# tut08sols - 2 The University of Sydney MATH2068 Number...

This preview shows pages 1–2. Sign up to view the full content.

The University of Sydney MATH2068 Number Theory and Cryptography (http://www.maths.usyd.edu.au/u/UG/IM/MATH2068/) Semester 2, 2008 Lecturer: R. Howlett Tutorial 8 Let n be a positive integer. A complex number ζ is called an n -th root of 1 if ζ n = 1. If in addition ζ k = 1 for all positive integers k less than n then ζ is called a primitive n -th root of 1. 1. By factorizing x n - 1, find all the n th roots of 1 for n = 1, n = 2, n = 3, n = 4 and n = 6. In each case determine which of the n -th roots of 1 are primitive n -th roots of 1. Solution. Obviously the only solution of x 1 - 1 = 0 is x = 1; so 1 is the only 1st root of 1. And there are no positive integers less than 1 satisfying 1 k = 1, because there are no positive integers less than 1 at all; hence 1 is a primitive 1st root of 1. Since x 2 - 1 = ( x - 1)( x + 1) we see that 1 and - 1 are the only 2nd roots of 1. Now ( - 1) k = 1 for all positive integers k < 2, since ( - 1) 1 = 1, and so - 1 is a primitive 2nd root of 1. But 1 1 = 1; so 1 is not a primitive 2nd root of 1. (In general, ζ can only be a primitive n th root of 1 for one value of n , and when ζ = 1 that value is 1, not 2.) Since x 3 - 1 = ( x - 1)( x 2 + x +1) we see that the 3rd roots of 1 are 1 and the two complex solutions of x 2 + x +1 = 0, namely ω 1 = - 1 2 + 3 2 i and ω 2 = - 1 2 - 3 2 i . We know that 1 is not a primitive 3rd root of 1 since it is a 1st root of 1; moreover, ω 1 and ω 2 are not 1st or 2nd roots of 1 since (as we have seen) the only 1st and 2nd roots of 1 are ± 1. So ω 1 and ω 2 are primitive 3rd roots of 1. Since x 4 - 1 = ( x - 1)( x + 1)( x 2 + 1) we see that the 4th roots of 1 are ± 1 and ± i . Since i and - i are not 1st, 2nd or 3rd roots of 1 they are primitive 4th roots of 1. The other two are 2nd roots of 1, hence not primitive 4th roots of 1. Since x 6 - 1 = ( x 3 - 1)( x 3 + 1) = ( x - 1)( x 2 + x + 1)( x + 1)( x 2 - x + 1), the 6th roots of 1 are the roots of the polynomials x - 1, x + 1, x 2 + x + 1 and x 2 - x + 1. The first three of these give, respectively, the primitive 1st, 2nd and 3rd roots of 1, namely 1, - 1, ω 1 , and ω 2 . These are the 6th roots of 1 that are not primitive 6th roots of 1. The two complex roots of x 2 - x = 1 = 0 are ζ 1 = 1 2 + 3 2 i and ζ 2 = 1 2 - 3 2 i , and we know that these are not 1st, 2nd, 3rd or 4th roots of 1 (since we have found all of those and they did not include ζ 1 or ζ 2 ). It is clear that ζ 5 1 = 1, since ζ 5 1 = 1 would give ζ 6 1 = ζ 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern