, 2023

Solution:
Let $x = 2000$, so the numerator is
$$x^{5} + x^{4} + 1 = (x^{2} + x + 1)(x^{3} - x + 1).$$
(This latter factorization can be noted by the fact that plugging in $\omega$ or $\omega^{2}$ into $x^{5} + x^{4} + 1$ gives $0$.) Then $x^{2} + x + 1 = 4002001$ divides the numerator. However, it can easily be checked that $101$ doesn't divide $4002001$ (since, for example, $101 \nmid 1 - 20 + 0 - 4$), so $4002001$ is one of the primes. Then the other one is
$$\frac{2000^{3} - 2000 + 1}{101} \approx \frac{2000^{3}}{101} > 2000^{2} \approx 4002001$$
so $4002001$ is the smaller of the primes.