IMO 1996 SL A7

Origin: ARM | Category: Algebra

Problem

Let $f$ be a function from the set of real numbers $\mathbb{R}$ into itself such that for all $x\in\mathbb{R}$, we have $|f(x)|\leq 1$ and

$$f\left(x+\frac13\right)+f(x)=f\left(x+\frac16\right)+f\left(x+\frac17\right).$$

Prove that $f$ is a periodic function (that is, there exists a nonzero real number $c$ such that $f(x+c)=f(x)$ for all $x\in\mathbb{R}$).

Solution

We are given that

$$f(x+a+b)-f(x+a)=f(x+b)-f(x),$$

where

$$a=\frac16,\qquad b=\frac17.$$

Summing up these equations for

$$x,\ x+b,\ \ldots,\ x+6b,$$

we obtain

$$f(x+a+1)-f(x+a)=f(x+1)-f(x).$$

Summing up the new equations for

$$x,\ x+a,\ \ldots,\ x+5a,$$

we obtain that

$$f(x+2)-f(x+1)=f(x+1)-f(x).$$

It follows by induction that

$$f(x+n)-f(x)=n[f(x+1)-f(x)].$$

If $f(x+1)\neq f(x)$, then $f(x+n)-f(x)$ will exceed in absolute value an arbitrarily large number for a sufficiently large $n$, contradicting the assumption that $f$ is bounded.

Hence

$$f(x+1)=f(x)$$

for all $x$.