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

Solution - Let $P(x)=Q(x)(1-x^2)^2+ax^3+bx^2+cx+d$. Solving $P(1)=P(-1), P(\sqrt{2})=P(0), P(\frac{\sqrt 3 +1}2)=P(\frac{\sqrt 3 -1}2)$ We get $a=b=c=0$ and hence \begin{align} P(x)=Q(x)(1-x^2)^2+d \end{align} Now we show that $Q(x)$ also satisfies the problem conditions. Define $\displaystyle f(\theta)=Q($cos$\theta +$sin$\theta)-Q($cos$\theta -$sin$\theta)$ Then note that from (1) we have that $f(\theta)=0$ whenever $\theta\neq (2n+1)\frac{\pi}{2}$ But it is obvious that $f$ is continuous . Hence $f(x)=0 \forall x\in \mathbb R$. So we are done by induction.

A similar problem which I have seen before Let, $P(x)$ be a real valued polynomial. It is given that $P(\sin x)=P(\cos x)~\forall x$. Show that, $\exists ~ Q(x) \in \mathbb{R}(x)$ such that $P(x)=Q(x^4-x^2)$

BOBTHEGR8 wrote: Tooo easy Solution- $\theta=\frac{\pi}{2} \implies P(1)=P(-1)$ Now let $P(x)=(1-x^2)Q(x)+ax+b , x=1,x=-1 \implies a+b=-a+b \implies a=0 \implies P(x)=(1-x^2)Q(x)+b$ Continuing we get required result. How we continue?

First, we show that $P$ is even. Indeed, for $\theta \in \left(0,\frac{\pi}{2}\right)$ letting $\cos \theta=t$ we have that $P(t+\sqrt{1-t^2})=P(t-\sqrt{1-t^2})$ and for $\varphi \in \left(0,\frac{\pi}{2}\right)$ letting $\sin \varphi =t$(the same $t$ as above) we have $P(\sqrt{1-t^2}+t)=P(\sqrt{1-t^2}-t)$, combining these two we get $P(t-\sqrt{1-t^2})=P(-t+\sqrt{1-t^2})$ , and since the image of $t-\sqrt{1-t^2}$ is uncountable, we have that $P$ is even. Since $P$ is even there exists a polynomial $Q$ such that $P(x)=Q(x^2)$. Now Taylor expanding $Q$ we have $$P(x)=Q(x^2)=Q(1+(x^2-1))=Q(1)+Q'(1) (x^2-1)+\dots +\frac{Q^{(k)}(1)}{k!}(x^2-1)^k$$where $k$ is the degree of $Q$. Now, plugging this polynomial in the identity $P(t+\sqrt{1-t^2})=P(t-\sqrt{1-t^2})$ we see that $Q^{(i)}(1)$ should be zero for odd $i$'s.

anantmudgal09 wrote: 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$$P(x)=a_0+a_1(1-x^2)+a_2(1-x^2)^2+\dots+a_n(1-x^2)^n$$for some real numbers $a_0, a_1, \dots, a_n$ and non-negative integer $n$. Proposed by C.R. Pranesacher Hey dude fix the typo; the powers of $1-x^2$ are all even.

The given implies $P(\sqrt{2} \cos \theta)=P(\sqrt{2} \sin \theta).$ Let $P(1)=a_0=P(-1).$ Then define $Q(x):=P(x)-a_0.$ I showed that $1,-1$ are double roots of $Q(x)$ by differentiating to get $(1-x^2)^2 \mid Q(x).$ The rest is routine (there are technical details after this too). Did anyone else get a similar solution? Also, this loosely reminded me of USA TSTST 2014/3, although in retrospect, they are not very similar.

Wizard_32 wrote: The given implies $P(\sqrt{2} \cos \theta)=P(\sqrt{2} \sin \theta).$ Let $P(1)=a_0=P(-1).$ Then define $Q(x):=P(x)-a_0.$ I showed that $1,-1$ are double roots of $Q(x)$ by differentiating to get $(1-x^2)^2 \mid Q(x).$ The rest is routine. But how would that prove that there is no $(1-x^2)^3$ term?

