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Let $S = 1^N + 2^N + \dots + N^N$. Show that $S \ \text{mod} \ N = 0$ for any odd $N$. What would be a good way to start this problem?

ironman123
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  • What means $S \mod N = 0$? Maybe you wanted to write $S\equiv 0\mod N$? You can start to see how the residues are structured. – Masacroso Mar 31 '16 at 03:30

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Hint: Write $S=0^N + 1^N + \cdots +N^N$ and consider pairs $(k,N-k)$ for $0 \le k < N/2$.

lhf
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