Polynomial Problem

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 :

$x_{1}=\phi$ and $x_{2}=1-\phi$ . where $\phi$ is the Golden Ratio.

But since our second polynomial contains high degrees it would be better to transform the second root into $\frac{-1}{\phi}$ .Therefore,

$x_{1}=\phi$ and $x_{2}=\frac{-1}{\phi}$ .

Now,after substituting $x$ we get the two equations:

$p\phi^{17}+q\phi^{16}=-1$

$\frac{q}{\phi^{16}}-\frac{p}{\phi^{17}}=-1$

Multiplying the second equation by $\phi^{32}$ then from the first unchanged equation substract the second one to get:

$p(\phi^{17}+\phi^{15})=\phi^{32}-1$ . And finally:

$\boxed{p=\frac{\phi^{32}-1}{\phi^{17}+\phi^{15}}}$ .

Substitute $\phi=\frac{1+\sqrt{5}}{2}$ to get $p=\boxed{987}$ .One can also solve this ,i think using Fibonacci numbers because the answer is equal to to $F_{16}$ where $F_{n}$ denotes the $n$ -th fibonacci number.

The roots of $x^{2}-x-1=0$ are $\alpha=\frac{1+\sqrt{5}}{2}$ and $\beta=\frac{1-\sqrt{5}}{2}$ .

These roots are also the roots of $px^{17}+qx^{16}+1=0$ .

So $p\alpha^{17}+q\alpha^{16} = -1 \space\space\_\_\_\_\_\_(1)$ ,

and $p\beta^{17}+q\beta^{16} = -1 \space\space\_\_\_\_\_\_(2)$ .

By multiplying $(1)$ by $\beta^{16}$ and using the fact $\alpha\beta = -1$ ,

$p\alpha+q = -\beta^{16}$ .

By multiplying $(2)$ by $\alpha^{16}$ and using the fact $\alpha\beta = -1$ ,

$p\beta+q = -\alpha^{16}$ .

By subtracting,

$p(\alpha-\beta) = \alpha^{16}-\beta^{16}$ .

So $p = \frac{\alpha^{16}-\beta^{16}}{\alpha-\beta}$ .

Note that this is the Binet's formula for the 16th Fibonacci number.

Finding the 16th Fibonacci number: $1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,...$ .

Therefore, $p = F_{16} = 987$ .

Alternative Solution:

$p = \frac{\alpha^{16}-\beta^{16}}{\alpha-\beta} = (\alpha^{8}+\beta^{8})(\alpha^{4}+\beta^{4})(\alpha^{2}+\beta^{2})(\alpha+\beta)$ .

Since

$\alpha+\beta = 1$ ,

$\alpha^{2}+\beta^{2} = (\alpha+\beta)^{2}-2\alpha\beta = 1+2 = 3$ ,

$\alpha^{4}+\beta^{4} = (\alpha^{2}+\beta^{2})^{2}-2\alpha^{2}\beta^{2} = 9-2 = 7$ ,

$\alpha^{8}+\beta^{8} = (\alpha^{4}+\beta^{4})^{2}-2\alpha^{4}\beta^{4} = 49-2 = 47$ ,

it follows that $p = 47 \times 7 \times 3 \times 1 = 987$ .

Let $f(x) = x^2-x-1$ , $g(x) = px^{17}+qx^{16}+1$ and the roots of $f(x)$ be $\varphi$ . If $f(x)$ is a factor of $g(x)$ , then $\varphi$ are also roots of $g(x)$ or $g(\varphi) = 0$ .

$f(\varphi) = \varphi^2 - \varphi\ -1 = 0\quad \Rightarrow \varphi^2 = \varphi +1$

$\Rightarrow \varphi^4 = (\varphi +1)^2 = \varphi^2 + 2\varphi +1 = 3\varphi +2$

$\Rightarrow \varphi^8 = (3\varphi +2)^2 = 9 \varphi^2 + 12\varphi +4 = 21 \varphi + 13$

$\Rightarrow \varphi^{16} = (21 \varphi + 13)^2 = 441 \varphi^2 + 546 \varphi + 169 = 987 \varphi + 610$

$\Rightarrow\varphi^{17} = \varphi (987 \varphi + 610) = 987 (\varphi + 1) + 610 \varphi = 1597 \varphi + 987$

Therefore, $g(\varphi) = p\varphi^{17}+q\varphi^{16}+1 = p(1597 \varphi + 987) + q(987 \varphi + 610) + 1$ $\quad \quad = (1597p + 987q) \varphi + 987p + 610q + 1 = 0$

$\Rightarrow \begin {cases} 1597p + 987q = 0 \\ 987p + 610q = -1 \end {cases} \quad \Rightarrow \begin {cases} 610(1597p + 987q) = 0 \\ 987(987p + 610q) = -987 \end {cases}$

$\Rightarrow \begin {cases} 974170p+ 602070q =0 \\ 974169p+602070q=-987 \end {cases} \quad \Rightarrow p = \boxed {987}$

Yes, $\varphi = \frac {1\pm \sqrt{5}} {2}$ is the golden ratio, and $\varphi^n = F(n)\varphi + F(n-1)$ , where $F(n)$ is the $n^{th}$ Fibonaci number.

$g(x)=x^2-x-1=0$ $\Rightarrow x=\dfrac{1\pm\sqrt{5}}{2}$ let $\varphi=\dfrac{1+\sqrt{5}}{2}, \psi=\dfrac{1-\sqrt{5}}{2}$ $f(x)=px^{17}+qx^{16}+1$ Since $g(x)$ is factor of $f(x)$ , so $\varphi$ and $\psi$ are also the roots of $f(x)$ $f(\varphi)=p\varphi^{17}+q\varphi^{16}+1=0$ $f(\psi)=p\psi^{17}+q\psi^{16}+1=0$ The above equations are linear equations in terms of $p$ and $q$ . so we use cross-multiplication to solve them. $\Rightarrow \dfrac{p}{\varphi^{16}-\psi^{16} }= \dfrac{1}{\varphi^{17}\psi^{16} - \varphi^{16}\psi^{17}}$ $\Rightarrow p=\dfrac{\varphi^{16}-\psi^{16}}{\varphi^{16}\psi^{16}(\varphi-\psi)}$ We know that $\psi^{16}\varphi^{16}=(-1)^{16}=1$ and also that $F_{n}=\dfrac{\varphi^{n}-\psi^{n}}{\varphi-\psi}$ and $F_{16}=987$ Substituting these values, we get $\boxed{p=987}$

try reverse synthetic division :D