IMO 1999 · Problem 6

Determine all functions \(f:\mathbb{R}\longrightarrow\mathbb{R}\) such that \[ f(x-f(y))=f(f(y))+xf(y)+f(x)-1 \] for all real numbers \(x,y\).

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

Answer

The unique solution is \[ \boxed{f(x)=1-\frac{x^2}{2}\qquad(x\in\mathbb{R}).} \]

Solution

Let \(S=f(\mathbb{R})\) and \(c=f(0)\). The given equation says that \[ f(x-t)=f(t)+xt+f(x)-1 \qquad(x\in\mathbb{R},\ t\in S). \] Setting \(x=t\), we obtain \[ c=2f(t)+t^2-1, \] so \[ f(t)=\frac{c+1-t^2}{2}\qquad(t\in S). \] Consequently, \[ f(x-t)=f(x)+xt-\frac{t^2}{2}+\frac{c-1}{2}. \]

Define \[ g(x)=f(x)+\frac{x^2}{2}, \qquad d=\frac{c-1}{2}. \] The preceding identity becomes \[ g(x-t)=g(x)+d \qquad(x\in\mathbb{R},\ t\in S). \tag{1} \] In particular, \(g(x+s)=g(x)-d\) for every \(s\in S\). Thus, for any \(s,t\in S\), \[ g(x+s-t)=g(x+s)+d=g(x). \tag{2} \]

We now prove that every real number is a difference of two elements of \(S\).

First, \(f\) cannot be constant: if \(f\equiv k\), the original equation with \(x=-1\) would give \(k=k-1\). Hence we can choose distinct \(s_0,t_0\in S\). Put \(h=s_0-t_0\ne0\). By (2), \(g(x+h)=g(x)\), and therefore \[ f(x+h)-f(x) =-\frac{(x+h)^2-x^2}{2} =-hx-\frac{h^2}{2}. \] As \(x\) ranges over \(\mathbb{R}\), the right-hand side takes every real value because \(h\ne0\). Both \(f(x+h)\) and \(f(x)\) belong to \(S\), proving the claim.

It now follows from (2) that \[ g(x+r)=g(x)\qquad(x,r\in\mathbb{R}), \] so \(g\) is constant. Equation (1) then forces \(d=0\), whence \(c=1\). Since \(g(0)=c=1\), we conclude that \[ f(x)=1-\frac{x^2}{2} \qquad(x\in\mathbb{R}). \]

Finally, this function satisfies the equation. Indeed, writing \(t=f(y)\), we have \[ f(t)+xt+f(x)-1 =\left(1-\frac{t^2}{2}\right)+xt+ \left(1-\frac{x^2}{2}\right)-1 =1-\frac{(x-t)^2}{2} =f(x-f(y)). \] Thus the displayed function is exactly the required solution.

The solution was checked against Evan Chen's published IMO solution notes.