Problem:
Find all functions $f : R → R$ such that
$$f(x + y)f(x − y) =
(f(x) + f(y))^2
− 4x^2
f(y)$$for all $x, y ∈ R$, where R denotes the set of all real numbers.
My approach: I substituted $x=y=0$ which gave $f(0)=0$. I then substituted $x=y=t$ which gave $f(2t)f(0) = 4f(t)(f(t)-t^2) => f(t) = 0, f(t) = t^2 $. Thus the only solutions are $f(x)= 0$ and $f(x) = x^2$.
But this is an olympiad problem (INMO 2011, India). So, there must be something more complicated than this that I am probably missing. My answer matches but the proof given in the official solution is quite long. Here
Best Answer
Your deduction revealing that for each $x$ either $f(x)=0$ or $f(x)=x^2$ is correct. What you are missing is that there are still infinitely many functions remaining. You may have $f(1)=1$ or $f(1)=0$, you may have $f(2)=4$ or $f(2)=0$. That's four combinations already. Are all the alternatives possible? You haven't even touched that question, yet.