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

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Let <math>\cos \theta + \sin \theta= c</math> and <math>\cos \theta - \sin \theta = d</math>.
 
Let <math>\cos \theta + \sin \theta= c</math> and <math>\cos \theta - \sin \theta = d</math>.
 
Then <math>1-c^2=d^2-1</math>.
 
Then <math>1-c^2=d^2-1</math>.
−
Let <cmath>P(x)\neq a_0+a_1(1-x^2)^2+a_2(1-x^2)^4+\dots+a_n(1-x^2)^{2n}</cmath> for <math>x\in</math> {<math>c,d</math>}. Then, clearly <cmath>P(c)\neq a_0+a_1(1-c^2)^2+
+
Let <cmath>P(x)= a_0+a_1(1-x^2)^2+a_2(1-x^2)^4+\dots+a_n(1-x^2)^{2n}</cmath> for <math>x\in</math> {<math>c,d</math>}. Then, clearly <cmath>P(c)=a_0+a_1(1-c^2)^2+
 
...a_n(1-c^2)^{2n}= a_0+a_1(d^2-1)^2+
 
...a_n(1-c^2)^{2n}= a_0+a_1(d^2-1)^2+
−
...a_n(d^2-1)^{2n}\neq P(d)</cmath> as <math>(1-c^2=d^2-1)</math> or <math>P(c) \neq P(d)</math> which is a contradiction. Thus,  <cmath>P(x)=a_0+a_1(1-x^2)^2+
+
...a_n(d^2-1)^{2n}=P(d)</cmath> as <math>(1-c^2=d^2-1)</math> or <math>P(c) = P(d)</math>. Thus,  <cmath>P(x)=a_0+a_1(1-x^2)^2+
 
...a_n(1-x^2)^{2n}</cmath> for infinite values of <math>x</math> (or particularly for <math>x\in</math> {<math>c,d</math>}). Next, let <cmath>f(x)=P(x)-(a_0+a_1(1-x^2)^2+a_2(1-x^2)^4+\dots+a_n(1-x^2)^{2n}</cmath>) where <math>f(x)</math> must be a polynomial of degree <math>k</math>.
 
...a_n(1-x^2)^{2n}</cmath> for infinite values of <math>x</math> (or particularly for <math>x\in</math> {<math>c,d</math>}). Next, let <cmath>f(x)=P(x)-(a_0+a_1(1-x^2)^2+a_2(1-x^2)^4+\dots+a_n(1-x^2)^{2n}</cmath>) where <math>f(x)</math> must be a polynomial of degree <math>k</math>.
  
 
Then, <math>f(x)=0</math> for all values of <math>x</math> because a nonzero polynomial of degree <math>k</math> can have at most <math>k</math> roots, which is only possible when <cmath>P(x)=a_0+a_1(1-x^2)^2+a_2(1-x^2)^4+\dots+a_n(1-x^2)^{2n}</cmath> for all x. ~Faraz0007
 
Then, <math>f(x)=0</math> for all values of <math>x</math> because a nonzero polynomial of degree <math>k</math> can have at most <math>k</math> roots, which is only possible when <cmath>P(x)=a_0+a_1(1-x^2)^2+a_2(1-x^2)^4+\dots+a_n(1-x^2)^{2n}</cmath> for all x. ~Faraz0007

Latest revision as of 11:45, 26 August 2025

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

$\textbf{Solution 2:}$ Let $\cos \theta + \sin \theta= c$ and $\cos \theta - \sin \theta = d$. Then $1-c^2=d^2-1$. Let \[P(x)= a_0+a_1(1-x^2)^2+a_2(1-x^2)^4+\dots+a_n(1-x^2)^{2n}\] for $x\in$ {$c,d$}. Then, clearly \[P(c)=a_0+a_1(1-c^2)^2+ ...a_n(1-c^2)^{2n}= a_0+a_1(d^2-1)^2+ ...a_n(d^2-1)^{2n}=P(d)\] as $(1-c^2=d^2-1)$ or $P(c) = P(d)$. Thus, \[P(x)=a_0+a_1(1-x^2)^2+ ...a_n(1-x^2)^{2n}\] for infinite values of $x$ (or particularly for $x\in$ {$c,d$}). Next, let \[f(x)=P(x)-(a_0+a_1(1-x^2)^2+a_2(1-x^2)^4+\dots+a_n(1-x^2)^{2n}\]) where $f(x)$ must be a polynomial of degree $k$.

Then, $f(x)=0$ for all values of $x$ because a nonzero polynomial of degree $k$ can have at most $k$ roots, which is only possible when \[P(x)=a_0+a_1(1-x^2)^2+a_2(1-x^2)^4+\dots+a_n(1-x^2)^{2n}\] for all x. ~Faraz0007

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