IMO 2019 · Problem 1

Let \(\mathbb{Z}\) be the set of integers. Determine all functions \(f:\mathbb{Z}\to\mathbb{Z}\) such that, for all integers \(a\) and \(b\), \[ f(2a)+2f(b)=f(f(a+b)). \]

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

Answer

The solutions are \[ \boxed{f(n)=0\text{ for all }n\in\mathbb Z} \] and \[ \boxed{f(n)=2n+c\text{ for all }n\in\mathbb Z,\quad c\in\mathbb Z.} \]

Solution

Let \(c=f(0)\). Setting \(a=0\) in the given equation gives \[ f(f(b))=c+2f(b). \] Consequently, the original equation becomes \[ f(2a)+2f(b)=c+2f(a+b). \tag{1} \] Taking \(b=0\) in (1), we obtain \[ f(2a)=2f(a)-c. \] Substituting this back into (1) and dividing by \(2\) yields \[ f(a+b)=f(a)+f(b)-c \qquad(a,b\in\mathbb Z). \]

Define \(g(n)=f(n)-c\). Then \(g(0)=0\) and \[ g(a+b)=g(a)+g(b). \] Write \(k=g(1)\in\mathbb Z\). Induction gives \(g(n)=kn\) for every nonnegative integer \(n\). Also, \[ g(n)+g(-n)=g(0)=0, \] so the same formula holds for negative integers. Thus \[ f(n)=kn+c\qquad(n\in\mathbb Z). \]

Substituting this expression into the original equation gives \[ 2k(a+b)+3c=k^2(a+b)+(k+1)c. \] Taking \(a+b=0\) gives \((k-2)c=0\). Taking \(a+b=1\) and subtracting the preceding equality gives \(k(k-2)=0\). Hence either:

  • \(k=0\), in which case \(c=0\) and \(f\equiv0\);
  • \(k=2\), in which case \(c\) is any integer.

Finally, \(f\equiv0\) plainly satisfies the equation. For \(f(n)=2n+c\), \[ f(2a)+2f(b)=4(a+b)+3c =f\bigl(2(a+b)+c\bigr) =f(f(a+b)). \] Therefore every listed function is a solution, and there are no others.

The solution was checked against the official IMO shortlist solutions.