# Functions for non-negative integers [duplicate]

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Let $\Bbb{Z}^+$be the set of all non-negative integers where $n$ and $k$ are given natural numbers. We consider the following non-decreasing function,

$$f:\Bbb{Z}^+ \to \Bbb{Z}^+$$ such that

\begin{align} f\left(\sum_{i=1}^{n} a_{i}^{n}\right) =\frac{1}{k}\sum_{i=1}^{n} \left(f(a_{i})\right)^{n}\\ \end{align}

Found all functions for $n=2014$ and $k=2014^{2013}$

Found, how many functions and which satisfy the condition of the problem depending on the values ​​of parameters $n$ and $k$.

PS. I've tried to make some operations with this function and found $f(0)$, but I can't imagine what can I do with it

## marked as duplicate by Hamou, apnorton, Hakim, MathOverview, Mark BennetOct 9 '14 at 21:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• This looks like a math contest problem? – Hagen von Eitzen Oct 7 '14 at 8:22
• @Hagen von Eitzen , yes it's problem from the math competition that was held in may – Antony Oct 7 '14 at 8:33

## 1 Answer

Letting all $a_i=0$, we find $$f(0)=\frac nk f(0)^n$$hence either $f(0)=0$ or $f(0)=\sqrt{2014^{2012}}$, but the latter is not $\in\mathbb Z^+$. We conclude $f(0)=0$. Letting all $a_i=0$ except $a_1=1$, we find $$f(1)= \frac1kf(1)^n$$ hence either $f(1)=0$ or $f(1)=2014$. In the latter case, we find $f(2)=\frac{2014^{2014}+2014^{2014}}{2014^{2013}}=2\cdot 2014$ and in fact $f(a)=2014a$ for $0\le a\le n$. In the first case we immediately find $f(2)=f(3)=\ldots = f(2014)=0$.

In fact, we immediately check that the two functions $f(a)=0$ and $f(a)=2014 a$ are valid solutions (in other words, $f(a)=af(1)$ and $f(1)$ is one of the two possible values found above). We suspect that there is no third solution and observe that we did not make use of the growth condition yet. Let $$S=\{\,a\in \mathbb Z^+\mid f(a)=af(1)\,\}$$Then $0,1,\ldots, 2014\in S$ as seen above. Moreover, if $a_1,\ldots, a_n\in S$ then $\sum a_i^n\in S$, hence $S$ is not bounded from above. If $f(1)=0$ this gives us immediately that $f(a)=0$ for all $a$ because $f$ is nondecreasing. Thus we may assume from now on that $f(1)=2014$.

For $a\in\mathbb N$ let $q(a) =\frac{f(a)}{2014a}$. From $f(ra^n)=\frac1k{rf(a)^n}=\frac{rq(a)^na^n2014^n}{2014^{2013}}=2014q^nra^n$ we see $q(ra^n)=q(a)^n$ for $1\le r\le 2014$. Hence, unless $f$ is linear, these $q(a)$ become arbitrarily big or arbitrarily small. Show that the set $S$ is "sufficiently" dense to disallow very small/big values of $q(a)$.