I seek a proof that xⁿ -1 = 0 has a primitive root.

I don't know what you are looking for. All the roots are trivially expressible, and the proof is that they satisfy the equation.

 

[latex]x_k=e^{\frac{2\pi ik}{n}},k=[0,...,n-1][/latex]

Edited September 29, 201510 yr by mathematic

Try induction on [math]x^2-1=0[/math] which has roots [math]\sqrt{1}[/math]. Clearly [math]-1[/math] is primitive since it is not a root for [math]x^1-1=0[/math]

 

i.e. assume for your induction hypothesis that the roots for [math]x^{n-1}-1=0[/math] are [math]^{n-1}\sqrt{1}[/math]

 

BUT beware of multiplicities!

Thanks for responses. I was thinking about a proof which did not require the explicit identification of a root, but I accept that

cos 2pi/n + i sin 2pi/n is indeed a primitive root (of xⁿ = 1).

Looking at my original post (exponential form), it is easy to see that for every n, the roots for k=1 and k=n-1 are primitive. In general roots where k/n is in lowest terms (as a fraction), kth root is primitve.