# 1994 IMO Problems/Problem 5

## Problem

Let $S$ be the set of real numbers strictly greater than $-1$. Find all functions $f:S \to S$ satisfying the two conditions:

1. $f(x+f(y)+xf(y)) = y+f(x)+yf(x)$ for all $x$ and $y$ in $S$;

2. $\frac{f(x)}{x}$ is strictly increasing on each of the intervals $-1 and $0.

## Solution

The only solution is $f(x) = \frac{-x}{x+1}.$

Setting $x=y,$ we get

$$f(x+f(x)+xf(x)) = x+f(x)+xf(x).$$

Therefore, $f(s) = s$ for $s = x+f(x)+xf(x).$

Note: If we can show that $s$ is always $0,$ we will get that $x+f(x)+xf(x) = 0$ for all $x$ in $S$ and therefore, $f(x) = \frac{-x}{x+1}.$

Let $f(s)=s.$ Setting $x=y=s,$ we get

$$f(s + f(s) + sf(s)) = f(2s + s^2) = 2s+s^2.$$

If $t=2s+s^2,$ we have $f(t)=t$ as well.

Consider $s > 0.$ We get $t = 2s + s^2 > s.$ Since $\frac{f(x)}{x}$ is strictly increasing for $x>0$ and $t > s$ in this domain, we must have $\frac{f(t)}{t} > \frac{f(s)}{s}$ but since $f(s) = s$ and $f(t) = t,$ we also have that $\frac{f(s)}{s} = 1 = \frac{f(t)}{t}$ which is a contradiction. Therefore $s \leq 0$

Consider $-1 < s < 0.$ Using a similar argument, we will get that $t = 2s + s^2 < s$ but also $\frac{f(s)}{s} = 1 = \frac{f(t)}{t},$ which is a contradiction.

Hence, $s$ must be $0.$ Since $s=0,$ we can conclude that $x+f(x)+xf(x) = 0$ and therefore, $f(x) = \frac{-x}{x+1}$ for all $x$.