Hl! I have problems with demostration
Let be a primitive nth root unity
For numbers and , prove that
and if is odd , that
Hint: Set if
I do not know how to start or how to use the hint
$a^n - b^n = b^n \left(\left(\dfrac a b\right)^n - 1\right)$
$x = \dfrac a b,~b \neq 0$
$a^n - b^n = b^n (x^n - 1)$
Consider the polynomial $x^n-1$
It's roots are precisely the $n$ roots of unity and further if $\zeta = e^{2\pi i \frac \theta n}$ then these roots are given by $\zeta^k,~k \in 0,~1,~2,\dots n-1$
thus we can write
$x^n - 1 = \displaystyle \prod_{k=1}^{n-1} \left(x-\zeta^k\right)$
$a^n - b^n = b^n \displaystyle \prod_{k=1}^{n-1} \left(x-\zeta^k\right)$
and you should be able to take it from here.