IMO 1986 · Problem 5

Find all functions \(f\), defined on the non-negative real numbers and taking non-negative real values, such that:

\[ \begin{aligned} \text{(i)}\quad & f(xf(y))\cdot f(y)=f(x+y)\quad\text{for all }x,y\ge 0,\\ \text{(ii)}\quad & f(2)=0,\\ \text{(iii)}\quad & f(x)\ne 0\quad\text{for }0\le x<2. \end{aligned} \]

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

Answer

The unique solution is \[ \boxed{ f(x)= \begin{cases} \dfrac{2}{2-x},&0\le x<2,\\[4pt] 0,&x\ge 2. \end{cases}} \]

Solution

Putting \(y=2\) in (i), we obtain \[ f(x+2)=f(xf(2))f(2)=0\qquad(x\ge0). \] Together with (iii), this shows that \[ f(t)=0\quad\Longleftrightarrow\quad t\ge2 \qquad(t\ge0). \]

Fix any \(y\in[0,2)\), and write \(a=f(y)>0\). Taking \(x=2-y\) in (i) gives \[ f((2-y)a)\,a=f(2)=0. \] Thus \(f((2-y)a)=0\), so the zero characterization above yields \[ (2-y)a\ge2, \qquad\text{hence}\qquad a\ge\frac{2}{2-y}. \]

On the other hand, taking \(x=2/a\) in (i) gives \[ f\left(y+\frac2a\right)=f(2)\,a=0. \] Consequently, \[ y+\frac2a\ge2, \qquad\text{hence}\qquad a\le\frac{2}{2-y}. \] Therefore \(f(y)=2/(2-y)\) for every \(0\le y<2\). This proves uniqueness.

It remains to verify the proposed function. Conditions (ii) and (iii) hold immediately. For (i), let \(x,y\ge0\).

If \(y\ge2\), then \(f(y)=0\) and \(x+y\ge2\), so both sides of (i) are zero.

If \(y<2\), then \[ xf(y)=\frac{2x}{2-y}. \] When \(x+y\ge2\), this argument is at least \(2\), so again both sides of (i) are zero. When \(x+y<2\), the argument is less than \(2\), and \[ f(xf(y))f(y) = \frac{2}{\,2-\frac{2x}{2-y}\,}\cdot\frac{2}{2-y} = \frac{2}{2-x-y} = f(x+y). \] Thus (i) holds in every case, completing the proof.

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