Prove that $\sum_{k=1}^{p-1} k^i$ is divisible by $p$

Question

Let $p$ be an odd prime. Prove that $$1^i + 2^i + \cdots + (p-1)^i \equiv 0 \pmod{p}$$ for all $i$, $1 \le i \le (p-2)$.

If $i$ is odd, then we are done, since $j^i + (p-j)^i \equiv 0 \pmod{p}$ for every $j$. But how can we prove this if $i$ is even? Any ideas? Thanks for your help.

https://math.stackexchange.com/questions/1777526/if-p-is-an-odd-prime-and-k-an-integer-with-0kp-1-then-1k-2k-ldot — lab bhattacharjee, Dec 25 '19 at 10:47
Does this answer your question? $p$ divides $\sum\limits_{k=1}^{p-1} (k^p)^n $ or this — rtybase, Dec 25 '19 at 11:49
@rtybase Neither of these solve the problem. The answers of the first post aren’t all that great, and the answers of the second post solve only the case where $i$ is odd. — ViHdzP, Dec 26 '19 at 00:19
@rtybase I correct myself: the answers in the second post solve only the even more trivial case $i=p$. — ViHdzP, Dec 26 '19 at 00:31
@rtybase Lubin states that the problem is false when $p-1\mid i$. This problem deals only with $1\leq i\leq(p-2)$. — ViHdzP, Dec 26 '19 at 00:35
@rtybase I hadn’t seen that one: this question is definitely a dupe of that. — ViHdzP, Dec 26 '19 at 00:47

score 1 · Accepted Answer · edited Dec 25 '19 at 09:09

1

Take $\xi$ a Primitive Root $\text{mod }p$. In particular, $\xi^i\not\equiv1\pmod{p}$. Therefore, $$1^i+2^i+\ldots+(p-1)^i\equiv\xi^0+\xi^i+\xi^{2i}+\ldots+\xi^{(p-2)i}\equiv\left(\xi^{(p-1)i}-1\right)\left(\xi^i-1\right)^{-1}\equiv0\pmod{p}.$$ $\blacksquare$

edited Dec 25 '19 at 09:09

Vann

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answered Dec 25 '19 at 08:35

ViHdzP

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Prove that $\sum_{k=1}^{p-1} k^i$ is divisible by $p$

1 Answers1