Functional-equation for strictly increasing functions Let $n,c$-given natural numbers .Let $f(x)$ - strictly increasing function , domain of definition and set of values of ​​which are non-negative integers ,$f(0)=0 , f(1)=c$  and 
 \begin{align}
f\left(\sum_{i=1}^{n} a_{i}^{n}\right) 
=\frac{1}{c^{n-1}}\sum_{i=1}^{n} \left(f(a_{i})\right)^{n}\\
\end{align}
Prove that $f(x)=cx$
 
 A: Part I - Let us first study the case $c = 1$.
Notice $f(0) = 0, f(1) = c = 1$, we have
$$\begin{align}
 & f(2) = f(2 \cdot 1^n +  (n-2) \cdot 0^n) = 2 f(1)^n + (n-2) f(0)^n = 2\\
\implies & f(2^n) = f(2^n + (n-1)\cdot 0^n) = f(2)^n + (n-2) f(0)^n = 2^n\\
\implies & f(2^{n^2}) = f((2^n)^n + (n-1)\cdot 0^n) = f(2^n)^n + (n-2)f(0)^n = 2^{n^2}\\
\vdots\\
\implies & f(2^{n^s}) = 2^{n^s}
\end{align}
$$
For any $x > 0$, pick a $s$ such that $y = 2^{n^s} \ge x$. 
Since $f(1) = 1$, $f(y) = y$ and $f(t)$ is strictly increasing for $1 \le t \le y$.
By Pigeon hole principle, we have $f(t) = t$ for all such $t$. In particular, this
means our original $x$ satisfies $f(x) = x$.
Since this is true for all $x$, we find $f(x) = x$ identically.
Part II - Let us switch to the case $c > 1$.
Factorize $c$ into its prime factors $c = p_1^{e_1} p_2^{e_2} \cdots p_r^{e_r}$.
Pick integers $\alpha_i$, $\beta_i$ such that
$e_i = \alpha_i n + \beta_i$ with $0 \le \alpha_i$ and $0 \le \beta_i < n$.
Since $f(0) = 0$, we find for any $x > 0$,
$$
c^{n-1} f(x^n) = c^{n-1} f(x^n + (n-1) 0^n ) = f(x)^n + (n-1) f(0)^n = f(x)^n
$$
This leads to
$$
\prod_{i=1}^r p_i^{(\alpha_i n + \beta_i)(n-1)} \bigg| f(x)^{n}
\quad\implies\quad
\prod_{i=1}^r p_i^{\left\lceil \frac{(\alpha_i n + \beta_i)(n-1)}{n}\right\rceil } = \prod_{i=1}^r p_i^{\alpha_i(n-1)+\beta_i} \bigg| f(x)
$$
As a result, there is a function $g: \mathbb{N} \to \mathbb{N}$ such that
$$f(x) = g(x)\prod\limits_{i=1}^r p_i^{\alpha_i(n-1)+\beta}$$
It is easy to see $g(x)$ satisfies an equation similar to that of $f(x)$:
$$
g\left(\sum_{i=1}^{n} a_{i}^{n}\right) =\frac{1}{\tilde{c}^{n-1}}\sum_{i=1}^{n} g(a_{i})^n
\quad\text{ with }\quad
\tilde{c} = \prod\limits_{i=1}^r p_i^{\alpha_i}
$$
As long as $\tilde{c} > 1$, we can repeat above procedures to reduce this constant again. 
After a finite number of steps, we can reduce this constant to $1$. What this means is 
there is a function $h : \mathbb{N} \to \mathbb{N}$ such that
$$f(x) = c h(x)
\quad\text{ and }\quad
h\left(\sum_{i=1}^{n} a_{i}^{n}\right) = \sum_{i=1}^{n} h(a_{i})^n
$$
By part I, we known $h(x) = x$. From this, we can conclude $f(x) = cx$.
