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 $x^{2022}+\frac{1}{x^{2022}}$?

The solutions you got are  (primitive) eighth roots of unity, $\zeta_8=e^{2\pi i/8},\dfrac 1{\zeta_8}=\bar {\zeta _8}$.
Now $$\zeta _8^{2022}=(\zeta _8^8)^{252}\cdot \zeta _8^6=1\cdot \zeta _8^6=\zeta_8^6$$.  And $\dfrac 1{\zeta_8^{2022}}=\zeta_8^{-6}$.
So  $$\zeta_8^6+\zeta_8^{-6}=-i+i=0$$
