Find the function $ f$: $\mathbb R^+$ $\to$ $\mathbb R^+$ satisfying: $ f(2f(x)+2y) = f(2x+y) +y $ Find the function $f : \mathbb R^+ \to \mathbb R^+$ satisfying: $f(2f(x) + 2y) = f(2x+y) + y.$
I think this problem is difficult in the set of determination and the set of values. If the deterministic set is $\mathbb R$, we can easily solve the problem:
Choose $y = 2x - 2f(x)$
$\Rightarrow 2f(x)+2y = 2x+y $
$\Rightarrow f(2f(x)+2y) = f(2x+y)$
$\Rightarrow y= 0$
$\Rightarrow f(x)=x$ (Try again and see if it is correct.)
That's all I did. I hope to get help from everyone. Thank you very much.

$f(2f(x) + 2y) = f(2x+y) + y$
Assuming that $f ( a ) = f (b)$
$\Rightarrow$ $ f(2a+b) = f(2b+y)$
If $ a {\neq} b {\Rightarrow} $ f is a periodic function with period s.
Notation P(a;b) : Replace x by a, y by b into the original equation .
$P(x;y+s)$ and $P(x;y)$ $\Rightarrow$ $s=0$ $\Rightarrow$ $a=b$
$P(x;y-f(x)+f(t))$ $\Rightarrow$ $f(2y+2f(t)) = f(2x+y-f(x)+f(t)) +y-f(x)+f(t) $
$\Rightarrow$ $f(2t+y) + y = f(2x+y-f(x) + f(t)) + y- f(x) + f(t) $
Set $2t-2x+f(x)-f(t) =s $
$\Rightarrow$ $( 2t+y)-(2x+y-f(x)+f(t)) =s $
Replace $y$ by $ 2x+y-f(x)+f(t)$ $\Rightarrow$ $f(y+s) +f(x) - f(t) = f(y) $.
Set $ f(x) - f(t) = r$
$\Rightarrow$ $f(y+s) +r = f(y)$
$\Rightarrow$ $f(2f(x)+2y)-r = f(2f(x)+2y+s) $
$\Rightarrow$ $f(2f(x)+2y)-2r = f(2f(x)+2y+2s) $
In a similar way we have $ f(2f(x)+2y)-2r = f(2f(x)+2y-2r)$
$\Rightarrow$ $2f(x)+2y+2s = 2f(x)+2y-2r$
$\Rightarrow$ $ s=-r $
$\Rightarrow$ $ 2t-2x+f(x)- f(t) = -f(x)+f(t) $
$\Rightarrow$ $f(x)-x = f(t)-t = C $
$\Rightarrow$ $f(x) = x+C $
It is easy to see that $C = 0$. $\Rightarrow$ $f(x) = x$
