For which values of the real parameter $a$ does the equation

$$
\sqrt{x}-\sqrt{x-a}=2
$$

have a solution in the set of real numbers?
I. solution. Since the right side is positive, due to the monotonicity of the square root function, it is necessary that $a>0$. The quantities in the equation are meaningful if $x \geq a$.

Rearrange $\sqrt{x-a}$ to the right side, then square both sides, since both sides have the same sign, this is an equivalent transformation:

$$
\begin{aligned}
\sqrt{x} & =2+\sqrt{x-a} \\
x & =4+4 \sqrt{x-a}+x-a
\end{aligned}
$$

Rearranging again:

$$
a-4=4 \sqrt{x-a}
$$

Since the right side is non-negative, it is necessary that $a \geq 4$ holds. In this case, squaring again (which is also an equivalent transformation) we get a solution to the equation:

$$
\begin{aligned}
a^{2}-8 a+16 & =16(x-a) \\
x & =\frac{a^{2}+8 a+16}{16}=\left(\frac{a+4}{4}\right)^{2}
\end{aligned}
$$

(The condition $x \geq a$ is satisfied because $x-a=\frac{(a-4)^{2}}{16}$.) Therefore, the equations have a solution when $a \geq 4$.

II. solution. Similarly to the previous solution, we can determine that $a>0$ and $x \geq a$.

Multiply and divide the left side by $(\sqrt{x}-\sqrt{x-a})$, which is positive:

$$
\begin{aligned}
\frac{(\sqrt{x}-\sqrt{x-a})(\sqrt{x}+\sqrt{x-a})}{(\sqrt{x}+\sqrt{x-a})} & =2 \\
\frac{a}{\sqrt{x}+\sqrt{x-a}} & =2
\end{aligned}
$$

The function on the left side is strictly monotonically decreasing, its value at $x=a$ is $\sqrt{a}$, and it approaches 0 as $x$ approaches $+\infty$. Since it is continuous, by the Bolzano theorem, its range is the interval $(0, \sqrt{a}]$. This interval contains $2$ precisely when $\sqrt{a} \geq 2$, i.e., $a \geq 4$.

Therefore, the equation has a solution when $a \geq 4$.