You can easily prove that \(n = 0\) by working mod 7.

Already posted: see here.

Ankoganit wrote: Already posted: see here. I know, but I re-posted without any errors. I need to add it to contest collection.

Solution : This post: http://www.artofproblemsolving.com/community/c260h1368546p7531805

PSY-Math wrote: Suppose $n \ge 0$ is an integer and all the roots of $x^3 + \alpha x + 4 - ( 2 \times 2016^n) = 0$ are integers. Find all possible values of $\alpha$. $n=0$ yields $\alpha=-3$. Now assume $n \geq 1$. Let the roots be $a,b,c$. Then $$a+b+c=0,abc=2(2016^{n})-4$$Hence $a^{3}+b^{3}+c^{3}=3abc=3\left(2(2016^{n})-4\right) \equiv 3(2(1)-4) \equiv 4 \pmod{5}$. Also, $a+b+c \equiv 0 \pmod{5}$. Hence we can easily check from the second condition that the only possibilities are $(a,b,c) \equiv (1,4,0),(2,3,0) \pmod{5}$ (and its permutations). But it is easy to see that neither of these two triples satisfy $a^{3}+b^{3}+c^{3} \equiv 4 \pmod {5}$. Thus, $\alpha=-3$ is the only possibility.

Wizard_32 wrote: PSY-Math wrote: Suppose $n \ge 0$ is an integer and all the roots of $x^3 + \alpha x + 4 - ( 2 \times 2016^n) = 0$ are integers. Find all possible values of $\alpha$. $n=0$ yields $\alpha=-3$. Now assume $n \geq 1$. Let the roots be $a,b,c$. Then $$a+b+c=0,abc=2(2016^{n})-4$$Hence $a^{3}+b^{3}+c^{3}=3abc=3\left(2(2016^{n})-4\right) \equiv 3(2(1)-4) \equiv 4 \pmod{5}$. Also, $a+b+c \equiv 0 \pmod{5}$. Hence we can easily check from the second condition that the only possibilities are $(a,b,c) \equiv (1,4,0),(2,3,0) \pmod{5}$ (and its permutations). But it is easy to see that neither of these two triples satisfy $a^{3}+b^{3}+c^{3} \equiv 4 \pmod {5}$. Thus, $\alpha=-3$ is the only possibility. What about $(a,b,c)\equiv (1,1,3)\pmod 5$?

TLP.39 wrote: What about $(a,b,c)\equiv (1,1,3)\pmod 5$? Oops! Missed that case! My bad So this problem boils down to mod 7 then, 5 would not work.

I worked mod 2 4 8 and got the solution

Here's my approach: Let $P(x) =^{Def} x^3 + \alpha x + 4 - 2\times 2016^n$. Denote by $a,b,c$ the roots of $P(x)$. By definition, $a,b,c \in \mathbb{Z}$. Thus, by Vieta , we have: \begin{align*} \sum a &= 0\\ \sum ab &= \alpha\\ abc &= 2\cdot 2016^n - 4 \end{align*}We consider $P(x)$ in$\pmod 7$. $\implies x^3+\alpha x +4 - 2\times 2016^n \equiv x^3 + \alpha x + 4 \pmod 7$ Note that the cubic residues modulo $7$ are $ 0,1,-1$.It is clear that $7$ does not divide $a,b,c$, otherwise, we have that $7 \mid 4$. Now since $ a+b+c = 0 \implies \sum a^3 = 3abc$ So $\sum a^3 = 3abc = 2 \pmod 7$ which is not possible. Thus $n=0$. So we have $a+b+c = 0$ and $ abc=-2$. From here it is easy to find that $\boxed{ (a,b,c)_{perm} = (1,1,-2)}$

$\color{red}\textbf{Claim:-}$ The only possible value of $\alpha$ is $-3.$ $\color{blue}\textbf{Proof:-}$ First thing we have to prove that there is no solution for $n \ge 1.$ Now let's assume the roots of $x^3+\alpha x+4-(2 \times 2016^n)=0$ are $a,b,c.$ By Vieta's Relation we get, $$a+b+c=0\implies a^3+b^3+c^3=3abc\equiv 3\times 3\equiv 2(\text{mod 7})$$$$ab+bc+ca=\alpha$$$$abc=(2\times 2016^n)-4\equiv -4\equiv 3(\text{mod 7})$$The possible remainder when any number is divided by $7$ are $$N\equiv 0,1,2,3,4,5,6 (\text{mod 7})$$Cubing we get, $$N^3\equiv 0,-1,1(\text{mod 7})\implies \boxed{a^3+b^3+c^3=0,-1,1(\text{mod 7}).}\implies \text{contradiction.}$$Now Let's assume $n=0$ we get, $$a+b+c=0 \hspace{2mm} \text{and} \hspace{2mm} abc=-2\implies \boxed{(a,b,c)=(1,1,-2).}$$Therefore the possible value of $\alpha$ is $$ab+bc+ca=\alpha\implies \boxed{\alpha=-3.}$$

Modulo 9 is also good after you got a³+b³+c³=3abc