Question: Let $f(x)=x^3+ax^2+bx+c$ and $g(x)=x^3+bx^2+cx+a$ where $a,b,c\in\mathbb{Z}$, $c\neq 0$. Suppose $f(1)=0$ and roots of $g(x)=0$ are squares of the roots of $f(x)=0$. Find the value of $a^{2020}+b^{2020}+c^{2020}$.  My approach: Let the other roots of $f(x)$ except $1$ be $\alpha$ and $\beta$. This implies that the roots of $g(x)$ are $1,\alpha^2$ and $\beta^2$. Thus, we have $1+\alpha+\beta=-a,$  $\alpha+\beta+\alpha\beta=b,$ $\alpha\beta=-c$ and $1+\alpha^2+\beta^2=-b,$ $\alpha^2+\beta^2+\alpha^2\beta^2=c,$ $\alpha^2\beta^2=-a.$ Note that thus we have $c^2=-a$. Again, $$a^2=(1+\alpha+\beta)^2=1+\alpha^2+\beta^2+2(\alpha+\beta+\alpha\beta)=-b+2b=b.$$ Also, since $f(1)=0\implies a+b+c=-1$. Now since we have $a=-c^2$ and $b=a^2\implies b=c^4,$ which in turn implies that $$-c^2+c^4+c+1=0\implies c^4-c^3+c^3-c^2+c-1=-2\\\implies (c^3+c^2+1)(c-1)=-2.$$ Thus, we either have $$\begin{cases}c^3+c^2+1=1\\ c-1=-2\end{cases} \text{ or }\begin{cases}c^3+c^2+1=-1\\ c-1=2\end{cases} \text{ or }\begin{cases}c^3+c^2+1=2\\ c-1=-1\end{cases} \text{ or }\begin{cases}c^3+c^2+1=-2\\ c-1=1\end{cases}.$$ Note that only the first diophantine equation yields a solution, that is $c=-1$. Thus, $a=-1$ and $b=1$. Therefore, $$a^{2020}+b^{2020}+c^{2020}=1+1+1=3.$$ Is this solution correct and rigorous enough and is there any other way to solve the problem? Also, does anyone know the original source of this problem?
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 $f(1) = g(1) = 0$
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