An algebra problem by Anandmay Patel

Let us first examine the nature of the function. Let $x=y=0:$ $f(0)^2=f(0)^2+f(0)^2$ ,which implies that $f(0)=0$ .

Now let $y=-x:$ $f(x-x)^2=f(0)^2=f(x)^2+f(-x)^2$

But,the square of a real number is always non-negative and 2 squares,when summed,can produce 0 iff each square(and thus the number squared) is 0.That is, $f(x)=f(-x)=0$ ,so, $f(x)^2+f(-x)^2=0^2+0^2=0$ .

Let us gather what we have obtained $:-$ $(1.)$ $f(0)=0$ $(2.)$ $f(x)=0$ and, $(3.)$ $f(-x)=0$ This can be evidently achieved iff $f(x)=0$ for all real $x$ . So it is a constant function.