Math 115B Homework 1sol - 2 x 1 = 10 or-10 Thus x = 58 or x...

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MAT 115B HOMEWORK 1 SOLUTION GK 1.1 Suppose that n = a 2 + b 2 . Since a 2 , b 2 0 , 1 (mod 4), their sum a 2 + b 2 can only be 0 , 1 , 2 (mod 4). GK 1.2 Use the previous problem and the fact that n (mod 4) is determined by the last two digits of n . GK 1.3 First suppose that ax b (mod n ). Then ax = ny + b for some integer y . Let d = gcd( a, n ). Since d | a, d | n , and b = ax - ny , we have d | b , and thus d | gcd( b, n ). Now suppose that gcd( a, n ) | gcd( b, n ). We first consider the special case when gcd( a, n ) = 1, and we would like to show that ax b (mod n ) has a unique solution. This is easy. In general, let d = gcd( a, n ) and d | b by hypothesis. We deduce from ax b (mod n ) that a 0 x b 0 (mod n 0 ), where a 0 = a/d , b 0 = b/d , and n 0 = n/d . This latter equation has a unique solution x (mod n 0 ) since gcd( a 0 , n 0 ) = 1. There are n/n 0 ways to lift x (mod n 0 ) to a residue (mod n ). Hence the original equation has n/n 0 = d solutions. GK 1.4 In Z / 107, 4 x 2 + 4 x + 8 = 0, we have (2 x + 1) 2 = - 7 = 100.
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Unformatted text preview: 2 x + 1 = 10 or-10. Thus, x = 58 or x = 48. GK 1.5 We are trying to solve the congruence x 2 ≡ -1 (mod 5 k ) for k = 1 , 2 , ··· , 6. For k = 1, one solutions is x ≡ 2 (mod 5). Lift this to a solution for k = 2 using Hensel’s lemma. Continue this process five times. We have x = ( ... 431212) 5 or x = ( ... 013233) 5 . GK 1.6 Let t ∈ F be such that t 2 =-1. We claim that 1 + ti ∈ F [ i ] is not invertible, and thus F [ i ] is not a field. In fact, suppose that (1 + ti )( a + bi ) = 1 for some a,b ∈ F . Then ( a-bt ) + ( at + b ) i = 1 and thus a-bt = 1 , at + b = 0 . This gives t = t ( a-bt ) = at-bt 2 = at + b = 0 . This is a contradiction. 1...
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