caledonia Posted September 28, 2015 Share Posted September 28, 2015 I seek a proof that xn -1 = 0 has a primitive root. Link to comment Share on other sites More sharing options...
mathematic Posted September 29, 2015 Share Posted September 29, 2015 (edited) 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, 2015 by mathematic Link to comment Share on other sites More sharing options...
Xerxes Posted September 29, 2015 Share Posted September 29, 2015 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! Link to comment Share on other sites More sharing options...
caledonia Posted September 30, 2015 Author Share Posted September 30, 2015 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). Link to comment Share on other sites More sharing options...
mathematic Posted September 30, 2015 Share Posted September 30, 2015 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. Link to comment Share on other sites More sharing options...
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