Find all prime number $p$ such that $\{1, 2, \dots, p-1\}$ can be partitioned into several $p$-good sets.

Solution. Clearly, we must have $p \equiv 1 \pmod 6$. To show that all such $p$ satisfies the condition, choose $g$ to be a primitive root modulo $p$. Let $x = g^{(p-1)/6}$. Can verify that $x^3 \equiv -1 \pmod p$ thus $x^2 - x + 1 \equiv 0 \pmod p$. So $g^n, x^2g^n, xg^n$ is $p$-good for every $n$. Pick $n \in \{0, 1, 2, \dots, \frac{p-1}{6} - 1\} \cup \{k, k+1, k+2, \dots, k+\frac{p-1}{6} - 1\}$ where $k = \frac{p-1}{2}$, and we have the desired partition.