Solve the following equation:

$$
x=\sqrt{-3+4 \sqrt{-3+4 \sqrt{-3+4 x}}}
$$
Solution. Let us introduce the sequence defined by the following recursion: $x_{1} \geq 0.75$ is a real number, and if $n>1$, then

$$
x_{n}=\sqrt{-3+4 x_{n-1}}
$$

The task is to find the values of $x_{1}$ for which $x_{1}=x_{4}$.

Let us examine the sequence $x_{n}$ in terms of monotonicity. The sequence $x_{n}$ is strictly increasing if for every $n$, $x_{n}>x_{n-1}$, that is, if

$$
\sqrt{-3+4 x_{n-1}}>x_{n-1}
$$

From the rule of sequence formation, it follows that every term is positive, so we can square both sides of the inequality:

$$
\begin{gathered}
-3+4 x_{n-1}>x_{n-1}^{2} \\
0>x_{n-1}^{2}-4 x_{n-1}+3 \\
0>\left(x_{n-1}-1\right)\left(x_{n-1}-3\right)
\end{gathered}
$$

This means that the sequence is strictly increasing precisely when for every $n>1$, $1<x_{n-1}<3$. It is sufficient to require this condition for $n=2$; if for some $k$, $1<x_{k-1}<3$ holds, then

$$
1=\sqrt{-3+4 \cdot 1}<\sqrt{-3+4 x_{k-1}}=x_{k}<\sqrt{-3+4 \cdot 3}=3
$$

also holds. Therefore, the sequence is strictly increasing if and only if $1<x_{1}<3$. Similarly, it can be shown that if $0.75 \leq x_{1}<1$ or $3<x_{1}$, then the sequence is strictly decreasing (or in the former case, it may strictly decrease for a while, then stop due to the next $-3+4 x_{t}$ becoming negative, meaning the subsequent terms are not defined). Therefore, if $x_{1}=x_{4}$, this can only be true for $x_{1}=1$ or $x_{1}=3$.

If $x_{1}=1$, then $x_{2}=\sqrt{-3+4 \cdot 1}=1$, which means that, due to the rule of sequence formation, every term of the sequence is 1. If $x_{1}=3$, then $x_{2}=\sqrt{-3+4 \cdot 3}=3$, so every term of the sequence is 3. Therefore, the solutions to the equation are: $x=1$ and $x=3$.

Remark: It follows from the derivation that equations of the form

$$
x=\sqrt{-3+4 \sqrt{-3+4 \sqrt{\ldots \sqrt{-3+4 x}}}}
$$

(in which $k$ square root symbols appear) also have only $x=1$ and $x=3$ as solutions.

Balázs Poronyi (Pécs, Janus Pannonius Gymnasium, 10th grade)