Dare to solve

Since $x^{2}-x-1$ is a factor of the first polynomial $px^{17}+qx^{16}+1$ it means that the roots of the first equation are also the roots of the second polynomial.We can solve the first equation by quadratic formula to find :

$x_{1}= \dfrac{1 + \sqrt{5}}{2} =a , x_{2} = \dfrac{1 - \sqrt{5}}{2} = b$

$f(x) = zx^{20} + yx^{19} + 1$

$f(a) = za^{20} + ya^{19} + 1$

$f(b) = zb^{20} + yb^{19} + 1$

Both are linear equations , thus applying cross multiplication method

$\dfrac{z}{a^{19} - b^{19}} = \dfrac{y}{b^{20} - a^{20}} = \dfrac{1}{a^{20}b^{19} - b^{20}a^{19}}$

$\dfrac{y}{a^{20} - b^{20}} = \dfrac{1}{a^{19}b^{19}(a - b)}$

Now,

$\dfrac{1 + \sqrt{5}}{2} \times \dfrac{1 - \sqrt{5}}{2} = -1$

$a^{19}b^{19} = -1$

$y = \dfrac{a^{20} - b^{20}}{(a - b)}$

$F_{n}=\dfrac{\varphi^{n}-\psi^{n}}{\varphi-\psi}$ (where $\varphi =a , \psi = b$ )

$F_{n} = F_{n-1} + F_{n-2}~ ,~~ F_{1}=1 = F_{2}$

Thus we have to find $F_{20} ~ which~is~equal~to~6765$

$\boxed{y = 6765}$

$Fibonacci ~ Series - 1,1,2,3,5,8,13,21,34.....$

How x^2=x+1 then x^20 = 6765x+4181 and x^19 = 4181x+2584. Replacing the espression given found : x(6765z+4181y)+4181z+2584y+1.

We have the zero the expressions " 6765z+4181y" and " 4181z+2584y+1", which brings us y =6765.