If $a{x}^{17}+b{x}^{16}+1$ is divisible by $x^2-x-1,$ then integral value of a is 

Let roots of $x^2-x-1=0$ are $\alpha ,\beta$  
$\Rightarrow \alpha +\beta =1,\alpha \beta =-1$ 
$also\{\begin{array}{l}a{\alpha }^{17}+b{\alpha }^{16}+1=0\\ a{\beta }^{17}+b{\beta }^{16}+1=0\end{array}$ 
$\Rightarrow a\alpha +b=-\frac{1}{{\alpha }^{16}}$ 
$a\beta +b=-\frac{1}{{\beta }^{16}}$     
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$\Rightarrow a(\alpha -\beta )=\frac{{\alpha }^{16}-{\beta }^{16}}{{\left(\alpha \beta \right)}^{16}}$ 
$\Rightarrow a=({\beta }^8+{\alpha }^8)({\beta }^4+{\alpha }^4)({\beta }^2+{\alpha }^2)(\beta +\alpha )$ 
$=\mathrm{47.7.3.1}=987$