Graph being a parabola

Question

Determine the polynomial $P\in \mathbb{R}[x]$ for which there exists $n\in \mathbb{Z}_{>0}$ such that for all $x\in \mathbb{Q}$ we have: $$P\left(x+\frac1n\right)+P\left(x-\frac1n\right)=2P(x)$$

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The solution given in the book says: If there is such a polynomial of degree $m ≥ 2$,differentiating the given relation $m−2$ times we find that there is a quadratic polynomial that satisfies the given relation. But then any point on its graph would be the vertex of the parabola, which of course is impossible. Hence only linear and constant polynomials satisfy the given relation. My doubt: Why do they say that the graph can't be a parabola? They didn't provide any information about it

Answer 1

Intuitive explanation : $$P\left(x+\frac1n\right)+P\left(x-\frac1n\right)-2P(x)=0$$

can be read as "second (discrete) derivative" is $0$ (see here for example). Therefore, no wonder that it is only first degree polynomials that are solutions.

Answer 2

If $P(x)=ax^2+bx+c$, then $$ 0=2P(x)-\left(P\left(x+\frac1n\right)+P\left(x-\frac1n\right)\right)=-\frac{2a}{n^2}, $$ and therefore $a=0$.