Saudi Booklet

Suppose firstly that there exists a $k \ge 2$ such that $a_{k+1} > a_k$. Then by induction we have $a_{n+1} > a_n$ for all $n \ge k$ (when taking out the modulus the expression does not change sign) and so $a_{n+2} = 4a_{n+1} - 3a_n$ for all $n \ge k$. Solving the characteristic equation $t^2 - 4t + 3 = 0$ implies that
$$a_n = C_1 + C_2 \cdot 3^n, \forall n \ge k.$$
Since the sequence is strictly increasing, we must have $C_2 > 0$ and in particular any prime divisor $p$ of some $a_m \ge 2$ will also be a prime divisor of $a_{m+k(p-1)}$ for any $k \ge 0$ as $3^{m+k(p-1)} \equiv 3^m \pmod{p}$ by Fermat's Little theorem. In particular, infinitely many terms would be composite.

Therefore $a_{k+1} \le a_k$ for all $k \ge 2$. Since $a_k > 0$, we must have $a_n = p$ where $p$ is a prime for all $n \ge n_0$ with some positive integer $n_0$. Thus
$$p = |4p - 3a_{n-1}| \implies a_{n-1} \in \{p, 3p\}.$$
The former case gives us $(b, c) = (p, p)$ which indeed satisfies the condition. Let's consider the latter case: $a_{n-1} = 3p$, so we can find out that $a_{n-2} = 4p \pm \frac{p}{3}$. We consider some cases:

* If there is no $a_{n-2}$, then $(b, c) = (3p, p)$ are the first two terms.
* If $p$ is divisible by $3$, thus $p = 3$, $a_{n-1} = 9$, $a_{n-2} = 11$ or $13$, which implies $a_{n-3}$ being not integer.

So, finally we get $(b, c) \in \{(11, 9), (13, 9), (p, p), (3p, p)\}$ where $p$ is any prime. $\square$