1. In the set of natural numbers, solve the equation $\sqrt{x}+\sqrt{y}=\sqrt{48}$.
First method:

From $\sqrt{x}+\sqrt{y}=\sqrt{48}$, it follows that $\sqrt{x}=\sqrt{48}-\sqrt{y}$, and after squaring, we get $x=48-2 \sqrt{48 y}+y$.

After partial rooting, we obtain $x=48+y-8 \sqrt{3 y}$.

Since $x$ and $y$ are natural numbers, the number $3 y$ must be the square of some natural number, i.e., it must be $y=3 m^{2}$, where $m$ is a natural number.

By analogous reasoning, we get $y=48+x-8 \sqrt{3 x}$, i.e., $x=3 n^{2}$, where $n$ is a natural number.

The given equation can be written in the form $\sqrt{3 m^{2}}+\sqrt{3 n^{2}}=\sqrt{48}$, i.e., $m \sqrt{3}+n \sqrt{3}=4 \sqrt{3}$. After dividing the equation by the number $\sqrt{3}$, we get the equation $m+n=4$. Its solutions are listed in the table:

| $m$ | 1 | 2 | 3 |
| :--- | :--- | :--- | :--- |
| $\boldsymbol{n}$ | 3 | 2 | 1 |

From $x=3 n^{2}$ and $y=3 m^{2}$, the solutions to the given equation are:

| $x$ | 27 | 12 | 3 |
| :--- | :--- | :--- | :--- |
| $y$ | 3 | 12 | 27 |