Find all polynomials $P(x)$ with real coefficients that satisfy
$$
P(x \sqrt{2})=P\left(x+\sqrt{1-x^{2}}\right)
$$
for all real numbers $x$ with $|x| \leq 1$.
The answer is any polynomial of the form $P(x)=f(U(x / \sqrt{2}))$, where $f \in \mathbb{R}[x]$ and $U$ is the unique polynomial satisfying $U(\cos \theta)=\cos (8 \theta)$.
Let $Q(x)=P(x \sqrt{2})$, then the condition reads
$$
Q(\cos \theta)=Q\left(\frac{1}{\sqrt{2}}(\cos \theta+\sin \theta)\right)=Q\left(\cos \left(\theta-45^{\circ}\right)\right) \quad \forall 0 \leq \theta \leq 180^{\circ} .
$$

We call a polynomial good if it satisfies this functional equation.
Lemma
The minimal (by degree) good nonconstant polynomial is $U$.

Proof. Since $U$ works, it suffices to show that $\operatorname{deg} Q \geq 8$. Note that:
$$
\begin{aligned}
Q\left(\cos 136^{\circ}\right) & =Q\left(\cos 91^{\circ}\right)=Q\left(\cos 46^{\circ}\right)=Q\left(\cos 1^{\circ}\right)=Q\left(\cos -44^{\circ}\right) \\
& =Q\left(\cos 44^{\circ}\right)=Q\left(\cos 89^{\circ}\right)=Q\left(\cos 134^{\circ}\right)=Q\left(\cos 179^{\circ}\right) .
\end{aligned}
$$

Hence $Q$ is equal at eight distinct values (not nine since $\cos -44^{\circ}=\cos 44^{\circ}$ is repeated), so $\operatorname{deg} Q \geq 8$ (unless $Q$ is constant).

Now, we claim $Q(x) \equiv f(U(x))$ for some $f \in \mathbb{R}[x]$. Indeed, if $Q$ is good, then by minimality the quotient $Q \bmod U$ must be constant, so $Q(x)=\widetilde{Q}(x) \cdot U(x)+c$ for some constant $c$, but then $\widetilde{Q}(x)$ is good too and we finish iteratively.