[ Systems of nonlinear algebraic equations ]

Given the system of equations:

$$
\left\{\begin{aligned}
& x_{1} x_{2} \cdot \ldots \cdot x_{1962}=1 \\
& x_{1}-x_{2} x_{3} \cdot \cdots \cdot x_{1962}=1 \\
& x_{1} x_{2}-x_{3} x_{4} \cdot \cdots \cdot x_{1962}=1 \\
& \cdots \cdots \cdots \cdots \cdots \\
& x_{1} x_{2} \cdot \ldots \cdot x_{1961}-x_{1962}=1
\end{aligned}\right.
$$

What values can $x_{25}$ take?

#
Note that the product $x_{1} \ldots x_{k}$ satisfies the equation $x+1 / x^{\prime}=1 \Rightarrow x^{2}-x-1=0$. Therefore, $x_{1} \ldots x_{k}=$ $\frac{1 \pm \sqrt{5}}{2}$. On the other hand, for the same reason, $x_{1} \ldots x_{k+1}=\frac{1 \pm \sqrt{5}}{2}$. From this, we get that $x_{k+1}=\frac{1+\varepsilon_{1} \sqrt{5}}{1+\varepsilon_{2} \sqrt{5}}$, where $\varepsilon_{1,2}= \pm 1$. Therefore, $x_{25}=1$ or $-\frac{3 \pm \sqrt{5}}{2}$. In the first case, the example fits:

$$
x_{1}=\frac{1+\sqrt{5}}{2}, x_{2}=x_{3}=\ldots=x_{1961}=1, x_{1962}=\frac{\sqrt{5}-1}{2} . \text { In the second case, the example fits }
$$

$$
x_{1}=-x_{1962}=\frac{1 \pm \sqrt{5}}{2}, x_{2}=x_{3}=\ldots=x_{24}=x_{26}=\ldots=x_{1961}=1, x_{25}=-\frac{3 \mp \sqrt{5}}{2}
$$

## Answer

1 and $-\frac{3 \pm \sqrt{5}}{2}$.