APMO 1989

Denote by $f_n$ the $n$th iterate of $f$, that is, $f_n(x)=\underset{n\text{ times }}{\underbrace{f(f(\dots f}}(x)))$. Plug $x\to f_{n+1}(x)$ in (2): since $g\left(f_{n+1}(x)\right)=g\left(f\left(f_n(x)\right)\right)=f_n(x)$, $$f_{n+2}(x)+f_n(x)=2f_{n+1}(x),$$ that is, $$f_{n+2}(x)-f_{n+1}(x)=f_{n+1}(x)-f_n(x).$$ Therefore $f_n(x)-f_{n-1}(x)$ does not depend on $n$, and is equal to $f(x)-x$. Summing the corresponding results for smaller values of $n$ we find $$f_n(x)-x=n(f(x)-x).$$ Since $g$ has the same properties as $f$, $$g_n(x)-x=n(g(x)-x)=-n(f(x)-x).$$ Finally, $g$ is also increasing, because since $f$ is increasing $g(x)>g(y)⟹f(g(x))>f(g(y))⟹x>y$. An induction proves that $f_n$ and $g_n$ are also increasing functions. Let $x>y$ be real numbers. Since $f_n$ and $g_n$ are increasing, $$x+n(f(x)-x)>y+n(f(y)-y)⟺n[(f(x)-x)-(f(y)-y)]>y-x$$ and $$x-n(f(x)-x)>y-n(f(y)-y)⟺n[(f(x)-x)-(f(y)-y)]<x-y.$$ Summing it up, $$|n[(f(x)-x)-(f(y)-y)]|<x-y\text{ for all }n\in {\mathbb{Z}}_{>0}.$$ Suppose that $a=f(x)-x$ and $b=f(y)-y$ are distinct. Then, for all positive integers $n$, $$|n(a-b)|<x-y,$$ which is false for a sufficiently large $n$. Hence $a=b$, and $f(x)-x$ is a constant $c$ for all $x\in \mathbb{R}$, that is, $f(x)=x+c$. It is immediate that $f(x)=x+c$ satisfies the problem, as $g(x)=x-c$.