I believe that the 8th Chebyshev polynomial works?

I am too lazy to post a full proof so here is a general outline: (1) Note that the monic polynomial of minimal degree to satisfy the equation is P(x) = x^2 * (x-1)^2 * (x+1)^2 * (x^2 - 2). Denote this polynomial by "X." (2) It is now clear that every polynomial "in" X work, and to prove that these are the only solutions, take an arbitrary polynomial, and reduce it modulo X. This must yield a constant, or else it contradicts the minimality of X. So every polynomial that works is a multiple of X plus a constant. Since taking a polynomial that works, subtracting the relevant constant and dividing it by X yields another polynomial that works, by induction the result follows. The hardest part of this problem is step 1 (I personally have rarely seen an answer so ugly). There are many different motivations... Chebyshev polynomials (this itself is motivated by the apparent trig sub), root-hunting, and complex numbers all lead to the correct result. However, what I personally did on the test was to just trudge through computation.... to get the square root out of x + sqrt(1-x^2) one must square it, subtract 1, and square it again. Performing the same operation to xsqrt(2) we have two fourth degree polynomials, and subtracting their average from one of them and squaring it we get the polynomial that we want (up to the constant term).

Another motivation is to start by taking the derivative of both sides, so we get $\sqrt{2}P'(x \sqrt{2})=(1-\frac{x}{\sqrt{1-x^2}})P'(x+\sqrt{1-x^2})$. Taking the limit of both sides as $x$ approaches $1$, we find we must have $P'(1)=0$. This also leads to $P'(\sqrt{2})=0$. Plugging in $0$ for $x$, we find $P'(0)=0$. Plugging in $-\frac{\sqrt{2}}{2}$ for $x$, we must have $P'(-1)=0$.

If anyone was interested, here is a full proof: Consider the monic polynomial $ f(x) $ of minimal degree that satisfies the equation. Now, assume WLOG that $ f(0) = 0 $. Letting $ x = 0 $ we have that $ f(1) = 0 $. Letting $ x = 1 $ we have that $ f(\sqrt{2}) = 0 $. Letting $ x = -\frac{\sqrt{2}}{2} $ we have that $ f(-1) = 0 $. Letting $ x = -1 $ we have that $ f(-\sqrt{2}) = 0 $. Now, differentiating both sides we have that $ \sqrt{2}f'(x\sqrt{2}) = (1 - \frac{x}{\sqrt{1 - x^2}})f'(x+\sqrt{1-x^2}) $. Letting $ x = \frac{\sqrt{2}}{2} $ we have that $ f'(1) = 0 $. Letting $ x = 1 $ we have that $ f'(0) = 0 $. Letting $ x = -\frac{\sqrt{2}}{2} $ we have that $ f'(-1) = 0 $. This implies that $ x^2(x - 1)^2(x + 1)^2(x^2 - 2) $ is a factor of $ f $ and upon checking we find that this eight degree polynomial (with any real constant added to it) is the minimal solution to the equation. It is also clear that any polynomial in $ f $ is a solution. I claim that these are the only solutions. Consider an arbitrary polynomial in $ \mathbb{R}[x] $ that is a solution to this equation. After subtracting off its constant term, it is clear that this polynomial must be a multiple of $ f $ or else we could reduce it modulo $ f $ and contradict the minimality of $ f $. After subtracting the constant term and dividing by $ f $, we get a new polynomial that also is a solution to the equation. After repeating this process, we inductively obtain the desired result.

If $0 <\theta<90$, then we easily see that$ P(\sqrt{2}\sin\theta)= P(\sqrt{2}\sin(\theta+45))$. If $z=e^{i\theta}$, and $w$ is the eighth root of unity, then we have $P(\frac{1}{\sqrt{2}i}(z-1/z))=P(\frac{1}{\sqrt{2}i}(zw-1/zw))$, for an infinite number of values of $z$. Then it must be true for all $z$. Repeating yields that $P(\frac{1}{\sqrt{2}i}(z-1/z))=P(-\frac{1}{\sqrt{2}i}(z-1/z))$, so $P(\sqrt{2}x)=Q(1-2x^2)$, for some polynomial $Q$. Then $Q(\cos2\theta)=Q(-\sin2\theta)$, for an infinite number of $\theta$. Again, $Q(x)$ must have only even exponents. It is easily seen that the minimal polynomial cannot be of degree 2, so the minimal polynomial must be $Q(x)=x^2(1-x^2)$, which yields $P(x)=x^2(x^2-1)^2(x^2-2)$ minimal.

