Find all functions $f: \mathbb{R} \to \mathbb{R}$ which satisfies $f(x)-y+f(f(y))=f(x+y)+f(f(x))$.

Question

Find all functions $f: \mathbb{R} \to \mathbb{R}$ which satisfies $f(x)-y+f(f(y))=f(x+y)+f(f(x))$.

My Attempt

\begin{align} &P(x, x): f(x)-x+f(f(x))=f(2x)+f(f(x)). \\ &\Rightarrow f(2x)=f(x)-x. \\ \ \\ &P(x, y): f(x)-y+f(f(y))=f(x+y)+f(f(x)). \\ &P(y, x): f(y)-x+f(f(x))=f(x+y)+f(f(y)). \\ &\therefore f(x)+f(y)-x-y+f(f(x))+f(f(y))=2f(x+y)+f(f(x))+f(f(y)). \\ &\therefore f(2x)+f(2y)=2f(x+y). \\ &x=\frac {a+b}{2}, y=\frac{a-b}{2} \Rightarrow f(a+b)+f(a-b)=2f(a) \end{align}

Answer 1

There are no such functions.

Set $y=0$. Then $$\require{cancel}\cancel{f(x)}+f(f(0))=\cancel{f(x)}+f(f(x))$$ So $f^2$ is constant. Now set $x=0$. Then $$f(0)-y+\cancel{f(f(y))}=f(y)+\cancel{f(f(0))}$$ (The canceled terms are equal because $f^2$ is constant.)

But then $$f(f(y))=f(f(0)-y)=\cancel{f(0)}-\cancel{f(0)}+y$$ which is not constant.

Answer 2

I think you've almost done everything.

Since $P(a,b):f(a+b)+f(a-b)=2f(a)$, then $P(x,x):f(2x)+f(0)=2f(x)$.

And we know $f(2x)=f(x)-x$, so $2f(x)=f(2x)+f(0)=f(x)+f(0)-x$.

$f(x)=f(0)-x$. I'll let $f(0)=c$, then $f(x)=c-x$ form.

Let's put this back into the first given condition.

$f(x)-y+f(f(y))=(c-x)-y+y=c-x$, and $f(x+y)+f(f(x))=c-(x+y)+x=c-y$.

This makes contradiction, so there are no such function.