2002 IMO Problems/Problem 5

Problem

Find all functions such that

for all real numbers .

Solution

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• • Problem of Functions with Parameters**

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.

1. **Initial Exploration:**

   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.

2. **Determining \( f(1) \):**

   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 \( f(1) = 1 \).

3. **Manipulating the Original Equation:**

   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.

4. **Using Parity:**

   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 \( m = 1 \).

5. **Final Conclusion:**

   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.

See Also

2002 IMO (Problems) • Resources
Preceded by Problem 4	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 6
All IMO Problems and Solutions