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

#Algebra #AlgebraicManipulation

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Comments

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

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

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

Try this similar problem too

U Z - 6 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$