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 $x,y$.
 A: 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. 
A: $f(0)=0$ By choosing $x=y=0$. Also by choosing $x=y$ we can see $f(2x)=4f(x)$. Moreover choose $x=0$ and use $f(0)=0$, then you have $f(y)=f(-y)$.
Now by strong induction one can prove that $f(nx)=n^2f(x)$. 
Then choose $x$ as $\frac{x}{n}$ in preceding equation, then you get:
$$
f(x)={n^2}f(\frac{x}{n})\rightarrow\frac{1}{n^2}f(x)=f(\frac{x}{n})
$$
And finally we have:
$$
f(\frac{mx}{n})=\frac{1}{n^2}f(mx)=\frac{m^2}{n^2}f(x).
$$
Now assume that $f(1)=a$ for some $a\in\mathbb{R}$, then you get
$$
f(x)=ax^2
$$
for all rational $x=\frac{m}{n}$.
A: Alternatively, one could, upon getting that $f(2x)=4f(x)$, keep differentiating both sides with respect to $x$;
$\quad 2f'(2x)=4f'(x), \\\quad 4f''(2x)=4f''(x), \\\quad 8f'''(2x)=4f'''(x),\\ \qquad \qquad \vdots \\ 2^nf^{(n)}(2x)=4f^{(n)}(x), \\ \qquad \qquad \vdots \\$
Then setting $x=0$, we have;
$\quad 2f'(0)=4f'(0), \\\quad 4f''(0)=4f''(0), \\\quad 8f'''(0)=4f'''(0),\\ \qquad \qquad \vdots \\ 2^nf^{(n)}(0)=4f^{(n)}(0),\qquad(1) \\ \qquad \qquad \vdots $
From this, it can be seen that $f''(0)$ can have an arbitrary value; with all the other derivatives being zero.
Using the Maclaurin's expansion for $f(x)$, we get the same answer as above;
$f(x)=ax^2\qquad$ (for some rational $a$)
