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Difference between revisions of "2020 INMO Problems/Problem 2"

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Latest revision as of 13:02, 5 November 2020

PROBLEM

Suppose $P(x)$ is a polynomial with real coefficients, satisfying the condition $P(\cos \theta+\sin \theta)=P(\cos \theta-\sin \theta)$, for every real $\theta$. Prove that $P(x)$ can be expressed in the form\[P(x)=a_0+a_1(1-x^2)^2+a_2(1-x^2)^4+\dots+a_n(1-x^2)^{2n}\]for some real numbers $a_0, a_1, \dots, a_n$ and non-negative integer $n$.

$\emph{Proposed by C.R. Pranesacher}$

SOLUTION(1)

Assume to the contrary. Suppose $P$ satisfies $P(\cos \theta + \sin \theta)=P(\cos \theta - \sin \theta)$ for all real $\theta$, and is of minimal degree and not of the prescribed form.

$\textbf{Claim:}$ For some $c \in \mathbb{R}$, we have $(1-x^2)^2 \mid P(x)-c$.

$\emph{Proof.}$ Note that $\theta=\frac{\pi}{2} \implies P(1)=P(-1)$. Set $c=P(1)$. Then $(1-x^2) \mid P(x)-c$ as the latter vanishes at both $\pm 1$. Now let $P(x)-c=(1-x^2)Q(x)$ for some $Q \in \mathbb{R}[x]$.

Then $Q(\cos \theta+\sin \theta)=-Q(\cos \theta-\sin \theta)$ holds for all $\theta$, by plugging $P(x)=(1-x^2)Q(x)$ in the original equation, since we have the identities $1-(\cos \theta + \sin \theta)^2=-\sin 2\theta$ and $1-(\cos \theta - \sin \theta)^2=\sin 2\theta$.

(Subtlety for beginners: while the equation in $Q$ only holds for $\theta$ away from roots of $\sin 2\theta=0$, since these form a discrete subset of $\mathbb{R}$, the equation extends to these as $Q$ is continuous.)

In particular, plugging $\theta=0, \pi$ we get $Q(1)=-Q(1)$ and $Q(-1)=-Q(-1)$ so $Q(\pm 1)=0$, hence $(1-x^2) \mid Q(x)$. Thus, $(1-x^2)^2 \mid P(x)-c$ as desired. $\blacksquare$

Finally, we see that $P(x)=c+(1-x^2)^2h(x)$ and $\text{deg} h<\text{deg} P$ so $h$ has the prescribed form. But then $P$ also has the prescribed form, and our result follows. $\blacksquare$ ~anantmdgal09

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