Difference between revisions of "2002 IMO Problems/Problem 5"
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==Solution== | ==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. | 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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| − | + | 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 <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. | |
==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 
such that
![\[(f(x)+f(z))(f(y)+f(t))=f(xy-zt)+f(xt+yz)\]](http://latex.artofproblemsolving.com/c/9/2/c92c29ce2db1e17848b4da5f6f018c0f03021faa.png)
for all real numbers 
.
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 
and 
, 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 | ||
