2021/075) If $x^2-x-1$ divides $ax^{17}+bx^{16} + 1$ find a-b

If $x^2-x-1$ divides $ax^{17}+bx^{16} + 1$ then $x^2=x+1$ =>  $ax^{17}+bx^{16} + 1 = 0$
The above is so because x =t which is root of $x^2= x+1$ is also a root of $ax^{17}+bx^{16} + 1= 0$

We have $x^2=x+1$
Putting $y=\frac{1}{x}$ we get $\frac{1}{y^2} =  \frac{1}{y} +1$
or $y^2 = 1 - y$
 $=>y^4 = (1-y)^2 = 1-2y +y^2 = (1-2y) + (1-y) = 2-3y$
$=>y^8 = (2-3y)^2 = 4-12y +9y^2 = (4-12y) + 9(1-y) =13-21y$
$=>y^{16} = (13-21y)^2 = 169-546y +441y^2 = (169-546y) + 441(1-y) = 610-987y$
$=>y^{17} = 610y-987y^2 = 610y - 987(1-y) = 1597y - 987$
Putting back $x=\frac{1}{y}$ we get
$\frac{1}{x^{17}} =  \frac{1597}{x} - 987$
Or $987x^{17} - 1597x^{16} +1 $ = 0

comparing with above we get a = 987, b = -  1597 so a - b = 2584