Solve the equation

$$
x=\sqrt{-3+4 \sqrt{-3+4 \sqrt{-3+4 x}}}
$$
The equation can only have a solution if $x \geqq \frac{3}{4}$. Consider the function

$$
f(x)=\sqrt{-3+4 x}
$$

This function is meaningful and strictly increasing for $x \geqq \frac{3}{4}$. The equation of the problem can be written in the form

$$
x=f(f(f(x)))
$$

If $x > f(x)$, then due to the strict monotonicity of $f$, we have $f(x) > f(f(x))$ and $f(f(x)) > f(f(f(x)))$, so $x > f(x) > f(f(x)) > f(f(f(x)))$. Therefore, such an $x$ cannot be a solution to the equation. Similarly, it can be shown that $x$ is also not a solution to the equation if $x < f(x)$.

The only remaining case is when $x = f(x)$, i.e., $x = \sqrt{-3 + 4x}$. Squaring and rearranging gives $0 = x^2 - 4x + 3 = (x-1)(x-3)$, so we get two possible roots:

$$
x_1 = 1, \quad x_2 = 3
$$

It can be verified by substitution that these are indeed solutions to the equation.

Peták Attila (Budapest, Berzsenyi D. Gymnasium, 4th grade)