Post Number:#2 by Thanic Nur Samin » Sun Apr 02, 2017 9:50 pm
The solutions are $2-x$ and $x$. They satisfy the FE.
Let $P(x,y)$ denote the FE.
$$P(0,0)\Rightarrow f(f(0))=0$$
$$P(0,f(0))\Rightarrow f(0)=0,2$$.
Case $1$: $f(0)=2$
$$P(0,x)\Rightarrow f(f(x))-f(x)=2(x-1)$$
Which implies $f(x)=x$ is only possible when $x=1$.
$$P(x,1)\Rightarrow f(x+f(x+1))=x+f(x+1)$$
So $x+f(x+1)=1$ and so $f(x)=2-x$ for all $x$.
Case $2$: $f(0)=0$.
$$P(-1,1)\Rightarrow f(-1)=-1$$
$$P(1,-1)\Rightarrow f(1)=1$$
$$P(x,0)\Rightarrow f(x+f(x))=x+f(x)$$
$$P(0,x)\Rightarrow f(f(x))=f(x)$$
$$P(x-1,1)\Rightarrow f(x-1+f(x))=x-1+f(x)$$
$$P(1,x-1+f(x))\Rightarrow f(x+1+f(x))=x+1+f(x)$$
$$\Rightarrow f(x+f(x-1))=x+f(x-1)$$
$$P(x,-1)\Rightarrow f(x)=-f(-x)$$
$$P(x,-x)\Rightarrow f(x)-f(x^2)=x-xf(x)$$
$$P(-x,x)\Rightarrow -f(x)-f(x^2)=-x-xf(x)$$
Substracting the last equation from the second last,
$f(x)=x$ For all real $x$.
Hammer with tact.
Because destroying everything mindlessly isn't cool enough.