8.50 Find the integer solutions of the equation $\underbrace{\sqrt{x+\sqrt{x+\cdots+\sqrt{x+\sqrt{x}}}}}_{1964 \text { }}=y$.
(4th All-Russian Mathematical Olympiad, 1964)
[Solution] Let $x$ and $y$ be a set of integer solutions that satisfy the equation.
By continuously squaring and rearranging the given equation, we get
$$
\sqrt{x+\sqrt{x}}=m .
$$

where $m$ is an integer.
In this case, we also have
$$
\sqrt{x}=m^{2}-x \text {. }
$$

Therefore, $\sqrt{x}$ is also an integer. Let $\sqrt{x}=k, k \geqslant 0$.
Then we have
$$
m^{2}=k^{2}+k \text {. }
$$

If $k>0$, then
$$
k^{2}<k^{2}+k<m^{2}<(k+1)^{2} \text {. }
$$

This implies that $m$ is not an integer when $k>0$, so (1) cannot hold, meaning $m$ is not an integer.
Thus,
$$
k=0 \text {. }
$$

At this point,
$$
x=y=0 .
$$

This is the only set of integer solutions that satisfy the given equation.