For what value of the parameter $a$ does the equation

$$
x+\sqrt{a+\sqrt{x}}=a
$$

have a solution?
I. Solution. The task was not to solve the equation; those who tried this way certainly did not get far.

From the definition of the square root, $\sqrt{x} \geq 0$, and

$$
\sqrt{a+\sqrt{x}} \geq 0, \quad \text { i.e., } \quad x \geq 0
$$

Rearrange the equation: $\sqrt{a+\sqrt{x}}=a-x$, and plot the corresponding functions in the coordinate system. The right side is a line that intersects both axes at the points $(0 ; a)$ and $(a ; 0)$. The left side is a strictly increasing function, whose graph starts from the point $\sqrt{a}$ on the $y$-axis. Thus, a solution exists precisely when $\sqrt{a} \leq a$, i.e., $a \leq a^{2}$, rearranging gives $a^{2}-a=a(a-1) \geq 0$, i.e., either $a=0$ or $a \geq 1$.

Therefore, the equation has a solution if $a=0$ or $a \geq 1$.

II. Solution. Let $a+\sqrt{x}=b^{2},(b \geq 0)$, then $a=b^{2}-\sqrt{x}$, and our equation takes the form:

$$
x+b=a, \quad \text { i.e., } \quad x+b=b^{2}-\sqrt{x}
$$

Rearranging gives

$$
b^{2}-x-(b+\sqrt{x})=0
$$

Using the fact that $b^{2}-x=b^{2}-(\sqrt{x})^{2}=(b-\sqrt{x})(b+\sqrt{x})$, (1) can be rewritten as a product:

$$
(b+\sqrt{x})(b-\sqrt{x}-1)=0
$$

A product is zero precisely when one of its factors is zero.

Thus:

1. $b+\sqrt{x}=0$; this can only hold if $a-x=b=\sqrt{x}=0$, i.e., $a=x=0$.
2. $b-\sqrt{x}-1=0$; substituting $b=a-x$ into this, we get a quadratic equation in $\sqrt{x}$, $x+\sqrt{x}+1-a=0$, one of whose roots must be non-negative:

$$
\sqrt{x}=-\frac{-1+\sqrt{1-4+4 a}}{2} \geq 0
$$

Since the sum of the roots here is -1, negative, the equation can only have a positive solution if the product of the roots $(1-a)$ is negative, i.e., $a \geq 1$. And in this case, the value of $x$ derived from (2) can be shown to indeed be a solution to the original equation.