Can one deduce that $P(x)$ is a polynomial of degree $0$ or $1$ from the following condition 
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.
 A: Let
$$Q(x)=P\bigg(x+\frac{2^k}{n}\bigg) -P(x)$$
Then, $Q(x)$ is a polynomial.
The given relation gives
$$Q(x)=Q(x-\frac{2^k}{n}) \ \ \forall x $$
from which it is easy to deduce that $Q$ is constant.
Let $C$ be the constant such that
$$C= Q(x)=P\bigg(x+\frac{2^k}{n}\bigg) -P(x)$$
Now, you can prove by induction that for all $m \in \mathbb N$ you have
$$P(m\frac{2^k}{n})=Cm+P(0)$$
Let $P(0)=a,\ \  b=C \frac{n}{2^k}$. Let $R(x)=bx+a$.
Then
$$P(m\frac{2^k}{n})-R(m\frac{2^k}{n})= 0 \ \ \forall m \in \mathbb N$$
Thus $P=R$.
A: Yes, even for a single fixed $k$.
Taking $n=2$ and $k=1$ as an example, the given equation implies $P(x+2)-2P(x+1)+P(x)=0$ for all $x$; the left-hand side is the result of twice applying the finite difference operator to $P(x)$. While formulas for finite differences of polynomials are not quite as slick as their derivatives, it is still the case that the finite difference of a degree-$d$ polynomial is a degree-$(d-1)$ polynomial when $d\ge1$. It follows that the only polynomials whose second-order finite difference vanishes are linear polynomials.
Other values of $n$ and $k$ follow immediately just by rescaling the variable $x$ by an appropriate constant, or using finite differences with an offset different from $1$.
A: Assuming that $P$ is a polynomial to begin with, the property transfers to the derivative of $P$. If $P$ has degree $n\ge2$, after $n-2$ differentiations we arrive at a polynomial of degree $2$. If $Q(x)=ax^2+bx+c$ satisfies the condition, then, with $r=2^k/n\ne0$,
$$
2ax^2+2bx+2c=a(x-r)^2+b(x-r)+c+a(x+r)^2+b(x+r)+c
$$
yielding
$$
ar^2=0
$$
Therefore $a=0$, a contradiction.
A: I've just come up with this solution:
Assuming $$P(x) = a_mx^m+a_{m-1}x^{m-1}+a_{m-2}x^{m-2}+...+a_0$$
we can find the simplified form for the RHS $(k=0)$:
$$P(x+\frac{1}{n})+P(x-\frac{1}{n})=2a_mx^m+2a_{m-1}x^{m-1}+\bigg(2a_{m-2}+a_m\frac{m(m-1)}{n^2}\bigg)x^{m-2}+Q(x)$$ where $\deg Q\le m-3$. The identity
$$2P(x)=P(x+\frac{1}{n})+P(x-\frac{1}{n})$$ gives
$$a_m\frac{m(m-1)}{n^2}=0$$ Since $a_m$ is a leading coefficient, $a_m\not =0$. So we must have $$m=0  \ \ \text{ or }  \ \ m=1$$ which completes the proof.
Even if it is correct (verifications would be awesome), I'm still open for solutions with different approaches like one from @GregMartin.
A: The 'shift' does not matter at all. Given any real $r≠0$ and polynomial function $P : \mathbb{R}→\mathbb{R}$ such that $2·P(x) = P(x+r) + P(x-r)$ for $x∈\mathbb{R}$, it must be that $P$ is linear.
Simply note that for each $k∈\mathbb{N}$ we have $P((k+1)·r)-P(k·r) = P(k·r)-P((k−1)·r)$, so a simple induction yields $P((k+1)·r)-P(k·r) = c$ where $c = P(r)-P(0)$, and another simple induction yields $P(k·r)-P(0) = k·c$. Let $f : \mathbb{R}→\mathbb{R}$ be defined by $f(x) = P(0)+x/r·c-P(x)$ for $x∈\mathbb{R}$. Then $f(k·r) = 0$ for every integer $k$. $f$ is a polynomial with infinitely many (distinct) zeros, which is only possible if $f$ is zero everywhere.
If you do not know the proof of the last fact used above, it is instructive to attempt to prove it. Hint: show by induction that a polynomial with degree $d$ that has $d+1$ zeros must be zero everywhere, by factorizing it using one of the zeros.
