Problem (Algebra):
Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ such that

$$
f(x+f(y))=f(x+y)+f(y)\tag{1}
$$

for all $x, y \in \mathbb{R}^{+}$. (Symbol $\mathbb{R}^{+}$denotes the set of all positive real numbers.)

Solution:
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 $f(y)<y$ for some $y$, then setting $x=y-f(y)$ we get

$$
f(y)=f((y-f(y))+f(y))=f((y-f(y))+y)+f(y)>f(y),
$$

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

$$
g(t+g(y))=g(t)+y \quad \text { for all } t>y>0 \tag{2}
$$

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,

$$
g(t+g(u)+g(v))=g(t+g(u))+v=g(t)+u+v=g(t+g(u+v)) \text {. }
$$

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

$$
g(u)+g(v)=g(u+v)\tag{3}
$$

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 $x<g(x)$ then $g(x)<g(g(x))=x$. Both cases lead to contradiction, so there exists no such $x$.

We have proved that $g(x)=x$ and therefore $f(x)=g(x)+x=2 x$ for all $x \in \mathbb{R}^{+}$. This function indeed satisfies the functional equation (1).

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

$$
f(t+n g(y))=f(t)+n f(y) \quad \text { for all } t>y>0, n \in \mathbb{N} \tag{4}
$$

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,

$$
\begin{gathered}
f\left(t+k g(y)-\ell_{k} g(z)\right)+\ell_{k} f(z)=f(t+k g(y))=f(t)+k f(y), \\
0<\frac{1}{k} f\left(t+k g(y)-\ell_{k} g(z)\right)=\frac{f(t)}{k}+f(y)-\frac{\ell_{k}}{k} f(z) .
\end{gathered}
$$

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 $f(x)=2 x$.

Final answer: $f(x)=2 x$