I propose the following solution for \[ f(f(f(n)))+6f(n)=3f(f(n))+4n+2001, \quad f: \mathbb{N}\to \mathbb{N}. \] Set $ x_{k+1}=f(x_k)$, then the previous equation becomes the linear recursive equation \[ x_{k+3}-3x_{k+2}+6x_{k+1}-4x_k=2001. \] Under the change $ y_k= x_k-667\times k$, the recursive equation reduces to the homogeneous equation: \[ y_{k+3}-3y_{k+2}+6y_{k+1}-4y_k=0. \] The associated characteristic equation being: \[ Y^3-3Y^2+6Y-4=(Y-1)(Y^2-2Y+4)=0, \] with roots in $ \mathbb{C}$ : $ 1, \exp(\pi/3),\exp(-\pi/3)$. The solution, in $ \mathbb{R}$, of the homogeneous equation is given by \[ y_k= a(n) + 2^k \left( b(n) \cos\frac{k\pi}{3}+ c(n)\sin\frac{k\pi}{3}\right), \] where $ a,b,c$ do not depend on $ k$. Because $ b(n)\cos\frac{k\pi}{3}+ c(n)\sin\frac{k\pi}{3}$ cannot be defined as a function from $ \mathbb{N}\to \mathbb{N}$ for all $ k$, the solution in $ \mathbb{N}$ of the homogeneous equation corresponds to $ b=c=0$ and $ y_k=a(n)$ with $ a: \mathbb{N}\to \mathbb{N}$, and is independant of $ k$. It follows : \[ x_k= a(n) + 667\times k,. \] Let $ n \in \mathbb{N}$ , for $ k=0,\, n= x_0=a(n)$ and for $ n=1,\,f(n)= x_1=n+667$. Finally, the solution in $ \mathbb{N}$ is unique and given by: \[ f: n\mapsto n+667. \]

Find all functions $ f:\mathbb{N}\to\mathbb{N} $ such that for all $ n\in\mathbb{N} $ and $ k=1,2,3, ... $ \[ f(f(f(n)))+6f(n)=3f(f(n))+4n+3k. \] k=1 _ Romania 1999