13. If $a, b$ are integers, and $x^{2}-x-1$ is a factor of $a x^{17}+b x^{10}+1$, try to find the value of $a$.


13. ( 987 )
The roots of $x^{2}-x-1=0$ are $p=\frac{1}{2}(1+\sqrt{5})$, $q=\frac{1}{2}(1-\sqrt{5})$. They must be the roots of $\boldsymbol{a} x^{17}$
$+b x^{18}+1=0$.
$$
\therefore a p^{17}+b p^{16}=-1, a q^{17}+b q^{18}=-1 \text {. }
$$

Multiplying the first equation by $q^{16}$, and the second equation by $p^{16}$, and using $p q=-1$, we get
$$
a p+b=-q^{18}, a q+b=-p^{10} .
$$

Subtracting the two equations, we get
$$
\begin{array}{l}
a(p-q)=p^{18}-q^{18} . \\
\therefore a=\frac{p^{16}-q^{16}}{p-q}=\left(p^{8}+q^{8}\right)\left(p^{4}+q^{4}\right) \\
\text { - }\left(p^{2}+q^{2}\right)(p+q) \text {. } \\
\because p+q=1 \text {, } \\
p^{2}+q^{2}=(p+q)^{2}-2 p q=1+2=3 \text {. } \\
p^{4}+q^{4}=\left(p^{2}+q^{2}\right)^{2}-2(p q)^{2}=9-2 \\
=7 \text {, } \\
p^{8}+q^{8}=\left(p^{4}+q^{4}\right)^{2}-2(p q)^{4}=49-2 \\
\end{array}
$$
$=47$.