Problem - 3653
For any real numbers $x$ and $y$, the following holds $$[f(x+y)]^2 = [f(x)]^2 + [f(y)]^2$$
Find the exact form of $f(x)$.
Let $x=y=0$: $$[f(0+0)]^2 = [f(0)]^2 + [f(0)]^2 \implies f(0)=0$$
Let $y=-x$: $$[f(x-x)]^2 = [f(x)]^2 + [f(-x)]^2$$
$$\therefore\quad 0=[f(x)]^2+[f(-x)]^2 \implies f(x)=f(-x)=0$$
It follows that for any real number $x$, $f(x)$ is a constant, $$f(x)=0$$