Let $f(x) = x^3 + ax^2 + bx + c$ and $g(x) = x^3 + bx^2 + cx + a$ , where $a, b, c$ are integers with $c \ne 0$ .
Suppose that the following conditions hold:
$f(1) = 0$
the roots of $g(x)$ are squares of the roots of $f(x)$ .
Find the value of $S=a^{2016}+b^{2016}+ c^{2016}$ .
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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 $S=a^{2016} + b^{2016} + c^{2016} = \boxed{3}$ .