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}
1.Find all functions for $n=2014$ and $k=2014^{2013}$
2.Find, how many functions and which satisfy the condition of the problem depending on the values of parameters $n$ and $k$.
I have some ideas for this problem .
I know how to found $f(0)$ (just take $a_1=a_2=\cdots=a_n=0$) .
I also can found $f(1)$ (take $a_1=1$ and $a_2=\cdots=a_n=0$)
Then if $f(0)=0$ I've get that $f(a_n)+f(b_n)=f(a_n+b_n)$.
I also think thas all functions will be as $f(x)=kx$ but I can't prove it .
Can you help me with some partical solution or some ideas which can help to find it.
And can someone tell me how in first example ($n=2014$ and $k=2014^{2013}$) found f(2015) or $f(2016)$ using that $f(i^n)=i\cdot f(1)$ ( for $2015>i>1$)and $f(1)=0$ or $f(1)=2014$
I do not know how to define all solutions for all values, but I know how to find all solutions on $\{0,1,\dots,n\}$.
Depending on the choice of the parameters, there are up to $4$ possible solutions in $\{0,1,\dots,n\}$.
Defining $f(0)$
If all $a$s $= 0$, we have $f(0)=\frac{n}{k}f^n(0)$, thus either $f(0)=0$ or $f(0)=\left(\frac{k}{n}\right)^{\frac{1}{n-1}}$ if this number happens to be integer. For your numerical example, this number is not an integer.
Defining $f(1)$ given $f(0)$
If $a_1=1$ and the rest is $0$, we have $f(1)=\frac{n-1}{k}f^n(0)+\frac{1}{k}f^n(1)=\frac{n-1}{n}f(0)+\frac{1}{k}f^n(1)$. If $f(0)=0$, we have either $f(1)=0$ or $f(1)=k^{1/(n-1)}$; for your numerical example, the latter value is $2014$. If $f(0)\neq 0$, we can have between $0$ and $2$ possible values for $f(1)$ - we can use them as long as they are integer.
Defining $f(2), \dots, f(n)$ given $f(1)$ and$f(0)$
Once we have decided on the values for $f(0)$ and $f(1)$, we can define $f(m)$, $1<m\leq n$, by choosing $a_1=\dots=a_m=1$, and the rest is $0$.
We then have:
$f(m)=\frac{1}{k}\left(m f^n(1) + (n-m) f^n(1)\right) = \frac{1}{k}\left(m f^n(1) + (n-m) f^n(0)\right)=m\left(f(1)-\frac{n-1}{n}f(0)\right) + \frac{n-m}{n}f(0)=mf(1)+(1-m)f(0)$
This holds for any $f(1)$ and $f(0)$.
For your parameters, we have $f(m)=2014m$ and $f(m)=0$