3 questions: 1) Why we have $P(\sqrt{2} \cos \theta)=P(\sqrt{2} \sin \theta).$ 2) Why we have $1,-1$ are double roots of $Q(x)$ 3) The next step of the proof please.

Arhaan wrote: But how would that prove that there is no $(1-x^2)^3$ term? Write $P(x)=a_0+c(1-x^2)^2Q(x)$ and repeat for $Q(x).$

I hope this works. anantmudgal09 wrote: 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 $n$. Proposed by C.R. Pranesacher We have \[P(\sin \theta - \cos \theta) = P(\sin \theta + \cos \theta) =P(\cos \theta + \sin \theta) = P(\cos \theta - \sin \theta).\]Therefore $P(x) = P(-x)$ for infinitely many $x$, so $P(x) = P(-x)$ for all $x$. We write $P(x) = Q(x^2)$. The problem statement becomes \[Q(1+\sin 2\theta) = P(\cos \theta + \sin \theta) = P(\cos \theta - \sin \theta) = Q(1-\sin 2\theta).\]Therefore $Q(1 - x) = Q(1 + x) := R(x)$ for all $x$. Therefore \[R(x) = c + x^2S(x^2) \implies Q(x) = c + (x-1)^2S((x-1)^2)\implies P(x) = c + (x^2-1)^2S((x^2-1)^2).\]

TheDarkPrince wrote: I hope this works. anantmudgal09 wrote: 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 $n$. Proposed by C.R. Pranesacher Clearly $P(x) = P(-x)$ for all $x$, so write $P(x) = Q(x^2)$. The problem statement becomes \[Q(1+\sin 2\theta) = P(\cos \theta + \sin \theta) = P(\cos \theta - \sin \theta) = Q(1-\sin 2\theta).\]Therefore $Q(1 - x) = Q(1 + x) := R(x)$ for all $x$. Therefore \[R(x) = c + x^2S(x^2) \implies Q(x) = c + (x-1)^2S((x-1)^2)\implies P(x) = c + (x^2-1)^2S((x^2-1)^2).\] I did something similar. Not exact though.

TheDarkPrince wrote: I hope this works. anantmudgal09 wrote: 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 $n$. Proposed by C.R. Pranesacher Clearly $P(x) = P(-x)$ for all $x$ .... How!?

How much marks for solving by comparing coefficients

hellomath010118 wrote: TheDarkPrince wrote: I hope this works. anantmudgal09 wrote: 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 $n$. Proposed by C.R. Pranesacher Clearly $P(x) = P(-x)$ for all $x$ .... How!? Did he proved that $P(x)$ is an even function or is it obvious conclusion?

aops29 wrote:

I appreciate it, because at priority you first proved that it's even function then arguments followed consequently.

Proving that it was even was pretty important I guess, deserved to be emphasised. @below, i would like to see your solution, could you post it?

Chemisorption wrote: How much marks for solving by comparing coefficients Anyone?

For how RMO 2019 Corrections went 0.

Very similar to TSTST 2014/3. Obviously, shifting $P$ by a constant does not affect the truth of the given statement. Hence, shift so that $P(1) = 0$. Then, by substituting $x = \frac{\pi}{2}$, we get $P(1) = P(-1) = 0$. Then, taking the derivative of the given relation yields \begin{align*} (-\sin\theta + \cos\theta)P'(\cos\theta + \sin\theta) = (-\sin\theta - \cos\theta)P'(\cos\theta - \sin\theta). \end{align*}Substituting $\theta = 0$ into this relation gives \begin{align*} P'(1) = -P'(1), \end{align*}so $1$ is a root of $P'$. Similarly, $-1$ is a root of $P'$. This implies that $1, -1$ are double roots of $P$, so $(1 - x^2)^2 \mid P$. Additionally, note that $P = (1 - x^2)^2$ satisfies the given equation. Hence, we may divide out the factor of $(1 - x^2)^2$ without affecting the truth of the condition. Thus, by repeating this argument until we arrive at $P \equiv 0$, we find that the only polynomials $P$ which work are the ones described in the problem, so we are done. $\Box$

