TITLE: Number theory, prime numbers
QUESTION [6 upvotes]: prove that for all prime numbers $p>2000$ the sum $$1+2^{2000}+3^{2000}+...+(p-1)^{2000}$$ is divisible by $p$
I tried this with primes smaller then $2000$, but I can't find a general rule.

REPLY [5 votes]: Pick any non-zero $a$ such that $a^{2000} \not \equiv 1 \pmod p$ (if no such $a$ exist then every number in the sum is congruent to one and hence the sum is divisible by $p$). Now if we multiply all the numbers $1 \ldots p-1$ by $a$ we'll get some permutation of $1 \ldots p-1$. So $(1^{2000} + \cdots + (p-1)^{2000}) \cdot a^{2000} = (1^{2000} + \cdots + (p-1)^{2000})$. Since $a^{2000} \not \equiv 1 \pmod p$ we must have $1^{2000} + \cdots + (p-1)^{2000} \equiv 0 \pmod p$.
Alternatively, you can pick any primitive root $g$ modulo $p$ and then the sum is equal to $g^{0} + g^{1 \cdot 2000} + g^{2\cdot 2000} + ... + g^{(p-2) \cdot 2000} = \frac{g^{2000(p-1)} - 1}{g^{2000}-1}$ which is also divisible by $p$.