All the above proofs are too smart! Here's the solution that I wrote during the actual exam. (ksun48 had the same solution) I will explicitly find 8 values of $P$ that are equal. Set $x = \frac{\sqrt{6}-\sqrt{2}}{4}$ to get that $P \left(\frac{\sqrt{3}-1}{2} \right) = P \left( \frac{\sqrt{6}}{2} \right)$. Set $x = \frac{\sqrt{3}}{2}$ to get that $P \left( \frac{\sqrt{6}}{2} \right) = P \left( \frac{\sqrt{3}+1}{2} \right)$. Set $x = \frac{1}{2}$ to get that $P \left( \frac{\sqrt{2}}{2} \right) = P \left( \frac{\sqrt{3}+1}{2} \right)$. Set $x = -\frac{1}{2}$ to get that $P \left( -\frac{\sqrt{2}}{2} \right) = P \left( \frac{\sqrt{3}-1}{2} \right)$. Set $x = -\frac{\sqrt{6}-\sqrt{2}}{4}$ to get that $P \left(-\frac{\sqrt{3}-1}{2} \right) = P \left( \frac{\sqrt{2}}{2} \right)$. Set $x = -\frac{\sqrt{3}}{2}$ to get that $P \left( -\frac{\sqrt{6}}{2} \right) = P \left( -\frac{\sqrt{3}-1}{2} \right)$. Set $x = -\frac{\sqrt{6}+\sqrt{2}}{4}$ to get that $P \left( -\frac{\sqrt{3}+1}{2} \right) = P \left( -\frac{\sqrt{2}}{2} \right)$. So the above 8 values in the equations are equal! Call them $v_1, v_2, \dots, v_8$. Say that they all equal some constant $c$. Then \[ P(x)-c = Q(x) \prod_{i = 1}^8 (x-v_i). \] Another direct computation shows that $\prod_{i = 1}^8 (x-v_i)$ in fact satisfies the polynomial equation! Substituting $P(x) = Q(x) \prod_{i = 1}^8 (x-v_i) + c$ into the equation, and cancelling, you get that $Q(x)$ satisfies the equation too! So you induct down, and all polynomials of $\prod_{i = 1}^8 (x-v_i)$ work. What a beautiful solution, and beautiful problem. In case you doubt that this solution works, in you expand everything, you actually get that $\prod_{i = 1}^8 (x-v_i)$ differs from $x^2(x^2-1)^2(x^2-2)$ by a constant only!

