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Unformatted text preview: Midterm the Second: Solutions 1a. Find the number of solutions to the congruence x 13 2 mod 53 , if any. Notice that (13 , 52) = 13. Using theorem 2.37, we check: 2 (53 1) / (13 , 53 1) = 2 52 / 13 = 2 4 = 16 6 1 mod 53 and we see that there are no solutions. 1b. Find the number of solutions to the congruence x 20 13 mod 61 , if any. Notice that (20 , 60) = 20. Using theorem 2.37, we check: 13 (61 1) / (20 , 61 1) = 13 60 / 20 = 13 3 = 2197 1 mod 61 and we see there are 20 solutions. 1c. Let p be a prime, and p 2 mod 3 . Show that, for an integer a such that ( a,p ) = 1 , the congruence x 3 a mod p always has a unique solution. Notice that (3 ,p 1) = (3 , 3 k +2 1) = (3 , 3 k +1) = 1. Now, again applying theorem 2.37, we have: a ( p 1) / 1 = a p 1 1 mod p so x 3 a mod p always has a solution, and as (3 ,p 1) = 1, we know that there is only one solution, thus it is unique....
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This note was uploaded on 06/15/2009 for the course MATH 115 taught by Professor Mok during the Fall '07 term at University of California, Berkeley.
 Fall '07
 MOK
 Number Theory, Congruence

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