Why is it the case that if $a$ is a primitive root of $x^p=1$, then $\frac{x^p-1}{x-1}=(x-a^2)(x-a^4)...(x-a^{2(p-1)})=1+x+x^2+...+x^{p-1}$?

Question

If $p$ is an odd prime. Why is it the case that if $a$ is a primitive root of $x^p=1$, then $\frac{x^p-1}{x-1}=(x-a^2)(x-a^4)...(x-a^{2(p-1)})=1+x+x^2+...+x^{p-1}$? I can see why $\frac{x^p-1}{x-1}=1+x+x^2+...+x^{p-1}$, but I’m unsure why the even powers of the primitive root are the roots of such things.

Answer 1

Note that as $ a^p= 1$. So in the factorisation after $ p-1/2$ terms you have $ a^{p+1} ,a^{p+3} .... a^{2p-2} $ which are $ a^3,a^5,...a^{p-2} $ and thus the expression on R.H.S turns out to be $$ \prod_{i=1}^{p-1} ( x -a^i) $$ which then gives you the factorisation that comes from fact that it is a primitve root