Notice that we have $P(\sqrt{2}x)=P(x+\sqrt{1-x^2})=P((\sqrt{1-x^2})+\sqrt{1-(\sqrt{1-x^2})^2})=P(\sqrt{2}\sqrt{1-x^2})$ and so if $x^2+y^2=2$ then $P(x)=P(y)$. Therefore since $(-x)^2=x^2$ we have $P$ is even. Let $Q$ be the polynomial that is the same as $P$ except divide all the exponents by $2$, and so $P(x)=Q(x^2)$ and $Q \in \mathbb{R}[X]$. The original equation becomes $Q(2x^2)=Q(1+2x\sqrt{1-x^2})$ and notice $Q(2x^2)=Q(1+2x\sqrt{1-x^2})=Q(2(1-x^2))$. Let $R(x)=Q(x+1)$, then $R(2x^2-1)=R(2(1-x^2)-1)$ and so $R(a)=R(-a)$ for any $a \in [-1,1]$ and therefore $R$ is even. The original equation becomes $R(2x^2-1)=R(2x\sqrt{1-x^2})$. Now let $S$ be the polynomial that is the same as $R$ except divide all exponents by $2$, so $R(x)=S(x^2)$. The original equation becomes $S((2x^2-1)^2)=S(4x^2(1-x^2))$ if $|x| \le 1$. Therefore for any $0 \le a \le 1$ we have $S((2a-1)^2)=S(4a(1-a))$ by letting $a=x^2$. Now let $b=4a^2-4a$, which can take on infinitely many values if $0 \le a \le 1$. We have $S(b+1)=S(-b)$ and therefore since this is true for infinitely many $b$, we have $S(x+1)=S(-x)$ for all $x$. Now let $T(x)=S\left( x+\frac{1}{2} \right)$. The above equation gives $T(x)=T(-x)$ for all $x$ and therefore $T$ is even. Let $U$ be the polynomial that is the same as $T$ except divide all exponents by $2$, so $T(x)=U(x^2)$. Notice $U \in \mathbb{R}[X]$. Now notice $P(x)=Q(x^2)=R(x^2-1)=S((x^2-1)^2)=T \left( (x^2-1)^2-\frac{1}{2} \right)=U \left( \left( (x^2-1)^2-\frac{1}{2} \right) ^2 \right)$. For the original equation to be satisfied we need, for $|x|\le 1$, $P(\sqrt{2}x)=P(x+\sqrt{1-x^2}) \Leftrightarrow T \left( (2x^2-1)^2-\frac{1}{2} \right) = T \left( ((x+\sqrt{1-x^2})^2-1)^2-\frac{1}{2} \right) \Leftrightarrow T \left( 4x^4-4x^2+\frac{1}{2} \right) = T \left( 4x^2-4x^4-\frac{1}{2} \right)$, which is a direct consequence of $T$ being even. Now notice $\left( (x^2-1)^2-\frac{1}{2} \right)^2 = x^8-4x^6+5x^4-2x^2+\frac{1}{4}$. Therefore the answer is $\boxed{P(x)=U\left( x^8-4x^6+5x^4-2x^2+\frac{1}{4} \right)}$ for any $U \in \mathbb{R}[X]$.

First we prove that Lemma: For a polynomial $P$, note there exists a unique sequence $(c_0, c_1,...)$ such that $P(X)=\sum_{i\ge 0} c_i T_i(X)$ where $T_n$ is the $n$th Chebyshev polynomial (meaning $T_n (\cos \theta)=\cos (n\theta)$). Denote this as the C-sequence of $P$. To prove existence and uniqueness, simply note we can induct on the degree of $P$ since $T_n$ has degree $n$ for all $n$. $\blacksquare$ Note that $|x|\le 1$ implies that if we take $\theta = \cos^{-1}(x)\in [0,\pi]$, we can substitute in to get $P(\cos (\theta) \sqrt{2})=P(\cos (\theta -\pi/4)\sqrt{2})$. Write $Q(X)=P(X\sqrt{2})$. We have the identity $Q(\cos \theta)=Q(\cos (\theta -\pi/4))$ for $\theta \in [0,\pi]$. Since $\cos \theta$ is even, note that we can extend this easily to all $\theta$ in $[0,2\pi]$. Now consider the C-sequence of $Q$. We have $\sum c_i T_i(\cos \theta)=\sum c_i T_i(\cos (\theta -\pi/4))$. This rearranges to $\sum c_i \cos (i\theta)=\sum c_i [\cos (i\theta)\cos (i\pi/4)+\sin (i\theta)\sin (i\pi/4)]$. But then by taking $\theta$ and then $-\theta$, it follows that $\sum c_i\sin (i\theta)\sin (i\pi/4)=0$; thus $\sum c_i(1-\cos (i\pi/4))\cos (i\theta)=0$. Because this is true for all $\theta$, it follows that if $R(X)$ has the C-sequence $(c_i(1-\cos (i\pi/4))$ then $R(X)=0$ for all $X$. But then this implies that, by uniqueness, $c_i(1-\cos (i\pi/4))=0$ for all $i$. Thus $c_i\neq 0\rightarrow 8\mid i$. In particular, because $T_{8n}(X)=T_n(T_8(X))$ we have $Q(X)=U(T_8(X))$ for $U\in\mathbb{R}[X]$ and $P(X)=Q(X/\sqrt{2})$ for some $U(X)$. Such solutions clearly work.

