Let $\alpha,\beta$ be the roots of the equation $x^2-ax-b=0$ with $\operatorname{Im}(\alpha)

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $P_{n+2}=aP_{n+1}+bP_n$. From $P_5$ and $P_6$ (dividing by $\sqrt7\,i$): $11=-3a-5b$ and $45=11a-3b$, so $b=-4$, $a=3$. Then $\alpha+\beta=3$, $\alpha\beta=4$, $\alpha^2+\beta^2=1$ and $\alpha^4+\beta^4=1-2\cdot16=-31$. The modulus is $31$. Answer: 31.

Q. Let $\alpha,\beta$ be the roots of the equation $x^2-ax-b=0$ with $\operatorname{Im}(\alpha)<\operatorname{Im}(\beta)$. Let $P_n=\alpha^n-\beta^n$. If $P_3=-5\sqrt7\,i$, $P_4=-3\sqrt7\,i$, $P_5=11\sqrt7\,i$ and $P_6=45\sqrt7\,i$, then $|\alpha^4+\beta^4|$ is equal to

Explanation

Reviewed by Dr. Abubacker

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