Taking $a = b = c = 0$, we have $3(f(0))^2 = 0$, which implies $f(0) = 0$.

Taking $a = 1, b = 0, c = 0$, we have $2f(1) = (f(1))^2 + (f(-1))^2$. Thus, $(f(1) - 1)^2 + (f(-1))^2 = 1$. This means either $f(1) = 1$ and $f(-1) = \pm 1$, or $f(-1) = 0$ and $f(1) = 0$ or $2$.

Taking $a = 1, b = 1, c = 0$, we have $2f(2) - 2f(1) = (f(1))^2 + (f(-1))^2 = 2f(1)$.
Hence, $f(2) = 2f(1)$.

Taking $a = 1, b = 0, c = -1$, we have $2f(2) - 2f(-1) = (f(2))^2 + 2(f(-1))^2$. Thus, $(f(2) - 1)^2 + 2(f(-1) + \frac{1}{2})^2 = 1.5$. This implies $f(2) = 0$ or $2$, and $f(-1) = 0$ or $-1$.

Combining the above, we conclude that either $f(1) = 1$ and $f(-1) = -1$, or $f(1) = 0$ and $f(-1) = 0$.

Let $g(k) = f(k)^2 - f(-k)^2$. We have $g(-k) = -g(k)$ and $g(0) = g(1) = 0$.
When we replace $(a, b, c)$ with $(b, c, a)$ in the original equation, the left side remains the same, and the change in the right side is exactly $g(a-b) + g(b-c) + g(c-a) = 0$. Taking $(a, b, c) = (k, 1, 0)$, we have $g(k-1)+g(1)+g(-k) = 0$, which implies $g(k) = g(k-1)+g(1)$. Therefore, we have $g(k) = 0$ for all $k \in \mathbb{Z}$, which means $f(-k) = \pm f(k)$ for all $k \in \mathbb{Z}$.

Taking $(a, b, c) = (k, -1, 0)$ and $(a, b, c) = (k, 1, 0)$ in the original equation and comparing the two resulting equations, we have
$$(6) \qquad 2(f(k) - f(-k)) = (f(k+1))^2 - (f(k-1))^2$$

Consider the first case where $f(1) = f(-1) = 0$. We will prove by mathematical induction that $f(k) = f(-k) = 0$ for all non-negative integers $k$. The base cases $k = 0$ and $k = 1$ have been established. Assume that $f(\pm k) = 0$ and $f(\pm(k-1)) = 0$ hold. From equation (6), we obtain $(f(k+1))^2 = 0$, which implies $f(k+1) = 0$ and $f(-k-1) = \pm f(k+1) = 0$. Thus, the induction hypothesis holds.

Therefore, in the first case, we have $f(m) = 0$ for all $m \in \mathbb{Z}$.

Now, consider the second case where $f(1) = 1$ and $f(-1) = -1$. We have $f(2) = 2f(1) = 2$.

By substituting $(a, b, c) = (k, 1, -1)$ and $(a, b, c) = (k, 1, 1)$ into the original equation and comparing the resulting equations, we obtain
$$2f(2k + 1) - 2f(-1) = (f(-1 - k))^2 - (f(1 - k))^2 + (f(2))^2.$$
$$(7) \qquad f(2k + 1) = \frac{1}{2}(f(k + 1))^2 - \frac{1}{2}(f(k - 1))^2 + 1$$

By substituting $(a, b, c) = (k, 2, 0)$ and $(a, b, c) = (k, -2, 0)$ into the original equation and comparing the resulting equations, we have
$$(8) \qquad 2f(2k) - 2f(-2k) = (f(k + 2))^2 - (f(k - 2))^2$$

By substituting $k = 1$ into equation (7), we have $f(3) = 3$.
By substituting $k = 2$ into equation (8), we obtain $(f(4))^2 = 2(f(4) - f(-4))$. If $f(4) = 0$, then substituting $k = 3$ into equation (6) yields $f(3) - f(-3) = -2$, which implies $f(-3) = 5 \neq \pm f(3)$, leading to a contradiction. Therefore, $f(4) \neq 0$. Since $f(-4) = \pm f(4)$, specifically $f(-4) = -f(4)$, we have $(f(4))^2 = 4f(4)$, which implies $f(4) = 4$.

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We will now prove by mathematical induction that $f(m) = m$ for all non-negative integers $m$. The base cases $m = 0, 1, 2, 3, 4$ have been established. Assume that $f(k) = k$ holds for $k = 0, 1, 2, \dots, m-1$.
For $m = 2k + 1 \ge 5$, using equation (7), we obtain $f(2k + 1) = \frac{1}{2}((f(k+1))^2 - (f(k-1))^2) + 1 = 2k + 1$.
For $m = 2k \ge 6$, utilizing equation (8), we have $f(2k) - f(-2k) = \frac{1}{2}((f(k+2))^2 - (f(k-2))^2) = 4k$. Furthermore, $f(-2k) = \pm f(2k)$, which implies $2f(2k) = 4k$ and $f(2k) = 2k$.

Hence, we conclude that $f(m) = m$ for $m \in \mathbb{Z}_+$. For positive integers $m$, substituting $k = m$ into equation (6) gives us
$$2(f(m) - f(-m)) = (f(m + 1))^2 - (f(m - 1))^2 = 4m, \Rightarrow f(-m) = -m.$$
Therefore, in the second case, we have $f(m) = m$ for all $m \in \mathbb{Z}$.

Upon verification, we find that both solutions, $f(m) = 0$ and $f(m) = m$, satisfy the given conditions. $\square$