<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] The trick with this question is to try finding $x^2+\frac{1}{x^2}$, via squaring $x+\frac{1}{x}$:
$(x+\frac{1}{x})^2=x^2+\frac{1}{x^2}+2$
$2=x^2+\frac{1}{x^2}+2$
$0=x^2+\frac{1}{x^2}$
The fact that this equals 0 is key to solving the problem, since subtracting $\frac{1}{x^2}$ from both sides and then putting both sides to the power of 1011 solves the problem:
$x^2=-\frac{1}{x^2}$
$(x^2)^{1011}=(-\frac{1}{x^2})^{1011}$
$x^{2022}=-\frac{1}{x^{2022}}$
$x^{2022}+\frac{1}{x^{2022}}=0$
So the answer is 0 and we are done!
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