Find the value of $a^{2020}+b^{2020}+c^{2020}$. 
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?

Another way to reach the solution is this:
Since $a=-c^2$ and $b=c^4$ you have $f(x) =x^3-c^2x^2+c^4x+c$ and since $f(1)=0$ you obtain $1-c^2+c^4+c=0$ i.e. $c^2(c^2-1)+c+1=(c+1)(c^3-c^2+1)=0$, by observing that $c^3-c^2+1=0$ can't have neither odd or even solution, you obtain $c=-1$ and your solution.
