Problem 1428

1. Given the system of equations:
$$\begin{cases} x = \sqrt{2y + 3} \\ y = \sqrt{2z + 3} \\ z = \sqrt{2x + 3} \end{cases}$$
We start by squaring each equation to eliminate the square roots:
$$\begin{cases} x^2 = 2y + 3 \\ y^2 = 2z + 3 \\ z^2 = 2x + 3 \end{cases}$$

2. Rearrange each equation to isolate the terms involving $x$, $y$, and $z$:
$$\begin{cases} x^2 - 2y = 3 \\ y^2 - 2z = 3 \\ z^2 - 2x = 3 \end{cases}$$

3. Notice that if $x = y = z$, then substituting $x$ for $y$ and $z$ in the original equations should satisfy all three equations. Let's test $x = y = z$:
$$x = \sqrt{2x + 3}$$
Squaring both sides:
$$x^2 = 2x + 3$$
Rearrange to form a quadratic equation:
$$x^2 - 2x - 3 = 0$$
Factor the quadratic equation:
$$(x - 3)(x + 1) = 0$$
Thus, the solutions are:
$$x = 3 \quad \text{or} \quad x = -1$$

4. Since $x$, $y$, and $z$ must be real and non-negative (as they are under square roots in the original equations), we discard $x = -1$. Therefore, we have:
$$x = y = z = 3$$

5. Verify the solution $x = y = z = 3$ by substituting back into the original equations:
$$\begin{cases} 3 = \sqrt{2 \cdot 3 + 3} = \sqrt{6 + 3} = \sqrt{9} = 3 \\ 3 = \sqrt{2 \cdot 3 + 3} = \sqrt{6 + 3} = \sqrt{9} = 3 \\ 3 = \sqrt{2 \cdot 3 + 3} = \sqrt{6 + 3} = \sqrt{9} = 3 \end{cases}$$
The solution $x = y = z = 3$ satisfies all three equations.

6. To check for any other possible solutions, consider the case where none of $x$, $y$, or $z$ is equal to 3. Multiply the three equations:
$$(x^2 - 2y)(y^2 - 2z)(z^2 - 2x) = 27$$
This simplifies to:
$$(x^2 - 2y)(y^2 - 2z)(z^2 - 2x) = 27$$
Since $x, y, z \geq 0$, the left-hand side must be non-negative. However, the product of three non-negative terms equaling 27 implies that each term must be at least 3, which is not possible unless $x = y = z = 3$.

Therefore, the only solution is $x = y = z = 3$.

The final answer is $\boxed{x = y = z = 3}$.