IMO 1992 · Problem 2

Let \(\mathbf{R}\) denote the set of all real numbers. Find all functions \(f:\mathbf{R}\to\mathbf{R}\) such that \[ f\left(x^2+f(y)\right)=y+(f(x))^2 \qquad\text{for all }x,y\in R. \]

Official text of the problem. Official problem paper (PDF)

Answer

The unique solution is \[ \boxed{f(x)=x\quad\text{for every }x\in\mathbf{R}.} \]

Solution

Put \(c=f(0)\). Setting \(x=0\) in the given equation gives \[ f(f(y))=y+c^2 \qquad (y\in\mathbf{R}). \tag{1} \] Consequently, \(f\) is injective: if \(f(u)=f(v)\), applying \(f\) and using (1) yields \(u=v\). It is also surjective, since the right-hand side of (1) takes every real value.

We first prove that \(c=0\). By surjectivity, choose \(a\in\mathbf{R}\) with \(f(a)=0\). Substituting \((x,y)=(a,0)\) and \((-a,0)\) into the original equation gives \[ f(a^2+c)=f(a)^2=0, \qquad f(a^2+c)=f(-a)^2. \] Thus \(f(-a)=0=f(a)\). Injectivity implies \(-a=a\), so \(a=0\), and hence \(c=f(0)=0\). Equation (1) therefore becomes \[ f(f(y))=y \qquad (y\in\mathbf{R}). \tag{2} \]

Next, we show that \(f\) is nondecreasing. Let \(s\le t\). In the original equation, take \[ x=\sqrt{t-s},\qquad y=f(s). \] Using (2), we obtain \[ f(t)=f(s)+\left(f\left(\sqrt{t-s}\right)\right)^2\ge f(s). \]

Finally, fix any \(u\in\mathbf{R}\). If \(f(u)>u\), monotonicity and (2) imply \[ u=f(f(u))\ge f(u)>u, \] a contradiction. If \(f(u)<u\), they imply \[ u=f(f(u))\le f(u)<u, \] again a contradiction. Therefore \(f(u)=u\) for every real \(u\).

Conversely, \(f(x)=x\) satisfies the equation, since \[ f(x^2+f(y))=x^2+y=y+(f(x))^2. \] This proves uniqueness and existence.

The solution was checked against John Scholes' IMO solution archive (Kalva).