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Find all functions $f:\mathbb{N} \to \mathbb{N}$ such that $$f(n) + f\big(f(n)\big) = 6n$$ for every $n\in\mathbb{N}$.

I don't know any formal way to solve this. I just tried to use some functions of the form $f(n) =\alpha n$ and found that $f(n) = 2n$ is the answer because $2n + 2(2n) = 6n$

I don't know how to correctly solve this type of questions and just used simple try-error method, So any help on solving this in mathematical way is appreciated.

I am not also sure about the tags I selected. It looks like a number theory problem to me that is also related to recurrence relations. Sorry if the tag is inappropriate.

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Prove that $$f^{k}(n)+f^{k-1}(n)-6\,f^{k-2}(n)=0$$ for every $n\in\mathbb{N}$ and $k\in\mathbb{Z}_{\geq 2}$. Here, $f^0:=\text{id}_\mathbb{N}$ and $f^k:=f\circ f^{k-1}$ for $k\in\mathbb{Z}_{\geq 1}$. Use a result from recursive sequences to show that $$f^k(n)=2^k\,A(n)+(-3)^k\,B(n)$$ for some functions $A,B:\mathbb{N}\to\mathbb{R}$. Verify that $B$ is identically zero.

Anyhow, it would be a very interesting problem to find all $f:\mathbb{Z}\to\mathbb{Z}$ such that $$f\big(f(n)\big)+f(n)-6n=0$$ for all $n\in\mathbb{Z}$. There are at least two solutions $f(n)=2n$ for each $n\in\mathbb{Z}$, and $f(n)=-3n$ for each $n\in\mathbb{Z}$. We have seen so far that $$f^{k}(n)=2^k\,\left(\frac{3n+f(n)}{5}\right)+(-3)^k\,\left(\frac{2n-f(n)}{5}\right)$$ for all $n\in\mathbb{Z}$ and $k\in\mathbb{Z}_{\geq 0}$. Are there more such functions?

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  • $\begingroup$ How do you find your initial equality without assuming $f$ is linear? $\endgroup$ Oct 12 '19 at 22:05
  • $\begingroup$ I do not apply $f$ on the whole equation (this is what I believe what you were referring to). I simply replace $n$ by, for example, $f(n)$. $\endgroup$ Oct 12 '19 at 22:06
  • $\begingroup$ Ahh... fair enough. $\endgroup$ Oct 12 '19 at 22:07
  • $\begingroup$ @Aqua Did you read the hidden box, where I gave explicitly $A(n)$ and $B(n)$ in terms of $f(n)$ and $n$? $\endgroup$ May 3 '20 at 7:20
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    $\begingroup$ @Aqua If not $(-3)^k\,B(n)$ will overcome $2^k\,A(n)$ for some large $k$, making $f^k(n)$ negative. $\endgroup$ May 3 '20 at 7:36

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