This is too hard FE Also is the TSTST the team selection test selection test... That's weird. Just wondering

Let $Q(x) = x^2(x^2 - 1)^2(x^2 - 2)$. We claim that the only polynomials which are work are those of the form $P(x) = a_nQ(x)^n + a_{n - 1}Q(x)^{n - 1} + \cdots + a_0$, where the $a_i$ are real numbers. We first rewrite the given relation in terms of trigonometry. Substitute $x = \sin\theta$, for $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$. We have \begin{align*} x + \sqrt{1 - x^2} &= \sin\theta + \cos\theta \\ &= \sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right), \end{align*}where the second equality is due to sum to product formula. Hence, the given relation rewrites as \begin{align} P(\sqrt{2}\sin\theta) &= P\left(\sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right)\right) \tag{$\spadesuit$} \end{align}where $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$. Furthermore, for $\frac{\pi}{2} \leq \theta \leq \frac{3\pi}{2}$, we have \begin{align*} P(\sin\theta) &= P(\sin(\pi - \theta)) \\ &= P\left(\sin\left(5\frac{\pi}{4} - \theta\right)\right) \\ &= P\left(\sin\left(\theta + \frac{\pi}{4}\right)\right), \end{align*}where the second equality holds because $-\frac{\pi}{2} \leq \pi - \theta \leq \frac{\pi}{2}$, whence $(\spadesuit)$ can be applied. As a result, $(\spadesuit)$ is in fact true for all $\theta$. Note that adding a constant to $P$ does not change the truth of $(\spadesuit)$; thus, we may WLOG assume that $P(0) = 0$. Now, by $(\spadesuit)$, \begin{align*} P\left(\sqrt{2}\sin\frac{-\pi}{2}\right) = P\left(\sqrt{2}\sin\frac{-\pi}{4}\right) &= P(\sqrt{2}\sin0) = P\left(\sqrt{2}\sin\frac{\pi}{4}\right) = P\left(\sqrt{2}\sin\frac{\pi}{2}\right) \\ \implies P(\pm\sqrt{2}) &= P(\pm 1) = P(0). \end{align*}Therefore, $0, \pm 1, \pm\sqrt{2}$ are all roots of $P$. Now, we claim that $0$ and $\pm 1$ are double roots of $P$. Indeed, differentiating $(\spadesuit)$, we obtain (by chain rule) \begin{align*} \sqrt{2}\cos\theta P'(\sqrt{2}\sin\theta) &= \sqrt{2}\cos\left(\theta + \frac{\pi}{4}\right)P'\left(\sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right)\right) \\ \cos\theta P'(\sqrt{2}\sin\theta) &= \cos\left(\theta + \frac{\pi}{4}\right)P'\left(\sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right)\right). \end{align*}Hence, we have \begin{align*} \cos\frac{-\pi}{4} P'\left(\sqrt{2}\sin\frac{-\pi}{4}\right) = \cos 0 P'\left(\sqrt{2}\sin 0\right) &= \cos\frac{\pi}{4} P'\left(\sqrt{2}\sin\frac{\pi}{4}\right) = \cos\frac{\pi}{2} P'\left(\sqrt{2}\sin\frac{\pi}{2}\right) \\ \frac{\sqrt{2}}{2}P'(\pm 1) &= P'(0) = 