Let $G = \mathbb{Z}/n\mathbb{Z}$ be the cyclic group of order $n$, and let $f : G \to \mathbb{R}$ be a function. Let $\Gamma$ be the Cayley graph of $G$ with generator $1$. Define the discrete derivation of $f$ by
$$Df(x) = f(x+1) - f(x).$$Let $c \in \mathbb{R}$ be a constant and suppose that $Df = c$, i.e.
$$f(x+1) - f(x) = c \quad \text{for all } x \in G.$$The question is: what constraints does the periodicity of $f$ impose on $c$?
Main Observation
We interpret $Df$ as a flow on the edges of $\Gamma$, and the condition $Df = c$ as a condition of constant flow. The graph $\Gamma$ is a cycle graph of order $n$. Summing the relation around the cycle gives
$$\sum_{x \in G} (f(x+1) - f(x)) = \sum_{x \in G} c = nc.$$The left-hand side telescopes to $0$, hence
$$nc = 0.$$Thus $c = 0$, and consequently $f$ is constant.
Proposition
Proposition. Let $G=\mathbb{Z}/n\mathbb{Z}$ and let $f:G\to\mathbb{R}$ satisfy $Df(x)=c$ for all $x \in G$. Then $c=0$, and hence $f$ is constant.
Generalization
Let $\phi : G \to \mathbb{R}$ be a function. Consider the equation
$$Df(x) = \phi(x).$$Since $G = \mathbb{Z}/n\mathbb{Z}$ is cyclic, any function defined by successive differences must be compatible with traversing the cycle and returning to the starting point.
This raises the question:
For which functions $\phi : G \to \mathbb{R}$ does there exist a function $f : G \to \mathbb{R}$ satisfying $$Df(x) = \phi(x) \, ?$$