Find the solutions of the equation

$$
\sqrt{x+\sqrt{4 x+\sqrt{16 x+\sqrt{\ldots+\sqrt{4^{n} x+3}}}}}=1+\sqrt{x}
$$
Solution

Let's raise the given equation to the power of $n$ times, obtaining

$$
\begin{aligned}
x+\sqrt{4 x+\sqrt{16 x+\sqrt{\ldots+\sqrt{4^{n} x+3}}}} & =1+2 \sqrt{x}+\chi \\
\sqrt{4 x+\sqrt{16 x+\sqrt{\ldots+\sqrt{4^{n} x+3}}}} & =1+2 \sqrt{x} \\
4 x+\sqrt{16 x+\sqrt{\ldots+\sqrt{4^{n} x+3}}} & =1+4 \sqrt{x}+4 x \\
\sqrt{16 x+\sqrt{\ldots+\sqrt{4^{n} x+3}}} & =1+4 \sqrt{x} \\
\cdots & =\cdots \\
\sqrt{4^{n-1} x+\sqrt{4^{n} x+3}} & =1+2^{n-1} \sqrt{x} \\
4^{n-1} x+\sqrt{4^{n} x+3} & =1+2^{n} \sqrt{x}+2^{2 n-2} x \\
\sqrt{4^{n} x+3} & =1+2^{n} \sqrt{x} \\
4^{n} x+3 & =1+2^{n+1} \sqrt{x}+2^{2 n x} x \\
3 & =1+2^{n+1} \sqrt{x}
\end{aligned}
$$

Therefore, $\sqrt{x}=2^{-n}$, that is, $x=2^{-2 n}$.