2. (51st Czech and Slovak Mathematical Olympiad (Open Book) Question) For a natural number $n(n \geqslant 2)$, for what value of the parameter $p$ does the system of equations
$$
\left\{\begin{array}{l}
x_{1}^{4}+\frac{2}{x_{1}^{2}}=p x_{2}, \\
x_{2}^{4}+\frac{2}{x_{2}^{2}}=p x_{3}, \\
\cdots \cdots \\
x_{n-1}^{4}+\frac{2}{x_{n-1}^{2}}=p x_{n}, \\
x_{n}^{4}+\frac{2}{x_{n}^{2}}=p x_{1}
\end{array}\right.
$$

have at least two real solutions?
2. Clearly $p \neq 0$, and if $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ is a solution to the original system of equations, then changing $p$ to $-p$ makes $\left(-x_{1},-x_{2}, \cdots,-x_{n}\right)$ also a solution to the original system of equations.

Assume without loss of generality that $p>0$, and $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ is a solution to the original system of equations, then $x_{i}>0, i=1,2, \cdots, n$. Since $x_{i}^{4}+\frac{2}{x_{i}^{2}} \geqslant 2 \sqrt{2} x_{i}$, we have $p^{\prime \prime} x_{1} x_{2} \cdots x_{n} \geqslant(2 \sqrt{2})^{n} x_{1} x_{2} \cdots x_{n}$, which means $p \geqslant 2 \sqrt{2}$.
When $p=2 \sqrt{2}$, from $x_{i}^{4}=\frac{2}{x_{i}^{2}}$, we get $x_{i}=\sqrt[6]{2}$, which gives a unique solution, contradicting the problem statement.
When $p>2 \sqrt{2}$, the original system of equations has at least two solutions. For example, if $x_{1}=x_{2}=\cdots=x_{n}=x$, then $x=\left(\frac{p \pm \sqrt{p^{2}-8}}{2}\right)^{\frac{1}{3}}$.
In summary, the range of $p$ is
$$
(-\infty,-2 \sqrt{2}) \cup(2 \sqrt{2},+\infty)
$$