0, \end{align*}which is enough to imply that $\pm 1$ and $0$ are roots of $P'$ as well. Therefore, as they are roots of both $P'$ and $\pm 1$, it follows that $\pm 1$ and $0$ are double roots of $P$. Hence, we have \begin{align*} P(x) &= Q(x)R(x), \end{align*}for some polynomial $R(x)$. Finally, note that \begin{align*} Q(\sqrt{2}\sin\theta) &= 2\sin^2\theta(2\sin^2\theta - 1)^2(2\sin^2\theta - 2) \\ &= -4\sin^2\theta\cos^2(2\theta)\cos^2\theta \\ &= -(2\sin\theta\cos\theta\cos(2\theta))^2 \\ &= -(\sin(2\theta)\cos(2\theta))^2 \\ &= -\frac{1}{4}\sin^2(4\theta). \end{align*}In particular, this implies \begin{align*} Q(\sqrt{2}\sin\theta) &= -\frac{1}{4}\sin^2(4\theta) \\ &= -\frac{1}{4}\sin^2\left(4\left(\theta + \frac{\pi}{4}\right)\right) \\ &= Q\left(\sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right)\right). \end{align*}Hence, we may rewrite $(\spadesuit)$ as \begin{align*} P(\sqrt{2}\sin\theta) &= P\left(\sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right)\right) \\ Q(\sqrt{2}\sin\theta)R(\sqrt{2}\sin\theta) &= Q\left(\sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right)\right)R\left(\sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right)\right) \\ R(\sqrt{2}\sin\theta) &= R\left(\sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right)\right) \end{align*}as long as $Q(\sqrt{2}\sin\theta) \neq 0$. By continuity, the above equality holds for all $\theta$. Thus, $R$ satisfies the given condition as well, so we may repeat the above argument on $R$. By repeating this argument until we eventually reach a constant polynomial, we find that $P(x)$ is of the form $P(x) = a_nQ(x)^n + a_{n - 1}Q(x)^{n - 1} + \cdots + a_0$, for some reals $a_0, \ldots, a_n$. Finally, we will check that all such $P$ satisfy $(\spadesuit)$, which is equivalent to the problem. We have \begin{align*} Q(\sqrt{2}\sin\theta) &= Q\left(\sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right)\right) \\ \implies Q(\sqrt{2}\sin\theta)^m &= Q\left(\sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right)\right)^m \end{align*}for any nonnegative integer $m$, which is enough to imply that \begin{align*} P(\sqrt{2}\sin\theta) &= a_nQ(\sqrt{2}\sin\theta)^n + a_{n - 1}Q(\sqrt{2}\sin\theta)^{n - 1} + \cdots + a_0 \\ &= a_nQ\left(\sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right)\right)^n + a_{n - 1}Q\left(\sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right)\right)^{n - 1} + \cdots + a_0 \\ &= P\left(\sqrt{2}\sin\left(\theta + \frac{\pi}{4}\right)\right). \end{align*}Thus, all such $P$ do indeed satisfy $(\spadesuit)$, so we are done.

