Let $a$, $b$, and $x_0$ all be positive integers. Sequence $\{x_n\}$ is defined as $x_{n+1}=ax_n + b$ where $n \ge 1$. Show that $x_1$, $x_2$, $\cdots$ cannot be all prime.
If $x_1$ is not a prime, then the claim holds. Otherwise, suppose $x_1 = p$ is a prime. Then $p > a$. Meanwhile, at least two of the next $(p+1)$ numbers, $x_2$, $x_3$, $\cdots$, $x_{p+2}$, are congruent to each other modulo $p$. Suppose $x_m \equiv x_n \pmod{m}$ where integers $m > n > 1$. Then we have $$x_m - x_n \equiv (ax_{m-1}+b) - (ax_{n-1}+b) \equiv a(x_{m-1} - x_{n-1})\equiv 0$$
$$\therefore x_{m-1}-x_{n-1}\equiv 0\pmod{p}$$
Continue this substitution will eventually lead to $$x_{m-n+1}-x_1\equiv 0\pmod{p}$$
This means $p \mid x_{m-n+1}$. Meanwhile, $\{x_n\}$ is a strictly increasing sequence because both $a$ and $b$ are positive integers. Therefore $x_{m-n+1} > x_1 =p$. Hence we conclude $x_{m-n+1}$ is a composite.