# Degree 17 by Degree 2

Find integers $$a$$ and $$b$$ such that $$x^2 - x - 1$$ divides $$ax^{17} + bx^{16} + 1 = 0$$

This is the question that I wasn't able to solve in 17th KVS JMO

3 years, 9 months ago

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$g(x)=x^2-x-1=0$

$\Rightarrow x=\dfrac{1\pm\sqrt{5}}{2}$

let $\varphi=\dfrac{1+\sqrt{5}}{2}, \psi=\dfrac{1-\sqrt{5}}{2}$

$f(x)=ax^{17}+bx^{16}+1$

Since $$g(x)$$ is factor of $$f(x)$$, so $$\varphi$$ and $$\psi$$ are also the roots of $$f(x)$$

$f(\varphi)=a\varphi^{17}+b\varphi^{16}+1=0$

$f(\psi)=a\psi^{17}+b\psi^{16}+1=0$

The above equations are linear equations in terms of $$p$$ and $$q$$. so we use cross-multiplication to solve them.

$\Rightarrow \dfrac{a}{\varphi^{16}-\psi^{16} }= \dfrac{1}{\varphi^{17}\psi^{16} - \varphi^{16}\psi^{17}}$

$\Rightarrow a=\dfrac{\varphi^{16}-\psi^{16}}{\varphi^{16}\psi^{16}(\varphi-\psi)}$

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

$\boxed{a=987}$

b can be similarly found @Akhilesh Prasad

- 3 years, 5 months ago

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