Is there any general form of even func. Or odd func...? Sorry for asking...I am a beginner

Puspi12345 wrote: Is there any general form of even func. Or odd func...? Sorry for asking...I am a beginner If $f(-x) = -f(x)$ then $f(x)$ is odd function and If $f(-x) = f(x)$ then $f(x)$ is an even function

The problem condition implies $P(\sqrt{2} \sin (\theta+\frac{\pi}4))=P(\sqrt{2} \sin (\frac{\pi}4 - \theta)) \implies P(\sqrt{2} \sin x)=P(\sqrt{2} \cos x)$. Now note that $P(\sqrt{2} \sin x)=P(\sqrt{2} \cos x)=P(\sqrt{2} \cos (-x)) =P(- \sqrt{2} \sin x) \implies P(y)=P(-y)$ for infinitely many y. Thus $P$ is even $\implies P(x)=Q(x^2) \implies Q(2 \sin ^2 x)=Q(2 \cos ^2 x)$. Hence we get $Q(1-\cos 2x)=Q(1+ \cos 2x)$. Letting $R(x)=Q(1+x)$, we get $R(x)=R(-x)$ for infinitely many $x \implies$ R is even $\implies R(x)=T(x^2)$. Thus, $P(x)=Q(x^2)=R(x^2-1)=T((1-x^2)^2)$, where $P,Q,R,T$ are all polynomials, and hence we're done.

First we observe, as others have done, $P(x)=Q(x^2)$ for some polynomial $Q(x)=b_0+b_1(1-x)+b_2(1-x)^2+\cdots b_n(1-x)^n$.We need to show $b_i=0$ for odd $i$. So its enough to show that $Q^{(i)}(1)=0$ for odd $i$. But this is very easy from $Q(1+\sin{2\theta})=Q(1-\sin{2\theta})$ (just differentiate desired number of times ans put $\theta=0$)

anantmudgal09 wrote: 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 $n$. Proposed by C.R. Pranesacher Pretty easy. Call a polynomial \(P\) diabetic if it satisfies the problems hypothesis. Also, we say that a polynomial \(Q\) is sugary if \[Q(\sin\theta+\cos\theta)+Q(\sin\theta-\cos\theta)=0\]Note that for any diabetic polynomial \(P\), we have \[(1-x^2)\mid P(x)-P(1)\]because \(P(1)=P(-1)\) and both \(1\) and \(-1\) are roots of \(1-x^2\). Therefore, \[P(x)=(1-x^2)Q(x)+c\]where \(c=P(1)\). This immediately tells us that \(Q\) is sugary by plugging \(x=\sin\theta+\cos\theta\) and \(\sin\theta-\cos\theta\) in the equation. We now prove a claim regarding sugary polynomials. Claim. If \(Q\) is a sugary polynomial, then \(1-x^2\) divides \(Q(x)\). Proof. This follows because plugging \(x=\frac{\pi}{2}\) gives us \(Q(1)=0\) and also plugging \(x=0\) gives us \(Q(-1)=0\), as desired. Now, let \(Q(x)=(1-x^2)R(x)\). Then, we see that \[P(x)=(1-x^2)^2R(x)+c\]implying that \(R\) is diabetic with \(\deg R<\deg P\). Continuing this procedure of diabetic polynomials, we see that \(P(x)\) has the desired form, and hence the result follows.

Chemisorption wrote: How much marks for solving by comparing coefficients Around 3/17 (according to a corrector).

