# $(x^{2022}+1)(1+x^2+x^4+...+x^{2020})=2022\cdot x^{2021}$

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.

• 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 $y:=x^{2022}$.
– Dave
Aug 10, 2019 at 19:25
• @NoChance- Common ratio is $x^2$. Aug 10, 2019 at 19:30
• Thanks for correcting this, now I have my glasses on... Aug 10, 2019 at 19:33
• 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.
– Joe
Aug 10, 2019 at 19:34
• @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 $x=1$. Aug 10, 2019 at 19:39

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.

• Doesn't Descartes' rule of signs give an odd number of roots, counting multiplicities? Aug 10, 2019 at 19:50
• No, you have a sign sequence $+,...,+,-,+,...,+$, this only has 2 sign variations, thus 2 or zero roots. Aug 10, 2019 at 19:52
• 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. Aug 10, 2019 at 19:56
• Yes, that's what I meant. I understand now. Aug 10, 2019 at 19:57
• @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. Aug 11, 2019 at 10:11

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 $$(x^{2022}+1)(1+x^2+x^4+\ldots+x^{2020})=1+x^2+x^4+\ldots+x^{4042}\geq \\ \geq 2022\sqrt{x^{0+2+\ldots+4042}}=2022x^{2021}.$$ 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 $$x=1$$.

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 $$0. Let $$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. Consequently $$x=1$$ is a only thing solution.