4. Let $n \geqq 2$ be a given natural number. For which values of the real parameter $p$ does the system of equations

$$
\begin{aligned}
& x_{1}^{4}+\frac{2}{x_{1}^{2}}=p x_{2}, \\
& x_{2}^{4}+\frac{2}{x_{2}^{2}}=p x_{3}, \\
& \cdots \ldots \ldots \ldots \ldots \ldots, \\
& x_{n-1}^{4}+\frac{2}{x_{n-1}^{2}}=p x_{n}, \\
& x_{n}^{4}+\frac{2}{x_{n}^{2}}=p x_{1}
\end{aligned}
$$

have at least two solutions in the set of real numbers?
SOLUTION. Since the given system (1) is very complex and there is probably no method to express all its solutions in a finite algebraic form, we will, on the one hand, think about the conditions for the solvability of this system, and on the other hand, look for some of its special solutions.

First, notice that the system (1) has no solutions for the value $p=0$, because the values of the left sides of the equations are positive numbers. Another observation, which we now state, is obvious: the $n$-tuple of numbers $\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ is a solution of the system (1) with the value of the parameter $p$, if and only if the $n$-tuple of opposite numbers $\left(-x_{1},-x_{2}, \ldots,-x_{n}\right)$ is a solution of the system (1) with the opposite value of the parameter $-p$. The values of the left and right sides of all equations in the system do not change when all values $x_{i} \mapsto-x_{i}$ and $p \mapsto-p$ are changed, because for any $x \neq 0$ and $p$ we have

$$
(-x)^{4}+\frac{2}{(-x)^{2}}=x^{4}+\frac{2}{x^{2}} \quad \text { and } \quad(-p)(-x)=p x
$$

Therefore, the system (1) with the value of the parameter $p$ has exactly as many solutions as the system (1) with the value of the parameter $-p$. We will therefore look for all positive numbers $p$ for which the system (1) has at least two solutions (in the answer, we will add all opposite numbers $-p$ to them.)

Until the end of the solution, we will therefore consider only positive values of the parameter $p$ of the system (1). From the positivity of its left sides, it follows that all right sides $p x_{i}$ must also be positive, and therefore (with respect to the assumption $p>0$) it must hold that $x_{i}>0$ for each $i$. Any solution $\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ of the system (1) is thus composed of $n$ positive numbers.

Now assume that for a given $p>0$ some solution $\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ of the system (1) exists, and multiply all $n$ equations among themselves. For positive numbers $x_{1}, x_{2}, \ldots, x_{n}$, we then get the equality

$$
\left(x_{1}^{4}+\frac{2}{x_{1}^{2}}\right)\left(x_{2}^{4}+\frac{2}{x_{2}^{2}}\right) \ldots\left(x_{n}^{4}+\frac{2}{x_{n}^{2}}\right)=p^{n} x_{1} x_{2} \ldots x_{n} .
$$

Each factor on the left side we estimate from below according to the known inequality

$$
u+v \geqq 2 \sqrt{u v}
$$

which holds for any positive numbers $u$ and $v$, with equality occurring if and only if $u=v$ (it is essentially the inequality between the arithmetic and geometric means of the numbers $u$ and $v$, easily derived from the obvious inequality $\left.(\sqrt{u}-\sqrt{v})^{2} \geqq 0\right)$. Therefore, for each index $i$ we have

$$
x_{i}^{4}+\frac{2}{x_{i}^{2}} \geqq 2 \sqrt{x_{i}^{4} \cdot \frac{2}{x_{i}^{2}}}=\left|x_{i}\right| \cdot 2 \sqrt{2}=x_{i} \cdot 2 \sqrt{2} .
$$

As a result of equality (2), we thus get the inequality

$$
\left(x_{1} 2 \sqrt{2}\right)\left(x_{2} 2 \sqrt{2}\right) \ldots\left(x_{n} 2 \sqrt{2}\right) \leqq p^{n} x_{1} x_{2} \ldots x_{n}
$$

from which, after canceling (the positive) product $x_{1} x_{2} \ldots x_{n}$, we get the condition on the number $p$ in the form

$$
p^{n} \geqq(2 \sqrt{2})^{n}, \quad \text { or } \quad p \geqq 2 \sqrt{2} \text {. }
$$

Let's summarize what we have just found: if the system (1) has at least one solution for a fixed $p>0$, then for this number $p$ the estimate $p \geqq 2 \sqrt{2}$ holds.

For the "extreme" value $p=2 \sqrt{2}$, we will now completely solve the system (1), i.e., find all its solutions. If $\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ is any solution of the system (1) with the value $p=2 \sqrt{2}$, then according to the considerations in the previous paragraph, equality must occur in inequality (4), which is possible only if equality occurs in all multiplied inequalities (3). Therefore, for each index $i$ we have

$$
x_{i}^{4}=\frac{2}{x_{i}^{2}}, \quad \text { or } \quad x_{i}^{6}=2, \quad \text { i.e., } \quad x_{i}=\sqrt[6]{2} .
$$

For the value $p=2 \sqrt{2}$, the system (1) thus has only one (!) solution

$$
\left(x_{1}, x_{2}, \ldots, x_{n}\right)=(\sqrt[6]{2}, \sqrt[6]{2}, \ldots, \sqrt[6]{2})
$$

From the results of the previous two paragraphs, it follows: if the system (1) has at least two solutions for a fixed $p>0$, then for this number $p$ the strict inequality $p>2 \sqrt{2}$ holds. If we therefore find two solutions of the system (1) with any value of the parameter $p>2 \sqrt{2}$, we will know the answer to the question in the problem. The mentioned two solutions we will look for among $n$-tuples $\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ composed of $n$ identical numbers; such an $n$-tuple $(x, x, \ldots, x)$ is clearly a solution of the system (1) if and only if the number $x$ is a solution (the only one) of the equation

$$
x^{4}+\frac{2}{x^{2}}=p x, \quad \text { or } \quad x^{6}-p x^{3}+2=0 .
$$

The last equation is quadratic with respect to the unknown $y=x^{3}$ and has two different solutions

$$
y_{1,2}=\frac{p \pm \sqrt{p^{2}-8}}{2}
$$

for each of the values of $p>2 \sqrt{2}$ we are considering, because for them $p^{2}-8>0$ holds. For each such $p$, the original system (1) thus has two solutions

$$
\left(x_{1}, \ldots, x_{n}\right)=\left(\sqrt[3]{y_{1}}, \ldots, \sqrt[3]{y_{1}}\right) \text { and }\left(x_{1}, \ldots, x_{n}\right)=\left(\sqrt[3]{y_{2}}, \ldots, \sqrt[3]{y_{2}}\right)
$$

(We do not exclude that in addition to these solutions, there may be other solutions, namely such that $x_{i} \neq x_{j}$ for some $i \neq j$.)

Answer. All the sought values of $p$ form the set $(-\infty ;-2 \sqrt{2}) \cup(2 \sqrt{2} ; \infty)$.