Overpowered $x$ 's - 3

$\begin{aligned} x +\frac {1} {x} & =0 \\ x^2 +1&=0 \\ \Rightarrow x&= \pm i \\ \Rightarrow x^{128} + \frac{1} {x^{128}} & = (\pm i)^{4 \times 32} + \frac {1} {(\pm i)^{4 \times 32} } \\ & = 1^{32} + \frac {1} {1 ^{32}} = \boxed {2} \end{aligned}$

Relevant wiki: Newton's Identities

Let $a_n=x^n+\frac{1}{x^n}$ . The question gives us $a_1=0$ .

$a_1 a_{n+1}=0=(x+\frac{1}{x})(x^{n+1}+\frac{1}{x^{n+1}})=x^{n+2}+\frac{1}{x^{n+2}}+x^{n}+\frac{1}{x^{n}}=a_{n+2} +a_n$ .

So $a_{n+2}+a_n=0 \Rightarrow a_{n+2}=- a_n \Rightarrow a_{n+4}=-a_{n+2}=-(-a_n)=a_n$ .

$a_2=x^2+\frac{1}{x^2}=(x+\frac{1}{x})^2 -2=a_1 -2=-2$ so $a_4=-(-2)=2$ .

We can extend the result $a_{n+4}=a_n \Rightarrow a_{n+8}=a_{n+4}=a_n ...$ to $a_{n+4k}=a_n$ for an integer $k$ by induction.

This gives $a_{128}=a_{4+4 \times 31}=a_4=2$

Assume $x\in \mathbb C$

$x=\mathrm{cis}(\theta)$

$x+\displaystyle\frac{1}{x}=0$

$\Rightarrow 2\cos(\theta)=0$

$\Rightarrow \theta=\frac{\pi}{2}$

Similarly

$x^{128}+\frac{1}{\displaystyle x^{128}}=2\cos(128 \theta)$

$\Rightarrow 2\cos(\displaystyle\frac{128 \times \pi}{2})=2\cos(64\pi)$ $\displaystyle = \boxed{2}$

Note that $\large {(x + \frac {1} {x})} ^ {2} = x^2 + \frac {1} {x^2} +2$ .

If $\large x+ \frac {1} {x} = 0$ then the equation $\large {x} ^ {{2} ^ {n}}+{(\frac {1} {x})} ^ {{2} ^ {n}}= 2$ is valid for all integer $n \geq 2$ .

Proof:

Squaring both sides once gives $x^2+ \frac {1} {x^2} = -2$

Squaring again gives $x^4+ \frac {1} {x^4} = 2$ Note that from now on squaring will double the power of x, but not change the constant, as $2^2-2=2$

This gives the desired formula.

Thanks to @Calvin Lin for feedback.

On a side note, I like these problems.