http://artofproblemsolving.com/community/c6h508345p2855945 http://artofproblemsolving.com/community/c6h1145580

Let $P(x,y) : f(x+y)=f(x-y)+f(f(1-xy))$ . $P(x+1,-1) : f(x)=f(x+2)+f(f(x))$ . ( ) $P(0,x) : f(x)=f(-x)$ ( ) By ( ) $\to P(x-1,1) : f(x)=f(x+2)+f(f(x))$ ( ) By ( ) and ( ) we get $f(f(x)) = 0$ . ( ) By ( ) and $P(x,y)$ , we get $f(x+y)=f(x-y)$ . then $f(x)=c$ . $\blacksquare$

Other Idea but I cant prove second idea . We get $f(f(x))=0$ . By $P(x+1 , -1)$ we see $f(x)=f(x+2)$ . So ! . Now our function is periodic. Now what is the problem ?! If I prove $f$ is Increasing , we can say $f(x)=c$ . But I cant prove $f$ is Increasing .

IstekOlympiadTeam wrote: Find all functions $f:\mathbb{R}\to\mathbb{R}$ for which \[f(x+y)=f(x-y)+f(f(1-xy))\]holds for all real numbers $x$ and $y$ $\color{blue}\boxed{\textbf{Answer: }f\equiv 0}$ $\color{blue}\boxed{\textbf{Proof:}}$ $\color{blue}\rule{24cm}{0.3pt}$ $$f(x+y)=f(x-y)+f(f(1-xy))...(\alpha)$$In $(\alpha) x=0:$ $$\Rightarrow f(y)=f(-y)$$$$\Rightarrow f \text{ is even}...(I)$$In $(\alpha) x=x+1, y=1:$ $$\Rightarrow f(x+2)=f(x)+f(f(-x))$$$$\Rightarrow f(f(-x))=f(x+2)-f(x)$$By $(I):$ $$\Rightarrow f(f(x))=f(x+2)-f(x)...(II)$$$(I)$ in $(\alpha):$ $$\Rightarrow f(x+y)=f(x-y)+f(f(xy-1))...(\beta)$$$(II)$ in $(\beta):$ $$\Rightarrow f(x+y)=f(x-y)+f(xy+1)-f(xy-1)...(\theta)$$In $(I) x=x-1,y=-1:$ $$\Rightarrow f(x-2)=f(x)+f(f(x))$$$$\Rightarrow f(f(x))=f(x-2)-f(x)...(III)$$By $(II)$ and $(III):$ $$\Rightarrow f(x-2)=f(x+2)...(IV)$$In $(\theta) y=2:$ $$\Rightarrow f(x+2)=f(x-2)+f(2x+1)-f(2x-1)$$By $(IV):$ $$\Rightarrow f(2x+1)=f(2x-1)...(V)$$In $(V) x=\frac{x}{2}:$ $$\Rightarrow f(x+1)=f(x-1)...(VI)$$$(VI)$ in $(\theta):$ $$\Rightarrow f(x+y)=f(x-y)$$$$\Rightarrow f\text{ is constant}$$$$\Rightarrow f\equiv c$$In $(\alpha):$ $$\Rightarrow c=2c$$$$\Rightarrow c=0$$$$\Rightarrow \boxed{f\equiv 0 \text{ is the only solution}}_\blacksquare$$$\color{blue}\rule{24cm}{0.3pt}$

strong_boy wrote: Let $P(x,y) : f(x+y)=f(x-y)+f(f(1-xy))$ . $P(x+1,-1) : f(x)=f(x+2)+f(f(x))$ . ( ) $P(0,x) : f(x)=f(-x)$ ( ) By ( ) $\to P(x-1,1) : f(x)=f(x+2)+f(f(x))$ ( ) By ( ) and ( ) we get $f(f(x)) = 0$ . ( ) Sorry for answering late but it's wrong since they are the same equation

Can you explain how $f(x+y)=f(x-y)$ implies that $f$ is constant?

amogususususus wrote: Can you explain how $f(x+y)=f(x-y)$ implies that $f$ is constant? $x=x+y:$ $$\Rightarrow f(x+2y)=f(x)$$$y=\frac{y}{2}:$ $$\Rightarrow f(x+y)=f(x)$$$$\Rightarrow f\text{ is constant}$$

hectorleo123 wrote: amogususususus wrote: Can you explain how $f(x+y)=f(x-y)$ implies that $f$ is constant? $x=x+y:$ $$\Rightarrow f(x+2y)=f(x)$$$y=\frac{y}{2}:$ $$\Rightarrow f(x+y)=f(x)$$$$\Rightarrow f\text{ is constant}$$ thanks a lot brother