5-144 Find all non-zero real-coefficient polynomials \( P(x) \) that satisfy
$$
P(x) P\left(2 x^{2}\right) \equiv P\left(2 x^{3}+x\right), x \in R
$$
[Solution] If $P(x)=a$ (a constant), then $a^{2}=a, a=1$, clearly $P(x)=1$ satisfies the condition.

Assume the non-zero polynomial $P(x)=a_{n} x^{n}+\cdots+a_{1} x+a_{0}$ satisfies the given condition, where $a_{n} \neq 0$. By comparing the coefficients of $x^{3 n}$ and $x^{0}$ in the original identity, we get $a_{n}^{2}=a_{n}, a_{0}^{2}=a_{0}$, thus $a_{n}=1$. If $a_{0}=0$, then $P(x)=x^{l} P_{1}(x)$, where
$$
P_{1}(0) \neq 0, l \in \mathbb{N},
$$

and from $x^{l} P_{1}(x)\left(2 x^{2}\right)^{l} P_{1}\left(2 x^{2}\right) \equiv\left(2 x^{3}+x\right)^{l} P_{1}\left(2 x^{3}+x\right)$,
we get $2^{l} x^{2 l} P_{1}(x) P_{1}\left(2 x^{2}\right) \equiv\left(2 x^{2}+1\right)^{l} P_{1}\left(2 x^{3}+x\right)$,
thus $P_{1}(0)=0$, a contradiction. Therefore, $a_{0}=1$. Let $\alpha=\rho(\cos \phi+i \sin \phi)$ be a root of the polynomial $P(x)$. Since
$$
P\left(2 \alpha^{3}+\alpha\right)=P(\alpha) P\left(2 \alpha^{2}\right),
$$

$2 \alpha^{3}+\alpha$ is also a root. Furthermore,
$$
\left|2 \alpha^{3}+\alpha\right|=|\alpha| \cdot\left|2 \alpha^{2}+1\right| \geqslant|\alpha| \cdot\left(2|\alpha|^{2}-1\right) .
$$

So, if $|\alpha|=\rho>1$, then $\left|2 \alpha^{3}+\alpha\right|>|\alpha|$, thus the non-zero polynomial $P(x)$ has infinitely many distinct roots
$$
\beta_{1}=\alpha, \beta_{j+1}=2 \beta_{j}^{3}+\beta_{j}, j=1,2, \cdots,
$$

which is impossible, therefore the roots of the polynomial $P(x)$ have modulus no greater than 1, but by Vieta's formulas, the product of the roots has modulus 1, so there are no roots with modulus less than 1. Thus $\rho=1$, and by
$$
\begin{aligned}
1 & =\left|2 \alpha^{3}+\alpha\right|^{2}=\left|2 \alpha^{2}+1\right|^{2} \\
& =\left|2 \cos 2 \phi+1+2 \sin 2 \phi_{i}\right|^{2} \\
& =4 \cos ^{2} 2 \phi+4 \cos 2 \phi+1+4 \sin ^{2} 2 \phi \\
& =4 \cos 2 \phi+5
\end{aligned}
$$

we get $\phi=\frac{\pi}{2}+m \pi, m \in \mathbb{Z}$.
Thus $\alpha= \pm i$. Since $P(x)$ has real coefficients, we have
$$
P(x)=\left(x^{2}+1\right)^{k}, k \in \mathbb{Z}^{+} \text {. }
$$

Finally, it can be verified that all polynomials of this form satisfy the given condition.