Solved with goodbear, Th3Numb3rThr33. The answer is \(P(x)\equiv f(T_8(x/\sqrt2))\), where \(T_8\) is the eighth Chebyshev polynomial and \(f\) is any polynomial. The problem is better phrased in terms of \(Q(x)\equiv P(x\sqrt2)\). The condition is then written as \[Q(\cos\theta)=Q(\cos(\theta-\pi/4))\]for all \(\theta\). It is easy to check \(f\circ T_8\) is a valid solution to this functional equation. Now we check these are the only solutions. Claim: \(\deg Q\ge8\) or \(Q\) is constant. Proof. Observe that \(Q\) is equal at \(\cos1\), \(\cos(1+\pi/4)\), \ldots, \(\cos(1+7\pi/4)\). \(\blacksquare\) Hence \(T_8\) is the monic polynomial of minimal degree obeying the functional equation. For any solution \(Q\), let its remainder upon division by \(T_8\) be \(R\). Then \(\deg R\le7\) and \(R\) also obeys the functional equation, so \(R\) is constant, the end.

Man I wish I had done this before INMO!!Anyway here goes my not so different solution:- numbertheorist17 wrote: Find all polynomials $P(x)$ with real coefficients that satisfy \[P(x\sqrt{2})=P(x+\sqrt{1-x^2})\]for all real $x$ with $|x|\le 1$. The answer is all polynomials in $Q(x)=x^2(x^2-1)^2(x^2-2)$. It is easy to verify this indeed fits in the original assertion. Now to get to this polynomial we go through a lot of steps:- Plugging $x=0,1,\dfrac{-1}{\sqrt{2}}$ we see $P(0)=P(1)=P(-1)=P(\sqrt{2})=P(-\sqrt{2})$.Thus $$x(x^2-1)(x^2-2)\mid P(x)-P(0) \Longleftrightarrow P(x)=Q(x)x(x^2-1)(x^2-2)+P(0)$$Comparing to the original assertion we get $$Q(x\sqrt{2})(x\sqrt{2})(2x^2-1)(2x^2-2)=Q(x+\sqrt{1-x^2})(x+\sqrt{1-x^2})(2x\sqrt{1-x^2})(2x\sqrt{1-x^2}-1)$$After simplifying and plugging $x=0,1,-1$ we get $Q(1)=Q(-1)=Q(0)=0$ thus $Q(x)=R(x)(x)(x^2-1)$; Hence we get $P(x)=R(x)x^2(x^2-1)^2(x^2-2)+P(0)$ so plugging in original assertion :- $$P(x\sqrt{2})=P(x+\sqrt{1-x^2})\Longleftrightarrow R(x\sqrt{2})(2x^2)(2x^2-1)^2(2x^2-2)=R(x+\sqrt{1-x^2})(1+2x\sqrt{1-x^2})(2x\sqrt{1-x^2})^2(2x\sqrt{1-x^2}-1)\Longleftrightarrow R(x\sqrt{2})=R(x+\sqrt{1-x^2})$$Thus $R(x)$ is a polynomial satisfying the same condition as the assertion but $\text{deg(R)}=\text{deg(P)}-8$ so we can induct down and so the conclusion follows.$\blacksquare$

Let $T(x)$ be the $8$-th Chebyshev polynomial. We argue that $P$ works if and only if $P(x)=S(T(x/\sqrt{2}))$ for some polynomial $S$. Let $Q(x)=P(x/\sqrt{2})$ and sub $x=\cos\theta$ so \[Q(\cos\theta)=Q(\cos\theta/\sqrt{2}+\sin\theta/\sqrt{2})=Q(\cos(\theta-45^\circ)).\]At this point, it is clear that any such polynomial $P$ works. Take $Q$ modulo $T(x)$ to yield a polynomial with degree at most $7$. This polynomial is constant because \[Q(\cos 136^\circ)=Q(\cos 91^\circ)=Q(\cos 46^\circ)=Q(\cos 1^\circ)=Q(\cos (-44^\circ))=\]\[Q(\cos 44^\circ)=Q(\cos 89^\circ)=Q(\cos 134^\circ)=Q(\cos 179^\circ),\]that is, $Q$ is the same at $8$ values of $x$, so we are done.

This is a very mechanical solution. I won't write any more because you can probably figure out the details, but... Imported from Discord wrote: my sol was literally like this prove even to substitute P(x)=Q(x^2) substitute Q(x)=R(x-1) prove even to substitute R(x)=S(x^2) substitute S(x)=T(x-1/2) prove even to substitute T(x)=U(x^2) remark that once we substitute U(x) the original FE becomes a tautology, so all such U(x) work

