IMO 2015 · Problem 5

Let \(\mathbb{R}\) be the set of real numbers. Determine all functions \(f:\mathbb{R}\to\mathbb{R}\) satisfying the equation \[ f(x+f(x+y))+f(xy)=x+f(x+y)+yf(x) \] for all real numbers \(x\) and \(y\).

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

Answer

The only solutions are \[ \boxed{f(x)=x\quad\text{for all }x\in\mathbb{R}} \qquad\text{and}\qquad \boxed{f(x)=2-x\quad\text{for all }x\in\mathbb{R}}. \]

Solution

Write \(P(x,y)\) for the given identity, and put \(c=f(0)\). From \(P(0,t)\), \[ f(f(t))=f(t)+c(t-1). \tag{1} \] Also, \(P(x,1)\) gives \[ f(x+f(x+1))=x+f(x+1). \tag{2} \]

Suppose first that \(c\ne0\). If \(f(u)=u\), then (1) implies \[ u=u+c(u-1), \] so \(u=1\). Thus (2) forces \(x+f(x+1)=1\) for every real \(x\), yielding \[ f(t)=2-t. \]

Now suppose that \(c=0\). Equation (1) becomes \[ f(f(t))=f(t). \tag{3} \] The substitutions \(P(-1,1)\) and \(P(1,-1)\) give, respectively, \[ 2f(-1)=-1+f(-1),\qquad f(1)+f(-1)=1-f(1). \] Consequently, \[ f(-1)=-1,\qquad f(1)=1. \tag{4} \]

From \(P(u,-u)\), we obtain \[ f(-u^2)=u-(u+1)f(u). \] Comparing this with the same formula for \(-u\) yields \[ (u+1)f(u)+(u-1)f(-u)=2u \qquad(u\in\mathbb{R}). \tag{5} \]

We claim that whenever \(f(r)=r\), we also have \(f(r+1)=r+1\). Indeed, using (4), the substitutions \(P(1,r-1)\) and \(P(-1,r+1)\) give \[ f(r+1)+f(r-1)=2r \] and \[ f(r-1)+f(-r-1)=-2. \] Subtracting and setting \(u=r+1\), we get \[ f(u)-f(-u)=2u. \] Substituting \(f(-u)=f(u)-2u\) into (5) gives \[ 2u\bigl(f(u)-u\bigr)=0. \] If \(u\ne0\), this implies \(f(u)=u\); if \(u=0\), the same conclusion follows from \(f(0)=0\). The claim is proved.

By (3), every \(r=f(t)\) is a fixed point. The claim therefore implies \[ f(1+f(t))=1+f(t). \] Finally, \(P(1,t-1)\) reads \[ f(1+f(t))+f(t-1)=t+f(t). \] Hence \(f(t-1)=t-1\) for every real \(t\), proving \(f(x)=x\).

Both candidates satisfy the original equation: for \(f(x)=x\), both sides equal \(2x+y+xy\); for \(f(x)=2-x\), both sides equal \(2+y-xy\). Thus both are solutions, and there are no others.

The solution was checked against the official IMO shortlist solutions.