IMO 2002 · Problem 5

Find all real-valued functions on the reals such that \((f(x)+f(y))((f(u)+f(v))=f(xu-yv)+f(xv+yu)\) for all \(x,y,u,v\).

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

Answer

The solutions are \[ \boxed{f(x)\equiv 0,\qquad f(x)\equiv \frac12,\qquad f(x)=x^2.} \]

Solution

The duplicated opening parenthesis in the printed statement is understood as a typographical duplication: the left-hand side is \((f(x)+f(y))(f(u)+f(v))\).

Let \(c=f(0)\). Substituting \(x=y=u=v=0\) gives \[ 4c^2=2c, \] so \(c=0\) or \(c=\frac12\).

If \(c=\frac12\), setting \(y=u=v=0\) gives \[ (f(x)+c)\,2c=2c, \] hence \(f(x)=\frac12\) for every real \(x\).

Now suppose \(c=0\). Setting \(y=v=0\) yields \[ f(xu)=f(x)f(u) \qquad(x,u\in\mathbb R). \tag{1} \] In particular, \(f(1)^2=f(1)\). If \(f(1)=0\), then (1) gives \(f(x)=f(x)f(1)=0\) for every \(x\).

It remains to consider \(f(1)=1\). Setting \(u=0\), \(v=1\) in the original equation gives \[ f(x)+f(y)=f(-y)+f(x), \] so \(f\) is even. Consequently, by (1), \[ f(t)=f(|t|)=f\bigl(\sqrt{|t|}\bigr)^2\ge 0 \qquad(t\in\mathbb R). \tag{2} \]

Define \[ g(t)=f(\sqrt t)\qquad(t\ge 0). \] Then \(g(t)\ge0\), \(g(0)=0\), and \(g(1)=1\). For arbitrary \(s,t\ge0\), substitute \[ x=u=\sqrt s,\qquad y=\sqrt t,\qquad v=-\sqrt t \] into the original equation. Evenness and \(f(0)=0\) give \[ (g(s)+g(t))^2=f(s+t). \] By (1), \(f(s+t)=f(\sqrt{s+t})^2=g(s+t)^2\). Both sides being squares of nonnegative quantities, we obtain \[ g(s+t)=g(s)+g(t)\qquad(s,t\ge0). \tag{3} \]

Equation (3) and \(g(1)=1\) imply \(g(q)=q\) for every nonnegative rational \(q\): first \(g(m)=m\) for nonnegative integers \(m\), and then \[ n\,g(m/n)=g(m)=m \] for every positive integer \(n\).

Moreover, \(g\) is nondecreasing, since for \(b\ge a\ge0\), \[ g(b)=g(a)+g(b-a)\ge g(a). \] Fix \(t\ge0\). For each positive integer \(n\), put \(m=\lfloor nt\rfloor\). Monotonicity and the rational values give \[ \frac mn\le g(t)\le\frac{m+1}{n}. \] Since \(t\) lies in the same interval, \(|g(t)-t|\le1/n\) for every positive integer \(n\). Therefore \(g(t)=t\).

Thus, for every real \(x\), \[ f(x)=f(|x|)=g(x^2)=x^2. \]

Finally, both constant functions in the answer satisfy the equation directly. The function \(f(x)=x^2\) satisfies it by the identity \[ (x^2+y^2)(u^2+v^2)=(xu-yv)^2+(xv+yu)^2. \] Hence all three listed functions, and only those functions, are solutions.

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