Find all solutions to the following system of equations

$$
y=x+\sqrt{x+\sqrt{x+\sqrt{x+\ldots+\sqrt{x+\sqrt{y}}}}}, \quad x+y=6
$$

where there are 1974 square root signs in the equation.
$$
y=x+\sqrt{x+\sqrt{x+\sqrt{x+\ldots+\sqrt{x+\sqrt{y}}}}}, \quad \text { (2) } \quad x+y=6
$$

First, consider the simpler equation (3) instead of equation (1):

$$
y=x+\sqrt{y}
$$

We claim that the solution to the system (3), (2) also satisfies the system (1), (2). To solve this, we solve the system of equations (3) and (2)! Substituting (2) into (3):

$$
2 y-\sqrt{y}-6=0
$$

This is a quadratic equation in $\sqrt{y}$. The only non-negative solution is $\sqrt{y}=2$, from which, according to (2), the only solution to the system (2), (3) is

$$
y=4, \quad x=2
$$

This pair of values also satisfies equation (1), since each expression under the square root sign will be $2+2=4$.

We claim that the system (1), (2) has no other solution. To prove this, it suffices to show that every solution $(x, y)$ of the system satisfies equation (3). We prove this statement indirectly: we assume that there is a pair $(x, y)$ that satisfies equations (1) and (2) but not (3). This assumption leads to a contradiction.

If (3) does not hold, then either $y<x+\sqrt{y}$ or $y>x+\sqrt{y}$. Consider the first case. Since $x$ and $y$ are numbers that satisfy equation (1), $y \geq 0$ and $x+\sqrt{y} \geq 0$ also hold. Thus, in the case $y<x+\sqrt{y}$, we have a contradiction. Similarly, in the case $y>x+\sqrt{y}$, we also have a contradiction. Therefore, the only solution to the original system of equations is:

$$
x=2, \quad y=4
$$