8-4 Let the sequence of positive integers $\left\{a_{n}\right\}$ satisfy
$$a_{n+3}=a_{n+2}\left(a_{n+1}+2 a_{n}\right), n=1,2, \cdots$$

and $a_{6}=2288$. Find $a_{1}, a_{2}, a_{3}$.
［Solution］From the recursive relation, we get
$$\begin{aligned}
a_{6} & =a_{5}\left(a_{4}+2 a_{3}\right)=a_{4}\left(a_{3}+2 a_{2}\right)\left(a_{4}+2 a_{3}\right) \\
& =a_{3}\left(a_{2}+2 a_{1}\right)\left(a_{3}+2 a_{2}\right)\left[a_{3}\left(a_{2}+2 a_{1}\right)+2 a_{3}\right] \\
& =a_{3}^{2}\left(a_{3}+2 a_{2}\right)\left(a_{2}+2 a_{1}\right)\left(a_{2}+2 a_{1}+2\right) .
\end{aligned}$$

Since \(a_{6}=2288=2^{4} \times 11 \times 13\), and \(a_{2}+2 a_{1}+2\) differs from \(a_{2}+2 a_{1}\) by 2, and both are factors of 2288, it follows that
$$a_{2}+2 a_{1}=11, a_{2}+2 a_{1}+2=13$$

Thus, \(a_{3}^{2}\left(a_{3}+2 a_{2}\right)=2^{4}\),
which implies that \(a_{3}\) can only be 1 or 2. If \(a_{3}=1\), then \(a_{2}=\frac{12}{5}\) is not a natural number, leading to a contradiction! Therefore, \(a_{3}=2, a_{2}=1\). From \(a_{2}+2 a_{1}=11\), we get \(a_{1}=5\).