## Problem 2

Define $f_{1}(x)=\sqrt{ }\left(x^{2}+48\right)$ and $f_{n}(x)=\sqrt{ }\left(x^{2}+6 f_{n-1}(x)\right)$. Find all real solutions to $f_{n}(x)=2 x$.


## Solution

Answer: For each $\mathrm{n}, \mathrm{x}=4$ is the only solution.

Obviously $\mathrm{x}=4$ is a solution. Since $\mathrm{f}_{\mathrm{n}}(\mathrm{x})>=0$, any solution must be non-negative. So we restrict attention to $\mathrm{x} \geq 0$.

Suppose $x2 x$. For $n=1$, the claim is equivalent to $4 x^{2}x^{2}+12 x$. But $x2 \mathrm{x}$, as required.

An exactly similar argument shows that $\mathrm{f}_{\mathrm{n}}(\mathrm{x})4$. Hence $\mathrm{x}=4$ is the only solution.

