Thus,Originally Posted by lalitchugh
Express in polar form thus,
Because given,Originally Posted by malaygoel
We can then show that if,
Furthermore, all elements are distinct.
Thus, by the pigeonhole principle we have exactly elements in this set. But since a polynomial can have a most solution we see that all of the solution of, are found in this set.
I like to think this is as elegant as it gets.
Yes you can take other values for , but I like to follow this reasoning. Because, I am certain that all the solutions are found by starting at k=0 and going to k=n-1.