Assuming $P(x)$ is a polynomial, can one deduce that $P(x)$ is a polynomial of degree $0$ or $1$ from the following condition: $$2P(x) = P\bigg(x-\frac{2^k}{n}\bigg) +P\bigg(x+\frac{2^k}{n}\bigg), \ \ \forall x $$ where $n$ is a fixed positive integer and $k\in\mathbb{N}\cup\{0\}$.
The identity gives a feeling like the values of $P(x)$ is "equally distributed" and its graph illustrates a straight line since $$P(x) = \cfrac{P\bigg(x-\frac{2^k}{n}\bigg) +P\bigg(x+\frac{2^k}{n}\bigg)}{2}$$ so its value at $x$ is an arithmetic mean of its values at $x-\frac{2^k}{n}$ and $x+\frac{2^k}{n}$. I don't have a potential argument though.
Update: I've just come up with a solution (posted below). Even if it is correct (verifications would be awesome), I'm still open for solutions with different approaches. Thanks.
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