Let $x = \cos \theta$. Subbing back we get \[ P(\sqrt{2} \cos \theta) = P(\sin \theta + \cos \theta) = P(\sqrt 2 \cos(\theta-\pi/4)) \]Thus, if we let $P(x/\sqrt{2})=Q(x)$ then $Q(\cos \theta)=Q(\cos(\theta+\pi/4))$. Letting $T$ be the 8'th Chebyshev polynomial we have $f \circ T$ works for any polynomial $f$. Assume $g_0$ is a polynomial for which $\deg g_0 \geq 8$ and $g_0 \not = f \circ T$ for any polynomial $f$. We can write $g_0$ as $f_1 \circ T + g_1$ where $f_1$ and $g_1$ are polynomials satisfying $\deg g_0 > \deg g_1$. Notice that if $g_0$ is a valid $Q$ then so is $g_1$ since \[ f_1(T(\cos \theta))+g_1(\cos \theta) = f_1(T(\cos(\theta+\pi/4))+g_1(\cos(\theta+\pi/4)) \]then $g_1(\cos \theta)=g_1(\theta+\pi/4)$. Thus, if we proceed inductively, there exists $g_n$ satisfying $\deg g_n \leq 7$ and $g_n$ is a valid $Q$. Let us show no such $g_n$ exists Indeed, notice that \[ g_n(\cos \theta)=g_n(\cos(\theta+\pi/4))= \cdots =g_n(\cos(\theta+7\pi/4)):=C \]So, $g_n-C$ has at least eight roots for a good choice of $\theta$ and thus must have degree of at least 8 or degree 0. $g_n$ can't be of degree 8 so $g_n = c$. This implies $g_{n-1}$ is a composed polynomial with $T$. Inducting upwards yields $g_1$ to be a composed polynomial with $T$ which is contradiction. So, we get all solutions of $P$ to be \[ P(x)=Q(x\sqrt{2})=f(T(x\sqrt{2})) \]where $f$ is an arbitrary polynomial and $T$ is the 8'th Chebyshev polynomial.

The answer is $p(Q(n)) $, for any polynomial $p(x)\in \mathbb{R}[x]$, where \[Q(x) = x^2 (x-1)^2 (x+1)^2 (x^2 - 2)\]These may be checked to work. Now we show they are the only ones. Claim: $Q(x)$ divides $P(x) - P(0)$ for all reals $x$. Proof: WLOG by shifting $P(0) = 0$. We show that $Q(x)$ divides $P(x)$. Setting $x=0$ gives $P(0) = P(1) = 0$. Setting $x = 1$ gives $P(1) = P(\sqrt{2}) = 0$. Setting $x = -\frac{\sqrt{2}}{2}$ gives $P(-1) = P(0) = 0$. Setting $x=-1$ gives that $P(-\sqrt{2}) = P(-1) = 0$. Thus, $-\sqrt{2}, -1,0,1,\sqrt{2}$ are all roots of $P$. Now take the derivative of both sides. We get \[\sqrt{2}\cdot P'(x\sqrt{2}) = \frac{\sqrt{1-x^2} - x}{\sqrt{1-x^2}} \cdot P'(x + \sqrt{1-x^2})\]for $x\ne -1,1$. Setting $x = \frac{\sqrt{2}}{2}$ here gives $P'(1) = 0$. Setting $x =- \frac{\sqrt{2}}{2}$ here gives $\sqrt{2}P'(-1) = 2P'(0)$. Setting $x =0$ here gives $\sqrt{2}P'(0) = P'(1) = 0$. Thus, $P'(-1) = P'(0) = P'(1) = 0$. Since $-1,0,1$ are roots of $P'(x)$ and $P(x)$, they must be double roots of $P(x)$, so $Q(x)$ must divide $P(x)$. $\square$ We can check that $\frac{P(x) - P(0)}{Q(x)}$ also satisfies the problem conditions. Since $Q(x)$ is the polynomial of minimal positive degree satisfying the problem conditions, we get that $P(x) = p(Q(x))$ for some polynomial $p$.

