$(x^{2022}+1)(1+x^2+x^4+...+x^{2020})=2022\cdot x^{2021}$ 
Let $S$ denote the set of all real values of $x$ for which $(x^{2022}+1)(1+x^2+x^4+...+x^{2020})=2022\cdot x^{2021}$, then the number of elements in $S$ is $0/1/2/$infinite?

My attempt: $$(x^{2022}+1)(\frac{(x^2)^{1011}-1}{x^2-1})=2022\cdot x^{2021}$$
$$x^{4044}-1=2022\cdot x^{2023}-2022\cdot x^{2021}$$
Don't know how to proceed next.
 A: The left side is positive, thus any real root will be positive. $x=0$ is not a solution.
If you multiply everything out you find that the polynomial is symmetric under degree reversion, so with some root $x$ also $x^{-1}$ is a root.  
If you divide by $x^{2021}$, the left side will turn into a convex function with a minimum at $x=1$, while the right side is constant. 
By inspection, $x=1$ is a solution.
Being a minimizer makes the root at $x=1$ into a double root. There are no other roots.

The polynomial equation can also be written as 
$$
0=\sum_{k=0}^{1010}(x^{2k}-2x^{2021}+x^{4042-2k})=\sum_{k=0}^{1010}x^{2k}(x^{2021-2k}-1)^2
\\
=(x-1)^2\sum_{k=0}^{1010}x^{2k}(x^{2020-2k}+x^{2019-2k}+...+x+1)^2
$$
which again is twice the linear factor for $x=1$ and then a positive factor with no additional root.
A: It is clear that equation does not have negative roots (otherwise left side is positive and right side is negative). Now, note that for every $x\geq 0$ from the AM-GM inequality we obtain that
$$
(x^{2022}+1)(1+x^2+x^4+\ldots+x^{2020})=1+x^2+x^4+\ldots+x^{4042}\geq 
\\
\geq 2022\sqrt[2022]{x^{0+2+\ldots+4042}}=2022x^{2021}.
$$
Therefore, if $x$ is a solution to the equation then we have $1=x^2=\ldots=x^{4042}$. Hence, the equation has only one root $x=1$.
A: The number of elements in $S$ is $1$. Proof. From $(x^{2022}+1)(1+x^2+x^4+...+x^{2020})=2022\cdot x^{2021}$ we see that $x>0$. Let $g(x)=2x^{2023}+2021-2023x^{2}$. Then $\frac{dg(x)}{dx}=4046x(x^{2021}-1)$ and $g(1)=0$. Consequently $g(x)>0$ if $x>1$ and $g(x)<0$ if $0<x<1$. Let $f(x)=x^{4044}-1+2022(x^{2021}-x^{2023})$. Then $\frac{df(x)}{dx}=2022x^{2020}g(x)$ and $f(1)=0$.Consequently $f(x)>0$ if $x>1$ and $f(x)<0$ if $0<x<1$. 
Consequently $x=1$ is a only thing solution.
