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