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TODAY's PROBLEM

*If k is an odd positive integer, prove that for any integer $\mathbf{ n \ge 1 , 1^k + 2^k + \cdots + n^k }$ is divisible by $\mathbf{ \frac {n(n+1)}{2} }$

Solution:
We write the given expression in two ways:

$\mathbf{ S= 1^k + 2^k + \cdots + n^k }$
$\mathbf{ S =n^k + (n-1)^k + \cdots + 1^k }$

This implies

$\mathbf{ 2S = (1^k + n^k ) + (2^k + (n-1)^k ) + \cdots + (n^k + 1^k) }$
Since k is odd, we know $\mathbf{ a^k + b^k = (a+b)(a^{k-1} - a^{k-2} b + \cdots + b^{k-1}) }$ , that is $\mathbf{ a+b}$ divides $\mathbf{a^k + b^k }$

Applying this to $\mathbf{ 2S = (1^k + n^k ) + (2^k + (n-1)^k ) + \cdots + (n^k + 1^k) }$ we have (n+1) divides $\mathbf{ 1^k + n^k }$ , $\mathbf{ (2 + n-1) =(n+1) }$ divides $\mathbf{(2^k + (n-1)^k)}$ and so on. Hence we can take n+1 common from each bracket, leading us to the following expression:

$\mathbf{ 2S = (n+1)\times(\text{some constant}) \implies S = \frac{(n+1)}{2} \times(\text{some constant}) }$ . This implies $\mathbf{ \frac{(n+1)}{2} }$ divides S if n+1 is even, other wise n+1 divides S.

Now we show n (or n/2) divides S (when is odd or even respectively). To show this we write

S= $\mathbf{ 1^k + 2^k + \cdots + n^k }$
$\mathbf{ S =n^k + (n-1)^k + \cdots + 1^k }$

Again $\mathbf{ 2S =n^k + (1^k+ (n-1)^k) + (2^k + (n-2)^k) \cdots + ((n-1)^k + 1^k) + n^k }$
Since k is odd n divides $\mathbf{ a^k + (n-a)^k }$ for all a from 1 to n. Hence we can take n as common and have:

$\mathbf{ 2S = (n)\times(\text{some constant}) \implies S = \frac{(n)}{2} \times(\text{some constant}) }$ . This implies $\mathbf{ \frac{(n)}{2} }$ divides S if n is even, other wise n divides S.