Functional equation extended solution

The question is Find all functions $f:R \to R$ such that $$f(x+y)f(x-y)=(f(x)+f(y))^2-4x^2f(y)$$ Taking $x=y=0$, we get $f(0)^2=4f(0)^2 \implies f(0)=0$. Now take $x=y$ which immediately gives $$4f(x)^2=4x^2f(x)\\\implies f(x)(f(x)-x^2)=0\\\implies f(x)=0 \space or \space f(x)=x^2 \space \forall x \in \mathbb R$$ This was my solution. But I was stunned when I looked at the official solution.

Why do we need to continue to do anything after getting what I got, isn't it sufficient?