Using Fermat's Little Theorem Prove if $p$ is prime, prove $1^p + 2^p + 3^p +...+(p-1)^p \equiv 0 \bmod{p}$ Using Fermat's Theorem prove if $p$ is prime, prove $1^p + 2^p + 3^p +...+(p-1)^p \equiv 0 \bmod{p}$
The two definitions of Fermat's Little Theorem is $a^p \equiv a \bmod{p}$ and $a^{p-1} \equiv 1 \bmod{p}$ but I don't know how to use this solve the problem 
 A: Since $1^p \equiv 1$, and $2^p \equiv 2$, $\ldots$ $(p-1)^p \equiv p-1$, we have that
$$1^p + 2^p + \ldots (p-1)^p \equiv 1 + 2 + \ldots + p-1$$
However, we know this sum: $\sum_{i = 1}^{p-1} i = \frac{p (p-1)}{2}$, which for odd primes would be a multiple of $p$, and we are done by reducing modulo $p$, then we only must check the case for $2$, which doesn't hold, and it's awkward...
A: Hint: Use Fermat's theorem for each $k$ ($1\le k\le p-1$) and add all them up. Then use Gauss's formula for the sum $1+2+\cdots +(p-1)$.
A: $1^p + ... + (p-1)^p \equiv 1 + 2 + ... (p-1) \equiv (1 + (p-1)) + (2 + (p-2)) + ... +(\frac{p-1}{2} + \frac{p+1}{2}) \equiv p + ... + p \equiv 0 (\mod p) $
Edit: Assuming p is an odd prime, of course
A: Before you wonder if you know how to use Fermat's theorem to solve the problem, first wonder if you are able to use Fermat's theorem at all.
There are a lot of things raised to the $p$-th power, and one form of Fermat's theorem tells you something about things raised to the $p$-th power. So you see what Fermat's theorem tells you about them.
You don't do this because you know it will solve the problem: you do this because maybe it will tell you something useful.
As it turns out, the thing it tells you is that your problem is equivalent to a problem you already know how to solve.
A: There is no need of Fermat's theorem. If $p$ is odd, then $p$ divides
$$ k^p +(p-k)^p = (k+p-k)\left(k^{p-1}-\ldots+(p-k)^{p-1}\right)=p A, $$
hence by collecting the terms of your sum in couples (just like Gauss did when he was five years old in order to prove that $1+2+\ldots+n = \frac{(n+1)n}{2}$) you get that the sum is a multiple of $p$.
A: This equation doesn't hold for $p=2$. However, if $p \ge 3$, a quick observation shows that $i$ and $p-i$ are paired. Since $i = i \text{ mod } p$, $p-i = -i \text{ mod } p$, we have
$$i^p = i^p \text{ mod } p, (p-i)^p = -i^p\text{ mod } p$$
Thus
$$i^p + (p-i)^p = 0 \text{ mod } p$$
sum up $i = 1, 2, \cdots, \frac{p-1}{2}$ gives original equation.
Apparently, for any odd $1\le d \le p$, $\sum_{i = 1}^{p-1} i^d = 0 \text{ mod } p$.
Additionally, this equation holds for any even $1\le d < p-1$. The only special case is when $d = p-1$, we can see using Euler's theorem that
$$\sum_{i = 1}^{p-1} i^{p-1} = p-1 \text{ mod } p$$
