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$k$ is odd number.
show that for arbitrary $n\in N$ , $1^{k}+2^{k}+\cdots+n^{k}$ is not divisible by $n+2$

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As $k$ is odd, $r^k+(n+2-r)^k$ is divisible by $r+n+2-r=n+2$

Putting $r=1,2,\cdots,n,n+1$ and adding we get,

$2\{1^k+2^k+\cdots +n^k+(n+1)^k\}$ is divisible by $n+2$

If $(n+2)$ divides $(1^k+2^k+\cdots +n^k),$

it will divide $2\{1^k+2^k+\cdots +n^k+(n+1)^k\}-2(1^k+2^k+\cdots +n^k)=2(n+1)^k$

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This is just a suggestion, nowhere near a complete answer. For the special case $k=3$, $1^3+2^3+3^3+4^3+ \cdot \cdot \cdot=(1+2+3+4+ \cdot \cdot \cdot)^2$. Maybe you could prove it for this case and then use induction.

Every perfect cube is the difference of two perfect squares?

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