Theorem 2 of Darij here

Let $ g$ one primitive root modulo $ p$. Then, the numbers $ 1,2,...,p-1$ are covered modulo $ p$ by the powers of $ g$: $ g^{0},g^{1}, ..., g^{p-2}$. So, the sum is: $ S = 1+g^{k}+g^{2k}+...+g^{(p-2)k}$ If $ p-1|k$, the sum is p-1 modulo p. If $ p-1$ not divides k, then $ S=\frac{g^{(p-1)k} - 1}{g^{k}-1}$ Seeing S modulo p, we have that S is 0 modulo p. Hence, $ k$ isn't multiple of $ (p-1)$

I know that this is WAY outdated, but I just looked at this problem and solved it, so

yugrey, your solution is wrong. It is not hard to verify that when $r=2$ and $p=5$, $1^2 + 2^2 + 3^2 + 4^2 = 30 \equiv 0 \pmod{5}$. Your error is here: Quote: If k is even, then mod p everything is 1,2,...(p-1)/2, (p-1)/2,... 2,1, as (-n)^(2r)=n^(2r) for integers r. Your exponent $r$ seems to have magically disappeared. In any case, I have my own solution different from the ones provided above. We claim that $S_k \equiv 0 \pmod{p}$ if and only if $p-1 \, |\, k$. As $a^{p-1} \equiv 1 \pmod{p}$ for $a = 1, 2, \ldots, p-1$, $p | S_k$ if and only if $p | S_{k\pm (p-1)}$. Since $p \not | S_0$, it is therefore sufficient to show that $p | S_k$ for $k = 1, 2, \ldots, p-2$. Let $e_n$ be the $n$th elementary symmetric polynomial evaluated at the numbers $1,2,\ldots,p-1$, and let $P(x) = x^{p-1} - 1$. By Fermat's little theorem, $x-a$ is a root of $P(x)$ for $a = 1, 2, \ldots, p-1$. Therefore $P(x)$ factors as $(x-1)(x-2) \ldots (x-(p-1))$ modulo $p$, which expands as $x^{p-1} - e_1 x^{p-2} + e_2 x^{p-3} - \ldots + e_{p-1}$. Comparing coefficients, we find that $e_i \equiv 0 \pmod{p}$ for $1 \leq i \leq p-2$. By Newton's identities, we have $S_n + \sum_{i=1}^{n-1} (-1)^i e_i S_{n-i} + (-1)^n n e_n$ for all $n \geq 1$. As $e_i \equiv 0 \pmod{p}$ for $1 \leq i \leq p-1$, the result follows immediately.

primitive roots kills this we just need $k$ not multiple of $\phi(p)$

Let us assume that x is the primitive root mod p then we have S_k=$1+x^k+x^{2k}..... x^{(p-2)k}$(mod p) therefore when (p-1)|k then we have S_k=-1(mod p) So let us assume that (p-1) does not divide k then we have $x^{(p-1)k}$=1(mod p) then on applying sum of gp formula we get that all k such that p-1 does not divide k satisfy the problem

Let $g$ be a primitive root $\pmod{p}$. Then we make two cases: Case 1: $p-1\mid k$. Then $S_k \equiv p-1 \pmod{p}$, so it doesnt hold. Case 2: $p-1 \nmid k$. Then $$S_k \equiv \sum_{m = 0}^{p-2} g^{mk} = \frac{1-g^{(p-1)k}}{1-g^k} \equiv 0 \pmod{p},$$so it works. $\square$