Find $a^{2013} + b^{2013} + c^{2013}$ Problem Statement
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 $c\not=0$ 
Suppose that the following conditions hold:


*

*$f(1)=0$

*the roots of $g(x)=0$ are the squares of the roots of $f(x)=0$

$$\text{Find the value of} \: \: a^{2013} + b^{2013} + c^{2013}$$

My attempt
Let $f(x) = (x-p)(x-q)(x-r) \: \:$ and $g(x) = (x - p^2)(x-q^2)(x-r^2)$
Expanding we get $$f(x) = x^3 - (p + q + r)x^2 + (pq + pr + qr)x - pqr$$
$$g(x) = x^3 - (p^2 + q^2 + r^2)x^2 + (p^2q^2 + p^2r^2 + r^2q^2)x - p^2q^2r^2$$
And so we have $$a = -(p+q+r) = -p^2q^2r^2$$
$$b = pq + pr + qr = p^2 + q^2 + r^2$$
$$c = -pqr = p^2q^2 + p^2r^2 + r^2q^2$$
Then we get the following relations
$$a = -c^2$$ $$b = a^2 - 2b \implies 3b = a^2 \implies b = \frac{c^4}{3}$$
From $f(1) = 0$ we have $a + b + c = -1 \implies \dfrac{c^4}{3} + c - c^2 = -1$
However, upon graphing this function, I saw that there are no integer solutions for $c$. Have I done something wrong?
Edit
For future readers, my approach can simplified alot if you replace $r$ with $1$ since $f(1) = g(1) = 0$

If the roots of $f$ are $1,u,v$, then the roots of $g$ are $1, u^2,v^2$.
Thus we obtain by Vieta
$$ \begin{align}-1-u-v&=a=-u^2v^2\\
 u+v+uv&=b=-1-u^2-v^2\\
 -uv&=c=u^2+v^2+u^2v^2\end{align}$$
In particular, 
$$ \tag1a+b+c = (-1-u-v)+(u+v+uv)-(-uv)=-1$$
and
$$\tag2 a^2=1+u^2+v^2+2u+2v+2uv=-b+2b=b$$
and $$ \tag3c^2=u^2v^2=-a.$$
By eliminating $a$ and $b$ from $(1)$, we find
$$\tag4 -c^2+c^4+c=-1$$
which makes $c$ a divisor of $-1$, so $c=\pm1$. But only $c=-1$ is a solution of $(4)$. We conclude $c=-1$, $a=-c^2=-1$, and $b=a^2=1$.
Thus the cubic equations are
$$x^3 -x^2+x-1=0$$
(with roots $1,i,-i$) and 
$$x^3 +x^2-x-1=0$$
(with $1$ and twice $-1$ as root).
