Solving $f(2011)=2012$, $f(4xy)=2yf(x+y)+f(x-y)$

Question

How to find the all functions $f$ :$ \mathbb{R}\longrightarrow\mathbb{R}$ such that $f(2011)=2012$,for every $x,y\in\mathbb{R}$ then: $$f(4xy)=2yf(x+y)+f(x-y)$$

Answer 1

If such a function exists, then $f(x)=f(0) \ne 0$ for every $x$, therefore $f(0)=2yf(0)+f(0)$ for every $y$, i.e. $y=0$ for every $y$. Thus such a function does not exist.