Enough of quadratic equations guys let us try something new

We have

$\begin{aligned} & x^2 - x - 1 =0 \\ \implies & x = \dfrac{1 \pm \sqrt{5}}{2} \end{aligned}$

$\therefore \ x^2 -x - 1 = 0 \implies \left[ x - \left( \dfrac{1 + \sqrt{5}}{2} \right) \right] \left[ x - \left( \dfrac{1 - \sqrt{5}}{2} \right) \right]$

Let the roots of the above quadratic be denoted by $\phi = \dfrac{1 + \sqrt{5}}{2}$ and $\psi = \dfrac{1 - \sqrt{5}}{2}$ respectively. Observe that $\phi \cdot \psi = \left( \dfrac{1 + \sqrt{5}}{2} \right) \left( \dfrac{1 - \sqrt{5}}{2} \right) = - \dfrac{4}{4} = -1$ .

Now since the quadratic is a factor of $ax^{17} + bx^{16} + 1$ therefore it's roots must also be two of the possible $17$ roots of $ax^{17} + bx^{16} + 1$ .

Thus

$\begin{aligned} & \begin{cases} a\phi^{17} + b\phi^{16} + 1 = 0 \\ a\psi^{17} + b\psi^{16} + 1 = 0 \end{cases} \\ \\ \implies & \begin{cases} a\phi^{17} + b\phi^{16} + 1 = 0 \\ a {\left( - \dfrac 1 \phi \right)}^{17} + b{\left( - \dfrac 1 \phi \right)}^{16} + 1 = 0 \end{cases} \\ \\ \implies & \begin{cases} a\phi^{17} + b\phi^{16} + 1 = 0 \\ - a + b\phi + \phi^{17} = 0 \end{cases} \\ \\ \implies & \begin{cases} {\color{#3D99F6} a\phi^{17} + b\phi^{16} + 1 = 0} \\ {\color{#3D99F6} - a\phi^{15} + b\phi^{16} + \phi^{32} = 0} \end{cases} \end{aligned}$

Subtracting the final two equations gives

$a \phi^{17} + a \phi^{15} + 1 - \phi^{32} = 0 \implies a = \dfrac{\phi^{32} - 1}{\phi^{17} + \phi^{15}} = \boxed{987}$