Let prove by induction that $x^{n+1}-f_n x-f_{n-1}=(x^2-x-1)(f_0x^{n-1}+f_1x^{n-2}+f_2x^{n-3}+...+f_{n-1})$ For $n=1$ it is obvious for $n=2$: $x^3-2x-1=(x^2-x-1)(x+1)$ is also true let it is true for $n=k$ then $(x^2-x-1)(f_0x^{k}+f_1x^{k-1}+f_2x^{k-2}+...+f_{k})=(x^2-x-1)( x(f_0x^{k-1}+f_1x^{k-2}+f_2x^{k-3}+...+f_{k-1})+f_k)=x(x^{k+1}-f_kx-f_{k-1})+f_k(x^2-x-1)=x^{k+2}-f_{k+1}x-f_k$ so it is true for $n=k+1$ Now let prove that $f_0x^{2k}+...+f_{2k}=0$ has no roots $\sum_{i=0}^{i=2k} f_i x^{2k-i}=\frac{x^{2k}+f_{2k}}{2}+\sum_{i=0}^{k-1} \frac{f_{2i}x^{2k-2i}+2f_{2i+1}x^{2k-2i-1}+f_{2i+2}x^{2k-2i-2}}{2} = \frac{x^{2k}+f_{2k}}{2}+\sum_{i=0}^{k-1} \frac{x^{2k-2i-2}}{2}( f_{2i}x^{2}+2f_{2i+1}x+f_{2i+2})$ But as $f_{2i+1}^2=f_{2i}*f_{2i+2}-1$ then $f_{2i}x^{2}+2f_{2i+1}x+f_{2i+2}>0$ for every $x$ and natural $i$ and so $\sum_{i=0}^{i=2k} f_i x^{2k-i}>0$ for every $x$ and natural $k$ So roots of $x^{2k+2}=f_{2k+1}x+f_{2k}$ are roots of $x^2-x-1=0$ or $ \frac{1 \pm \sqrt{5}}{2}$