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 $f(1) = g(1) = 0$

The line
$$b = pq + pr + qr = p^2 + q^2 + r^2$$
is wrong.
$$b = pq + pr + qr =-( p^2 + q^2 + r^2)$$
