Let $n$ be an odd positive integer and $\varepsilon_0$, $\varepsilon_1$,..., $\varepsilon_{n-1}$ the complex roots of unity of order $n$. Prove that $$\prod_{k=0}^{n-1}(a+b\varepsilon^2_k)=a^n+b^n,$$ for all complex numbers $a$ and $b$.
I believe this problem comes from the 2000 RMO. I'm not too sure where to really start the problem, any hints?
If $b=0,$ the equation holds. Else, let $c=a/b$ and divide by $b^n$ so that we have to show $\prod\limits_{k=0}^{n-1} (c+\epsilon_k^2) = c^n + 1.$ The LHS is simply $\prod\limits_{k=0}^{n-1} (c+\epsilon_k)$ because the $\epsilon_k^2$ form a permutation of the roots of unity as you observed. Now do you see how to finish off with Vieta's Theorem?
Extra hint: What are the roots of $(x-c)^n - 1$?