Question

Show that \(1^p + 2^p +\cdots + (p-1)^p \equiv 0(mod p)\)
Please, help

Answer 1

I am looking for several methods. One of them is
p > 1, 2, ...., (p-1) , and p is prime, hence (p, 1) = 1, (p, 2) =1.... and so one
that is by Wilson's theorem, \(1^ p \equiv 1 (modp)\\2^p \equiv 2 (mod p)\\\cdots\\(p-1)^p \equiv p-1 (mod p)\)

then, \(1^p +2^p +\cdots +(p-1)^p \equiv 1+2 +\cdots +(p-1) (mod p)\)
But the right hand side is \(\sum_{i =1}^{p-1} i = \dfrac{(p-1)(p-1+1)}{2}= \dfrac{p(p-1)}{2}\)
and p is odd, hence 2| p-1, then \(\dfrac{p(p-1)}{2}\equiv 0(modp)\)

Answer 2

I'm kinda thinking we might be able to pair up the last and first terms:

\[[1^p+(p-1)^p] + [2^p+(p-2)^p] + \cdots \equiv 0 \mod p\]
Since distributing the binomials, and p is odd we will get:

\[(p-n)^p \equiv -n^p \mod p\]

so all the terms cancel each other out. I think if we wrote it out in summation notation it'd be more rigorous and also we could prove for p=2 independently since it's the only even prime.

Answer 3

Yes, it works to me. :) Thank you

Answer 4

Well I already typed this up, so here we go I guess haha. I guess since \(1 \cancel \equiv 0 \mod 2\) is false, p is a prime greater than 2 for this problem. Oh well, no big deal.

\[\sum_{n=1}^{p-1} n^p \mod p\]
\[\sum_{n=1}^{(p-1)/2} n^p + (p-n)^p \equiv 0 \mod p\]

Answer 5

Again, thank you so much.