A4. Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$such that
$$
f(x+f(y))=f(x+y)+f(y)
$$
for all $x, y \in \mathbb{R}^{+}$. (Symbol $\mathbb{R}^{+}$denotes the set of all positive real numbers.)
Answer. $f(x)=2 x$.
Solution 1. First we show that $f(y)>y$ for all $y \in \mathbb{R}^{+}$. Functional equation (1) yields $f(x+f(y))>f(x+y)$ and hence $f(y) \neq y$ immediately. If $f(y)f(y),
$$
contradiction. Therefore $f(y)>y$ for all $y \in \mathbb{R}^{+}$.
For $x \in \mathbb{R}^{+}$define $g(x)=f(x)-x$; then $f(x)=g(x)+x$ and, as we have seen, $g(x)>0$. Transforming (1) for function $g(x)$ and setting $t=x+y$,
$$
\begin{aligned}
f(t+g(y)) & =f(t)+f(y) \\
g(t+g(y))+t+g(y) & =(g(t)+t)+(g(y)+y)
\end{aligned}
$$
and therefore
$$
g(t+g(y))=g(t)+y \quad \text { for all } t>y>0 .
$$

Next we prove that function $g(x)$ is injective. Suppose that $g\left(y_{1}\right)=g\left(y_{2}\right)$ for some numbers $y_{1}, y_{2} \in \mathbb{R}^{+}$. Then by $(2)$
$$
g(t)+y_{1}=g\left(t+g\left(y_{1}\right)\right)=g\left(t+g\left(y_{2}\right)\right)=g(t)+y_{2}
$$
for all $t>\max \left\{y_{1}, y_{2}\right\}$. Hence, $g\left(y_{1}\right)=g\left(y_{2}\right)$ is possible only if $y_{1}=y_{2}$.
Now let $u, v$ be arbitrary positive numbers and $t>u+v$. Applying (2) three times,
$$
g(t+g(u)+g(v))=g(t+g(u))+v=g(t)+u+v=g(t+g(u+v)) .
$$

By the injective property we conclude that $t+g(u)+g(v)=t+g(u+v)$, hence
$$
g(u)+g(v)=g(u+v) .
$$

Since function $g(v)$ is positive, equation (3) also shows that $g$ is an increasing function.
Finally we prove that $g(x)=x$. Combining (2) and (3), we obtain
$$
g(t)+y=g(t+g(y))=g(t)+g(g(y))
$$
and hence
$$
g(g(y))=y
$$

Suppose that there exists an $x \in \mathbb{R}^{+}$such that $g(x) \neq x$. By the monotonicity of $g$, if $x>g(x)$ then $g(x)>g(g(x))=x$. Similarly, if $x0$ ) imply $g(x)=c x$. So, after proving (3), it is sufficient to test functions $f(x)=(c+1) x$.
Solution 2. We prove that $f(y)>y$ and introduce function $g(x)=f(x)-x>0$ in the same way as in Solution 1.
For arbitrary $t>y>0$, substitute $x=t-y$ into (1) to obtain
$$
f(t+g(y))=f(t)+f(y)
$$
which, by induction, implies
$$
f(t+n g(y))=f(t)+n f(y) \quad \text { for all } t>y>0, n \in \mathbb{N} .
$$

Take two arbitrary positive reals $y$ and $z$ and a third fixed number $t>\max \{y, z\}$. For each positive integer $k$, let $\ell_{k}=\left\lfloor k \frac{g(y)}{g(z)}\right\rfloor$. Then $t+k g(y)-\ell_{k} g(z) \geq t>z$ and, applying (4) twice,
$$
\begin{aligned}
f\left(t+k g(y)-\ell_{k} g(z)\right)+\ell_{k} f(z) & =f(t+k g(y))=f(t)+k f(y), \\
0<\frac{1}{k} f\left(t+k g(y)-\ell_{k} g(z)\right) & =\frac{f(t)}{k}+f(y)-\frac{\ell_{k}}{k} f(z) .
\end{aligned}
$$

As $k \rightarrow \infty$ we get
$$
0 \leq \lim _{k \rightarrow \infty}\left(\frac{f(t)}{k}+f(y)-\frac{\ell_{k}}{k} f(z)\right)=f(y)-\frac{g(y)}{g(z)} f(z)=f(y)-\frac{f(y)-y}{f(z)-z} f(z)
$$
and therefore
$$
\frac{f(y)}{y} \leq \frac{f(z)}{z}
$$

Exchanging variables $y$ and $z$, we obtain the reverse inequality. Hence, $\frac{f(y)}{y}=\frac{f(z)}{z}$ for arbitrary $y$ and $z$; so function $\frac{f(x)}{x}$ is constant, $f(x)=c x$.

Substituting back into (1), we find that $f(x)=c x$ is a solution if and only if $c=2$. So the only solution for the problem is $f(x)=2 x$.