(IMO SL 1992)(M-D) $a, b>0$ Determine the solutions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that $f(f(x))+$ $a f(x)=b(a+b) x$.
We fix an $x$ and define $x_{0}=x$ and $x_{n+1}=f\left(x_{n}\right)$, and then $x_{n+2}=-a x_{n+1}+b(b+a) x_{n}$, so we write $x_{n}=A a^{n}+B(-a-b)^{n}$ but since $x_{n} \geqslant 0$ for all $n$ we must have $B=0$ and then we get $x_{1}=f(x)=b x$ which is indeed a solution.