Find all pair of integers $(a,b)$ so that $x^2-x-1 \mid ax^{17} + bx^{16} + 1$

Question

Find all pair of integers $(a,b)$ so that $x^2-x-1 \mid ax^{17} + bx^{16} + 1$.

This problem is from a book. In the book says that there is only one solution! That is $(a,b) = (987, -1597)$, i.e., $(F_{16}, -F_{17}), F_n = n-th\; Fibonacci\; number$.

But I did this way. As $x^2-x-1=0$ has roots $\phi$ and $1-\phi$. So they must be root of $ax^{17} + bx^{16} + 1$ also.
So we have $$a\phi ^ {17} + b\phi ^{16} + 1 = 0\; \text{and}\\a(1-\phi)^{17}+b(1-\phi)^{16} + 1 = 0.$$ Subtracting these two and dividing by $\sqrt{5}$ gives $$a \frac{\phi ^ {17}-(1-\phi)^{17}}{\sqrt{5}} + b \frac{\phi ^ {16}-(1-\phi)^{16}}{\sqrt{5}} = 0\\ \implies aF_{17} + bF_{16} = 0.$$
$F_{17}$ and $F_{16}$ are coprime integers so this has solutions $(a,b) = (kF_{16}, -kF_{17}), \forall k \in \mathbb{Z}$.
Where I am mistaken?

Source: 101 Problems in Algebra written by Titu Andrescu.
I'd also like to know other approaches to solve this problem.

Answer 1

Just use $x^2=x+1$ four times and solve the system.

Finely, $a=987$ and $b=-1597$.

Answer 2

Let $N=17$. Note that $\phi^n=F_n\phi+F_{n-1}$ for every $n$, by induction. It follows that $a\phi^N+b\phi^{N-1}+1=(aF_{N}+bF_{N-1})\phi+(aF_{N-1}+bF_{N-2}+1)$. So you need $aF_{N}+bF_{N-1}=0$ and $aF_{N-1}+bF_{N-2}+1$. As you already noticed, the first equation gives $(a,b)=k(F_{N-1},-F_N)$ for some integer $k$. Reinjecting into the second equation, we obtain $kD_N=-1$ where $D_N=F_{N-1}^2-F_NF_{N-2}$.

Now, for every $n$ we have

$$ D_{n+1}=F_{n}^2-F_{n+1}F_{n-1}=F_n(F_{n-2}+F_{n-1})-(F_n+F_{n-1})F_{n-1}=-D_n $$

so that $D_n=(-1)^{n}$ by induction on $n$. It follows that $k=(-1)^{N+1}=1$.