Problem 2. Solve the system of equations $(n>2)$ :

$$
\begin{aligned}
& \sqrt{x_{1}}+\sqrt{x_{2}+\cdots+x_{n}}=\sqrt{x_{2}}+\sqrt{x_{3}+\cdots+x_{n}+x_{1}}=\cdots \\
& \cdots=\sqrt{x_{n}}+\sqrt{x_{1}+\cdots+x_{n-1}} ; \quad x_{1}-x_{2}=1
\end{aligned}
$$
Solution: By squaring the equation

$$
\sqrt{x_{1}}+\sqrt{x_{2}+\cdots+x_{n}}=\sqrt{x_{2}}+\sqrt{x_{3}+\cdots+x_{n}+x_{1}}
$$

and subtracting $x_{1}+x_{2}+\cdots+x_{n}$ from both sides of the resulting equation and squaring again, we arrive at

$x_{1}\left(x_{2}+\cdots+x_{n}\right)=x_{2}\left(x_{3}+\cdots+x_{n}+x_{1}\right) \Longleftrightarrow\left(x_{1}-x_{2}\right)\left(x_{3}+\cdots+x_{n}\right)=0$.

Since $x_{1}-x_{2}=1$, then $x_{3}+\cdots+x_{n}=0$. This means that $x_{3}=\cdots=x_{n}=0$, because the domain of the system is $x_{i} \geq 0$ for $i=1 . . n$.

If $x_{2} \neq 0$, then $x_{3}-x_{2} \neq 0$ and similarly we would get $x_{1}+x_{4}+\cdots+x_{n}=0$, from which $x_{1}=0$. But then $x_{2}=-1$, a contradiction.

Therefore, $x_{2}=0$ and $x_{1}=1$.