Root of unity: Is it true that $w_N = w_N^{(N-1)(N-1)}$ and why?

Question

Is it true that for the $N$th root of unity $w_N = w_N^{(N-1)(N-1)}$ and why?

score 6 · Accepted Answer · answered Jun 01 '14 at 14:50

6

Yes, because $(N-1)(N-1)=1\pmod{N}$.

answered Jun 01 '14 at 14:50

Did

Aaahhh, the return of downvotes on arithmetic answers... I wonder what this reminds me. – Did Jun 01 '14 at 16:06

score 4 · Answer 2 · answered Jun 01 '14 at 15:51

4

$$ w_N^{(N-1)(N-1)} = w_N^{N^2-2N+1} = w_N^{N^2-2N}w_N^{1} = (w_N^N)^{N-2}w_N = 1^{N-2}w_N = w_N. $$

answered Jun 01 '14 at 15:51

score 3 · Answer 3 · answered Jun 01 '14 at 15:51

Hint $\ $ If $\ w^N = 1\,$ then exponents on $\,w\,$ may be considered mod $\,N.\,$ More precisely

Lemma $\,\ \color{#c00}{w^N= 1},\,\ A\equiv B\pmod N\ \Rightarrow\ w^A = w^B$

Proof $\,\ A = B + N k\,\Rightarrow\, w^A = w^{B\,+\,Nk} = w^B (\color{#c00}{w^N})^k = w^B \color{#c00}1^k = w^B\ \ $ QED

Thus $\,\ {\rm mod}\ N\!:\,\ (N\!-\!1)^2 \equiv (-1)^2\equiv 1\,\Rightarrow\, w^{(N-1)^2} =\,w^1$

3 Answers3