Functional Equation (no. of solutions): $f(x+y) + f(x-y) = 2f(x) + 2f(y)$ Find all the functions $f\colon\mathbb{Q} \to \mathbb{Q}$ such that
$f(x+y) + f(x-y) = 2f(x) + 2f(y)$, for all rationals $x,y$.

That equation, which is often called Appolonius' formula or the parallelogram law, is a characterization of homogeneous quadratic forms on a vector space (without solutions of $v+v=0$).  
In plane geometry, with $f(\mathbb{x})=|\mathbb{x}|^2$ the squared Pythagorean distance, the formula is the relation proved by Appolonius between the diagonals and sides of a parallelogram.  In a vector space, for any dot product [symmetric bilinear form] $B$, the associated 'squared distance' function [the restriction to a quadratic form] $f(x)=B(x,x)$ satisfies the equation, due to
$B(x+y,x+y)=B(x,x) + B(y,y) + 2B(x,y)$.
Conversely, any solution of the equation comes from this construction, and $2B(x,y) = f(x+y) - f(x) - f(y) $ .  The posted problem is the case of a 1-dimensional vector space over the field of rational numbers. 
