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Question 4 from IMO 2008

By zieglerk

Problem: Find all functions f: \mathbb{R}^{+} \to \mathbb{R}^{+}, s.t.

\displaystyle \frac{f(w)^2 + f(x)^2}{f(y^2)+f(z^2)} = \frac{w^2+x^2}{y^2+z^2} \ \ \ \ \ (1)

whenever wx=yz. (Author: Hojoo Lee, South Korea)

Remark: Obvious solutions are \mathop{id}: x \mapsto x and \mathop{inv}: x \mapsto 1/x. We show, that those are also the only solutions.

Fact 1: f(1) = 1.

Proof: Let w=x=y=z=1.

Fact 2: f(x^2) = f(x)^2.

Proof: Let w=y=1 and x=z.

Using Fact 2, we can rewrite (1) as

\displaystyle \frac{f(w)+f(x)}{w+x}=\frac{f(y)+f(z)}{y+z} \ \ \ \ \ (2)

whenever wx=yz.

Fact 3: For every x \in \mathbb{R}^{+}

\displaystyle f(x) = x \text{ or } f(x) = 1/x .

Proof: Let w=x, y=1 \text{ and } z = x^2. Then (2) implies

\displaystyle \frac{2f(x)}{2x} = \frac{1+f(x)^2}{1+x^2}

which is a quadratic equation for f(x) with the two solutions x and 1/x.

Fact 4 (

Non-mixing Lemma): If f(x) = x for some x \neq 1, then for all x \in \mathbb{R}^{+}.

Proof: Assume f(x) = x, for some x \neq 1, but f(w) \neq w for some w \neq 1. By Fact 3, we know that in this case f(w) = 1/w. We will show, that now the two possibilities for f(xw), i.e. f(xw) = xw and f(xw) = 1/(xw) both lead to contradictions.

Let y=1 and z = x \cdot w. Then (2) reads as

\displaystyle \frac{1/w + x}{w+x} = \frac{1+f(xw)}{1+xw} \ \ \ \ \ (3).

If f(xw) = xw, this implies w= 1. If f(xw) = 1/(xw) this implies x =1. Either way, a contradiction.

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