Let $p$ be any prime number. Find all non-negative integer solutions to the equation

$$
\sqrt{x}+\sqrt{y}=\sqrt{p}
$$
If $x$ and $y$ are non-negative integers, then squaring the original equation, we get an equivalent equation:

$$
x+y+2 \sqrt{x y}=p
$$

Since $x$, $y$, and $p$ are integers, $2 \sqrt{x y}$ must also be an integer, hence $\sqrt{x y}$ must be rational, and thus $\sqrt{x y}$ must be an integer, meaning $x y$ must be a perfect square.

Let $d$ be the greatest common divisor of $x$ and $y$. Since $d \mid x$ and $d \mid y$, it follows that $d^{2} \mid x y$, thus $d \mid \sqrt{x y}$, and therefore $d$ must divide the left-hand side of (1), i.e., $d \mid p$. Since $p$ is prime, $d=1$ or $d=p$ must hold.

If $d=1$, then $x$ and $y$ are coprime. Since $x y$ is a perfect square, both $x$ and $y$ must be perfect squares, and from (1) we have $(\sqrt{x}+\sqrt{y})^{2}=p$, which is impossible since a prime number $p$ cannot be a perfect square. Therefore, only $d=p$ can hold, in which case one of $x$ and $y$ must be equal to $p$ and the other must be 0.

Thus, the only solutions to the equation are $x=0 ; y=p$ and $y=0 ; x=p$.

Based on the work of Dénes Koch (Linz, Akademisches Gymn., 10th grade)