How to find all the continuous functions satisfying an equation?

Question

The problem that I want to solve is:

"find all the continuous functions $f\colon \mathbb R\to \mathbb R$ such that for every $x$, $f(f(f(x))) = x $ , I know that f(x) = x is an answer but how can one find (and prove) all the continuous answers ? And more generally what are the steps to find similar problems ?

Answer 1

If $f(x)$ is a polynomial of degree $d \ge 2$, let $f(x) = ax^d+g(x)$ with degree of $g \lt d$. Then $f(f(x)) =a(ax^d+g(x))^d+g(ax^d+g(x)) =a(a^dx^{d^2}+...)+g(ax^d+g(x)) $ so $degree(f(f(x)) = d^2$ (leading term is $a^dx^{d^2}$) and $degree(f(f(f(x)))) =d^3$ so $f$ can not be a polynomial of degree $\ge 2$.

If $f(x) = ax+b$ then $f(f(x)) =a(ax+b)+b =a^2x+ab+b $ and $f(f(f(x))) =a(a^2x+ab+b)+b =a^3x+a^2b+ab+b $ so $a^3 = 1$ and $a^2b+ab +b= 0$.

If $b = 0$ then $a^3 = 1$ which has one real and two complex solutions.

If $b \ne 0$ then $0 =a^2+a+1 $ so $a =\dfrac{-1\pm\sqrt{-3}}{2} =a_1, a_2 $.

Therefore solutions are $f(x) = ax$ with $a^3 = 1$ (so $a = 1, a_1, a_2$) and, for any $b$, $f(x) = a_{1, 2}x+b $.

So, if the solution has to be a polynomial with real coefficients, it can only be $f(x) = x$.