tut11sols - 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 11 1. Given that 1081 = 23 × 47, find four solutions of x 2 2 ( mod 1081 ) with 1 x 1080. ( Hint: First solve x 2 1 2 ( mod 23 ) and x 2 2 2 ( mod 47 ) ( e.g. by evaluating 1 2 , 2 2 , 3 2 , etc. ) . By the Chinese Remainder Theorem there must exist an x such that x x 1 ( mod 23 ) and x x 2 ( mod 47 ) . ) Solution. Clearly ± 5 are two solutions of x 2 2 ( mod 23 ) and ± 7 are two solutions of x 2 2 ( mod 47 ) . So we seek x ’s in the range 1 x 1080 with x ≡ ± 5 ( mod 23 ) and x ≡ ± 7 ( mod 47 ) . Assume first that x 5 ( mod 23 ) , so that x = 23 k + 5 for some k . Then if x 7 ( mod 47 ) we deduce that 23 k 2 ( mod 47 ) . To solve this we need to find the inverse of 23 modulo 47, which can be done by use of the extended Euclidean Algorithm. In this case, though, the answer is almost obvious since 2 × 23 ≡ - 1 ( mod 47 ) : the inverse of 23 is - 2. So 23 k 2 gives k ≡ - 4 43 ( mod 47 ) . This gives x = 23 × 43 + 5 = 994 as one solution of x 2 2 ( mod 1081 ) . Returning to x = 23 k + 5, suppose now that x ≡ - 7 ( mod 47 ) . Then 23 k ≡ - 12 ( mod 47 ) , whence 46 k ≡ - 24, and k 24 ( mod 47 ) . So x = 23 × 24 + 5 = 557 is our next solution. Now assume that x ≡ - 5 ( mod 23 ) , so that x = 23 k - 5 for some k . If x 7 ( mod 47 ) we obtain 23 k 12 ( mod 47 ) , and k ≡ - 24 23 ( mod 47 ) . So x = 23 2 - 5 = 524 is our third solution. Finally, suppose that x = 23 k - 5 and x ≡ - 7 ( mod 47 ) . Then 23 k ≡ - 2, giving x 4 ( mod 47 ) . So our fourth solution is x = 23 × 4 - 5 = 87. We could have saved ourselves some work by realizing that two of the solutions are the negatives ( mod 1081 ) of the other two. 2. It was proved in lectures that if p is prime and d | ( p - 1 ) then x d 1 mod p has exactly d solutions mod p . Given that 2 is a primitive root mod 101, find the 10 solutions of x 10 1 ( mod 101 ) . ( Hint: Consider values of d for which 2 10 d 1 ( mod 101 ) . ) Solution. The 10 solutions are 2
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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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tut11sols - 2 The University of Sydney MATH2068 Number...

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