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.enter image description here

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

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    $\begingroup$ Because it may be the case that $f(x)=0$ for some $x$ and $f(x)=x^2$ for others. A priori $$f(x)=\begin{cases}0,& x<0\\x^2,& x\geq0\end{cases}$$ could have been a solution. In fact, this function is a solution of $f(x)(f(x)-x^2)=0$, but not of the original functional equation. $\endgroup$ – Pp.. Jan 24 '15 at 15:07

Solve $\sqrt{x - 1} = \sqrt{2x}$ for real $x$.

Square both sides. $x-1 = 2x$ so $x = -1$.

Done, right?

No! You have to verify that it actually satisfies the original.

In your case, the functional equation implies that $f(x) = 0$ or $f(x) = x^2$ for each $x$. For some $x$ it could be $0$, for some others it could be $x^2$.

So this actually describes an uncountable set of functions! Pick some subset of $\mathbb{R}$ where it is $0$. At the rest it is $x^2$. This subset could be arbitrary.

You now need to verify which of the uncountable functions actually satisfy the original equation.

That is the interesting part of the question...

  • $\begingroup$ But in my answer, the solution can be easily verified by plugging it into the given condition,...anyway I understood your point.... $\endgroup$ – Abhishek Bakshi Jan 24 '15 at 18:17
  • $\begingroup$ @AbhishekBakshi: Easily verified? No. It is not, IMO. That is why the official solutions goes through all those motions. Please read the paragraph in my answer which talks about it being $0$ at an arbitrary subset. How do you easily verify for every possible such subset? $\endgroup$ – pedant Jan 24 '15 at 18:19
  • $\begingroup$ No, I am not saying that my answer is complete but that $f(x)=x^2 \forall x \in \mathbb R $ and $f(x)=0 \forall x \in \mathbb R$ satisfies the given condition in the question... $\endgroup$ – Abhishek Bakshi Jan 25 '15 at 17:54

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