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Let $n$ be an odd positive integer. Show that the sum $1^n$ + $2^n$ + · · · + $n^n$ is divisible by $n^2$.

I tried induction on $n$ and thought of manipulating the terms by separating for example $3^{2n+3}$ into $3^{2n+1}\cdot 3^2$... then thought the sum of the squares of integers would play in to showing the inductive proof step for $n=2k+3$.

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Using the binomial theorem, note that $k^n+(n-k)^n$ is congruent to $k^n+(-1)^nk^n$ modulo $n^2$, and since $n$ is odd, this is zero. Thus, pair off all terms, except the last one which is divisible by $n^2$.

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    $\begingroup$ Great answer! :) $\endgroup$ – abc... Jun 18 '18 at 5:04
  • $\begingroup$ I still dont get it can u explain ? $\endgroup$ – Randin Jun 19 '18 at 5:01
  • $\begingroup$ @RandinMichaelDivelbiss $1^n \equiv -(n-1)^n \pmod {n^2}$, $2^n \equiv -(n-2)^n \pmod {n^2}$, and so on, because $n$ is odd, and because $(n-k)^n = (\text{terms having n with power }\ge 2) + (-1)^{n-1}n^2k^{n-1} + (-1)^nk^n$, due to which only the last term remains. So, all such pairs cancel out and only $n^n \pmod {n^2}$ is left, which is $0$. $\endgroup$ – GoodDeeds Jun 19 '18 at 5:10
  • $\begingroup$ Darn i still dont get it can u gmail me ? GoodDeeds12? $\endgroup$ – Randin Jun 19 '18 at 17:34
  • $\begingroup$ I got it ..should have posted this weeks ago ..right after i asked if u could email me it saw it ! 💡💡💡💡💡 $\endgroup$ – Randin Dec 20 '18 at 7:53

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