due to $b^n + 1$ taking the form ${b^2}^k + 1$ for $n=2^k$ and Fermat's little theorem but seemingly without any progress.






number-theory prime-numbers fermat-numbers







share|cite|improve this question











$endgroup$




I have been having problems finding a solution for this problem and honestly have no ideas left how to solve this, please help.



Assume that $b^n + 1= $ a prime number for some integers $b,n$ where $b>1$ and $n>1$. Prove that $n$ must take the form $n=2^k$ for some positive integers $k$



What i have been looking at are the generalized Fermat numbers due to $b^n + 1$ taking the form ${b^2}^k + 1$ for $n=2^k$ and Fermat's little theorem but seemingly without any progress.






number-theory prime-numbers fermat-numbers




number-theory prime-numbers fermat-numbers






share|cite|improve this question















share|cite|improve this question













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edited Jan 30 at 0:24









TPace

5111318




5111318










asked Jan 30 at 0:12









RyxoRyxo

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204












  • $begingroup$
    See this question, three days ago.
    $endgroup$
    – Dietrich Burde
    Jan 30 at 18:01


















  • $begingroup$
    See this question, three days ago.
    $endgroup$
    – Dietrich Burde
    Jan 30 at 18:01
















$begingroup$
See this question, three days ago.
$endgroup$
– Dietrich Burde
Jan 30 at 18:01




$begingroup$
See this question, three days ago.
$endgroup$
– Dietrich Burde
Jan 30 at 18:01










2 Answers
2






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If $n$ is an odd natural number then $-1$ is a root of the polynomial $P(x)=x^n+1$ and hence $x+1|x^n+1$.



Now suppose $b^n+1$ is prime and using prime decomposition write $n=2^rm$ when $m$ is odd. Then $b^n+1=(b^{2^r})^m+1$ and hence $b^{2^r}+1|b^n+1$. But we supposed $b^n+1$ is prime so from here we conclude that $b^{2^r}+1=b^n+1$. This implies $n=2^r$.






share|cite|improve this answer









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    1












    $begingroup$

    Suppose that $b^n+1$ is prime, $b>1$ and $n>1$. We shall show that $n$ has no odd factor except for $1$: this proves that $n$ is a power of $2$.



    So, let $n=st$ where $s,t$ are positive integers and $s$ is odd. You should know the factorisation
    $$x^s+1=(x+1)(x^{s-1}-x^{s-2}+cdots-x+1) .$$
    Substituting $x=b^t$ shows that $b^t+1$ is a factor of $b^n+1$. But $b^n+1$ is prime, so there are two options:
    $$b^t+1=1quadhbox{or}quad b^t+1=b^n+1 .$$
    The first is clearly impossible since $b>1$; and the second gives $t=n$, hence $s=1$. We have shown that that only possible odd factor of $n$ is $1$, and as explained above, this means that $n$ is a power of $2$.






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      2 Answers
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      2 Answers
      2






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      3












      $begingroup$

      If $n$ is an odd natural number then $-1$ is a root of the polynomial $P(x)=x^n+1$ and hence $x+1|x^n+1$.



      Now suppose $b^n+1$ is prime and using prime decomposition write $n=2^rm$ when $m$ is odd. Then $b^n+1=(b^{2^r})^m+1$ and hence $b^{2^r}+1|b^n+1$. But we supposed $b^n+1$ is prime so from here we conclude that $b^{2^r}+1=b^n+1$. This implies $n=2^r$.






      share|cite|improve this answer









      $endgroup$


















        3












        $begingroup$

        If $n$ is an odd natural number then $-1$ is a root of the polynomial $P(x)=x^n+1$ and hence $x+1|x^n+1$.



        Now suppose $b^n+1$ is prime and using prime decomposition write $n=2^rm$ when $m$ is odd. Then $b^n+1=(b^{2^r})^m+1$ and hence $b^{2^r}+1|b^n+1$. But we supposed $b^n+1$ is prime so from here we conclude that $b^{2^r}+1=b^n+1$. This implies $n=2^r$.






        share|cite|improve this answer









        $endgroup$
















          3












          3








          3





          $begingroup$

          If $n$ is an odd natural number then $-1$ is a root of the polynomial $P(x)=x^n+1$ and hence $x+1|x^n+1$.



