Taiwan IMO Selection Camp

Let $p_n$ be the largest prime divisor of $n^4 + n^2 + 1$, and let $q_n$ be the largest prime divisor of $n^2 + n + 1$.
It is easy to see that $p_n = q_{n^2}$. Also, from
$$n^4+n^2+1 = (n^2+1)^2-n^2 = (n^2-n+1)(n^2+n+1) = ((n-1)^2+(n-1)+1)(n^2+n+1)$$
we know that for all $n \ge 2$, $p_n = \max\{q_n, q_{n-1}\}$. Also since $n^2 - n + 1$ is odd,
$$\text{gcd}((n-1)^2+(n-1)+1, n^2+n+1) = \text{gcd}(2n, n^2-n+1) = \text{gcd}(n, n^2-n+1) = 1.$$
Hence $q_n \neq q_{n-1}$.
Returning to the original problem. Consider the set
$$S := \{n \in \mathbb{Z}_{\ge 2} \mid q_n > q_{n-1}, q_n > q_{n+1}\}.$$
Note that for all $n \in S$, we have
$$p_n = \max\{q_n, q_{n-1}\} = q_n = \max\{q_n, q_{n+1}\} = p_{n+1},$$
so the original problem is equivalent to proving that $S$ has infinitely many elements. Here note that $q_2 = 7 < 13 = q_3$ and $q_3 = 13 > 7 = q_4$, so $S$ is nonempty.
We now argue by contradiction: suppose $S$ has a largest element $m$. Since all $q_i$ are positive integers, it is impossible that $q_m > q_{m+1} > q_{m+2} > \dots$, so there must exist $k \ge m$ such that $q_k < q_{k+1}$ (note that it has already been shown that $q_k \ne q_{k+1}$). Also, note that since $q_{(k+1)^2} = p_{k+1} = \max\{q_k, q_{k+1}\} = q_{k+1}$, it is likewise impossible that $q_k < q_{k+1} < \dots$, so there must exist a smallest $l \ge k+1$ such that $q_l > q_{l+1}$.

But since $l$ is smallest, we must have $q_l > q_{l-1}$, and hence $l \in S$, contradicting the assumption that $m$ is the largest element. Contradiction! Therefore $S$ has infinitely many elements. This completes the proof.