Finding a function which satisfies an equation

Question

During my analysis course, I found this interesting problem: Let $f: A \to A$ a function such that $f(f(x))+3x=4f(x)$ for every $x \in A$, where A is a finite set of positive real numbers. Find the function f.

Answer 1

I claim that $f(x)=x$ for all $x\in A$.

Proof. The given functional equation can be written in the form $$f\bigl(f(x)\bigr)-f(x)=3\bigl(f(x)-x\bigr)\qquad(x\in A)\ .\tag{1}$$ Assume that there is an $x_0\in A$ such that $x_1:=f(x_0)\ne x_0$. Proceed recursively by putting $$x_{n+1}:=f(x_n)\qquad(n\geq0)\ .$$ Then from $(1)$ it follows that $$x_{n+2}-x_{n+1}=3(x_{n+1}-x_n)\qquad(n\geq0)\ ,$$ so that we obtain $x_{n+1}-x_n=3^n(x_1-x_0)$ for all $n\geq0$. This isincompatible with $A$ being finite.