shohvanilu wrote: Does existes a function $f:N->N$ and for all positeve integer n $f(f(n)+2011)=f(n)+f(f(n))$ Yes. Choose for example : Let $A=\{2011k$ $\forall$ positive integer $k\ge 3\}$ $\forall x\in A$ : $f(x)=\frac {x(x-2011)}{4022}$ $\forall x\in\mathbb N\setminus A$ : $f(x)=6033$

Dear Mr pco, could you explain why could you choose the such exemple?

Dear Mr Patrick,How did you get that ? Maybe it's your experience or a miracle ? Can you post a full solution or your approach about it ? Thank alot in advance !

nam wrote: Dear Mr Patrick,How did you get that ? Maybe it's your experience or a miracle ? Can you post a full solution or your approach about it ? Thank alot in advance ! No miracle. The question is to find one solution, not all. We have $f(x+2011)=x+f(x)$ $\forall x\in f(\mathbb N)$ With such an equation, the first trial is to look for a polynomial expression and then it's simple to think to something like $f(x)=ax^2+bx+c$ which gives $a=\frac 1{4022}$ and $b=-\frac 12$ and any $c$ And so for example $f(x)=\frac{x(x-2011)}{4022}$. In this attempt, we need $2011|x$ and $x>2011$ and so $x=2011k$ with $k\ge 2$; Note that this imply $2011\notin f(\mathbb N)$ But $x=2011\times 2$ would imply $f(x)=2011$ and so does not fit; If we decide that $k\ge 3$, then $f(2011\times 3)=2011\times 3$, which fits. So we can decide that $f(\mathbb N)\subseteq A=\{2011k$ $\forall$ positive integer $k\ge 3\}$ and $f(x)=\frac{x(x-2011)}{4022}$ $\forall x\in A$ for $x\notin A$, we can choose for $f(x)$ any value in $A$, so for example $f(x)=6033$ That's the method I used to get this example. Obviously, there are a lot of other solutions. But the problem was just to find one.