Solve the following equation in the set of integers:

$$
\underbrace{\sqrt{x+\sqrt{x+\sqrt{x+\cdots+\sqrt{x}}}}}_{1992 \text { square root signs }}=y
$$
We interpret the square root function only for non-negative numbers, and its value is also non-negative, so the equation can only be satisfied for $x, y \geq 0$. Let

$$
f_{1}(x)=\sqrt{x}, \quad f_{2}(x)=\sqrt{x+\sqrt{x}}, \ldots, f_{n}(x)=\underbrace{\sqrt{x+\sqrt{x+\ldots+x \sqrt{x}}}}_{\mathrm{n} \text { nested radicals }}
$$

Then our equation can be written as

$$
f_{1992}(x)=y
$$

Notice that

$$
\left(f_{n}(x)\right)^{2}-x=f_{n-1}(x)
$$

Therefore, if $x$ and $f_{n}(x)$ are integers, then $f_{n-1}(x)$ is also an integer, and similarly, $f_{n-2}(x), \ldots, f_{2}(x), f_{1}(x)$ are all integers. Thus, if

$$
f_{1992}(x)=y
$$

is an integer, then $f_{1}(x)=\sqrt{x}$ and $f_{2}(x)=\sqrt{x+\sqrt{x}}$ are also integers. Let their values be $z$ and $k$, respectively, so

$$
k^{2}=x+\sqrt{x}=z^{2}+z
$$

where $z$ and $k$ are non-negative integers.

If $z=0$, then $x=0$, and

$$
f_{1}(x)=f_{2}(x)=\cdots=f_{1992}(x)=y=0
$$

so $(x, y) = (0, 0)$ is a solution. If $z>0$, then

$$
z^{2}<z^{2}+z<z^{2}+2 z+1=(z+1)^{2}
$$

so $z^{2}+z$ cannot be a perfect square. Therefore, the only solution is the pair $(0,0)$.