$ a^{n}+b^{n} $ prime. How to show that n is a power of two?












3














Let $ a,b in mathbb{N} setminus {0,1}$ and $n in mathbb{N^{ast }}. $
Let $a^{n}+b^{n} $ prime. How to show that n is a power of two ? I know it works if we suppose that $n=2k(2p+1)$ and then show that it implies $ a^{n}+b^{n} = 0 $ but can't do it.



Sorry, I have a little bit of trouble with LaTex too.






elementary-number-theory arithmetic







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  • Hint: if $n=2^km$ where $m$ is odd then $a^n+b^n$ is divisible by $a^{2^k}+b^{2^k}$. Of course you have to handle the case where the quotient is $1$, as in $2=1^3+1^3$.
    – lulu
    Jul 8 '17 at 14:05
















3














Let $ a,b in mathbb{N} setminus {0,1}$ and $n in mathbb{N^{ast }}. $
Let $a^{n}+b^{n} $ prime. How to show that n is a power of two ? I know it works if we suppose that $n=2k(2p+1)$ and then show that it implies $ a^{n}+b^{n} = 0 $ but can't do it.



Sorry, I have a little bit of trouble with LaTex too.






elementary-number-theory arithmetic







share|cite|improve this question
























  • Hint: if $n=2^km$ where $m$ is odd then $a^n+b^n$ is divisible by $a^{2^k}+b^{2^k}$. Of course you have to handle the case where the quotient is $1$, as in $2=1^3+1^3$.
    – lulu
    Jul 8 '17 at 14:05














3












3








3


1





Let $ a,b in mathbb{N} setminus {0,1}$ and $n in mathbb{N^{ast }}. $
Let $a^{n}+b^{n} $ prime. How to show that n is a power of two ? I know it works if we suppose that $n=2k(2p+1)$ and then show that it implies $ a^{n}+b^{n} = 0 $ but can't do it.



Sorry, I have a little bit of trouble with LaTex too.






elementary-number-theory arithmetic







share|cite|improve this question















Let $ a,b in mathbb{N} setminus {0,1}$ and $n in mathbb{N^{ast }}. $
Let $a^{n}+b^{n} $ prime. How to show that n is a power of two ? I know it works if we suppose that $n=2k(2p+1)$ and then show that it implies $ a^{n}+b^{n} = 0 $ but can't do it.



Sorry, I have a little bit of trouble with LaTex too.






elementary-number-theory arithmetic




elementary-number-theory arithmetic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jul 8 '17 at 14:14









MathOverview

8,53643163




8,53643163










asked Jul 8 '17 at 13:56









ClipperClipper

192




192












  • Hint: if $n=2^km$ where $m$ is odd then $a^n+b^n$ is divisible by $a^{2^k}+b^{2^k}$. Of course you have to handle the case where the quotient is $1$, as in $2=1^3+1^3$.
    – lulu
    Jul 8 '17 at 14:05


















  • Hint: if $n=2^km$ where $m$ is odd then $a^n+b^n$ is divisible by $a^{2^k}+b^{2^k}$. Of course you have to handle the case where the quotient is $1$, as in $2=1^3+1^3$.
    – lulu
    Jul 8 '17 at 14:05
















Hint: if $n=2^km$ where $m$ is odd then $a^n+b^n$ is divisible by $a^{2^k}+b^{2^k}$. Of course you have to handle the case where the quotient is $1$, as in $2=1^3+1^3$.
– lulu
Jul 8 '17 at 14:05




Hint: if $n=2^km$ where $m$ is odd then $a^n+b^n$ is divisible by $a^{2^k}+b^{2^k}$. Of course you have to handle the case where the quotient is $1$, as in $2=1^3+1^3$.
– lulu
Jul 8 '17 at 14:05










2 Answers
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Try to show the following:



Let $n=l2^k$, where $l$ is co prime to $2$ (in other words, it's odd). Then, it is infact true that $a^n+b^n$ is a multiple of $a^{2^k} + b^{2^k}$.



If you can't then look at this box:




Let $d = a^{2^k}$ and $e=b^{2^k}$, then this is basically saying that $d^l + e^l$ is a multiple of $d+e$ whenever $l$ is odd.
Which is true, since $dfrac{d^l+e^l}{d+e} = displaystylesum_{k=0}^{l-1} (-1)^ka^kb^{l-1-k}$. Put back $d$ and $e$ in terms of $a$ and $b$ to get the proof.




Therefore, unless $l=1$, the expression $a^n+b^n$ always has a non-trivial factor. Infact, we can even figure out what it is from the above formula.



For example, $7^6+2^6 = 117713 = 53 times 2221$, and $53 = 7^2+2^2$.



This proves your statement.






share|cite|improve this answer































    1














    Hint If $n=(2k+1)m$ with $kgeq 1$ odd then
    $$a^n+b^n=(a^m)^{2k+1}+(b^m)^{2k+1}=(a^m+b^m)(a^{2km}-a^{(2k-1)m}b^m+...+b^{2km})$$






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

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      2














      Try to show the following:



      Let $n=l2^k$, where $l$ is co prime to $2$ (in other words, it's odd). Then, it is infact true that $a^n+b^n$ is a multiple of $a^{2^k} + b^{2^k}$.



