<s>[INST] if $x+\frac{1}{x}=\sqrt2$, then find the value of $x^{2022}+\frac{1}{x^{2022}}$? It is question of mathematical olympiad. kindly solve it guys!
I tried a bit.I am sharing this with u...
•$x+\frac{1}{x}=\sqrt{2}$
•$x^2+1=x\sqrt{2}$
•$x^2-x\sqrt{2}+1=0$
so, $x=\frac{\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}$ or $\frac{\sqrt{2}}{2}-\frac{i\sqrt{2}}{2}$ and
$\frac{1}{x}=\frac{\sqrt{2}}{2}-\frac{i\sqrt{2}}{2}$ or $\frac{\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}$
How can I find $x^{2022}+\frac{1}{x^{2022}}$?
 [/INST] $$x + \frac{1}{x} = \sqrt2 $$
squaring both sides,
$$x^2 + \frac{1}{x^2} = 0 $$
then,
$$x^{2022} + \frac{1}{x^{2022}} = (x^2 + \frac{1}{x^2})(x^{2020} + \frac{1}{x^{2020}}) - (x^{2018}+ \frac{1}{x^{2018}}) = - (x^{2018}+ \frac{1}{x^{2018}})$$
repeat the above process,
$$x^{2022} + \frac{1}{x^{2022}} = x^2 + \frac{1}{x^2} = 0$$
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