Question 82: Find all real-coefficient polynomials $P$ such that for any $x \in \mathbb{R}$, we have $P(x) P(x+1) = P(x^2 - x + 3)$.
The 82nd question:
Solution: Clearly, the zero polynomial satisfies the equation in the question.
Now assume that the polynomial $\mathrm{P}(\mathrm{x})$ is not the zero polynomial.
If $\mathrm{P}(\mathrm{x})$ has a root $\alpha$, then $\mathrm{P}\left(\alpha^{2}-\alpha+3\right)=\mathrm{P}(\alpha) \mathrm{P}(\alpha+1)=0$, so $\alpha^{2}-\alpha+3$ is also a root of $\mathrm{P}(\mathrm{x})$. Let $f(x)=x^{2}-x+3$, then $\alpha, f(\alpha), f^{(2)}(\alpha), \ldots$ are all roots of $P(x)$.

For any real number $x, f(x)-x=(x-1)^{2}+2>0$. If $\alpha$ is a real number, then $\alpha<f(\alpha)<f^{(2)}(\alpha)<\ldots$, thus $\mathrm{P}(\mathrm{x})$ would have infinitely many roots, which is impossible. Therefore, $\mathrm{P}(\mathrm{x})$ has no real roots, and its degree must be even.

Consider the leading coefficient $t$ of $P(x)$, comparing the leading coefficients of $P(x) P(x+1)$ and $P\left(x^{2}-x+3\right)$, we know $t^{2}=t$, hence $t=1$. Thus, $P(x)$ is a monic polynomial.

Let the degree of $P(x)$ be $2n$, and let $P(x)=\left(x^{2}-2 x+3\right)^{n}+Q(x)$. Since $P(x)$ is a monic polynomial of degree $2n$, we have $\deg \mathrm{Q}<2 \mathrm{n}$. According to the condition, we have:
$$
\begin{array}{l}
{\left[\left(x^{2}-2 x+3\right)^{n}+Q(x)\right]\left\{\left\{(x+1)^{2}-2(x+1)+3\right]^{n}+Q(x+1)\right\}} \\
=\left[\left(x^{2}-x+3\right)^{2}-2\left(x^{2}-x+3\right)+3\right]^{n}+Q\left(x^{2}-x+3\right)
\end{array}
$$

Simplifying, we get:
$$
\begin{array}{l}
Q(x) \cdot Q(x+1)+Q(x) \cdot\left[(x+1)^{2}-2(x+1)+3\right]^{n}+Q(x+1) \cdot\left(x^{2}-2 x+3\right)^{n} \\
=Q\left(x^{2}-x+3\right)
\end{array}
$$

If $Q(x)$ is not the zero polynomial, let the degree of $Q(x)$ be $q$, and the leading coefficient be $\mathrm{k}$. Since $\mathrm{q}<2 \mathrm{n}$, the highest degree term on the left side of (*) is $2 k^{2 n+q}$, while the highest degree term on the right side is $k x^{2 q}$. Clearly, $2 k x^{2 n+q} \neq k x^{2 q}$, which leads to a contradiction.
Therefore, $Q(x)$ must be the zero polynomial, meaning $P(x)$ can only be the zero polynomial, or $\left(x^{2}-2 x+3\right)^{n}, n \in \mathbb{N}$.