Finding a Root of a Polynomial in New Year's Eve - 2021

The key to solving this problem is to realize that we can represent the $n^{th}$ polynomial of the sequence by the formula $P_n(x) = \cos (2^n \arccos x),$ for all $x \in [-1,1].$ This is easy to prove by Mathematical Induction.

We can use the previous representation to find the real roots of $P_n(x).$ Indeed, we would need to solve the equation $\cos (2^n \arccos x)=0.$ Then we obtain that $2^n\arccos x =\frac{\pi}{2}+k \pi.\;\;\;\;\; (*)$
Since $0\leq \arccos x \leq \pi,$ then in the previous equation $k$ is any integer number satisfying $0\leq k \leq 2^n-1.$ Solving the equation $(*)$ for $x,$ we obtain $2^n$ distinct real solutions given by the formula $x_k=\cos(\frac{\pi}{2^{n+1}}+k \frac{\pi}{2^n})$ where $k$ is any integer number satisfying $0\leq k \leq 2^n-1.$

It is easy to see that the $x_k's$ introduced before are the $2^n$ distinct real roots of $P_n.$ Notice that the degree of $P_n$ is $2^n.$ Besides that, we can also see that $x_0> x_1>x_2>...>x_{2^n-1}.$

Now , assume that $n=11.$ Then $P_{11}$ has $2^{11}$ distinct real roots given by $x_k= \cos(\frac{\pi}{2^{12}}+k \frac{\pi}{2^{11}}),$ where $0\leq k \leq m=2^{11}-1=2047.$ Therefore $m \geq 2021$ and $x_{2021}=\cos(\frac{\pi}{2^{12}}+\frac{2021\pi}{2^{11}})=-0.999173... .$ So, the answer for this problem is $\boxed{999173}.$