Why for $k$ odd, $(1+2+…+n)$ divides $(1^k+2^k+…n^k)$? [duplicate]

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  • Showing that $1^k+2^k + dots + n^k$ is divisible by $n(n+1)over 2$

    3 answers



The question asks how to prove this statement. I have been unable to prove by contradiction and by induction for the same $k$.




elementary-number-theory summation




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marked as duplicate by Dietrich Burde, 6005, barto, Xander Henderson, Key Flex 20 mins ago


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  • Indeed, the RHS is a polynomial in the LHS for odd $k$.
    – mvw
    6 hours ago










  • @mvw Really? Do you know where I can find a proof of that?
    – Batominovski
    6 hours ago










  • Look for Faulhaber polynomials.
    – mvw
    6 hours ago














up vote
3
down vote

favorite
2













This question already has an answer here:



  • Showing that $1^k+2^k + dots + n^k$ is divisible by $n(n+1)over 2$

    3 answers



The question asks how to prove this statement. I have been unable to prove by contradiction and by induction for the same $k$.




elementary-number-theory summation




share|cite|improve this question













marked as duplicate by Dietrich Burde, 6005, barto, Xander Henderson, Key Flex 20 mins ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • Indeed, the RHS is a polynomial in the LHS for odd $k$.
    – mvw
    6 hours ago










  • @mvw Really? Do you know where I can find a proof of that?
    – Batominovski
    6 hours ago










  • Look for Faulhaber polynomials.
    – mvw
    6 hours ago












up vote
3
down vote

favorite
2









up vote
3
down vote

favorite
2






2






This question already has an answer here:



  • Showing that $1^k+2^k + dots + n^k$ is divisible by $n(n+1)over 2$

    3 answers



The question asks how to prove this statement. I have been unable to prove by contradiction and by induction for the same $k$.




elementary-number-theory summation




share|cite|improve this question














This question already has an answer here:



  • Showing that $1^k+2^k + dots + n^k$ is divisible by $n(n+1)over 2$

    3 answers



The question asks how to prove this statement. I have been unable to prove by contradiction and by induction for the same $k$.





This question already has an answer here:



  • Showing that $1^k+2^k + dots + n^k$ is divisible by $n(n+1)over 2$

    3 answers





elementary-number-theory summation





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited 6 hours ago









6005

34.4k749122




34.4k749122









asked 6 hours ago









Alexandre Tourinho

1305




1305




marked as duplicate by Dietrich Burde, 6005, barto, Xander Henderson, Key Flex 20 mins ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Dietrich Burde, 6005, barto, Xander Henderson, Key Flex 20 mins ago


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.













  • Indeed, the RHS is a polynomial in the LHS for odd $k$.
    – mvw
    6 hours ago










  • @mvw Really? Do you know where I can find a proof of that?
    – Batominovski
    6 hours ago










  • Look for Faulhaber polynomials.
    – mvw
    6 hours ago
















  • Indeed, the RHS is a polynomial in the LHS for odd $k$.
    – mvw
    6 hours ago










  • @mvw Really? Do you know where I can find a proof of that?
    – Batominovski
    6 hours ago










  • Look for Faulhaber polynomials.
    – mvw
    6 hours ago















Indeed, the RHS is a polynomial in the LHS for odd $k$.
– mvw
6 hours ago




Indeed, the RHS is a polynomial in the LHS for odd $k$.
– mvw
6 hours ago












@mvw Really? Do you know where I can find a proof of that?
– Batominovski
6 hours ago




@mvw Really? Do you know where I can find a proof of that?
– Batominovski
6 hours ago












Look for Faulhaber polynomials.
– mvw
6 hours ago




Look for Faulhaber polynomials.
– mvw
6 hours ago










1 Answer
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Since $1+2+...+n = dfracn(n+1)2$, you can prove separately that $S_k = 1^k+2^k+...+n^k$ is divisible by both $n$ and $dfracn+12$ for $n$ odd, $n/2$ and $n+1$ for $n$ even.



For instance, when $n$ is odd, prove that $n+1|a^k+(n-a+1)^k$ and then partition $S_k$ in such a way that every summand is divisible by $n+1.$ The rest is pretty much similar for the different cases.



You can also use a more powerful apparatus called Bernoulli numbers and how they are related to power sums. In fact, you can prove that $S_k$ is a polynomial of $S_1$ with rational coefficients.






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    1 Answer
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    1 Answer
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    active

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    active

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    active

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    up vote
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    Since $1+2+...+n = dfracn(n+1)2$, you can prove separately that $S_k = 1^k+2^k+...+n^k$ is divisible by both $n$ and $dfracn+12$ for $n$ odd, $n/2$ and $n+1$ for $n$ even.



    For instance, when $n$ is odd, prove that $n+1|a^k+(n-a+1)^k$ and then partition $S_k$ in such a way that every summand is divisible by $n+1.$ The rest is pretty much similar for the different cases.



    You can also use a more powerful apparatus called Bernoulli numbers and how they are related to power sums. In fact, you can prove that $S_k$ is a polynomial of $S_1$ with rational coefficients.






    share|cite|improve this answer



























      up vote
      2
      down vote













      Since $1+2+...+n = dfracn(n+1)2$, you can prove separately that $S_k = 1^k+2^k+...+n^k$ is divisible by both $n$ and $dfracn+12$ for $n$ odd, $n/2$ and $n+1$ for $n$ even.



      For instance, when $n$ is odd, prove that $n+1|a^k+(n-a+1)^k$ and then partition $S_k$ in such a way that every summand is divisible by $n+1.$ The rest is pretty much similar for the different cases.



      You can also use a more powerful apparatus called Bernoulli numbers and how they are related to power sums. In fact, you can prove that $S_k$ is a polynomial of $S_1$ with rational coefficients.






      share|cite|improve this answer

























        up vote
        2
        down vote










        up vote
        2
        down vote









        Since $1+2+...+n = dfracn(n+1)2$, you can prove separately that $S_k = 1^k+2^k+...+n^k$ is divisible by both $n$ and $dfracn+12$ for $n$ odd, $n/2$ and $n+1$ for $n$ even.



        For instance, when $n$ is odd, prove that $n+1|a^k+(n-a+1)^k$ and then partition $S_k$ in such a way that every summand is divisible by $n+1.$ The rest is pretty much similar for the different cases.



        You can also use a more powerful apparatus called Bernoulli numbers and how they are related to power sums. In fact, you can prove that $S_k$ is a polynomial of $S_1$ with rational coefficients.






        share|cite|improve this answer















        Since $1+2+...+n = dfracn(n+1)2$, you can prove separately that $S_k = 1^k+2^k+...+n^k$ is divisible by both $n$ and $dfracn+12$ for $n$ odd, $n/2$ and $n+1$ for $n$ even.



        For instance, when $n$ is odd, prove that $n+1|a^k+(n-a+1)^k$ and then partition $S_k$ in such a way that every summand is divisible by $n+1.$ The rest is pretty much similar for the different cases.



        You can also use a more powerful apparatus called Bernoulli numbers and how they are related to power sums. In fact, you can prove that $S_k$ is a polynomial of $S_1$ with rational coefficients.







        share|cite|improve this answer















        share|cite|improve this answer



        share|cite|improve this answer








        edited 6 hours ago


























        answered 6 hours ago









        dezdichado

        5,4411826




        5,4411826












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