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I am working in some questions, but there is one that I did partialy in a exam, and I need help to do it right.

The question is:

Let $n$ an odd number. Prove that $n$ divides $\displaystyle\sum_{i=1}^{n-1} i^n$.

I tried to look at the remains of the euclidian division of each $i^n$ for $n$. However, once $n$ isn't necessarily prime, i can't use Fermat's Little Theorem, and don't know how to procced.

Appreciate your help

user1176409
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Matheus Modesti
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  • Maybe you mean $k^n$. Anyway, edit the question and tell what you did, for a better explanations. – Rafa Budría Jan 16 '23 at 20:52
  • thank you for the corretion – Matheus Modesti Jan 16 '23 at 20:58
  • Yes, use the remainders. But it'll be clearer and easier using modular arithmetic, e.g $10= 3(mod)7$. Consider pairing the remainders into the sum: e.g. $1^n=1(mod)n$ $,(n-1)^n=-1(mod)n$ (because $n-1=-1(mod)n$ and n odd. – Rafa Budría Jan 16 '23 at 21:10
  • I hadn't thought about congruences. Once $1^n\equiv 1 \mod n$ and $(n-1)^n\equiv -1 \mod n$, when i sum both, there will be zero reminder. Doing this for all of them, since there is $(n-1)$ numbers, the sum will be zero. Is this? – Matheus Modesti Jan 16 '23 at 21:16
  • Yes, that's the easiest way to do it - exploiting the reflection symmetry $k\to -k\pmod{n},$ along with $i^n$ being an odd function - see the Hint in the linked dupe. – Bill Dubuque Jan 16 '23 at 21:34
  • Yes, symmetry and congruences. – Rafa Budría Jan 17 '23 at 12:27

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