2021 April MIMC 10 Problems/Problem 12

Given that $x^2-\frac{1}{x^2}=2$, what is $x^{16}-\frac{1}{x^{8}}+x^{8}-\frac{1}{x^{16}}$?

$\textbf{(A)} ~1120 \qquad\textbf{(B)} ~1180 \qquad\textbf{(C)} ~3780 \qquad\textbf{(D)} ~840\sqrt{2} \qquad\textbf{(E)} ~1260\sqrt{2}$

Solution

We know that $2=x^2-\frac{1}{x^2}=(x+\frac{1}{x})(x-\frac{1}{x})$. Squaring both sides, we can get that $(x+\frac{1}{x})^2\cdot(x+\frac{1}{x})^2=4$. Therefore, $(x^2+2+\frac{1}{x^2})(x^2-2+\frac{1}{x^2})=4$. Setting $a=x^2+\frac{1}{x^2}$, we can get that $(a+2)(a-2)=4$, and therefore $a^2-4=4$. So $x^2+\frac{1}{x^2}=\sqrt{8}=2\sqrt{2}$. In addition, $x^4-\frac{1}{x^4}=(x^2+\frac{1}{x^2})(x^2-\frac{1}{x^2})=2\cdot2\sqrt{2}=4\sqrt{2}$. $\frac{2}{x^2}=2\sqrt{2}-2$, so $\frac{1}{x^2}=\sqrt{2}-1$ and $\frac{1}{x^4}=3-2\sqrt{2}$. Therefore, $x^4+\frac{1}{x^4}=4\sqrt{2}+2(3-2\sqrt{2})=6$. Using a similar approach, we can calculate $x^8-\frac{1}{x^8}=4\sqrt{2}\cdot6=24\sqrt{2}$ and $\frac{1}{x^8}=(3-2\sqrt{2})^2=17-12\sqrt{2}$. Therefore, $x^8+\frac{1}{x^8}=24\sqrt{2}+2(17-12\sqrt{2})=34$. Also, $x^{16}-\frac{1}{x^{16}}=24\sqrt{2}\cdot34=816\sqrt{2}$. As a result, the answer of $x^{16}-\frac{1}{x^{8}}+x^{8}-\frac{1}{x^{16}}$ would be $816\sqrt{2}+24\sqrt{2}=\boxed{\textbf{(D)} 840\sqrt{2}}$.

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