$f(x + y)f(x − y) = ( f(x) + f(y) )^ 2− 4x^ 2 f(y)$ so why $f(2) \neq 2$?

Question

Let $f : \mathbb R → \mathbb R$ be a function that satisfies for all $x,y ∈ \mathbb R $ defined as $f(x + y)f(x − y) = ( f(x) + f(y) )^ 2− 4x^ 2 f(y)$. So, why $f(2) = 2$ is impossible?

Why $f(0) = 0, f(3) = 9, f(5) = 0$ are ok but not $f(2) = 2$?

Answer 1

With $x=y=0$, the relation shows you that $f(0)^2=4f(0)^2$, so $f(0)$ must equal $0$.

Then with $x=y$, the relation shows that $0=4f(x)^2-4x^2f(x)$, so for all $x$, $$f(x)(f(x)-x^2)=0$$

And so for each $x$, either $f(x)=0$ or $f(x)=x^2$.