## Task $4 / 74$

In a right-angled Cartesian coordinate system, a curve is given by the equation

$$
\sqrt{x}+\sqrt{y}=\sqrt{12}
$$

The task is to find all points on the curve with integer coordinates.
Obviously, the coordinates $x_{1}=0, y_{1}=12$ and $x_{2}=12, y_{2}=0$ are integer solutions of the given equation, so the points $\left.P_{1} / 0 ; 12\right)$ and $P_{2}(12 ; 0)$ satisfy the given condition. To find more points, we solve the equation for $y$.

$$
\begin{array}{r}
\sqrt{y}=\sqrt{12}-\sqrt{x} \\
y=12-4 \sqrt{3 x}+x
\end{array}
$$

From (1) it already follows that $0 \leq x ; y \leq 12$, and from (2) it follows (due to the required integer nature of $y$) that $\sqrt{3 x}$ must be an integer, so $3 x=k^{2}$ with $k \in N$. Then $k$ must be a multiple of 3. Therefore, only $k=0, k=3$, and $k=6$ are possible.

For $k>6$, $x>12$ would result. For $k=0$ and $k=6$, the already known points $P_{1}$ and $P_{2}$ are obtained, and for $k=3$ we get $x_{3}=3, y_{3}=3$ and thus $P_{3}(3 ; 3)$.