TITLE: $(x^{2022}+1)(1+x^2+x^4+...+x^{2020})=2022\cdot x^{2021}$
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: 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 $x=1$.