1
$\begingroup$

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.

$\endgroup$
9
  • 2
    $\begingroup$ math.stackexchange.com/questions/1777526/… $\endgroup$ – lab bhattacharjee Dec 25 '19 at 10:47
  • 2
    $\begingroup$ Does this answer your question? $p$ divides $\sum\limits_{k=1}^{p-1} (k^p)^n $ or this $\endgroup$ – rtybase Dec 25 '19 at 11:49
  • $\begingroup$ @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. $\endgroup$ – ViHdzP Dec 26 '19 at 0:19
  • $\begingroup$ @URL I seriously insist you look at the links again. $\endgroup$ – rtybase Dec 26 '19 at 0:29
  • $\begingroup$ @rtybase I correct myself: the answers in the second post solve only the even more trivial case $i=p$. $\endgroup$ – ViHdzP Dec 26 '19 at 0:31
1
$\begingroup$

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

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.