, 2014

a. By assumption $P(0) = a_0 \in \mathbb{Z}$, $P(-1) = a_0 - a_1 + a_2 \in \mathbb{Z}$ and $P(1) = a_0 + a_1 + a_2 \in \mathbb{Z}$. This implies that $a_1 + a_2$ and $a_2 - a_1$ are integers, hence also $2a_1$ and $2a_2$. From $P(n) = a_0 + a_1 n + a_2 n^2 = a_0 + a_1(n + n^2) + (a_2 - a_1)n^2$ we see now that $P(n) \in \mathbb{Z}$, as $n + n^2 = n(n + 1)$ is always even.

b. For integers $k \ge 0$ we define polynomials $Q_k$ as follows:
$$Q_0(x) = Q(x) \quad \text{and} \quad Q_{k+1}(x) = Q_k(x+1) - Q_k(x)$$
for all $k \ge 0$. Because for any polynomial $f$, the leading terms of $f(x+1)$ and $f(x)$ coincide, it follows that the degree of $Q_{k+1}$ is smaller than the degree of $Q_k$. As $Q_0$ was of degree three, it follows that $Q_3$ is a constant. The polynomials $Q_1$, $Q_2$, $Q_3$ can easily be determined, but their explicit form is not needed below.
The assumption that $Q(i)$, $Q(i+1)$, $Q(i+2)$ and $Q(i+3)$ are integers implies that $Q_1(i)$, $Q_1(i+1)$ and $Q_1(i+2)$ are integers. In turn we get that $Q_2(i)$ and $Q_2(i+1)$ are integers, which finally yields that $Q_3(i) \in \mathbb{Z}$. As $Q_3$ is constant, this shows that $Q_3(n) \in \mathbb{Z}$ for all $n \in \mathbb{Z}$.
We show next that $Q_{k+1}(n) \in \mathbb{Z}$ for all $n \in \mathbb{Z}$ implies that $Q_k(n) \in \mathbb{Z}$ for all $n \in \mathbb{Z}$, provided that $Q_k(i) \in \mathbb{Z}$ for at least one $i \in \mathbb{Z}$. For $n \ge i$ this follows by induction from the equation $Q_k(n+1) = Q_{k+1}(n) + Q_k(n)$. For $n \le i$ we use induction and the equality $Q_k(n-1) = Q_k(n) - Q_{k+1}(n-1)$.
Because we have seen that $Q_3(n) \in \mathbb{Z}$ and that $Q_2(i)$, $Q_1(i)$ and $Q_0(i)$ are integers, it follows now that $Q_0(n)$ is an integer for all $n \in \mathbb{Z}$.