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

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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

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edited 6 hours ago

6005

34.4k749122

asked 6 hours ago

Alexandre Tourinho

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.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  

  
  
  
  

add a comment | 

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

edited 6 hours ago

6005

34.4k749122

asked 6 hours ago

Alexandre Tourinho

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.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  

  
  
  
  

add a comment | 

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

edited 6 hours ago

6005

34.4k749122

asked 6 hours ago

Alexandre Tourinho

1305

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

edited 6 hours ago

6005

34.4k749122

asked 6 hours ago

Alexandre Tourinho

1305

share|cite|improve this question

share|cite|improve this question

edited 6 hours ago

6005

34.4k749122

edited 6 hours ago

6005

34.4k749122

edited 6 hours ago

6005

34.4k749122

34.4k749122

asked 6 hours ago

Alexandre Tourinho

1305

asked 6 hours ago

Alexandre Tourinho

1305

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
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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

add a comment | 

1 Answer
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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.

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edited 6 hours ago

answered 6 hours ago

dezdichado

5,4411826

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

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

edited 6 hours ago

answered 6 hours ago

dezdichado

5,4411826

add a comment | 

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

edited 6 hours ago

answered 6 hours ago

dezdichado

5,4411826

add a comment | 

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

edited 6 hours ago

answered 6 hours ago

dezdichado

5,4411826

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

edited 6 hours ago

answered 6 hours ago

dezdichado

5,4411826

share|cite|improve this answer

share|cite|improve this answer

edited 6 hours ago

edited 6 hours ago

edited 6 hours ago

answered 6 hours ago

dezdichado

5,4411826

answered 6 hours ago

dezdichado

5,4411826

answered 6 hours ago

dezdichado

5,4411826

5,4411826

add a comment | 

add a comment |