Mongolian Mathematical Olympiad Round 3

It is clear if $a$ and $b$ are not relatively prime, thus we assume that $a$ and $b$ are relatively prime.
For $n \ge 1$, let $a_n = a^n + b$. First we choose $l \ge 0$ such that $|a|^l > |b|$. Then for any $n \ge l+1$, we have $|a_n| < |a|^n + |a|^l \le 2|a|^n - |a|^l < |a_{n+1}|$.
In particular, for any $n \ge l+3$, we have $|a_n| \ge 2$. Let $m = l+3$ and let $N = 2022$. Each of $a_{m+1}, a_{m+2}, \dots, a_{m+N}$ has at least one prime divisor, and we choose a prime divisor for each: $p_1 \mid a_{m+1}$, $p_2 \mid a_{m+2}$, $\dots$, $p_N \mid a_{m+N}$. Clearly $(a, p_i) = 1$ for each $1 \le i \le N$.
Now let $k = (p_1-1)(p_2-1)\dots(p_N-1) \ge 1$. Then $a_{m+k+i} = a^{m+i}(a^k-1) + a_{m+i}$ is divisible by $p_i$ by Fermat's theorem for any $1 \le i \le N$. Moreover, $|a_{m+k+i}| > |a_{m+i}|$. Hence $a_{m+k+1}, a_{m+k+2}, \dots, a_{m+k+N}$ are all composite.