<s>[INST] 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?
 [/INST] 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.
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