Solve the equation $\underbrace{\sqrt{n+\sqrt{n+\ldots \sqrt{n}}}}_{1964 \text { times }}=m$ in integers.
Let's solve the more general equation $A_{y}(x)=z$ in integers, where $A_{y}(x)=\underbrace{\sqrt{x+\sqrt{x+\ldots \sqrt{x}}}}_{y \text { times }}$. Clearly, $x+A_{y-1}(x)=z^{2}$, or $A_{y-1}(x)=z^{2}-x$.

Thus, if the number $A_{y}(x)=z$ is an integer, then the number $A_{y-1}(x)=\left(z^{2}-x\right)$ is also an integer; but then the number $A_{y-2}(x)=\left(z^{2}-x\right)^{2}-x$ is also an integer, and $A_{y-3}$ is an integer, ..., and $A_{1}(x)=\sqrt{x}$ is an integer. Therefore, $x=t^{2}$ is an integer (where $t$ is an integer).

On the other hand, for any integer $t\left(t^{2}, 1, t\right)$ is a solution to our equation.

Let $y>1$. In this case, the numbers $A_{1}(x)=\sqrt{x}=t$ and $A_{2}(x)=\sqrt{x+\sqrt{x}}=\sqrt{t^{2}+t}=\sqrt{t(t+1)}$ must be integers. But the numbers $t$ and $t+1$ are coprime, which implies that $t$ and $t+1$ are perfect squares, i.e., $t=0$. Therefore, $x=0$ and $A_{y}(x)=0$ for any $y$.

## Answer

$(0,0)$.