2007 IMO Shortlist Problems/A4


(Thailand) Find all functions $f: \mathbb{R}^+ \to \mathbb{R}^+$ such that \[f(x+f(y)) = f(x+y) + f(y)\] for all $x,y \in \mathbb{R}^+$. (Symbol $\mathbb{R}^+$ denotes the set of all positive real numbers [sic].)


We will show that $f(x) = 2x$ is the unique solution to this equation. To this end, let $g(x) = f(x) - x$. The given condition then translates to \[g(x+y+ g(y)) + x+y+g(y) = g(x+y) + x+y + g(y) + y ,\] or \[g(x+y+g(y)) = g(x+y) + y .\]

Lemma 1. The function $g$ is injective.

Proof. Suppose $g(a) = g(b)$. Then \[a = g(b+a+g(a)) - g(b+a) = g(a+b+g(b)) - g(a+b) = b,\] as desired. $\blacksquare$

Lemma 2. If $a>b$, then $g(a+g(b)) = g(a) + b$.

Proof. Set $y= b$, $x=a-b$. $\blacksquare$

Lemma 3. For all $a,b$, $g(a+b) = g(a) + g(b)$.

Proof. Pick an arbitrary positive real $c > \max(g(a)+g(b), g(a))$. Then by Lemma 2, \[g(c+g(a)+g(b)) = g(c+g(a))+b = g(c) + a+b = g(c+g(a+b)) .\] Since $g$ is injective, it follows that $c+g(a)+g(b) = c + g(a+b)$. The lemma then follows. $\blacksquare$

Now, let $x$ be any positive real; pick some $y>x$. Then by Lemmata 3 and 2, \[g(x) + g(y) = g(x+y) = g(y) + x .\] Hence $g(x)=  x$ and $f(x) = g(x) + x = 2x$. Therefore the function $x \mapsto 2x$ is the only possible solution to the problem. Since this function evidently satisfies the problem's condition, it is the unique solution, as desired. $\blacksquare$


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