5-33 If $a, b$ are integers, and $x^{2}-x-1$ is a factor of $a x^{17}+b x^{16}+1$, try to find the value of $a$.


[Solution] The roots of the equation $x^{2}-x-1=0$ are
$$
p=\frac{1+\sqrt{5}}{2}, q=\frac{1-\sqrt{5}}{2},
$$

and $p+q=1, p q=-1$.
From the problem, we know that $p, q$ are also roots of the equation
$$
a x^{17}+b x^{16}+1=0
$$

Therefore,
$$
\begin{array}{l}
a p^{17}+b p^{16}+1=0 \\
a q^{17}+b q^{16}+1=0
\end{array}
$$
(1) $\times q^{16}$, we get $a p^{17} q^{16}+b p^{16} q^{16}+q^{16}=0$

Using $p q=-1$, we get
$$
a p+b=-q^{16}
$$

Similarly, (2) $\times p^{16}$, and using $p q=-1$, we get
$$
a q+b=-p^{16}
$$
(3) - (4) gives
$$
\begin{array}{l}
a(p-q)=p^{16}-q^{16} \\
a=\frac{p^{16}-q^{16}}{p-q} \\
\quad=\left(p^{8}+q^{8}\right)\left(p^{4}+q^{4}\right)\left(p^{2}+q^{2}\right)(p+q)
\end{array}
$$

Since
$$
\begin{array}{l}
p+q=1, \\
p^{2}+q^{2}=(p+q)^{2}-2 p q=1+2=3, \\
p^{4}+q^{4}=\left(p^{2}+q^{2}\right)^{2}-2(p q)^{2}=9-2=7, \\
p^{8}+q^{8}=\left(p^{4}+q^{4}\right)^{2}-2(p q)^{4}=49-2=47,
\end{array}
$$

Therefore, $a=47 \times 7 \times 3 \times 1=987$.