
2. Suppose $n \geq 0$ is an integer and all the roots of $x^{3}+\alpha x+4-\left(2 \times 2016^{n}\right)=0$ are integers. Find all possible values of $\alpha$.


Solution 1: Let $u, v, w$ be the roots of $x^{3}+\alpha x+4-\left(2 \times 2016^{n}\right)=0$. Then $u+v+w=0$ and $u v w=-4+\left(2 \times 2016^{n}\right)$. Therefore we obtain

$$
u v(u+v)=4-\left(2 \times 2016^{n}\right)
$$

Suppose $n \geq 1$. Then we see that $u v(u+v) \equiv 4\left(\bmod 2016^{n}\right)$. Therefore $u v(u+v) \equiv 1$ $(\bmod 3)$ and $u v(u+v) \equiv 1(\bmod 9)$. This implies that $u \equiv 2(\bmod 3)$ and $v \equiv 2(\bmod 3)$. This shows that modulo 9 the pair $(u, v)$ could be any one of the following:

$$
(2,2),(2,5),(2,8),(5,2),(5,5),(5,8),(8,2),(8,5),(8,8)
$$

In each case it is easy to check that $u v(u+v) \not \equiv 4(\bmod 9)$. Hence $n=0$ and $u v(u+v)=2$. It follows that $(u, v)=(1,1),(1,-2)$ or $(-2,1)$. Thus

$$
\alpha=u v+v w+w u=u v-(u+v)^{2}=-3
$$

for every pair $(u, v)$.
