<s>[INST] Find a such that $ax^{17}+bx^{16}+1$ is divisible by $x^2-x-1$. 
Find $a$ such that $ax^{17}+bx^{16}+1$ is divisible by $x^2-x-1$.

I tried taking the roots of the polynomial which are $\frac{1±\sqrt{5}}{2}$
And I got the equation $a(\frac{1±\sqrt{5}}{2})^{17}+b(\frac{1±\sqrt{5}}{2})^{16}+1=0$
Now I don't know what to do next.
Any help would be appreciated.
 [/INST] Hint. If $x=\frac{1±\sqrt{5}}{2}$ then $x^2=x+1$ and
\begin{align}
ax^{17}+bx^{16}+1 &=(ax+b)(x+1)^8+1 \\
                  &=(ax+b)(x^2+2x+1)^4+1\\
                  &=(ax+b)(3x+2)^4+1\\
                  &=(ax+b)(21x+13)^2+1.
\end{align}
Can you take it from here? Note that 3,2, 21,13 are all Fibonacci numbers.
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