Miquel-point wrote: 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: \[P\left(x+\frac1n\right)+P\left(x-\frac1n\right)=2P(x).\] If $P(x)$ fits, $P'(x)$ fits too. Easy to see that no degree $2$ polynomial can fit and so no polynomial with degree $\ge 2$ And since any $\boxed{ax+b}$ fits, these are all the polynomials.