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 $x^{2022}+\frac{1}{x^{2022}}$?
 A: 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!
A: 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
$$
\begin{align}
x^{2022}+1/x^{2022}&=x^{4\times505}x^2+{\bar x}^{4\times505}{\bar x}^2\\
&=(-1)^{505}(x^2+{\bar x}^2)\\
&=0
\end{align}
$$.
A: $$x + \frac{1}{x} = \sqrt2 $$
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,
$$x^{2022} + \frac{1}{x^{2022}} = x^2 + \frac{1}{x^2} = 0$$
A: 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  $$\zeta_8^6+\zeta_8^{-6}=-i+i=0$$
