How find this function $f(x)$ such $f(a+f(b))=f(a+b)+f(b)$ let function $f:R_{+}\to R_{+}$,and such
$$f(a+f(b))=f(a+b)+f(b),\forall a,b\in R_{+}$$
Find $f(x)$.
my try: let $a=b=1$,then 
$$f(1+f(1))=f(2)+f(1)$$
$a=1,b=2$,then
$$f(1+f(2))=f(3)+f(2)$$
then I can't find have any regular,so I can't.Thank you
Edit: This is IMO2007 SL, Problem A4.
 A: Here is a proof assuming that $D={\mathbb R}_{+}$ means $(0,+\infty)$ (and 
so does not include $0$).
Step 1. $f(x)\neq x$ for every $x\in D$.
Indeed, if $f(x)=x$, then $f(a+f(x))=f(a+x)+f(x)$ becomes $f(x)=0$, so $x=0$ which
is impossible.
Step 2. $f(x) > x$ for every $x\in D$.
Suppose by contradiction that $f(x)<x$ for some $x\in D$. Let $d=x-f(x)$.
Let $y > x$. Putting $a=y-f(x)$, $f(a+f(x))=f(a+x)+f(x)$ becomes
$f(y)=f(y+d)+(x-d)$, so that $f(y+d)=f(y)-(x-d)$. By induction on
$j\geq 1$, we have $f(y+dj)=f(y)-j(x-d)$ for every $j\geq 1$ and every
$y>x$. For large enough $j$, we will therefore have $f(y+dj)<0$ for large
enough $j$, an impossibility.
Now let us define a sequence $(x_n)_{n\geq 1}$ by
$$
x_1=x, x_{n+1}=f(x_n)-x_n \ (n\geq 1). \tag{1}
$$
Putting $a=y-x_n$ in the equation
$f(a+f(x_n))=f(a+x_n)+f(x_n)$ we see that
$$
f(y+x_{n+1})=f(y)+x_n+x_{n+1} \ \text{whenever} \ y > x_n \tag{2}
$$
Let $z > x_1$. Taking $y=z$ or $f(z)$ and $n=1$ in (2)  above, we see that
$$
\begin{array}{lcl}
f(z+x_2) &=& f(z)+x_1+x_2  \\
f(f(z)+x_2) &=& f(f(z))+x_1+x_2  \\
\end{array}\tag{3}
$$
So that the identity $f(x_2+f(z))=f(x_2+z)+f(z)$ becomes :
$$
f(f(z))=2f(z) \ \text{whenever} \ z > x_1 \tag{4}
$$
But given any $z\in D$, we can always find an $x\in D$ such that
$x<z$ (for example $x=\frac{z}{3}$), so that (4) holds in fact
for any $z\in D$ :
$$
f(f(z))=2f(z) \ \text{whenever} \ z > 0 \tag{5}
$$
Let $M={\sf max}(x_1,x_2,x_3)$ and $y > M$. It follows from (2) that
$f(f(y+x_3))=f(f(y)+x_2+x_3)$, or $2f(y+x_3)=f(f(y)+x_2+x_3)$. Now 
$$
f(f(y)+x_2+x_3)=
f(f(y)+x_2)+x_2+x_3=
f(f(y)))+x_1+2x_2+x_3=2f(y)+x_1+2x_2+x_3 \tag{6}
$$
$$
f(y+x_3)=f(y)+\frac{x_1+2x_2+x_3}{2} \ \text{whenever} \ y > M \tag{7}
$$
On the other hand, we know from (2) that $f(y+x_3)=f(y)+x_2+x_3$, so
that $\frac{x_1+2x_2+x_3}{2}=x_2+x_3$, and hence $x_3=x_1$. Revisiting (2)
for $n=1$ and $n=2$, we obtain that
$$
f(y+x_2)=f(y+x_1),  \text{whenever} \ y > M. \tag{8}
$$
If we put $T=|x_2-x_1|$, it follows from (8) that
$$
f(z+T)=f(z), \text{whenever} \ z > M-{\sf min}(x_1,x_2). \tag{9}
$$
By induction we have $f(z+Tj)=f(z)$ for any $j\geq 1$ and any $z > M-{\sf min}(x_1,x_2)$.
From step 2 we deduce $f(z)>z+Tj$, which is possible only if $T=0$, so $x_2=x_1$
or in other words 
$$
f(x)=2x
$$ 
A: We can let $a=0$ and see that:
$$f(f(b)) = f(b) + f(b)$$
$$f(f(b)) = 2.f(b)$$
This function $f$ has a domain that is the same as its range.
Substituting $f(b)$ for $x$:
$$f(x) = 2x $$
Which gives:
$$f:Range(f) \rightarrow Range(f)$$
$$f: x \mapsto 2x$$
This allows us to then define the range arbitrarily (so long as the range is closed under multiplication by 2). Since the question stipulates that $f : \mathbb{R_{+}} \rightarrow \mathbb{R_{+}}$, that is what we will choose.
Thus $f$ is:
$$f: \mathbb{R_{+}} \rightarrow \mathbb{R_{+}}$$
$$x \mapsto 2x$$
A: the hint is :
$f(f(2a))=f(a+f(a+f^{-1}(a)))=f(2a+f^{-1}(a))+a$
$f(2f(a))=f(f(a)+a)+f(a)=f(2a)+2a$
$\Longrightarrow$$f(2a)+a=f(2a+f^{-1}(a))$
$\Longrightarrow$$f(x)=2x$
