# 2002 IMO Problems/Problem 5

## Problem

Find all functions $f:\Bbb{R}\to \Bbb{R}$ such that

$$(f(x)+f(z))(f(y)+f(t))=f(xy-zt)+f(xt+yz)$$

for all real numbers $x,y,z,t$.

## Solution

Given the problem $$(f(x) + f(y))(f(u) + f(v)) = f(xu - yv) + f(xv - yu)$$, we aim to find a function that satisfies it.

   We start by considering the case when $$x = y = u = v = 0$$.


   This leads us to $$4f(0)^2 = 2f(0)$$, implying $$f(0) = 0$$ or $$f(0) = 1/2$$.


   If $$f(0) = 1$$, then putting $$x = y = u = 0$$ gives us $$f(u) = 1/2$$ for all $$u \in \mathbb{R}$$.


   On the other hand, if $$f(0) = 0$$, putting $$y = v = 0$$ gives us $$f(x)f(u) = f(xu)$$, indicating that $$f$$ is multiplicative.


   If $$f(0) = 0$$, we have $$f(1) = 0$$ or $$f(1) = 1$$.


   If $$f(1) = 0$$, then $$f(x) = f(x \cdot 1) = f(x)f(1) = 0$$ for all $$x \in \mathbb{R}$$.


   Disregarding constant solutions, we assume $$f(0) = 0$$ and $$f(1) = 1$$.


   Taking $$x = y = 1$$ in the original equation, we arrive at $$2f(u) + 2f(v) = f(u + v) + f(u - v)$$.


   Taking $$u = 0$$, we get $$f(v) = f(-v)$$, indicating that $$f$$ is an even function.


   Using parity and taking $$a = u$$ and $$b = -v$$ in the original equation, we get $$f(u^2 + v^2) = (f(u) + f(v))^2$$.


   This implies $$f(x) > 0$$ for all $$x > 0$$, allowing us to define an auxiliary function $$g$$ as $$g(x) = \sqrt{f(x)}$$.


   Then, taking $$a = u^2$$ and $$b = v^2$$, the equation rewrites as $$g(a+b) = g(a) + g(b)$$.


   This leads us to $$g$$ being additive, and therefore, there exists $$m \in \mathbb{N}$$ such that $$g(x) = mx$$ for all $$x > 0$$. Since $$g(1) = \sqrt{f(1)} = 1$$, we have $$m = 1$$.


   We will prove that $$f$$ is increasing on $$[0, \infty)$$. Given $$a > b \geq 0$$, we express $$a = u^2 + v^2$$ and $$b = u^2$$ for $$u, v \in \mathbb{R}$$.


   Then, $$f(u^2 + v^2) = (f(u) + f(v))^2 = f(u^2) + 2f(uv) + f(v^2) > f(u^2)$$, implying $$f(a) > f(b)$$, since $$f$$ is multiplicative.


   Therefore, the only solutions are $$f(x) = 0$$, $$f(x) = 1/2$$, and $$f(x) = x^2$$, which can be easily verified in the original equation.