Korean Mathematical Olympiad Final Round

Let $u:=x+\frac{1}{y}$, $v:=y+\frac{1}{x}$ for $x>0,y>0$. Then we have $$f(u+v)=f(u)+f(v)(1)$$ Here $uv=xy+2+\frac{1}{xy}\ge 4$. First we show that for any $u,v>0$ with $uv\ge 4$ can be represented as $$u=x+\frac{1}{y},v=y+\frac{1}{x}(2)$$ for some $x>0,y>0$. From (2), we deduce $$u=x+\frac{1}{y}⟹x=u-\frac{1}{y}=\frac{uy-1}{y}⟹v=y+\frac{1}{x}=\frac{u{y}^2}{uy-1}⟹u{y}^2-uvy+v=0.$$

The last quadratic equation has a real solution $y>0$ because $D=(uv{)}^2-4uv\ge 0$ and $u,v>0$. (Observe that there's no negative real solution for the equation.) Similarly, there exists real $x>0$, which satisfies (2) together with the above $y$. Therefore, (1) holds for all $u>0,v>0$ $uv\ge 4$. We verify that (1) is valid for all $u,v>0$. Indeed, for given $u,v>0$, we may choose a real number $w>0$ such that $$(u+v)w\ge 4,u(v+w)\ge 4,vw\ge 4.$$ Then from $$\begin{array}{rl}f(u+v+w)& =f(u+v)+f(w)\\ & =f(u)+f(v+w)\\ & =f(u)+f(v)+f(w).\end{array}$$ we may conclude that $f(u+v)=f(u)+f(v)$ for all $u,v\ge 0$. Now we put $h(x)=f(x)-f(1)x=f(x)-2008x$ and show that $h(x)$ is a zero function. It is clear that $h$ satisfies (1) and $$h(x+1)=h(x)+h(1)=h(x)(3)$$ for all $x\in {\mathbb{R}}^{+}$. It follows from the condition $|f(x)|\le x^2+1004^2$ that $$-(x+1004{)}^2\le h(x)\le (x-1004{)}^2,$$ which implies that $h(x)$ is bounded on $(0,1]$ and hence $h(x)$ is bounded on ${\mathbb{R}}^{+}$ according to (3). Since $h(nx)=nh(x)$ for all positive integer $n$, we must have $h\equiv 0$. Therefore we have $f(x)=2008x$. $\square$