tut02sols - 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 2 1. For each natural number n , let a n be the residue of 2 n modulo 13. Observe that a k +1 2 a k (mod 13) for each k N , and thus the a k are easy to compute recursively, starting at a 0 = 1 and repeatedly doubling and reducing mod 13 to get successive terms of the sequence. Compute the first 15 or 20 terms, and then compute a 2008 . Solution. 1, 2, 4, 8, 3, 6, 12, 11, 9, 5, 10, 7, 1 – and then it repeats. In other words, a 0 = a 12 = a 24 = ··· , and a 1 = a 13 = a 25 = ··· , etc. In general, a i = a j if and only if i j (mod 12). (In the terminology introduced in lectures, ord 13 (2) = 12.) Now 2004 4 (mod 12); so a 2008 = a 4 = 3. 2. Repeat Question 1 with 13 replaced by 3, 5, 7, 11, 31, 47. Solution. For 3 the sequence is just 1, 2, 1, 2, . . . ; that is, a i = 1 if i is even and a i = 2 if i is odd (and ord 3 (2) = 2). So a 2008 = 1. Using 5 we get 1, 2, 4, 3, 1, . . . . So ord 5 (2) = 4, and a 2008 = a 0 = 1 (since 2008 0 (mod 4)). Using 7 we get 1, 2, 4, 1, . . . . So ord 7 (2) = 3, and a 2008 = a 1 = 2 (since 2008 1 (mod 3)). Using 11 we get 1, 2, 4, 8, 5, 10, 9, 7, 3, 6, 1, . . . . So ord 11 (2) = 10, and a 2008 = a 8 = 3 (since 2008 8 (mod 10)). Using 31 we get 1, 2, 4, 8, 16, 1, . . . . So ord 31 (2) = 5, and a 2008 = a 3 = 8 (since 2008 3 (mod 5)). Using 47 we get 1, 2, 4, 8, 16, 32, 17, 34, 21, 42, 37, 27, 7, 14, 28, 9, 18, 36, 25, 3, 6, 12, 24, 1, . . . . So ord 47 (2) = 23, and a 2008 = a 7 = 34 (since 2008 7 (mod 23)). Observe that in all these case, ord
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tut02sols - 2 The University of Sydney MATH2068 Number...

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