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Find all functions $ f: \mathbb{N} \rightarrow \mathbb{N}\; $ satisfying

$ f(f(f(n))) + 6f(n) = 3f(f(n)) + 4n + 2001 , \forall n \in\mathbb{N} $

After some trial and error I assumed the solution to be of the form $ f(n) = n + r\;$ which gave me $ f(n) = n + 667\; $ as a solution but I am unable to come up with any other solutions. How would you guys solve this problem?

Thanks in advance.

1 Answers 1

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Given $n \in \mathbb{N}$, define $ g_n(d) = f^{(d)}(n) - 667d. $ Routine computation yields the recurrence equation $ g_n(d+3) = 3g_n(d+2) - 6g_n(d+1) + 4g_n(d). $ The characteristic polynomial is $ x^3 - 3x^2 + 6x - 4 = (x-1)(x^2-2x+4). $ The two roots of $x^2-2x+4$ are $2e^{\pm i \pi/3}$.

It follows from the general theory of recurrence relations that for some real $A,B$ and angle $\theta$, $g_n(d) = A + B2^d \cos (d\pi/3 + \theta).$ The cosine part is a periodic function in $d$ attaining $6$ fixed values, at least one of them strictly positive and at least one strictly negative. In particular, if $B \neq 0$ then for some $e \in \{0,\ldots,5\}$ we have $C \triangleq B\cos (e\pi/3 + \theta) < 0.$ This implies that $f^{(6d+e)}(n) = g_n(6d+e) + 4002d + 667e = A + 2^d C + 4002d + 667e $ is eventually negative. This contradicts the premise that $f \geq 0$, and so $B = 0$. Substituting $d = 0$ in the formula for $g_n$, we deduce that $g_n(d) = A = n$ and so $ f^{(d)}(n) = n + 667d. $ In particular, $f(n) = n + 667$.

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    As for choosing to look at iterates of $f$, this is suggested by the form of the functional equation. Moreover, lately I have been working on something that involves solving a non-homogeneous recurrence relation (though with non-constant coefficients), so this "solution pattern" suggested itself after a short while.2011-06-21