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

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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 6 = 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 6 = 1 for all positive integers k < 2, since ( - 1) 1 6 = 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
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This note was uploaded on 09/12/2009 for the course MATH 2068 taught by Professor Howlett during the One '08 term at University of Sydney.

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

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