USAMO or AIME or IMO?

If $f_n(x) = 2x$ is true for $n=1$ , then

$\begin{aligned} \sqrt{x^2+48} & = 2x \\ x^2+48 & = 4x^2 \\ \implies 3x^2 & = 48 \\ x & = 4 & \small \color{#3D99F6}{\text{Since }f_1(x) \ge 0} \\ \implies f_1(x) & = 2(4) = 8 \end{aligned}$

Let us check if the claim $f_n(x) = 8$ is true for all $n \ge 1$ by induction.

1. It is shown that the claim is true for $n=1$ .
2. Assuming that the claim is true for $n$ , then

$\begin{aligned} \quad f_{n+1}(x) & = \sqrt{x^2+6f_n(x)} = \sqrt{4^2+6(8)} = \sqrt{16+48} = 8 \end{aligned}$

$\quad$ Therefore, the claim $f_n(x) = 8$ is also true for $n+1$ and it is true for all $n \ge 1$ .

Therefore, the solution is $x=\boxed{4}$ .

For every integer there is only one solution. So let's take $n=1$

$f_1(x) = 2x \\ \sqrt[]{x^2+48} = 2x \\ x^2+48 = 4x^2 \\ 3x^2 = 48 \\ x^2 = 16 \\ x = \pm 4$

Since $x$ is positive, $x=4$