Inspired by Chung Kevin

Note that for an integer $n$ , $x^{2n} + \dfrac{1}{x^{2n}} =\left(x^n+\dfrac{1}{x^n}\right)^2 - 2$ .

So if $T_n=x^n+\dfrac{1}{x^n}$ then we have the recurrence relation $T_{2n} = T_{n}^2 - 2$ . So let us see what happens for different values of $n$ :

$\begin{aligned}T_{1} &= \sqrt{3} \\ T_2 &= (\sqrt{3})^2 - 2 = 3-2=1 \\ T_4 &=1^2-2 = -1 \\ T_8 &=(-1)^2-2=-1 \\ T_{16} &=(-1)^2-2=-1 \\T_{32} &=(-1)^2-2=-1 \\ \dots \end{aligned}$

What do we observe now? Since $(-1)^2-2=-1$ , we have another interesting relation : $\Large T_{2^y} = -1$ for all integers $y>1$ .

Hence , $\Large T_{2^{1729}} = \boxed{-1}$ .

Again, I will lift some of them from original. Notice that $\displaystyle x+\frac{1}{x} = 2\cos(\frac{\pi}{6})\Rightarrow x =\cos(\frac{\pi}{6})+i\sin(\frac{\pi}{6})$ Hence $\displaystyle x^{\pm 2^{1729}} = \cos(\frac{4^{864}\pi}{3})\pm i\sin(\frac{4^{864}\pi}{3}) = \cos(\frac{2\pi}{3})\pm i\sin(\frac{2\pi}{3}) = -\frac{1}{2}\pm\frac{\sqrt{3}i}{2}$ Therefore $x^{2^{1729}}+x^{-2^{1729}} = 2(\frac{-1}{2}) = -1$