, 2010

No such $k$ exists. For suppose that $k$ and $p$ are as described. Consider the number
$$A = \sum_{i=0}^{p-1} (i^3 - 1)^{3k}.$$
Because $p-1 = 6k$ is divisible by $3$, there are three cube roots of $1$ modulo $p$. Therefore, three terms in the sum are $0$ modulo $p$, and the others are $[(p-1)/2]$th powers of nonzero residues, hence are congruent to either $1$ or $-1$ modulo $p$. Consequently, $A$ is congruent to one of the residues $p-3, p-5, p-7, \dots, -(p-3)$ modulo $p$. In particular, $A$ cannot be congruent to $1$ or $-1$ modulo $p$.

On the other hand, applying the binomial theorem and changing the order of summation, we have
$$A = \sum_{i=0}^{p-1} \left( \sum_{j=0}^{3k} \binom{3k}{j} (-1)^j i^{3(3k-j)} \right) = \sum_{j=0}^{3k} \left( (-1)^j \binom{3k}{j} \sum_{i=0}^{p-1} i^{3(3k-j)} \right), \quad (1)$$
where we use the convention $0^0 = 1$ for the case $i = 0, j = 3k$.

Now, we claim that $\sum_{i=0}^{p-1} i^d \equiv 0 \pmod{p}$ if $p-1 \nmid d$. For this, choose a primitive root $g$ modulo $p$ and notice that $g^d \not\equiv 1 \pmod{p}$, while we have
$$(g^d - 1) \sum_{i=0}^{p-1} i^d \equiv \sum_{i=0}^{p-1} (g \cdot i)^d - \sum_{i=0}^{p-1} i^d \equiv 0 \pmod{p},$$
giving the claim. Further, notice that $\sum_{i=0}^{p-1} i^d \equiv 0 \pmod{p}$ when $d = 0$.

Now, because $3 \cdot (3k) < 2(p-1)$, the only value of $j$ such that $3 \cdot (3k-j)$ is a multiple of $p-1$ is $j = k$, where $3 \cdot (3k-j) = 6k = p-1$. This means that the sums $\sum_i i^{3 \cdot (3k-j)}$ are congruent to $0$ modulo $p$ unless $j = k$. Further, for $j = k$, the sum evaluates to
$$\sum_{i=0}^{p-1} i^{6k} \equiv \sum_{i=0}^{p-1} i^{p-1} \equiv \sum_{i=1}^{p-1} 1 \equiv -1 \pmod{p}.$$

Considering (1) in this light, we find that
$$A \equiv (-1)^k \cdot \binom{3k}{k} \cdot (-1) \pmod{p}.$$
Since we saw that $A$ is not congruent to $1$ or $-1$ modulo $p$, we conclude that $\binom{3k}{k} \not\equiv 1 \pmod{p}$, a contradiction.