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 $x,y$.

$f(0)=0$ 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 $x=\frac{m}{n}$.