      If you can't then look at this box:




      Let $d = a^{2^k}$ and $e=b^{2^k}$, then this is basically saying that $d^l + e^l$ is a multiple of $d+e$ whenever $l$ is odd.
      Which is true, since $dfrac{d^l+e^l}{d+e} = displaystylesum_{k=0}^{l-1} (-1)^ka^kb^{l-1-k}$. Put back $d$ and $e$ in terms of $a$ and $b$ to get the proof.




      Therefore, unless $l=1$, the expression $a^n+b^n$ always has a non-trivial factor. Infact, we can even figure out what it is from the above formula.



      For example, $7^6+2^6 = 117713 = 53 times 2221$, and $53 = 7^2+2^2$.



      This proves your statement.






      share|cite|improve this answer




























        2














        Try to show the following:



        Let $n=l2^k$, where $l$ is co prime to $2$ (in other words, it's odd). Then, it is infact true that $a^n+b^n$ is a multiple of $a^{2^k} + b^{2^k}$.



        If you can't then look at this box:




        Let $d = a^{2^k}$ and $e=b^{2^k}$, then this is basically saying that $d^l + e^l$ is a multiple of $d+e$ whenever $l$ is odd.
        Which is true, since $dfrac{d^l+e^l}{d+e} = displaystylesum_{k=0}^{l-1} (-1)^ka^kb^{l-1-k}$. Put back $d$ and $e$ in terms of $a$ and $b$ to get the proof.




        Therefore, unless $l=1$, the expression $a^n+b^n$ always has a non-trivial factor. Infact, we can even figure out what it is from the above formula.



        For example, $7^6+2^6 = 117713 = 53 times 2221$, and $53 = 7^2+2^2$.



        This proves your statement.






        share|cite|improve this answer


























          2












          2








          2






          Try to show the following:



          Let $n=l2^k$, where $l$ is co prime to $2$ (in other words, it's odd). Then, it is infact true that $a^n+b^n$ is a multiple of $a^{2^k} + b^{2^k}$.



          If you can't then look at this box:




          Let $d = a^{2^k}$ and $e=b^{2^k}$, then this is basically saying that $d^l + e^l$ is a multiple of $d+e$ whenever $l$ is odd.
          Which is true, since $dfrac{d^l+e^l}{d+e} = displaystylesum_{k=0}^{l-1} (-1)^ka^kb^{l-1-k}$. Put back $d$ and $e$ in terms of $a$ and $b$ to get the proof.




          Therefore, unless $l=1$, the expression $a^n+b^n$ always has a non-trivial factor. Infact, we can even figure out what it is from the above formula.



          For example, $7^6+2^6 = 117713 = 53 times 2221$, and $53 = 7^2+2^2$.



          This proves your statement.






          share|cite|improve this answer














          Try to show the following:



          Let $n=l2^k$, where $l$ is co prime to $2$ (in other words, it's odd). Then, it is infact true that $a^n+b^n$ is a multiple of $a^{2^k} + b^{2^k}$.



          If you can't then look at this box:




          Let $d = a^{2^k}$ and $e=b^{2^k}$, then this is basically saying that $d^l + e^l$ is a multiple of $d+e$ whenever $l$ is odd.
          Which is true, since $dfrac{d^l+e^l}{d+e} = displaystylesum_{k=0}^{l-1} (-1)^ka^kb^{l-1-k}$. Put back $d$ and $e$ in terms of $a$ and $b$ to get the proof.




          Therefore, unless $l=1$, the expression $a^n+b^n$ always has a non-trivial factor. Infact, we can even figure out what it is from the above formula.



          For example, $7^6+2^6 = 117713 = 53 times 2221$, and $53 = 7^2+2^2$.



          This proves your statement.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 4 at 5:50

























          answered Jul 8 '17 at 14:21









          астон вілла олоф мэллбэргастон вілла олоф мэллбэрг

          37.4k33376




          37.4k33376























              1














              Hint If $n=(2k+1)m$ with $kgeq 1$ odd then
              $$a^n+b^n=(a^m)^{2k+1}+(b^m)^{2k+1}=(a^m+b^m)(a^{2km}-a^{(2k-1)m}b^m+...+b^{2km})$$






              share|cite|improve this answer


























                1














                Hint If $n=(2k+1)m$ with $kgeq 1$ odd then
                $$a^n+b^n=(a^m)^{2k+1}+(b^m)^{2k+1}=(a^m+b^m)(a^{2km}-a^{(2k-1)m}b^m+...+b^{2km})$$






                share|cite|improve this answer
























                  1












                  1








                  1






                  Hint If $n=(2k+1)m$ with $kgeq 1$ odd then
                  $$a^n+b^n=(a^m)^{2k+1}+(b^m)^{2k+1}=(a^m+b^m)(a^{2km}-a^{(2k-1)m}b^m+...+b^{2km})$$






                  share|cite|improve this answer












                  Hint If $n=(2k+1)m$ with $kgeq 1$ odd then
                  $$a^n+b^n=(a^m)^{2k+1}+(b^m)^{2k+1}=(a^m+b^m)(a^{2km}-a^{(2k-1)m}b^m+...+b^{2km})$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jul 8 '17 at 14:53









                  N. S.N. S.

                  102k5110206




                  102k5110206






























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