## Task 1 - 291241

For each real number $a$, investigate whether the equation

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

(has at least) one real solution $x$, and determine all real solutions of equation (1).
a) Reducing (1) to a simpler equation:

I. If (1) has a real solution $x$ for a real $a$, then it is also a solution of the equation

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

Proof: If $a+\sqrt{x}>x$, then it would follow that

$$
a+\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{x}}}}>a+\sqrt{a+\sqrt{a+\sqrt{x}}}>\ldots>a+\sqrt{x}>x
$$

contradicting the assumption (1). The proof runs essentially the same for $a+\sqrt{x}<x$. 

In essence, the proof runs analogously for $a+\sqrt{x}<x$.

The solution to (2) is exactly the number

$$
x=0.5 \cdot (2a + 1 + \sqrt{4a + 1})
$$

which is a solution to (2), and thus to (1).

Adapted from $[1]$

Hint: In the original problem statement, the equation was given as

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

i.e., the $a+$ was missing. This made the equation (1*) practically unsolvable for the participants. In the recommended solution, the correct equation (1) was used.