, 2019

Substituting $a=0$, $b=n+1$ gives $f(f(n+1)) = f(0) + 2f(n+1)$. Substituting $a=1$, $b=n$ gives $f(f(n+1)) = f(2) + 2f(n)$.
In particular, $f(0) + 2f(n+1) = f(2) + 2f(n)$, and so $f(n+1) - f(n) = \frac{1}{2}(f(2) - f(0))$. Thus $f(n+1) - f(n)$ must be constant. Since $f$ is defined only on $\mathbb{Z}$, this tells us that $f$ must be a linear function; write $f(n) = M n + K$ for arbitrary constants $M$ and $K$, and we need only determine which choices of $M$ and $K$ work.

Now, (1) becomes
$$2M a + K + 2(M b + K) = M(M(a+b) + K) + K$$
which we may rearrange to form
$$(M-2)(M(a+b) + K) = 0$$
Thus, either $M = 2$, or $M(a+b) + K = 0$ for all values of $a+b$. In particular, the only possible solutions are $f(n) = 0$ and $f(n) = 2n + K$ for any constant $K \in \mathbb{Z}$, and these are easily seen to work.

Let $K = f(0)$.
First, put $a = 0$ in (1); this gives
$$\begin{equation*} f(f(b)) = 2f(b) + K \tag{2} \end{equation*}$$
for all $b \in \mathbb{Z}$.
Now put $b = 0$ in (1); this gives
$$f(2a) + 2K = f(f(a)) = 2f(a) + K,$$
where the second equality follows from (2). Consequently,
$$\begin{equation*} f(2a) = 2f(a) - K \tag{3} \end{equation*}$$
for all $a \in \mathbb{Z}$.
Substituting (2) and (3) into (1), we obtain
$$\begin{aligned} f(2a) + 2f(b) & = f(f(a+b)) \\ 2f(a) - K + 2f(b) & = 2f(a+b) + K \\ f(a) + f(b) & = f(a+b) + K . \end{aligned}$$
Thus, if we set $g(n) = f(n) - K$ we see that $g$ satisfies the Cauchy equation $g(a+b) = g(a) + g(b)$. The solution to the Cauchy equation over $\mathbb{Z}$ is well-known; indeed, it may be proven by an easy induction that $g(n) = M n$ for each $n \in \mathbb{Z}$, where $M = g(1)$ is a constant.
Therefore, $f(n) = M n + K$, and we may proceed as in Solution 1.