2002 IMO Problems/Problem 5

Problem

Find all functions such that

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 $[0,\mathrm{\infty })$. Given $a>b\ge 0$, 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