System of equations involving complex numbers I'm getting confused to figure this work out. The only thing that came into my head was using AM-GM inequality, but i just get stuck. Here's the problem: 
Let $a,b,c,$ be the complex numbers such that $abc=1$. and
\begin{cases}
a^{20}+b^{20} + c^{20} &= \frac{1}{a^{20}} + \frac{1}{b^{20}} + \frac{1}{c^{20}} \\
a^{17}+b^{17}+c^{17} &= \frac{1}{a^{17}} + \frac{1}{b^{17}} + \frac{1}{c^{17}}\\
a^{2017} + b^{2017}+c^{2017} &= \frac{1}{a^{2017}} + \frac{1}{b^{2017}} + \frac{1}{c^{2017}}\\
\end{cases}
show that $1 \in \{a,b,c\}$.
Please help me to solve this question, any thought would be helpful.
 A: Actually,
Lemma 1: if $abc=1$, and if $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, then one of $a,b,c$ is 1.
Proof: multiply by $abc=1$:
$$a+b+c=ab+bc+ac\Leftrightarrow abc-ab-ac-bc+a+b+c-1=0\Leftrightarrow (a-1)(b-1)(c-1)=0$$
Lemma 2: If $\gcd(u,v)=1$, and $a^u=a^v=1$, then $a=1$.
Proof: $1=mu+nv$ for some $m,n\in\mathbb Z$, and so $a=a^{mu+nv}=(a^u)^m(a^v)^n=1$.
We can now apply Lemma 1 to the triplets $(a^{17}, b^{17}, c^{17})$, $(a^{20}, b^{20}, c^{20})$ and $(a^{2017}, b^{2017}, c^{2017})$. Note also that $17, 20, 2017$ are pairwise coprime. We have two cases:


*

*Either the same variable ($a$, $b$ or $c$) raised to two different exponents is 1. WLOG, let's assume $a^{17}=1$ and $a^{20}=1$. Then $a=1$ by Lemma 2. The same proof goes if the exponents are $20,2017$ or $17,2017$.

*Or, it is all three variables. WLOG, assume $a^{17}=b^{20}=c^{2017}=1$. Thus $a^{17\cdot 20}b^{17\cdot 20}=1$ so $c^{17\cdot 20}=1$, but because also $c^{2017}=1$ and $\gcd(17\cdot 20, 2017)=1$ we then have $c=1$ by Lemma 2. 
A: Hint. If $abc=1$ then, for any integer $n$, the equation
$$a^{n}+b^{n} + c^{n}= \frac{1}{a^{n}} + \frac{1}{b^{n}} + \frac{1}{c^{n}}$$
is equivalent to
$$(1-a^n)(1-b^n)(1-c^n)=0.$$
Moreover note that $\gcd(20,17)=1$, $\gcd(2017,17)=1$, and $\gcd(2017,20)=1$.
