, 2015

a) Let $P(x) = c_nx^n + c_{n-1}x^{n-1} + \dots + c_1x + c_0 - 2015$, then $P(\alpha) = 0$. Suppose that when we divide $P(x)$ by $x^2 + x - 5$, we have a quotient $Q(x)$ and the remainder $R(x) = Ax + B$ with integers $A, B$. Then
$$0 = P(\alpha) = (\alpha^2 + \alpha - 5)Q(\alpha) + A\alpha + B = A\alpha + B.$$
Since $\alpha$ is irrational, it follows that $A = B = 0$. So
$$P(x) = Q(x)(x^2 + x - 5).$$
When $x = 1$, we have $P(1) = -3Q(1)$. This implies that
$$c_0 + c_1 + \dots + c_n \equiv 2015 - 3Q(1) \equiv 2 \pmod{3}.$$

b) Suppose that $(c_0, c_1, \dots, c_n)$ is the set of nonnegative integers such that
(1) $c_0 + c_1\alpha + \dots + c_n\alpha^n = 2015$; and
(2) $c_0 + c_1 + \dots + c_n$ is minimal.
We note that $0 \le c_i \le 4$ for all $i = 0, 1, \dots, n-2$, since otherwise, the set $(c_0, \dots, c_{i-1}, c_i-5, c_{i+1}+1, c_{i+2}+1, c_{i+3}, \dots, c_n)$ also satisfies (1) and has a smaller sum, which is a contradiction.
Let $Q(x) = a_{n-2}x^{n-2} + \dots + a_0$. Since $P(x) = Q(x)(x^2 + x - 5)$, we have
$$\begin{aligned} c_0 - 2015 &= -5a_0 \\ c_1 &= -5a_1 + a_0 \\ c_2 &= -5a_2 + a_1 + a_0 \\ c_3 &= -5a_3 + a_2 + a_1 \\\dots \end{aligned}$$
Since $c_i \in \{0, 1, 2, 3, 4\}$, it follows from the first line that $c_0 = 0$ and $a_0 = 403$. From the second line, we have $c_1 = 3, a_1 = 80$. In general, $c_{i+1} = \text{MOD}(a_i + a_{i-1}, 5)$ and $a_{i+1} = \text{DIV}(a_i + a_{i-1}, 5)$. Therefore, we have
$$\begin{aligned} \{a_0, a_1, \dots, a_{11}\} &= \{403, 80, 96, 35, 26, 12, 7, 3, 2, 1, 0, 0\} \\ \{c_0, c_1, \dots, c_{11}\} &= \{0, 3, 3, 1, 1, 1, 3, 4, 0, 0, 3, 1\}. \end{aligned}$$
Hence, the minimum value of $c_0 + \dots + c_n$ is
$$0 + 3 + 3 + 1 + 1 + 1 + 3 + 4 + 0 + 0 + 3 + 1 = 20.$$