## Task 1.

Let $\alpha$ be a real number. Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that for all $x, y \in \mathbb{R}$ the following holds:

$$
f(x+\alpha+f(y))=f(f(x))+f(\alpha)+y
$$
## Solution.

By substituting $x=y=-\alpha$ we get

$$
f(f(-\alpha))=f(f(-\alpha))+f(\alpha)-\alpha \quad \Longrightarrow \quad f(\alpha)=\alpha
$$

By substituting $x=-\alpha, y=\alpha$ we get

$$
\alpha=f(f(\alpha))=f(f(-\alpha))+\alpha+\alpha \quad \Longrightarrow \quad f(f(-\alpha))=-\alpha
$$

Let's denote $f(-\alpha)=a$. We know that $f(a)=-\alpha$. By substituting $x=\alpha, y=a$ we get

$$
\alpha=2 \alpha+a \quad \Longrightarrow \quad a=-\alpha
$$

Finally, by substituting $y=-\alpha$ we get

$$
f(x)=f(f(x)), \quad \forall x \in \mathbb{R}
$$

and by substituting $x=-\alpha$ we get

$$
f(f(y))=y, \quad \forall y \in \mathbb{R}
$$

It follows that

$$
f(x)=f(f(x))=x, \quad \forall x \in \mathbb{R}
$$

Verification shows that this is indeed a solution.