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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.

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    $\begingroup$ I don't know if this helps, but this is the first thing I saw: we can write the equation as $$\frac{dy}{dx}=\frac{y^2-1}{x^2-1}$$ where $y:=x^{2022}$. $\endgroup$
    – Dave
    Aug 10, 2019 at 19:25
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    $\begingroup$ @NoChance- Common ratio is $x^2$. $\endgroup$
    – aarbee
    Aug 10, 2019 at 19:30
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    $\begingroup$ Thanks for correcting this, now I have my glasses on... $\endgroup$
    – NoChance
    Aug 10, 2019 at 19:33
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    $\begingroup$ Well, it’s a polynomial equation, so we know it doesn’t have infinitely many solutions. Performing algebra to get a polynomial equal to zero, we see the degree of the polynomial is odd, so we know it has at least 1 solution (in fact, by guessing, we can see x=1 is a solution). But I’m not sure if there’s an easy way to confirm/deny the existence of another solution. $\endgroup$
    – Joe
    Aug 10, 2019 at 19:34
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    $\begingroup$ @Dave- Thanks to your input. I solved the integration, and got $y=x$ i.e.$x^{2021}=1$. $x$ being real, the only solution should be $x=1$. $\endgroup$
    – aarbee
    Aug 10, 2019 at 19:39

3 Answers 3

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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.

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    $\begingroup$ Doesn't Descartes' rule of signs give an odd number of roots, counting multiplicities? $\endgroup$
    – saulspatz
    Aug 10, 2019 at 19:50
  • $\begingroup$ No, you have a sign sequence $+,...,+,-,+,...,+$, this only has 2 sign variations, thus 2 or zero roots. $\endgroup$ Aug 10, 2019 at 19:52
  • $\begingroup$ If you mean the "compactified" polynomial from the question, this has indeed 3 sign variations, according to its triple root at $1$. Note that one of those is introduced by multiplying with $x^2-1$, so the original polynomial again has an even number of positive roots. $\endgroup$ Aug 10, 2019 at 19:56
  • $\begingroup$ Yes, that's what I meant. I understand now. $\endgroup$
    – saulspatz
    Aug 10, 2019 at 19:57
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    $\begingroup$ @Ramit : These counts are always with multiplicity, as the sign structure is invariant to small perturbations of the non-zero coefficients, and does thus not distinguish the generic case of simple roots from the exceptional case of roots with higher multiplicity. $\endgroup$ Aug 11, 2019 at 10:11
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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 $x=1$.

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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.

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