Solve $f(x+f(2y))=f(x)+f(y)+y$ 
Find all $f:\mathbb{R}^+\to \mathbb{R}^+$ such that for each $x$ and $y$ in $\mathbb{R}^+$, $$f(x+f(2y))=f(x)+f(y)+y$$

Note:
$f(x)=x+b$ is a solution for all  $b\in\mathbb{R}^+$ but I can not prove it.
 A: EDIT: I now have a complete solution.
$$f:\mathbb{R}^+\to\mathbb{R}^+$$
$$f(x+f(2y))=f(x)+f(y)+y$$
Note that $f(a+f(2x_1)+f(2x_2)+...+f(2x_n))=f(a)+f(x_1)+x_1+f(x_2)+x_2+...+f(x_n)+x_n$
Let us calculate the expression $f(x+f(2(y+f(2z)))$ in two ways:
$$f(x+f(2(y+f(2z)))=f(x)+f(y+f(2z))+y+f(2z)=f(x)+f(y)+f(z)+z+y+f(2z)$$
$$f(x+f(2(y+f(2z)))=f(x+f(2y+f(2z)+f(2z)))=f(x+f(2y)+2f(z)+2z)=f(x+2z)+f(y)+y+2f(\frac{z}{2})+2\cdot{}\frac{z}{2}$$
Thus, cancelling $f(y)+y+z$ we get
$$f(x)+f(z)+f(2z)=f(x+2z)+2f(\frac{z}{2})$$
i.e. (for $y=2z$)
$$f(x+y)=f(x)+f(y)+f(\frac{y}{2})-2f(\frac{y}{4})$$
Since the expression $f(x+y)-f(x)-f(y)$ is symmetric in $x$ and $y$, we get that $f(\frac{y}{2})-2f(\frac{y}{4})=f(\frac{x}{2})-2f(\frac{x}{4})$, i.e. $f(2y)-2f(y)=f(2x)-2f(x)$ for all $x,y$. so $f(2x)-2f(x)$ is some constant $-c$. So we obtained:
$$f(x+y)=f(x)+f(y)-c$$
Define $i=inf_{\mathbb{R}^+}(f(x))$. Clearly, $0\le{}i\le{}\infty$. as $x$ and $y$ go through all positive real values, the infimum of LHS is $i$ and the infimum of RHS is $2i-c$, therefore $c=i$. since $f(y)-i\ge{}0$, we get $f(x+y)\ge{}f(x)$, so $f$ is monotonically increasing.
Define $g:\mathbb{R}^+\to{}\mathbb{R}^+$ as $g(x)=f(x)-i$. from $f(x+y)=f(x)+f(y)-c$ we get that $g$ is additive, i.e. $g(x+y)=g(x)+g(y)$. Thus, by induction, $g(k)=kg(1)$ for all $k\in{}\mathbb{N}$. This argument can be expanded for all $k\in\mathbb{Q}^+$ (why?), so $g(q)=qg(1)$ for all $q\in{}\mathbb{Q}^+$. but since $f$ is monotonic we get that $g$ is monotonic, so $g(x)=xg(1)$ for all $x>0$. Thus $g$ is linear, so $f$ is linear. Substituting $f(x)=ax+b$ in the original equation yields the solutions:
$$a=1$$
$$a=-\frac{1}{2}, b=0$$
We are interested in solutions that give only positive values in the image of $f$. Thus $f(x)=x+b$ for some constant $b\ge{}0$. This function satisfies the equation for all $b\ge{}0$.
A: Another solution is
$f(x) =-x/2$.
Here,
at the end of my ramblings,
is how I found it.
Inspired by the comment
that a solution is
$x+b$,
let
$f(x) = x+g(x)$.
Then
$x+2y+g(2y)+g(x+2y+g(2y))
=x+g(x)+y+g(y)+y
$
or
$$g(2y)+g(x+2y+g(2y))
=g(x)+g(y).
$$
Letting $y=0$,
this is
$g(0)+g(x+g(0))
=g(x)+g(0)
$
or
$g(x) = g(x+g(0))$.
If $g(0) = 0$,
this tells nothing.
If $g(0) \ne 0$,
$g$ is periodic
with period $g(0)$.
In particular,
$g(0) 
= g(g(0))
= g(g(g(0)))
=...
$.
Setting $x=0$,
this is
$g(2y)+g(2y+g(2y))
=g(0)+g(y)
$.
This doesn't seem to help much.
If $x=2y$,
this is
$g(2y)+g(4y+g(2y))
=g(2y)+g(y)
$
or
$g(y)
=g(4y+g(2y))
$,
so
$g(y+g(0))
=g(4y+g(2y))
$.
If
$y+g(0)
=4y+g(2y)$,
then
$g(2y)
=g(0)-3y
$
or
$g(y)
=g(0)-3y/2
$.
This suggests trying
$f(x) = ax+b$.
Then
$a(x+2ay+b)+b
=a(x+y)+2b+y
$
or
$2a^2y+ab
=(a+1)y+b
$
or
$0
=(2a^2-a-1)y+b(a-1)
$.
Therefore
$0=(2a^2-a-1$
or
$a
=\dfrac{1\pm\sqrt{1+8}}{4}
=\dfrac{1\pm 3}{4}
=1, -\frac12
$.
If $a=1$
then any $b$ works.
If
$a=-\frac12$,
then
$b=0$,
so
$-x/2$
is a solution.
To check,
$f(x+f(2y))
=f(x-y)
=(y-x)/2
$
and
$f(x)+f(y)+y
=-x/2-y/2+y
=(y-x)/2
$.
That's all I have.
