## Task 4 - 331234

Determine the number of all pairs $(x, y)$ of integers $x$, y, for which $\sqrt{x}+\sqrt{y}=1993$ holds.
Given $x, y \geq 0$, it also holds that $x, y \leq 1993^{2}$. For $x$ and $y$, only the square numbers $0,1,4,9,16, \ldots, 1993^{2}$ are possible.

Proof:

If $x$ is not a square number, then $\sqrt{x}$ is irrational. If we solve the equation from the problem statement for $y$:

$$
\begin{gathered}
y=(1993-\sqrt{x})^{2} \\
y=1993^{2}+x-2 \cdot 1993 \sqrt{x}
\end{gathered}
$$

Then $y$ would be irrational, but it must be an integer, which is a contradiction. Therefore, if $x=n^{2}$ is a square number with $0 \leq n \leq 1993$, then $y=(1993-n)^{2}$. Thus, there are 1994 pairs of numbers that satisfy the equation from the problem statement.