IMO 1994 · Problem 5

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

Official text of the problem. Official problem paper (PDF)

Answer

The unique solution is \[ \boxed{f(x)=-\frac{x}{1+x}\qquad(x>-1).} \]

Solution

We first prove that every fixed point of \(f\) must be \(0\).

Suppose that \(f(a)=a\), and put \[ b=2a+a^2=(1+a)^2-1. \] Since \(a>-1\), we have \(b>-1\). Substituting \(x=y=a\) into the functional equation gives \[ f(b)=b. \] If \(a>0\), then \[ b-a=a(1+a)>0, \] so \(0<a<b\). If \(-1<a<0\), then \(b-a<0\), so \[ -1<b<a<0. \] Thus, whenever \(a\ne0\), the distinct numbers \(a,b\) lie in the same interval on which \(f(t)/t\) is strictly increasing. But \[ \frac{f(a)}a=\frac{f(b)}b=1, \] a contradiction. Consequently, every fixed point is \(0\).

Now take any \(x\in S\), and set \[ z=x+f(x)+xf(x). \] This number belongs to \(S\), because \[ 1+z=(1+x)(1+f(x))>0. \] The functional equation with \(y=x\) gives \[ f(z)=x+f(x)+xf(x)=z. \] Hence \(z\) is a fixed point and therefore \(z=0\). It follows that \[ x+(1+x)f(x)=0, \qquad\text{so}\qquad f(x)=-\frac{x}{1+x}. \]

It remains to verify this function. For every \(x>-1\), \[ 1+f(x)=\frac1{1+x}>0, \] so \(f\) maps \(S\) into \(S\). Moreover, for \(x,y\in S\), \[ x+f(y)+xf(y)=\frac{x-y}{1+y}, \qquad 1+\frac{x-y}{1+y}=\frac{1+x}{1+y}>0. \] Therefore \[ f(x+f(y)+xf(y)) =f\!\left(\frac{x-y}{1+y}\right) =\frac{y-x}{1+x} =y+f(x)+yf(x). \] Finally, for \(x\ne0\), \[ \frac{f(x)}x=-\frac1{1+x}. \] If \(-1<u<v\), then \[ -\frac1{1+v}+\frac1{1+u} =\frac{v-u}{(1+u)(1+v)}>0. \] Thus the required quotient is strictly increasing on each specified interval. Both conditions hold.

The solution was checked against John Scholes' IMO solution archive (Kalva).