<s>[INST] Show that if $m,n$ are positive integers, then $1^m+2^m+\cdots+(n-2)^m+(n-1)^m$ is divisible by $n$. Show that if $m,n$ are positive integers and $m$ is odd, then $1^m+2^m+\cdots+(n-2)^m+(n-1)^m$ is divisible by $n$.
(Hint: Let $s=1^m+2^m+\cdots+(n-2)^m+(n-1)^m$. Obviously $s=(n-1)^m+(n-2)^m+\cdots+2^m+1^m$.
Consider these relations as equivalent ${}\bmod n$ and add them.)
$$1^m+2^m+\cdots+(n-2)^m+(n-1)^m=((n-1)^m+(n-2)^m+\cdots+2^m+1^m)\bmod n$$
$$(n-1)^m+(n-2)^m+\cdots+2^m+1^m=(1^m+2^m+\cdots+(n-2)^m+(n-1)^m)\bmod n$$
By adding them we get:
$$(1^m+(n-1)^m)+(2^m+(n-2)^m)+\cdots+((n-2)^m+2^m)+((n-1)^m+1^m)=((n-1)^m+1^m)+((n-2)^m+2^m)+\cdots+(2^m+(n-2)^m)+(1^m+(n-1)^m))\bmod n$$
That means that:
$$n|[(1^m+(n-1)^m)+(2^m+(n-2)^m)+\cdots+((n-2)^m+2^m)+((n-1)^m+1^m)]-[((n-1)^m+1^m)+((n-2)^m+2^m)+\cdots+(2^m+(n-2)^m)+(1^m+(n-1)^m))]$$
Or not?? 
And that is equal to $$n\mid 0$$
Is this correct? How can I continue??
 [/INST] This is not true. Let $m=2$ and $n=3$. Then $3$ does not divide $1^2+2^2$. 
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