46. Solve the equation in integers

$$
\sqrt{\sqrt{x+\sqrt{x+\sqrt{x+\ldots+\sqrt{x}}}}}=y
$$
46. Answer: the only solution is $x=y=0$.

Let $\boldsymbol{x}$ and $\boldsymbol{y}$ be integers satisfying the condition. After a series of squaring, we are convinced that $\sqrt{x+\sqrt{x}}=\boldsymbol{m}$ and $\sqrt{x}=k$ are integers, and

$$
m^{2}=k(k+1)
$$

If $k>0$, then it must be that $k^{2}<m^{2}<(k+1)^{2}$, so $k<m<k+1$ and therefore $m$ is not an integer. Thus, $k=0$, i.e., $x=0$.

$\nabla$ Note that the contradiction of equation (*) also follows from problem 42.