Answer (E): Let $Q(m) = 8P(m) = 4m + 2m^2 + m^4 + m^8$. Because the coefficients of $Q$ are integers, it follows that if $a \equiv b \pmod{8}$, then $Q(a) \equiv Q(b) \pmod{8}$. It suffices to show that the 8 numbers $Q(-3)$, $Q(-2)$, $Q(-1)$, ..., $Q(4)$ are all divisible by 8. If $m$ is even, then each of the monomials of $Q(m)$ is divisible by 8. If $m = \pm 1$, then $Q(m) = \pm 4 + 4 \equiv 0 \pmod{8}$. If $m = \pm 3$, then $m^2 = 9 \equiv 1 \pmod{8}$, which implies that $m^4 \equiv 1 \pmod{8}$, and so also that $m^8 \equiv 1 \pmod{8}$. Hence $Q(\pm 3) \equiv \pm 12 + 2 + 1 + 1 \equiv 0 \pmod{8}$.

Therefore $8P(m)$ is divisible by 8 for all integers $m$, which implies that $P(m)$ is an integer for all $m$. In particular, all 4 of the given values of $P(m)$ are integers.

Let $Q(m)$ be defined as in the first solution, and note that $Q(m)$ is divisible by 8 if $m$ is even. To treat odd $m$, write
$$\begin{align*} Q(m) &= 8m + 4(m^2 - m) + 2m^2(m^2 - 1) + m^4(m^4 - 1) \\ &= 8m + 4m(m - 1) + 2m^2(m + 1)(m - 1) + m^4(m^2 + 1)(m + 1)(m - 1) \end{align*}$$
and note that because $m+1, m-1$, and $m^2+1$ are all even, each term has at least three factors of 2. The solution concludes as above.

Another way to see that $Q(m)$ is divisible by 8 when $m$ is odd in the second solution is to apply more general facts from number theory. Fermat's Little Theorem asserts that if $p$ is prime, then $a^p \equiv a \pmod p$ for all integers $a$. In particular, $m^2 \equiv m \pmod 2$. Euler's Totient Theorem asserts that if $\gcd(a, q) = 1$, then $a^{\phi(q)} \equiv 1 \pmod{q}$, where $\phi(q)$ is the number of positive integers less than $q$ that are relatively prime to $q$. Because $\phi(4) = 2$, it follows that $m^2 \equiv 1 \pmod{4}$ when $m$ is odd. Also, $\phi(8) = 4$, so $m^4 \equiv 1 \pmod{8}$ if $m$ is odd.