Polynomials are interesting!

The roots of of $x^2-x-1=0$ are the golden ratios $\varphi =\frac{1\pm\sqrt{5}}{2}$ . As $(x^2-x-1) | (ax^{17} + bx^{16} + 1)$ . This means that $\varphi$ are also those of $ax^{17} + bx^{16} + 1=0$ .

We note that:

\(\begin{array} {} & \quad \space \space \space \varphi = \varphi & \Rightarrow \varphi^1 = F_1 \varphi + F_0 & \\ \varphi^2 - \varphi - 1 =0 & \Rightarrow \varphi^2 = \varphi + 1 & \Rightarrow \varphi^2 = F_2 \varphi + F_1 \\ \varphi^3 = \varphi^2 + \varphi & \Rightarrow \varphi^3 = 2\varphi + 1 & \Rightarrow \varphi^3 = F_3 \varphi + F_2 \\ \varphi^4 = 2\varphi^2 + \varphi & \Rightarrow \varphi^4 = 3\varphi + 2 & \Rightarrow \varphi^4 = F_4 \varphi + F_3 \\ \varphi^5 = 3\varphi^2 + 2\varphi & \Rightarrow \varphi^3 = 5\varphi + 3 & \Rightarrow \varphi^5 = F_5 \varphi + F_4 \\ & & \color{blue} {\Rightarrow \varphi^n = F_n \varphi + F_{n-1}} \end{array} \)

Therefore,

$\begin{aligned} a\varphi^{17} + b\varphi^{16} + 1 & = 0 \\ a(F_{17} \varphi + F_{16} ) + b(F_{16} \varphi + F_{15}) + 1 & = 0 \\ (1597a + 987b)\varphi + 987a + 610b + 1 & = 0 \end{aligned}$

Since $\varphi$ is irrational, $\varphi \ne -\dfrac{987a + 610b}{1597a + 987b}$ , we have:

$\begin{cases} 1597a + 987b = 0 \\ 987a + 610b + 1 = 0 \end{cases}$

Solving the above equations, we get:

$a = \pm 987, b = \mp 1597 \quad \Rightarrow |a| + |b| = 987 + 1597 = \boxed{2584}$