Problem 920

Solution:
Take a generator $g \pmod{p}$ so that $g$ is not equivalent to anything of the form $-a / b$ for integers $a, b \leq 2023$. For big enough primes, such a $g$ must exist as there are at most a constant number of such fractions.

Now number the vertices with the residues $\bmod\ p$. For any residue $r \pmod{p}$, consider the sequence of vertices $P_{r}$ formed by the multiples of $r$, starting from $r$ and ending at $p r \equiv 0 \pmod{p}$. We claim that the concatenation of $P_{1}, P_{g}, P_{g^{2}}, \ldots$ works.

Clearly no $P_{r}$ contains a cycle. Thus, if a cycle were to exist, it would have to be formed by the end of one $P$ and the start of another. In other words, we would have $c g^{i+1} \equiv (p-d) g^{i} \pmod{p}$ for $c+d \leq 2023$. However, this is impossible by choice of $g$.