Please find the solution of this problem in the attachment. Hope it helps

Solution from my inmo mock today similar to the geo p1, when i had first seen this problem just when starting oly math, i was sure i would never solve it. glad i proved myself wrong First a small note: If two polynomials are equal in some non trivial interval, they are equal everywhere, since their difference then has infinite roots. $\theta \rightarrow \frac{\pi}2 - \theta \implies P(\cos \theta - \sin \theta)= P(\sin \theta - \cos\theta)$ so $P$ is even, and as it is a polynomial, this means that it has only even degree terms, so it is possible to find a polynomial $A(x) \in \mathbb{R[X]}$ such that $P(x)=A(x^2)$ Now, for $\theta \in [0,\frac{\pi}4]$, the original condition is equivalent to $P(\sqrt{1+\sin 2\theta})=P(\sqrt{1-\sin 2\theta}) \implies A(1+x)=A(1-x)$ Let $A(x)=B(1-x) \iff B(x)=A(1-x)$ So, $P(x)=A(x^2)=B(1-x^2)$ So, it suffices to prove that $B$ is even. But $B(x) = A(1-x)=A(1+x)=B(-x)$ So we are done (edit: I am attaching a photo of my scratch work, because idk why i really like looking at old rough work)

Here goes the proof: Notice that: $$P(\cos \theta - \sin \theta) = P(\cos \theta + \sin \theta) \implies P(\cos (90^\circ - \theta) - \sin(90^\circ - \theta) = P(\cos (90^\circ - \theta) + \sin(90^\circ - \theta)$$$$\implies P(\sin \theta -\cos \theta) = P(\sin \theta + \cos \theta)$$Thus, $P(\cos \theta - \sin \theta) = P(\sin \theta + \cos \theta)$, for all $\theta \in \mathbb R$. Thus, $P$ is an even polynomial. Let $P(x) = Q(1-x^2)$ for some polynomial $Q$. Thus, it suffice to prove that $Q$ is an even polynomial. Notice that: $$P(\cos \theta - \sin \theta) = Q(\sin 2 \theta)$$$$ P(\cos \theta + \sin \theta) = Q(-\sin 2 \theta)$$Thus, $Q$ is an even polynomial and we conclude.

nice!

Is this correct? First let's prove that $P$ is even, note that $cos \ \theta + sin \ \theta$ is surjective in the interval $[ -\sqrt{2}, \sqrt{2}]$ and only takes value in that interval, let $cos \ \theta + sin \ \theta=x $. Then we have $$P(x)=P(\sqrt{2-x^2}) \Rightarrow P(x)=P(-x)$$. Since $P(x)=P(-x)$ for infinitely many $x$, $P$ is even. So we can let $Q(1-x^2)=P(x)$ where $Q$ is a polynomial with real coefficients. It suffices to prove that $Q$ is even, indeed $$Q(1-x^2)=P(x)=P(\sqrt{2-x^2})=Q(x^2-1)$$for all $x \in [ -\sqrt{2}, \sqrt{2}]$, by the same reason $Q$ is even. Hence we can write $P$ as the desired.

Note that $\theta = -\pi/2 \implies P(-1) = P(1)$, so consider $P_1(x) = P(x)- P(1)$. Note that $1-x^2 \mid P_1(x)$. Therefore, let $P_1(x) = (1-x^2)Q(x)$. The condition is equivalent to $Q(\cos \theta+\sin \theta) = -Q(\cos \theta-\sin \theta)$ for $\theta \in \mathbb{R} \setminus \{\frac{k\pi}{2}: k \in \mathbb{Z}\}$. But since $f(x) =Q(\cos \theta+\sin \theta)+Q(\cos \theta-\sin \theta)$ is continuous in $\theta$, the statement holds for all $\theta$. From here, $\theta = 0 \implies Q(1) = -Q(1) \implies Q(1) = 0$, and $\theta = \pi/2 \implies P(-1) = -P(1) = 0$. Thus $1-x^2 \mid Q(x).$ Now let $Q(x) = (1-x^2) R(x)$. Note that from the same continuity argument, $R$ satisfies $R(\cos \theta+\sin \theta)=R(\cos \theta-\sin \theta)$, and is of degree less than that of $P$. So from here we induct on the degree of $P$ and win