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Unformatted text preview: 18.781, Fall 2007 Problem Set 6 Solutions to Selected Problems Problem 3.2.6 First note that 1009 is a prime number. We need to decide the value of ( 150 1009 ) . We can find that 150 1009 = 2 1009 3 1009 25 1009 . Because 1009 ≡ 1 (mod 8), we have ( 2 1009 ) = 1. By the theorem 3.5, with the fact that 1009 = 3 · 336 + 1, ( 3 1009 ) = ( 1 3 ) = 1. Since 25 is a square number, ( 25 1009 ) = 1. In conclusion, we have ( 150 1009 ) = 1. Therefore, the given equation is solvable. (Actually, 139 2 ≡ 150 (mod 1009).) 2 Problem 3.2.7 First, it is easily observed that x 2 ≡ 13 (mod p ) has a solution when p is 2 or 13. Now assume that p is neither 2 nor 13. Then p is an odd prime, and we have x 2 ≡ 13 (mod p ) has a solution. ⇔ 13 p = 1. ⇔ ( p 13 ) ( 1) 13 1 2 p 1 2 = ( p 13 ) = 1. By a little computation, we can easily verify that the quadric residues of 13 are { 1 , 3 , 4 , 9 , 10 , 12 } ....
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This note was uploaded on 04/28/2009 for the course MATH 18.781 taught by Professor Brubaker during the Spring '09 term at MIT.
 Spring '09
 BRUBAKER
 Number Theory

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