Any solutions guys?

Suppose, for some $k$ we have $a_k< a_{k+1}$. Then, $a_n<a_{n+1}\,,\, \forall n, n\geq k$ and $a_{n+2}= 3a_{n+1}-2a_n\,,\, n\geq k$. It means $a_{k+n}=C_1 + C_2 2^n\,,\, n=0,1,\dots$. In that case, there would be infinite number of composite terms $a_n$. Indeed, let $p$ be a prime divisor of $C_1+C_2=a_k$ (we can assume $a_k>1, p>2$). Then, $p\mid C_1 + C_2 2^{m(p-1)}\,,\, m=1,2,\dots$ Going further, it's impossible $a_{n+1}<a_n$ for all $n$, hence from some place on, $a_n=a$, where $a$ is a prime. Going backward, there are two possibilities: $a_{n-1}=a$, or $a_{n-1}=2a$. The former gives us $(b,c)=(a,a)\,,\, \forall a \in \mathbb{P}$ Let's consider the latter one: $ a_{n-1}=2a , a_{n-2}=3a\pm a/2$. Hence, either there is no $a_{n-2}$, meaning $(b,c)=(2a,a)\,,\, a\in \mathbb{P}$, or $a$ is even, therefore $a=2$, $a_{n-1}=4, a_{n-2}=5$ or $7$, which implies $a_{n-3}$ being not integer. So, finally we get: \[(b,c) \in \{(5,4), (7,4), (p,p), (2p,p) \}\,, p\in \mathbb{P} \]

Suppose firstly that there exists a $k \geq 2$ such that $a_{k+1} > a_k$. Then by induction we have $a_{n+1} > a_{n}$ for all $n\geq k$ (when taking out the modulus the expression does not change sign) and so $a_{n+2} = 3a_{n+1} - 2a_n$ for all $n\geq k$. Solving the characteristic equation $t^2 - 3t + 2 = 0$ implies that $a_n = C_1 + C_2 \cdot 2^n$ for any $n\geq k$. Since the sequence is strictly increasing, we must have $C_2 > 0$ and in particular any prime divisor $p$ of some $a_m \geq 2$ will also be a prime divisor of $a_{m+k(p-1)}$ for any $k\geq 0$ as $2^{m+k(p-1)} \equiv 2^m \pmod p$ by Fermat's Little theorem. In particular, infinitely many terms would be composite. Therefore $a_{k+1} \leq a_k$ for all $k\geq 2$. Since $a_k > 0$, from some point on we must have $a_n = a$, where $a$ is a prime. Going backward, there are two possibilities: $a_{n-1}=a$, or $a_{n-1}=2a$. The former gives us $(b,c)=(a,a)$. Let's consider the latter one: $ a_{n-1}=2a , a_{n-2}=3a\pm a/2$. Hence, either there is no $a_{n-2}$, meaning $(b,c)=(2a,a)$, or $a$ is even, therefore $a=2$, $a_{n-1}=4, a_{n-2}=5$ or $7$, which implies $a_{n-3}$ being not integer. So, finally we get: \[(b,c) \in \{(5,4), (7,4), (p,p), (2p,p) \} \]where $p$ is an arbitrary prime.