          Now suppose $b^n+1$ is prime and using prime decomposition write $n=2^rm$ when $m$ is odd. Then $b^n+1=(b^{2^r})^m+1$ and hence $b^{2^r}+1|b^n+1$. But we supposed $b^n+1$ is prime so from here we conclude that $b^{2^r}+1=b^n+1$. This implies $n=2^r$.






          share|cite|improve this answer









          $endgroup$



          If $n$ is an odd natural number then $-1$ is a root of the polynomial $P(x)=x^n+1$ and hence $x+1|x^n+1$.



          Now suppose $b^n+1$ is prime and using prime decomposition write $n=2^rm$ when $m$ is odd. Then $b^n+1=(b^{2^r})^m+1$ and hence $b^{2^r}+1|b^n+1$. But we supposed $b^n+1$ is prime so from here we conclude that $b^{2^r}+1=b^n+1$. This implies $n=2^r$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 30 at 0:24









          MarkMark

          10.4k1622




          10.4k1622























              1












              $begingroup$

              Suppose that $b^n+1$ is prime, $b>1$ and $n>1$. We shall show that $n$ has no odd factor except for $1$: this proves that $n$ is a power of $2$.



              So, let $n=st$ where $s,t$ are positive integers and $s$ is odd. You should know the factorisation
              $$x^s+1=(x+1)(x^{s-1}-x^{s-2}+cdots-x+1) .$$
              Substituting $x=b^t$ shows that $b^t+1$ is a factor of $b^n+1$. But $b^n+1$ is prime, so there are two options:
              $$b^t+1=1quadhbox{or}quad b^t+1=b^n+1 .$$
              The first is clearly impossible since $b>1$; and the second gives $t=n$, hence $s=1$. We have shown that that only possible odd factor of $n$ is $1$, and as explained above, this means that $n$ is a power of $2$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Suppose that $b^n+1$ is prime, $b>1$ and $n>1$. We shall show that $n$ has no odd factor except for $1$: this proves that $n$ is a power of $2$.



                So, let $n=st$ where $s,t$ are positive integers and $s$ is odd. You should know the factorisation
                $$x^s+1=(x+1)(x^{s-1}-x^{s-2}+cdots-x+1) .$$
                Substituting $x=b^t$ shows that $b^t+1$ is a factor of $b^n+1$. But $b^n+1$ is prime, so there are two options:
                $$b^t+1=1quadhbox{or}quad b^t+1=b^n+1 .$$
                The first is clearly impossible since $b>1$; and the second gives $t=n$, hence $s=1$. We have shown that that only possible odd factor of $n$ is $1$, and as explained above, this means that $n$ is a power of $2$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Suppose that $b^n+1$ is prime, $b>1$ and $n>1$. We shall show that $n$ has no odd factor except for $1$: this proves that $n$ is a power of $2$.



                  So, let $n=st$ where $s,t$ are positive integers and $s$ is odd. You should know the factorisation
                  $$x^s+1=(x+1)(x^{s-1}-x^{s-2}+cdots-x+1) .$$
                  Substituting $x=b^t$ shows that $b^t+1$ is a factor of $b^n+1$. But $b^n+1$ is prime, so there are two options:
                  $$b^t+1=1quadhbox{or}quad b^t+1=b^n+1 .$$
                  The first is clearly impossible since $b>1$; and the second gives $t=n$, hence $s=1$. We have shown that that only possible odd factor of $n$ is $1$, and as explained above, this means that $n$ is a power of $2$.






                  share|cite|improve this answer









                  $endgroup$



                  Suppose that $b^n+1$ is prime, $b>1$ and $n>1$. We shall show that $n$ has no odd factor except for $1$: this proves that $n$ is a power of $2$.



                  So, let $n=st$ where $s,t$ are positive integers and $s$ is odd. You should know the factorisation
                  $$x^s+1=(x+1)(x^{s-1}-x^{s-2}+cdots-x+1) .$$
                  Substituting $x=b^t$ shows that $b^t+1$ is a factor of $b^n+1$. But $b^n+1$ is prime, so there are two options:
                  $$b^t+1=1quadhbox{or}quad b^t+1=b^n+1 .$$
                  The first is clearly impossible since $b>1$; and the second gives $t=n$, hence $s=1$. We have shown that that only possible odd factor of $n$ is $1$, and as explained above, this means that $n$ is a power of $2$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 30 at 0:27









                  DavidDavid

                  69.7k668131




                  69.7k668131






























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