Problem by Arturo Presa - It is easier than it seems

Multiplying both sides of the equation $x^{100}+\frac{1}{x^{100}}=2$ by $x^{100}$ , subtracting $2 x^{100}$ from both sides and factoring we get $(x^{100}-1)^2=0.$ So we get that $x^{100}=1.$ Therefore $x= \cos(k\pi/50)+i\sin (k\pi/50)$ for $k\in \{0,1, 2, 3,..., 99\}$ . Now $x+\frac{1}{x}=\cos(k\pi/50)+i\sin (k\pi/50)+\frac{1}{\cos(k\pi/50)+i\sin (k\pi/50)}=$ $=\cos(k\pi/50)+i\sin (k\pi/50)+\cos(k\pi/50)- i\sin (k\pi/50)=2\cos(k\pi/50)$ So $a=2\cos(k\pi/50)$ . Since the sequence $\{2\cos(k\pi/50)\},$ where $k\in \{0,1, 2, 3,..., 99\},$ is decreasing , it attains its largest value at $k=0$ and that value is 2. The second largest value is the one corresponding to $k=1$ and it is $2\cos(\pi/50)$ which is approximately 1.996.

$f(X)= X^{100}+X^{-100}=2, \implies~X^{100}=1=e^{2*\pi*i} \\\therefore ~X=e^{\frac{n*2*\pi*i} {100} }, n~an~integer~from~1 ~to~100.\\If~n=1, f(X)=2, ~so~next~ best~n=2.\\ \large So~f(X)=X+X^{-1}=e^{\frac{2*2*\pi*I} {100}}+e^{\frac{-2*2*\pi*I} {100}} \\= ~~~~\Large \color{#D61F06} {1.996}$