Determine all functions from $\mathbb{Z}$ to $\mathbb{Z}$ satisfying:

$$
\forall(x, y) \in \mathbb{Z}^{2}, f(x+y)=f(x)+f(y)+2
$$
Let $y=1$. We then have: $f(x+1)=f(x)+(f(1)+2)$. Thus, the sequence $(f(n))_{\mathbb{N}}$ is an arithmetic sequence with common difference $f(1)+2$ and first term $f(0)$. Therefore, $\forall n \in \mathbb{N}, f(n)=n(f(1)+2)+$ $f(0)$. With $x=y=0$, we have $f(0)=-2$. Thus, $\forall n \in \mathbb{N}, f(n)=n(f(1)+2)-2$. Therefore, with $x \in \mathbb{N}$ and $y=-x$, we have: $f(0)=f(x)+f(-x)+2$ or $f(-x)=-x(f(1)+2)-2$. We verify that all functions $f: x \mapsto a x-2$ with $a \in \mathbb{Z}$ are solutions, these are therefore the only ones.