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Difference between revisions of "2002 IMO Problems/Problem 5"
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==Solution==
 
==Solution==
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{{solution}}
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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.
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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.
+
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 \( f(0) = 1/2 \).  
  
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1. **Initial Exploration:**
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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.  
  
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    We start by considering the case when \( x = y = u = v = 0 \).
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If \( f(0) = 0 \), we have \( f(1) = 0 \) or \( f(1) = 1 \).  
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    This leads us to \( 4f(0)^2 = 2f(0) \), implying \( f(0) = 0 \) or \( f(0) = 1/2 \).
 
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    If \( f(0) = 1 \), then putting \( x = y = u = 0 \) gives us \( f(u) = 1/2 \) for all \( u \in \mathbb{R} \).
 
−
   
 
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    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.
 
  
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2. **Determining \( f(1) \):**
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If \( f(1) = 0 \), then \( f(x) = f(x \cdot 1) = f(x)f(1) = 0 \) for all \( x \in \mathbb{R} \).
  
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    If \( f(0) = 0 \), we have \( f(1) = 0 \) or \( f(1) = 1 \).
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Disregarding constant solutions, we assume \( f(0) = 0 \) and \( f(1) = 1 \).  
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    If \( f(1) = 0 \), then \( f(x) = f(x \cdot 1) = f(x)f(1) = 0 \) for all \( x \in \mathbb{R} \).
 
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    Disregarding constant solutions, we assume \( f(0) = 0 \) and \( f(1) = 1 \).
 
  
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3. **Manipulating the Original Equation:**
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Taking \( x = y = 1 \) in the original equation, we arrive at \( 2f(u) + 2f(v) = f(u + v) + f(u - v) \).  
  
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    Taking \( x = y = 1 \) in the original equation, we arrive at \( 2f(u) + 2f(v) = f(u + v) + f(u - v) \).
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Taking \( u = 0 \), we get \( f(v) = f(-v) \), indicating that \( f \) is an even function.  
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    Taking \( u = 0 \), we get \( f(v) = f(-v) \), indicating that \( f \) is an even function.
 
  
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4. **Using Parity:**
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Using parity and taking \( a = u \) and \( b = -v \) in the original equation, we get \( f(u^2 + v^2) = (f(u) + f(v))^2 \).  
  
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    Using parity and taking \( a = u \) and \( b = -v \) in the original equation, we get \( f(u^2 + v^2) = (f(u) + f(v))^2 \).
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This implies \( f(x) > 0 \) for all \( x > 0 \), allowing us to define an auxiliary function \( g \) as \( g(x) = \sqrt{f(x)} \).  
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    This implies \( f(x) > 0 \) for all \( x > 0 \), allowing us to define an auxiliary function \( g \) as \( g(x) = \sqrt{f(x)} \).
 
−
   
 
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    Then, taking \( a = u^2 \) and \( b = v^2 \), the equation rewrites as \( g(a+b) = g(a) + g(b) \).
 
−
   
 
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    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 \( m = 1 \).
 
  
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5. **Final Conclusion:**
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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 <math>\boxed{f(x) = 0}, \boxed{ f(x) = 1/2 },</math> and <math>\boxed{f(x) = x^2}</math>, which can be easily verified in the original equation.
  
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    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.
 
−
   
 
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    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==
 
==See Also==
  
 
{{IMO box|year=2002|num-b=4|num-a=6}}
 
{{IMO box|year=2002|num-b=4|num-a=6}}

Latest revision as of 18:41, 14 April 2025

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 $\boxed{f(x) = 0}, \boxed{ f(x) = 1/2 },$ and $\boxed{f(x) = x^2}$, 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
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