2001 IMO Shortlist Problems/A4

Problem

Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ , satisfying

$f(xy)(f(x) - f(y)) = (x - y)f(x)f(y)$

for all $x,y$ .

Solution

Assume $f(1) = 0$ . Take $y = 1$ . We get $f^2(x) = 0,\ \forall x$ , so $f(x) = 0,\ \forall x$ . This is a solution, so we can take it out of the way: assume $f(1)\ne 0$ .

$y = 1\Rightarrow f(x)[f(x) - f(1)] = (x - 1)f(x)f(1)$ . We either have $f(x) = 0$ or $f(x) = f(1)x$ , so for every $x$ , $f(x)\in\{0,f(1)x\}$ . In particular, $f(0) = 0$ .

Assume $f(y) = 0$ . We get $f(x)f(xy) = 0,\ \forall x$ . This means that $f(a),f(b)\ne 0\Rightarrow f\left(\frac ab\right)\ne 0\ (*)$ ( $\frac ab$ is defined because $f(b)\ne 0\Rightarrow b\ne 0$ ). Assume now that $x\ne y$ and $f(x),f(y)\ne 0$ . We get $f(x) = f(1)x,\ f(y) = f(1)y$ , and after replacing everything we get $f(xy) = f(1)xy\ne 0$ , so $x\ne y,\ f(x),f(y)\ne 0\Rightarrow f(xy)\ne 0\ (**)$ . Assume now $f(x)\ne 0$ . From $(*)$ we get $f\left(\frac 1x\right)\ne 0$ , and after applying $(*)$ again to $a = x,b = \frac 1x$ we get $f(x^2)\ne 0\ (***)$ . We can now see that $(**),(***)$ combine to $f(x),f(y)\ne 0\Rightarrow f(xy)\ne 0\ (\#)$ .

Let $G = \{x\in\mathbb R|f(x)\ne 0\}$ . $(*)$ and $(\#)$ simply say that $(G,\ \cdot)$ is a subgroup of $(\mathbb R^{*},\ \cdot)$ .

Conversely, let $G$ be a subgroup of the multiplicative group $\mathbb R^*$ . Take $f(x) = \left\{\begin{array}{c}f(1)x,\ x\in G \\ 0,\ x\not \in G\end{array}\right\}$ . It's easy to check the condition $f(xy)[f(x) - f(y)] = (x - y)f(x)f(y)$ .

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2001 IMO Shortlist Problems/A4

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