Substituting $x=y=0$, we get $f(f(1))=0$. Knowing this, we substitute only $x=0$. We get $f(y)=f(-y)$. So $f(f(1))=f(f(-1))=0$. Substituting $y=\frac{2}{x}$, we get $f(x+\frac{2}{x})=f(x-\frac{2}{x})$. Though I cannot proceed further, I have a feeling its worth something. Peace. Faustus

Setting $y=1$ yields $f(x+1)-f(x-1)=f(f(1-x))$, so initial equation may be rewritten as $f(x+y)=f(x-y)+f(xy+1)-f(xy-1)$. I'm stuck here.

nsato wrote: Find all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ for which \[f(x + y) = f(x - y) + f(f(1 - xy))\] holds for all real numbers $x$ and $y$. Let $y=0$ then $ f(f(1))=0$ Let $x=0$ then $f(y)=f(-y)$(*) Let $y=1$ have $f(x+1)=f(x-1)+f(f(1-x))$ then $f(x+2)=f(x)+f(f(x))=f(2-x)=f(x-2)$ (with (*)) Let $y=2$ with (**) then $f(f(2x-1))=0$ for all x or $f(f(x)=0$ for all x then $f(x+y)=f(x-y)$ all x then $f(x)=const$ all x then $f(x)=0$ all x

vntbqpqh234 wrote: Let $y=0$ then $ f(f(1)=1$ Should be $f(f(1)) = 0$, but it seems like only a typo (you use it properly for $x=0$. vntbqpqh234 wrote: then $f(x+2)=f(x)+f(f(x))=f(2-x)$ How?

chaotic_iak wrote: vntbqpqh234 wrote: Let $y=0$ then $ f(f(1)=1$ Should be $f(f(1)) = 0$, but it seems like only a typo (you use it properly for $x=0$. vntbqpqh234 wrote: then $f(x+2)=f(x)+f(f(x))=f(2-x)$ How? You are right $f(f(1))=0$ I write fast but after I do $with f(f(1)=0$ vntbqpqh234 wrote: nsato wrote: Find all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ for which \[f(x + y) = f(x - y) + f(f(1 - xy))\] holds for all real numbers $x$ and $y$. Let $y=0$ then $ f(f(1)=1$ Let $x=0$ then $f(y)=f(-y)$(*) Let $y=1$ have $f(x+1)=f(x-1)+f(f(1-x))$ then $f(x+2)=f(x)+f(f(x))=f(2-x)=f(x-2)$ (with (*)) Let $y=2$ with (**) then $f(f(2x-1))=0$ for all x or $f(f(x)=0$ for all x then $f(x+y)=f(x-y)$ all x then $f(x)=const$ all x then $f(x)=0$ all x because $f(x+1)=f(x-1)+f(f(1-x))$ all x and $f(x)=f(-x)$ all x then $f(x+2)=f(x)+f(f(x))=f(-x)+f(f(-x))=f(2-x)=f(x-2)$

How is this true? $f(-x)+f(f(-x))=f(2-x)$

Just an idea: It is pretty obvious (putting y = 0) that f(f(1))=0. Now, $f(x+y) = f(x-y) + f(f(1-xy))$ or, $f(x+y) - f(x-y) = f(f(1-xy))$ or, $[f(x+y) - f(x-y)]/2y = [f(f(1-xy))]/2y$ With the assumption that f is differentiable at x (I will try to prove it later), if we let y tend to zero on both sides, then we have $f'(x) = lim(y -> 0) [f(f(1-xy))]/2y = -(x/2)*[f'(f(1))*f'(1)]$, as we have f(f(1)) = 0, so the numerator/denominator is in the form of 0/0; hence L'Hopital's Rule seems to be applicable (my only concern is whether f has to be differentiable at any x or not). I think this may be an outline to a solution; considering f'(f(1))*f'(1) as some constant c, we can write: f'(x) = -cx/2 or, f(x) = -cx^2/4 + d, where c and d are constant we need to determine. Faustus wrote: Substituting $x=y=0$, we get $f(f(1))=0$. Knowing this, we substitute only $x=0$. We get $f(y)=f(-y)$. So $f(f(1))=f(f(-1))=0$. Substituting $y=\frac{2}{x}$, we get $f(x+\frac{2}{x})=f(x-\frac{2}{x})$. Though I cannot proceed further, I have a feeling its worth something. Peace. Faustus Using this relation, $f(x+2/x) = f(x-2/x)$ or, (-c/4)[(x+2/x)^2] = (-c/4)[(x-2/x)^2] Quite clearly, $(x+2/x)^2$ is never equal to $(x-2/x)^2$ and so we have c = 0 as the only possible result. Hence, $f(x) = d$ (some constant) which we need to determine. But f(f(1)) = 0 = d, so d = 0. Hence, f(x) = 0 for all x. Really unsure whether this is the correct solution or not. Need help

Does anyone know the full solution of this?

for those who couldn't understand vntbqpqh234-s solution(think it's still same) - after $f(f(1))=0$,$f(x)=f(-x)$,$f(x+2)=f(x)+f(f(x))$ put $P(1-x,1)$ $\rightarrow$ $f(2-x)=f(-x)+f(f(-x))$ which means $f(x+2)=f(x-2)$.Put $P(x,2)$ $\rightarrow$ $f(x+2)=f(x-2)+f(f(1-2x))$ and since $f(x+2)=f(x-2)$ we deduce that $f(f(1-2x))=0=f(f(2x-1))$ and since $2x-1$ can get any real value,it means that $f(f(x))=0$ for all $x$ real.And if we put $P(\frac{x+y}{2},\frac{x-y}{2})$ $\rightarrow$ $f(x)=f(y)$ for any $x,y$ real..And from these,it's obvious that $f(x)=0$ for all $x$ real.

$y=0$ gives $f(f(1))=0$. $x=0$ gives $f(y)=f(-y)$ or that $f$ is even. Swapping sign on $y$ gives $f(x-y)=f(x+y)+f(f(1+xy))$. Hence, $$f(f(1+xy))+f(f(1-xy))=0\implies f(f(1+t))+f(f(1-t))=f(f(t+1))+f(f(t-1))=0$$Take $y=1$ to get $f(x+1)=f(x-1)+f(f(1-x))=f(x-1)+f(f(x-1))$. Replace $x$ with $x+2$ to get $f(x+3)=f(x+1)+f(f(x+1))$. Summing the 2 equations and using the above equation gives $f(x+3)=f(x-1)$, meaning $f$ has period 4. Now, let $y=4$ $$f(x+4)=f(x-4)+f(f(1-4x)),$$but $f(4+x)=f(x)=f(x-4)$, so we have $f(f(1-4x))=0$ for all $x$. Thus, $f(f(r))=0$ for all $r$. Putting this back in the original FE, we get $f(x+y)=f(x-y)$, which shows $f$ is constant. Thus, $\boxed{f=0}.$

Denote the assertion by $P(x,y).$ $P(0,0)\implies f(f(1))=0.$ $P(0,x)\implies f(x)=f(-x).$ $P(1+x,1)-P(1-x,1)\implies f(x+2)=f(2-x)=f(x-2).$ $P(x,2)\implies f(f(2x-1))=0\implies f(f(x))=0.$ $P(x,y)\implies f(x+y)=f(x-y)=\text{constant}=0.$ This works.