JuanOrtiz wrote: Notice that we have $P(\sqrt{2}x)=P(x+\sqrt{1-x^2})=P((\sqrt{1-x^2})+\sqrt{1-(\sqrt{1-x^2})^2})=P(\sqrt{2}\sqrt{1-x^2})$ and so if $x^2+y^2=2$ then $P(x)=P(y)$. Therefore since $(-x)^2=x^2$ we have $P$ is even. Let $Q$ be the polynomial that is the same as $P$ except divide all the exponents by $2$, and so $P(x)=Q(x^2)$ and $Q \in \mathbb{R}[X]$. The original equation becomes $Q(2x^2)=Q(1+2x\sqrt{1-x^2})$ and notice $Q(2x^2)=Q(1+2x\sqrt{1-x^2})=Q(2(1-x^2))$. Let $R(x)=Q(x+1)$, then $R(2x^2-1)=R(2(1-x^2)-1)$ and so $R(a)=R(-a)$ for any $a \in [-1,1]$ and therefore $R$ is even. The original equation becomes $R(2x^2-1)=R(2x\sqrt{1-x^2})$. Now let $S$ be the polynomial that is the same as $R$ except divide all exponents by $2$, so $R(x)=S(x^2)$. The original equation becomes $S((2x^2-1)^2)=S(4x^2(1-x^2))$ if $|x| \le 1$. Therefore for any $0 \le a \le 1$ we have $S((2a-1)^2)=S(4a(1-a))$ by letting $a=x^2$. Now let $b=4a^2-4a$, which can take on infinitely many values if $0 \le a \le 1$. We have $S(b+1)=S(-b)$ and therefore since this is true for infinitely many $b$, we have $S(x+1)=S(-x)$ for all $x$. Now let $T(x)=S\left( x+\frac{1}{2} \right)$. The above equation gives $T(x)=T(-x)$ for all $x$ and therefore $T$ is even. Let $U$ be the polynomial that is the same as $T$ except divide all exponents by $2$, so $T(x)=U(x^2)$. Notice $U \in \mathbb{R}[X]$. Now notice $P(x)=Q(x^2)=R(x^2-1)=S((x^2-1)^2)=T \left( (x^2-1)^2-\frac{1}{2} \right)=U \left( \left( (x^2-1)^2-\frac{1}{2} \right) ^2 \right)$. For the original equation to be satisfied we need, for $|x|\le 1$, $P(\sqrt{2}x)=P(x+\sqrt{1-x^2}) \Leftrightarrow T \left( (2x^2-1)^2-\frac{1}{2} \right) = T \left( ((x+\sqrt{1-x^2})^2-1)^2-\frac{1}{2} \right) \Leftrightarrow T \left( 4x^4-4x^2+\frac{1}{2} \right) = T \left( 4x^2-4x^4-\frac{1}{2} \right)$, which is a direct consequence of $T$ being even. Now notice $\left( (x^2-1)^2-\frac{1}{2} \right)^2 = x^8-4x^6+5x^4-2x^2+\frac{1}{4}$. Therefore the answer is $\boxed{P(x)=U\left( x^8-4x^6+5x^4-2x^2+\frac{1}{4} \right)}$ for any $U \in \mathbb{R}[X]$. I'm very glad that my solution is the same as yours! This is also the most basic solution in my opinion. It can avoid using derivatives and Chebyshev's polynomials

Replace $x$ with $\sin x$ so that $$P(\sqrt{2}\cos x)=P(\sin x+\cos x).$$ Then, we let $Q(x)=P(x\sqrt{2})$ so that this is $$Q(\cos x)=Q(\cos x-\pi/4).$$Let $T$ be the polynomial of degree 8 such that $T(\cos x)=\cos 8x$. Then, $Q=R(T(x))$ for any polynomial $R$ works, since $T(\cos x)=T(\cos x-\pi/4)$. We claim that these are the only $Q$ that work. By picking 8 equally spaced points around the unit circle that are not axis-aligned, we obtain 8 distinct values of $\cos x$ with $x$ spaced in intervals of $\pi/4$. Thus, by the periodicity, $Q$ is equal at $8$ distinct values, so it has degree at least 8 unless it is constant. Thus, noting that the solution set is closed under addition and subtraction, we can finish as follows: if $Q$ works, then the remainder when $Q$ is divided by $T$ also works, so it must be constant as its degree is less than 8. Thus, the quotient also works, since if the original function is periodic with period $\pi/4$ and $T(\cos x)$ also is, the quotient also is. Thus, we have that a function being good implies that its quotient when divided by $T$ works. This combined with the minimial degree solves the problem.

Can someone, please, explain in more details why the remainder of a solution of this polynomial equation when divided by the minimal solution would also be a solution of the initial equation.