Edit: Thanks, Bandera. Hopefully fixed now. Let $y=x$; we have $f(f(x)-x)=0$ or $f(x+f(x))=0$ so $f(f(0))=0$. Taking $S(0, f(0))$ gives $f(0)=0$. We can now either proceed as above or take $S(0,x)$ to get $\boxed{f(x)=x}$.

eisirrational wrote: Let $y=x$; we have $f(f(x)-x)=0$ or $f(x+f(x))=0$. Note that $f(f(0))=0$; then, for $y=f(x)-f(0)$, we must have $x= \pm y$. Thus, for every $x$, either $f(x)=x+c$ or $c-x$ for some $c$. Let $y=0$; we have $f(f(x)) \cdot f(x+f(0))=x^2$. Plugging in our choices, note $f(f(x))=2f(0)-x$ or $x$ and $f(x+f(0))=x+2f(0)$ or $2f(0)$. Note that for any $f(0) \neq 0$, there exists a value of $x$ so that none of the above products can equal $x^2$. Thus, $f(0)=0$, so $f(x)= \pm x$. Now suppose $\exists x,y \neq 0$ s.t. $f(x)=x$ and $f(y)=-y$. Then, $(x-y)(x-y)= \pm(x^2-y^2)$, which is impossible. Hence our two functions are $\boxed{ f(x)=x \text{ and } f(x)=-x}$. Something's wrong with your solution because $f(x) \equiv -x$ doesn't satisfy the equation.

from above f(0)=0 set y=f(x) 0=x^2-f(x)^2 f(x)=x or -x But f(x)=-x doesn't work so f(x)=x

$\LaTeX$ from above $f(0)=0$ set $y=f(x)$ $0=x^2-f(x)^2$ $f(x)={x}~\text{or}~{(-x)}$ But $f(x)=-x$ doesn't work so $f(x)=x$

Aritra12 wrote: Latex.from above $f(0)=0$ set $y=f(x)$ $0=x^2-f(x)^2$ $f(x)=\frac{x}{-x}$ But $f(x)=-x$ doesn't work so $f(x)=x$ sorry f(x)≠x/(-x) i wanted to write f(x) =x or -x. Thanks.

We denote with $c=f(0)$. First we plug in $x=y=0$, from here we have that $f(f(0))=0$ If we now plug in $y=c$, we have the following: $$f(f(x)-c)f(x)=x^2-c^2$$now we plug in $x=0$, from which we have: $$c.c=-c^2 \implies c=0$$ Now let's set $y=0$, from here we have: $$f(f(x))f(x)=x^2 \; \; \text{(1)}$$now we set $y=0$, from here we have: $$f(-y)f(f(y))=-y^2$$From here we see that: $$f(-x)=-f(x)$$Now if set $y \rightarrow -y$ and $x \rightarrow f(y)$, we have that: $$f(y)^2=y^2$$Now we have two cases here. Obviously $f(y)=-y$ isn't a solution (first case). Now we have that $f(y)=y$, but let's say $\exists k \; \text{such that} \; f(k)=-k$. Now we plug in $x=k$ into $\text{(1)}$ we get that $k=0$, thus the only solution is $f(x)=x \; \; \forall x \in \mathbb{R}$

Let $P(x,y)$ the assertion of the given F.E. $P(0,0)$ $$f(f(0))=0$$$P(0,f(0))$ $$f(0)^2=-f(0)^2 \implies f(0)=0$$$P(-f(x),x)$ $$f(x)^2=x^2 \implies f(x)=\pm x$$Now assume that there exists $a,b \ne 0$ such that $f(a)=a$ and $f(b)=-b$ $P(a,b)$ $$(a-b)^2=a^2-b^2 \implies 2b(b-a)=0 \implies b=0 \; \text{or} \; a=b=0 \; \text{contradiction!!}$$Hence or $f(x)=-x$ or $f(x)=x$ but since $f(x)=-x$ is not a solution we see that only $f(x)=x$ thus we are done

Let $P(x,y)$ be the assertion. $P(0,0)\Rightarrow f(f(0))=0$ $P(0,f(0))\Rightarrow f(0)=0$ $P(x,f(x))\Rightarrow f(x)^2=x^2$ Assume $f(a)=a$ and $f(b)=-b$ for some $a,b\ne0$. $P(a,b)\Rightarrow a=b$, contradiction. Hence $\boxed{f(x)=x}$ since $-x$ is not a solution.

$P(0,0)$ gives that $f(f(0))=0.$ Then $P(x, f(x)-f(0))$ yields $f(x)=\pm x + f(0),$ and from here it is easy to see that $f(x)\equiv x$.

JustKeepRunning wrote: $P(0,0)$ gives that $f(f(0))=0.$ Then $P(x, f(x)-f(0))$ yields $f(x)=\pm x + f(0),$ and from here it is easy to see that $f(x)\equiv x$. How u do the pointwise trap with that f(0)