Consider $P(x, y, z) = (y-x)^2 + (z-y)^2$. Then, $P(n, n+m, n+2m) = 2m^2$. Clearly, $r > 2$ does not work. We show that $r = 2$.

If there exist $(a_1, b_1, c_1)$, $(a_2, b_2, c_2)$ such that $P(a_1, b_1, c_1) > 0$ and $P(a_2, b_2, c_2) < 0$, draw a continuous path from $(a_1, b_1, c_1)$ to $(a_2, b_2, c_2)$, not passing through the line $x = y = z$.
Since $P$ is continuous, there exists $(a_3, b_3, c_3)$ on the path such that $P(a_3, b_3, c_3) = 0$. This is a contradiction as the condition that $a_3 = b_3 = c_3$ is not satisfied. Hence, either all $P \le 0$ or $P \ge 0$.

Let $u = y - x$, $v = z - y$. $P(x, y, z) = Q(x, u, v) = uvQ_1(x, u, v) + u^2Q_2(x, u, v) + v^2Q_3(x, u, v) + uR_1(x) + vR_2(x) + R_3(x)$. Substituting $u, v = 0$, $x = a$, we have $R_3(a) = 0$ for all real $a \Rightarrow R_3(x) = 0$. Substituting $v = 0$, $x = a$, we have $Q(a, u, 0) = u^2Q_2(a, u, 0) + uR_1(a)$. However, since $P \ge 0$ or $P \le 0$, for any fixed real $a$, there must be a double root at $u = 0$. Hence, $R_1(a) = 0$ for all $a \Rightarrow R_1(x) = 0$. Similarly, $R_2(x) = 0$.

Since $m \mid u, v$, we have that $m^2$ divides $P(n, n+m, n+2m) = Q(n, m, m) = m^2(Q_1(n, m, m) + Q_2(n, m, m) + Q_3(n, m, m))$.