48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions

First we show that $f(y)>y$ for all $y\in {\mathbb{R}}^{+}$. Functional equation (1) yields $f(x+f(y))>f(x+y)$ and hence $f(y)\ne y$ immediately. If $f(y)<y$ for some $y$, then setting $x=y-f(y)$ we get $$f(y)=f((y-f(y))+f(y))=f((y-f(y))+y)+f(y)>f(y),$$ contradiction. Therefore $f(y)>y$ for all $y\in {\mathbb{R}}^{+}$. For $x\in {\mathbb{R}}^{+}$ define $g(x)=f(x)-x$; then $f(x)=g(x)+x$ and, as we have seen, $g(x)>0$. Transforming (1) for function $g(x)$ and setting $t=x+y$, $$\begin{array}{rl}f(t+g(y))& =f(t)+f(y)\\ g(t+g(y))+t+g(y)& =(g(t)+t)+(g(y)+y)\end{array}$$ and therefore $$\begin{array}{c}g(t+g(y))=g(t)+y\text{ for all }t>y>0\end{array}\text{\hspace{1em}(2)}$$ Next we prove that function $g(x)$ is injective. Suppose that $g\left(y_1\right)=g\left(y_2\right)$ for some numbers $y_1,y_2\in {\mathbb{R}}^{+}$. Then by (2), $$g(t)+y_1=g\left(t+g\left(y_1\right)\right)=g\left(t+g\left(y_2\right)\right)=g(t)+y_2$$ for all $t>\max \{y_1,y_2\}$. Hence, $g\left(y_1\right)=g\left(y_2\right)$ is possible only if $y_1=y_2$. Now let $u,v$ be arbitrary positive numbers and $t>u+v$. Applying (2) three times, $$g(t+g(u)+g(v))=g(t+g(u))+v=g(t)+u+v=g(t+g(u+v))$$ By the injective property we conclude that $t+g(u)+g(v)=t+g(u+v)$, hence $$\begin{array}{c}g(u)+g(v)=g(u+v)\end{array}\text{\hspace{1em}(3)}$$ Since function $g(v)$ is positive, equation (3) also shows that $g$ is an increasing function. Finally we prove that $g(x)=x$. Combining (2) and (3), we obtain $$g(t)+y=g(t+g(y))=g(t)+g(g(y))$$ and hence $$g(g(y))=y$$ Suppose that there exists an $x\in {\mathbb{R}}^{+}$ such that $g(x)\ne x$. By the monotonicity of $g$, if $x>g(x)$ then $g(x)>g(g(x))=x$. Similarly, if $x<g(x)$ then $g(x)<g(g(x))=x$. Both cases lead to contradiction, so there exists no such $x$. We have proved that $g(x)=x$ and therefore $f(x)=g(x)+x=2x$ for all $x\in {\mathbb{R}}^{+}$. This function indeed satisfies the functional equation (1).

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Alternative solution.

We prove that $f(y)>y$ and introduce function $g(x)=f(x)-x>0$ in the same way as in Solution 1. For arbitrary $t>y>0$, substitute $x=t-y$ into (1) to obtain $$f(t+g(y))=f(t)+f(y)$$ which, by induction, implies $$\begin{array}{c}f(t+ng(y))=f(t)+nf(y)\text{ for all }t>y>0,n\in \mathbb{N}.\end{array}\text{\hspace{1em}(4)}$$ Take two arbitrary positive reals $y$ and $z$ and a third fixed number $t>\max \{y,z\}$. For each positive integer $k$, let ${\ell }_k=\lfloor k\frac{g(y)}{g(z)}\rfloor$. Then $t+kg(y)-{\ell }_{k}g(z)\ge t>z$ and, applying (4) twice, $$\begin{array}{rl}f\left(t+kg(y)-{\ell }_{k}g(z)\right)+{\ell }_{k}f(z)& =f(t+kg(y))=f(t)+kf(y)\\ 0& <\frac{1}{k}f\left(t+kg(y)-{\ell }_{k}g(z)\right)=\frac{f(t)}{k}+f(y)-\frac{{\ell }_k}{k}f(z)\end{array}$$ As $k\to \infty$ we get $$0\le \underset{k\to \infty }{\lim }\left(\frac{f(t)}{k}+f(y)-\frac{{\ell }_k}{k}f(z)\right)=f(y)-\frac{g(y)}{g(z)}f(z)=f(y)-\frac{f(y)-y}{f(z)-z}f(z)$$ and therefore $$\frac{f(y)}{y}\le \frac{f(z)}{z}$$ Exchanging variables $y$ and $z$, we obtain the reverse inequality. Hence, $\frac{f(y)}{y}=\frac{f(z)}{z}$ for arbitrary $y$ and $z$; so function $\frac{f(x)}{x}$ is constant, $f(x)=cx$. Substituting back into (1), we find that $f(x)=cx$ is a solution if and only if $c=2$. So the only solution for the problem is $f(x)=2x$.