IMO 1995- Problem 4

Problem: The positive real numbers $x_0,x_1,...,x_{1995}$ satisfy $x_0=x_{1995}$ and $x_{i-1}+\frac{2}{x_{i-1}}=2x_i+\frac{1}{x_i}$ for $i=1,2,...,1995$. Find the maximum value that $x_0$ can have.
Solution: The given condition is equivalent to $(2x_i^2-x_{i-1})(x_i-\frac{1}{x_{i-1}})=0$
which yield $x_i=\frac{x_{i-1}}{2}$ and $x_i=\frac{1}{x_{i-1}}$. We shall prove by induction that $x_i=2^{k_i}x_0^{\epsilon_{i}}$, where $|k_i|\le i$ and $\epsilon_i=(-1)^{k_i+i}$. This is true for $i=0$, with $k_0=0,\epsilon_0=1$. Assume that it is true for $i-1$. Then for $x_i=\frac{1}{2}x_{i-1}$, we have $k_i=k_{i-1}-1$ and $\epsilon_i=\epsilon_{i-1}$. For $x_i=\frac{1}{x_{i-1}}$, we have $k_i=-k_{i-1}$ and $\epsilon_{i}=-\epsilon_{i-1}$. In each case, it is immediate that $|k_i|\le i$ and $\epsilon_i=(-1)^{k_i+i}$. Thus $x_0=x_{1995}=2^{k}x_0^{\epsilon}$ where $k=k_{1995}$ and $\epsilon=\epsilon_{1995}$. Thus $|k|\le 1995$ and $\epsilon=(-1)^{1995+k}$. If $k$ is odd then $\epsilon=1$ and this is impossible. Thus $k$ is even and $\epsilon=-1$ and $x_0^2=2^k$. Since $|k|\le 1995$, we have $k\le 1994$. Hence $x_0\le 2^{997}$. The value $x_0=2^{997}$ can be attained by choosing $x_i=\frac{1}{2}x_{i-1}$ for $i=1,2,...,1994$ and $x_{1995}=\frac{1}{x_{1994}}$.
Source: https://sms.math.nus.edu.sg/Simo/IMO_Problems/95.pdf