## Subject II. (20 points)

Determine the real numbers $x, y, z, t$ that satisfy the relations:

$$
x-\sqrt{y}=y-\sqrt{z}=z-\sqrt{t}=t-\sqrt{x}=2 .
$$

Prof. Gheorghe Lobonț, National College "Mihai Viteazul" Turda
## Subject II.

a) If $xy$ follows the idea from point a) and we obtain: $x>y>z>t>x$, thus a series of inequalities that cannot hold. It follows that $x=y=z=t$ and we have $x-\sqrt{y}=2 \Leftrightarrow x-\sqrt{x}=2 \Leftrightarrow x-2=\sqrt{x}$. The conditions for existence are $\left\{\begin{array}{c}x-2 \geq 0 \\ x \geq 0\end{array}\right.$, which means $x \in[2,+\infty)$. Performing the calculations, we obtain the equation $x^{2}-5 \cdot x+4=0$, which admits the solutions $x_{1}=1$ and $x_{2}=4$. Given that $x \in[2,+\infty)$, only $x=4$ is a solution to the irrational equation. The solution to the problem is $x=y=z=t=4$.

(10 points)