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

Your answers for $x=(1\pm i)/\sqrt2$, $1/x=(1\mp i)/\sqrt2$ are correct, now notice that
$$
{1\over x}={\bar x},
$$
the "complex conjugate", and
$$
x^2+{\bar x}^2={1+2i-1+1-2i-1\over 2}=0
$$
and $x^4={\bar x}^4=-1$,
so
$$
\begin{align}
x^{2022}+1/x^{2022}&=x^{4\times505}x^2+{\bar x}^{4\times505}{\bar x}^2\\
&=(-1)^{505}(x^2+{\bar x}^2)\\
&=0
\end{align}
$$.
