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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

Note by Akhilesh Prasad
3 years ago

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

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


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



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


and \(F_{16}=987\) Substituting these values, we get


b can be similarly found @Akhilesh Prasad

Try this similar problem too

U Z - 2 years, 9 months ago

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