Find all function $ f:\Bbb R\to\Bbb R$ such that $ f(x+y)f(x-y)=(f(x)+f(y))^2-4x^2f(y)$.

Question

While doing some INMO questions, one entry went this way:

Find all function $ f:\Bbb R\to\Bbb R$ such that $f(x+y)f(x-y)=(f(x)+f(y))^2-4x^2f(y)$.

I made an approach similar to this :

On Putting $x=0, y=0$ we get

$$f(0+0)f(0-0)=(f(0)+f(0))^2-4\times0^2f(0)$$ $$f(0)^2=(2\times f(0))^2$$Which gives us $f(0)=0$.

Then, on putting $x=1, y=1$, $$f(1+1)f(1-1)=(f(1)+f(1))^2-4\times1^2f(1)$$ $$f(2)f(0)=(2\times f(1))^2+-4\times f(1)$$ $$4\times f(1)=4\times f(1)^2$$

Which gives us $f(1)=0 $ or $f(1)=1$.

From here, I can't go further. I think that method I m working on is quite right and will take me to the right answer. But the problem is that I can't find that right answer. I shall be thankful if you can provide me a hint or a complete solution. Thanks.

SIDE NOTE: I am using this method for a while (I got this one from then answer of a post). Now I m thinking to switch, If you know some other method to solve such functional equations, please try to give your answer by that method.

Answer 1

HINT:

First:

$$x=y=0 \Rightarrow f^{2}(0)=4f^{2}(0) \Rightarrow f(0)=0$$ Second:

$$x=y \Rightarrow 0=4f^{2}(x)-4x^{2}f(x) \Rightarrow f(x)[f(x)-x^{2}]=0 (1)$$

If we check, we can see that $f(x)=0$ and $f(x)=x^{2}$ are solutions.

Now you have to prove that those are the only ones. For that some ideas can help.

$$x=0 \Rightarrow f(y)[f(-y)-f(y)]=0 (2) $$ $$y=-x \Rightarrow 0=[f(x)+f(-x)]^{2}-4x^{2}f(-x)(3)$$

Suppose that we have a solution $f(x)$ such that for some $x_0$ we have $f(x_0) \ne 0$, then backing to $(1)$ and $(2)$ we get $f(x_0)=f(-x_0)=x_{0}^{2}$. You have to prove that $x_0=0.$