(Cauchy)

Find all functions $f: \mathbb{Z} \mapsto \mathbb{Z}$ such that for all $x, y \in \mathbb{Z}$,

$$
f(x+y)=f(x)+f(y)
$$
We notice that $f(0)=0$ by setting $x=y=0$, then we use this information to obtain $f(x)=-f(-x)$ by setting $y=-x$ in the equation from the statement. Be careful here, if the domain of definition of $f$ was $\mathbb{N}$, we would not be allowed to perform this last substitution. $f$ is therefore odd, and it suffices to solve over $\mathbb{N}$. For this, we will prove by induction that for all $n \geq 0$,

$$
f(n)=n \cdot f(1) \text {. }
$$

The base case is clear since $f(0)=0$. Now suppose for the inductive step that for some $n \geq 0, f(n)=n f(1)$. Then from the initial equation with $x=n$ and $y=1$, we get

$$
f(n+1)=f(n)+f(1)=(n+1) f(1)
$$

using the inductive hypothesis. We conclude the analysis by oddness, the only potential solutions are the functions defined by $f(x)=a x$, where $a \in \mathbb{Z}$. It is easy to verify that these functions (the linear functions) are solutions.