(Try to find the number of real numbers $a$ such that for each $a$, the cubic equation $x^{3}=a x+a+1$ has an even root satisfying $|x|<1000$. (Xiong Bin))

---

The above text has been translated into English, preserving the original text's line breaks and format.
Solution 1: Let $x_{0}=2 n, n$ be an integer, and $|2 n|<1000$, i.e., $|n| \leqslant 499$, so at most $2 \times 499+1=999$ numbers can be taken, i.e., $n \in\{-499,-498, \cdots, 0, 1, \cdots, 499\}$. Substituting $x_{0}=2 n$ into the original equation gives $a=\frac{8 n^{3}-1}{2 n+1}$.

Let $f(n)=\frac{8 n^{3}-1}{2 n+1}$. For any $n_{1}, n_{2} \in\{-499,-498, \cdots, 0, 1, \cdots, 499\}$, when $n_{1} \neq n_{2}\left(n_{1}, n_{2} \in \mathbf{Z}\right)$, if $f\left(n_{1}\right)=f\left(n_{2}\right)$, set $n_{1}=\frac{x_{1}}{2}, n_{2}=\frac{x_{2}}{2}$, where $x_{1}, x_{2}$ are two roots of the equation $x^{3}-a x-a-1=0$, and let the other root be $x_{3}$. By the relationship between roots and coefficients,
$$
\text { i.e., } \begin{array}{l}
\left\{\begin{array}{l}
x_{3}=-\left(x_{1}+x_{2}\right), \\
x_{1} x_{2}+x_{2} x_{3}+x_{3} x_{1}=-a, \\
x_{1} x_{2} x_{3}=a+1,
\end{array}\right. \\
\left\{\begin{array}{l}
4 N_{1}=-a, \\
8 N_{2}=a+1,
\end{array}\right.
\end{array}
$$

where $N_{1}=-\left(n_{1}^{2}+n_{2}^{2}+n_{1} n_{2}\right), N_{2}=-n_{1} n_{2}\left(n_{1}+n_{2}\right)$, i.e., $4 N_{1}+8 N_{2}=1$, contradiction!

Therefore, for different $n_{1}, n_{2} \in\{-499,-498, \cdots, 0,1, \cdots, 499\}$, we have $f\left(n_{1}\right) \neq f\left(n_{2}\right)$. Thus, there are exactly 999 real numbers $a$ that satisfy the condition.

Solution 2: For any $|x| \leqslant 998, x$ being even, the values of $a=\frac{x^{3}-1}{x+1}$ are all different.

Proof by contradiction: If there exist $x_{1} \neq x_{2}$ such that $\frac{x_{1}^{3}-1}{x_{1}+1}=\frac{x_{2}^{3}-1}{x_{2}+1}$, where $x_{1}, x_{2}$ are even, then
$$
\left(x_{1}-x_{2}\right)\left(x_{1}^{2} x_{2}+x_{1} x_{2}^{2}+x_{1}^{2}+x_{2}^{2}+x_{1} x_{2}+1\right)=0,
$$

Since $x_{1} \neq x_{2}$, then $x_{1}-x_{2} \neq 0$. Also, because $x_{1}^{2} x_{2}+x_{1} x_{2}^{2}+x_{1}^{2}+x_{2}^{2}+x_{1} x_{2}$ is even, we have
$$
\left(x_{1}-x_{2}\right)\left(x_{1}^{2} x_{2}+x_{1} x_{2}^{2}+x_{1}^{2}+x_{2}^{2}+x_{1} x_{2}+1\right) \neq 0,
$$

contradiction. Therefore, there are 999 values of $a$ that satisfy the condition.