How find this function such $f(2010f(n)+1389)=1+1389+1389^2+1389^3+\cdots+1389^{2010}+n$ Question:

Find all function:
$f:N\to N$, such  that
$$f(2010f(n)+1389)=1+1389+1389^2+1389^3+\cdots+1389^{2010}+n,\forall n\in N$$

Maybe this is 2010 Mathematical olympiad problem.But I can't find it.since
$$1+1389+1389^2+\cdots+1389^{2010}=\dfrac{1-1389^{2011}}{1-1389}$$
so
$$f(2010f(n)+1389)=\dfrac{1-1389^{2011}}{1-1389}+n,\forall n\in N$$
then follow I can't it.Thank you
Now I have google this problem have post this PDF 116 problems - Scribd But I can't find solution
 A: Hint
Show $f$ is injective. Further show that the image of the A.P. $1389+2010k$ includes all numbers above a certain $C$. So there are not enough numbers to map the rest of naturals, a contradiction. 

P.S.  Adding details.  For ease of writing, let $C = 1+1389+1389^2+\dots + 1389^{2010}$ and $g(k)=2010k+1389$. So the condition we have is $f\circ g\circ f(n) = C+n$.
1. $f$ is injective 

 $f(a)=f(b) \implies f\circ g\circ f(a)=f\circ g\circ f(b) \implies C+a=C+b \implies a=b$, so $f$ is injective.

2. Image of $g(k)$ covers all numbers above C 

 As $k$ runs through all natural numbers, $g(k)$ runs through an arithmetic progression which must contain all numbers of form $g\circ f(n)$, and hence $f\circ g(k)$ must include all numbers of form $C+n$.

3. The contradiction 

 The natural numbers, after removing the A.P. $g(k)$, still contains infinite numbers. As $f$ is injective they must map to a set of infinite cardinality.  However the only possibilities left are natural numbers $\le C$, a finite set.  Hence no such $f$ can exist.

