Solving the functional Equation $f(f(x))=f(x)+x$

Question

Let $f$ be continuous on $\mathbb{R}$. Then how to find all continuous functions satisfying $f(f(x))=f(x)+x$

Answer 1

This one is a problem from a journal or from competitions at the level of the Putnam contest (see reference below).

Hint: $g(x) = x + Af(x)$ satisfies $g(f^n(x))=A^ng(x)$ when $A^2 = A + 1$; consider the cases $n \to \pm \infty$.

Source for a similar problem, with solution: http://books.google.com/books?id=-CNbGp2ZFXUC&pg=PA21

Answer 2

In fact this belongs to a functional equation of the form http://eqworld.ipmnet.ru/en/solutions/fe/fe1220.pdf.

Let $\begin{cases}x=u(t)\\f=u(t+1)\end{cases}$ ,

Then $u(t+2)=u(t+1)+u(t)$

$u(t+2)-u(t+1)-u(t)=0$

$u(t)=C_1(t)\left(\dfrac{1+\sqrt{5}}{2}\right)^t+C_2(t)\left(\dfrac{1-\sqrt{5}}{2}\right)^t$ , where $C_1(t)$ and $C_2(t)$ are arbitrary periodic functions with unit period

$\therefore\begin{cases}x=C_1(t)\left(\dfrac{1+\sqrt{5}}{2}\right)^t+C_2(t)\left(\dfrac{1-\sqrt{5}}{2}\right)^t\\f=C_1(t)\left(\dfrac{1+\sqrt{5}}{2}\right)^{t+1}+C_2(t)\left(\dfrac{1-\sqrt{5}}{2}\right)^{t+1}\end{cases}$ , where $C_1(t)$ and $C_2(t)$ are arbitrary periodic functions with unit period