2016 is absolutely awesome 15

Note that $g(1) = f(1) = 0$ , so $1$ is a root of both $f(x)$ and $g(x)$ .

Let $p$ and $q$ be the other two roots of $f(x)$ , so $p^2$ and $q^2$ are the other two roots of $g(x)$ .

We then get $pq = -c$ and $p^2q^2 = -a$ , so $a = -c^2$ .

Also, $(-a)^2= (p+q + 1)^2 = p^2 +q^2 + 1 + 2(pq +p+q) = -b+ 2b = b$ .

Therefore $b = c^4$ .

Since $f(1) = 0$ we therefore get $1 + c - c^2 + c^4 = 0$ .

Factorising, we get $(c + 1)(c^3 - c^2 + 1) = 0$ .

Note that $c^3 - c^2 + 1 = 0$ has no integer root and hence $c = -1, b = 1, a = -1$ .

Therefore $S=a^{2016} + b^{2016} + c^{2016} = \boxed{3}$ .

Though Khang Nguyen Thanh's solution is the best,here is another approach which is a little TROLL and objective.

We know that any integer to the power 2016 is always positive.

If any of $a,b,c$ is either $\le-2$ or $\ge2$ then $a^{2016}+b^{2016}+c^{2016}=2^{2016} + something positive$

Since such an huge answer cannot be entered in Brilliant answer box,we now have to check for the values $0,1,-1$ . But since $c\ne0$ , $c^{2016}=1$ .

Hence this rules out $0$ as an answer.Now the answers can only be $1,2,3$ which can be easily done since we get $3$ tries.

Thus when you enter $3$ as your answer,you get CORRECT