17. Find all pairs of positive integers $(b, c)$ such that the sequence $a_{1}=b, a_{2}=c, a_{n+2}=\left|3 a_{n+1}-2 a_{n}\right|(n \geqslant 1)$, has only finitely many composite numbers.
(2002, Bulgaria National Mathematical Olympiad (Final))
Let $(b, c)$ be the pair of numbers that satisfy the conditions.
If $a_{k}=a_{k+1}$, and $k$ is the smallest index that satisfies this condition. If $k \geqslant 4$, since $a_{k+1}=\left|3 a_{k}-2 a_{k-1}\right|$, and $a_{k} \neq a_{k-1}$, then $a_{k+1}=2 a_{k-1}-3 a_{k}$, i.e., $a_{k-1}=2 a_{k}$.

Also, because $a_{k}=13 a_{k-1}-2 a_{k-2} \mid$, so, $2 a_{k-2}=5 a_{k}$ or $2 a_{k-2}=7 a_{k}$, which means $a_{k}$ can be divided by 2.

On the other hand, it is easy to know that $a_{n}=a_{k}$, where $n \geqslant k$. Let $P$ be the set of all prime numbers and 1, then $a_{k} \in P$. Therefore, we know $a_{k}=2$, $a_{k-1}=4, a_{k-2}=5$ or 7, which contradicts $4=a_{k-1}=\left|3 a_{k-2}-2 a_{k-3}\right|$. Because the right side is odd, and the left side is even, so, $k \leqslant 3$.

When $k=3$, $a_{3}=2, a_{2}=4, a_{1}=5$ or 7, i.e., $(b, c)=(5,4)$ or $(7,4)$.

When $k=2$, $a_{2}=p, a_{1}=2 p$, where $p \in P$, i.e., $(b, c)=(2 p, p)$.

When $k=1$, $a_{1}=p, p \in P, a_{2}=p$, i.e., $(b, c)=(p, p)$.

If for all $n$, $a_{n} \neq a_{n+1}$, since $a_{n} \geqslant 0$, there must exist a positive integer $l$, such that $a_{i+1}>a_{l}$. By mathematical induction, when $n \geqslant l$, we have $a_{n+1}>a_{n}$. Using $a_{n+2}=3 a_{n+1}-2 a_{n}$, by mathematical induction, we get
$$
a_{n}=a_{l+2}+2\left(2^{n-1-2}-1\right)\left(a_{l+2}-a_{l+1}\right).
$$

If $a_{l+2}$ is even, then $a_{n} \geqslant 4$ is even, where $n \geqslant l+3$, which is a contradiction.
If $a_{l+2}$ is odd, then $a_{l+2} \geqslant 3$, and $\left(a_{l+2}, 2\right)=1$.
By Euler's theorem, $a_{l+2}$ can divide $2^{\varphi\left(a_{l+2}\right)}-1$, where $\varphi$ is the Euler function, i.e., $\varphi(m)$ represents the number of positive integers less than $m$ and coprime with $m$.

Let $n=l+2+i \varphi\left(a_{l+2}\right), i=1,2, \cdots$, then $a_{l+2}$ can divide $a_{i}\left(a_{n}>a_{l+2}\right)$, which is a contradiction.

In summary, $(b, c)=(5,4)$ or $(7,4)$ or $(2 p, p)$ or $(p, p)$, where $p \in P$.