22.11. (Jury, NRB, 79). Find all non-zero polynomials $P(x)$ with real coefficients, satisfying the identity

$$
P(x) P\left(2 x^{2}\right) \equiv P\left(2 x^{3}+x\right), \quad x \in \mathbf{R}
$$
22.11. Let a non-zero polynomial $P(x)=a_{n} x^{n}+\ldots+a_{1} x+a_{0}$, where $a_{n} \neq 0$, satisfy the conditions of the problem. Then, by comparing the coefficients of $x^{3 n}$ and $x^{0}$ in the original identity, we obtain the equalities $a_{n}^{2}=a_{n}, a_{0}^{2}=a_{0}$, from which $a_{n}=1$ and $a_{0}=1$ (if $a_{0}=0$, then $P(x)=x^{\prime} P_{1}(x)$, where $P_{1}(0) \neq 0, l \in \mathrm{N}$, and from the identity

$$
x^{l} P_{1}(x) \cdot\left(2 x^{2}\right)^{l} P_{1}\left(2 x^{2}\right)=\left(2 x^{3}+x\right)^{l} P_{1}\left(2 x^{3}+x\right)
$$

we obtain the identity

$$
2^{l} x^{2 l} P_{1}(x) P_{1}\left(2 x^{2}\right)=\left(2 x^{2}+1\right)^{l} P_{1}\left(2 x^{3}+x\right) ;
$$

from here $P_{1}(0)=0$ - a contradiction). Let $\alpha=\rho(\cos \varphi+i \sin \varphi)$ be an arbitrary root of the polynomial $P(x)$; then the number $2 \alpha^{3}+\alpha$ is also its root, since $P\left(2 \alpha^{3}+\alpha\right)=P(\alpha) P\left(2 \alpha^{2}\right)$. Further,

$$
\left|2 \alpha^{3}+\alpha\right|=|\alpha| \cdot\left|2 \alpha^{2}+1\right| \geqslant|\alpha| \cdot\left(2|\alpha|^{2}-1\right),
$$

therefore, if $|\alpha|=\rho>1$, then $\left|2 \alpha^{3}+\alpha\right|>|\alpha|$, and the polynomial $P(x) \not=$ $\not \equiv 0$ has infinitely many distinct roots $\beta_{i}=\alpha, \beta_{j+1}=2 \beta_{j}^{3}+\beta_{j}$ for $j=1,2, \ldots$, which is impossible. Therefore, the modulus of any root of the polynomial $P(x)$ does not exceed 1, but since the product of all roots of this polynomial is 1 (by Vieta's theorem), none of its roots can have a modulus less than 1. Thus, $\rho=1$ and from the equality

$$
1=\left|2 \alpha^{3}+\alpha\right|^{2}=\left|2 \alpha^{3}+1\right|^{2}=4 \cos 2 \varphi+5
$$

we have $\varphi=\frac{\pi}{2}+\pi m, m \in \mathrm{Z}$, and $\alpha= \pm i$. Since the polynomial $P(x)$ has real coefficients, then $P(x)=\left(x^{2}+1\right)^{k}$, where $k \in \mathbf{Z}^{+}$. Finally, verification shows that all polynomials of this form satisfy the conditions of the problem.