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

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## Comments

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TopNewest\[g(x)=x^2-x-1=0\]

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

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

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