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

Question

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

Answer 1

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.

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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.