Some problem

Suppose $z$ is a solution to the equation $z^{2014133742} = 1$ . Since $2014133742$ is even, $-z$ also is a solution. Hence the sum of all solutions is $0$ .

EDITED:

We have:

$x^{2014133742}-1=0$

By Vieta's, the sum of the roots is the coefficient of $x$ . Thus the sum of the roots is $\boxed{0}$ .

For any integer n > 1, the sum of the roots of x^n - 1 = 0 is the negative of the coefficient of x^(n-1), which will always be 0 for n > 1.

The large number was intended to intimidate solvers. Don't get intimidated! The sum of the $n^{\text{th}}$ roots is 0.