Solve the following equation:

$$
x-2=\sqrt{4-3 \sqrt{4-3 \sqrt{10-3 x}}}
$$
Solution. Let's substitute $\sqrt{1} x$ with $(a+2)-\mathrm{t}$:

$$
a=\sqrt{4-3 \sqrt{4-3 \sqrt{10-3 a-6}}}=\sqrt{4-3 \sqrt{4-3 \sqrt{4-3 a}}}
$$

Let $f(x)=\sqrt{4-3 x}$. We are looking for the value of $a$ such that $a=f(f(f(a)))$.

If $a=f(a)$, then this is clearly true. $a=f(a)$ holds if, under the conditions $a \geq 0, a \leq \frac{4}{3}$, $a^{2}+3 a-4=0$, from which $a=1$, so $x=3$.

We need to check if there is any other $a$ that satisfies the equation $a=f(f(f(a)))$.

$a$ cannot be negative because it is an element of the range of $f$, so it is sufficient to look for further solutions of $a$ in the intervals $[0 ; 1)$ and $\left(1 ; \frac{4}{3}\right]$.

$f$ maps elements from the interval $[0 ; 1)$ to the interval $(1 ; 2]$, and elements from the interval $\left(1 ; \frac{4}{3}\right]$ to the interval $[0 ; 1)$.

Therefore, if $a \in[0 ; 1)$, then $f(f(f(a))) \in(1 ; 2]$; if $a \in(1 ; 2]$, then $f(f(f(a))) \in[0 ; 1)$ (assuming the expression is meaningful). This means that the equation has no other solutions.

Remark. The reasoning is incomplete if it claims that

$$
a=f(f(\ldots(f(a) \ldots)))
$$

is only possible if $a=f(a)$. Several submissions simplified the solution of the problem in this way, which is incorrect. This is not necessarily true even for monotonic functions. The function $f(x)=-x$ is monotonic, yet it is not true that $f(f(x))=x$ only holds if $x=f(x)$ (since $f(f(x))=x$ for all $x$). Similarly, it is conceivable that $a \neq f(a), a \neq f(f(a))$, but $a=f(f(f(a)))$. This is about functions that are not monotonic, as discussed in problem B. 3541 (KöMaL 2003/2, p. 89), where the function $f(x)=\frac{x-3}{x-2}$ is mentioned in the remark, and a method is shown for constructing such first-degree rational functions. The function $f(x)=-\frac{2 x+7}{x+3}$, examined in problem B. 3481, is also such, and the solution of the problem relies precisely on this property (KöMaL 2002/1, p. 40).[^0]


[^0]:    ${ }^{1}$ With this substitution, we obtained problem B. 3579, whose solution can be found on p. 153 of the 2003/3 issue of KöMaL.