Solution: Clearly $ p=q=2$ satisfies to the conditions.Now assume that $ p,q\geq 3$. As we know $ \frac{(p^2+8)(q^2+8)}{pq}$ is an integer. It follows that $ \frac{8(p^2+q^2+8)}{pq}$, recalling that $ p,q$ are odd, it can be inferred that $ \frac{p^2+q^2+8}{pq}$ is an integer as well. So the problem reduces to finding all odd solutions $ (p_n,q_n)$, that are less than $ 2005$ and satisfy to condition, that: \[ \frac{p^2+q^2+8}{pq}=k \] for some fixed $ k$. Assume that $ (p_0,q_0)$ is the solution, so that the sum $ p+q$ reaches its minimum, and $ p_0\geq q_0$. If $ p_0=q_0$, then, obviously $ p_0=q_0=1$. Now assume $ p_0>q_0$. It follows that $ p_0^2-p_0(q_0k)+q_0^2+8=0$ As we know, by Vieta's formula: $ p'=\frac{q_0^2+8}{p_0}<\frac{p_0^2-2p_0+9}{p_0}<p_0$, as long as, $ p_0\geq 5$, so this case leads to contradiction. So we might assume that $ p_0\leq 4$. If $ p_0=3$, then $ q_0=1$, but this case is invalid. So if $ (p_0,q_0)$ is a solution than $ p_0=q_0=1$ and $ k=10$. All solutions determined by this method are in the sequence $ q_{n+1}=p_n, p_{n+1}=10q_n-p_n$ are $ (1,1),(9,1),(89,9),(881,89)$. Now let's prove that all solutions of the equation $ p^2-10pq+q^2+8=0$ are in the described sequence. Let $ (p',q')$ be the odd solution with minimal sum that is not in the sequence. By the explicitly described method, we could proceed to the another solution with less sum and that will either lead to the contradiction of minimality or to the contradiction of the assumptation that this solution is not in the sequence. So all solutions are: \[ (2,2),(881,89),(89,881) \]

you missed $ (3,17)$ and $ (17,3)$...

well, it is because my assumptation about the case $ p_0 = 3, q_0 = 1$ and consequently $ k = 6$ was wrong and this case is valid. I have eventually realized that but it was too late to fix the solution. is it clear now?

yes, it's clear now..