A game on equiangular polygons Let $n\geq 3$ be a natural number and  Consider the following game: 
Correspond an integer to each vertices of an equiangular polygon 
(at least two of the numbers are unequal). 
(1) Replace the number of each vertices with the number obtained by the sum of 
its neighborhood numbers minus its own number. (Edit: Do this for all vertices at once, not one by one)
This gives us a new equiangular polygon with an integer corresponded to each of its vertices. 
Again do as in (1) and Continue the progress. 
Let $A$ be the set containing the numbers corresponded to vertices of this polygon
in this (infinite) process. Now the question is: 
what are the possible values of $n$ if we want $A$  to be a bounded set of integers!?
 A: For any $n$ there is always a (certainly non-constant) initial choice of integer labels $v=(v_1,\dots,v_n)$ of the vertices of the $n$-gone for which the set of numbers $A$ is unbounded. 
The question is more subtle if you ask for which $n$ it is possible to find a  non-constant  integer choice of labels for which the set of numbers $A$ is bounded. The  answer is then: it is possible if and only if $n$ is either a multiple of $4$ or a multiple of $6$.
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The iteration you consider is described by a simple symmetric circulant matrix $L$ of order $n$  (in the notation of the link, the non-zero coefficients are $c_0=-1$, $c_1=c_{n-1}=1$). It has therefore $n$ real eigenvalues (repeated according to their multiplicity, which is either $1$ or $2$)
$$\lambda_k=2\cos(2\pi k/n) -1,\qquad k=1,\dots, n\ . $$
In particular, the spectral radius of $L$ is larger than $1$, corresponding to $\lambda_{ \lfloor n/2\rfloor} $ . Since $\mathbb{Z}^n$ spans linearly $\mathbb{R}^n$, some element $v$ of $\mathbb{Z}^n$ must have a non-zero component w.r.to $\lambda_{ \lfloor n/2\rfloor} $ in the spectral decomposition, implying that $A=\{L^k v\}_{k\ge0}$ is unbounded. Note that $v$ is certainly  not a constant vector, as required. Also note that for even $n$, $\lambda_{n/2}=-3$ corresponds to the  example in Barry Cipra's comment. 
For what $n$ it is possible to find non-constant integer labels $v=(v_1,\dots,v_n)$ of the vertices for which $A=\{L^k v\}_{k\ge0}$ is bounded? This is certainly the case if $n$ is a multiple of $4$: then we have the eigenvector $(1,0,-1,0,\dots)$ of the eigenvalue  $-1$.
If $n$ is a multiple of $6$, $L$ is singular with a $2$ dimensional kernel spanned by the $6$-periodic vectors $(1,0,-1,-1,0,1,\dots)$ and  $(1,1,-1,-1,1,1,\dots)$. Of course these generate bounded orbits.
Conversely, assume that $n$ is such that there exists a non-constant $v\in \mathbb{Z}^n$ 
that generates a bounded orbit $\{L^k v\}_k $ . If $L$ is singular, $n$ must be a multiple of $6$, as it follows from the above expression for the eigenvalues. If $L$ is not singular the orbit $\{L^k v\}_k $ is periodic, that is, $L^m v = v$  for some $m$. So $1$ is one of the eigenvalues of $L^m$, which are of the form  $\lambda_k^m$, and this is only possible if $\lambda_k=-1$ or  $\lambda_k=1$. In the former case, $k=0$ and $v$ is a  simple eigenvalue of $L$ with constant coordinates, which is not the case. In the latter case, $\cos(2\pi k/n)=0$, which implies that $n=4k$.
