InternetPerson10 wrote: Find all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that $$f(2f(x)) = f(x - f(y)) + f(x) + y$$for all $x, y \in \mathbb{R}$. Let $P(x,y)$ be the assertion $f(2f(x))=f(x-f(y))+f(x)+y$ $f(x)$ is injective (look at $y$) $P(x,-f(0)-f(x)+f(2f(x)))$ $\implies$ $f(x-f(-f(0)-f(x)+f(2f(x))))=f(0)$ $\implies$ (injectivity) $f(-f(0)-f(x)+f(2f(x)))=x$ And so $f(x)$ is surjective. Let then $u$ such that $f(u)=0$ : $P(x,u)$ $\implies$ $f(2f(x))=2f(x)+u$ $\implies$ (surjectivity) $f(x)=x+u$ Plugging this back in original equation, we get $u=0$ and so $\boxed{f(x)=x\quad\forall x}$, which indeed fits.

Solution without using surjectivity $f(2f(x))=f(x-f(y))+f(x)+y$ it is obvious that $f$ is injective $P(0,-f(0))$ $2f(0)=-f(-f(0))$ $P(0,0)$ $f(2f(0))=-f(0)$ $P(0,2f(0))$ $ff(0)=-4f(0)$ $P(f(0),0)$ $f(-8f(0))=-3f(0)$ $P(f(0),-f(0))$ $f(-8f(0))=-7f(0)$ Thus $f(0)=0$ $P(x,-f(x))$ $2f(x)=x-f(-f(x))$ $P(0,x)$ $f(-f(x))=-x$ So that $2f(x)=2x$ and $f(x)=x$

First observe $f(0)=0$ Then at $P(0,x)$ $f(-f(y))=-y$ We have an involution This implies function is bijective At $P(x,0)$ $f(2f(x))=2f(x)$ Now it is trivial and so $\boxed{f(x)=x\quad\forall x}$, which indeed fits