MATH10111 - SOLUTIONS TO EXERCISES 8. 8.1 (i), (ii), (iv) and (v) are true, (iii)
and (vi) are false. 8.2 (i) 9; (ii) 9; (iii) 0; (iv) 12; (v) 4; (vi) 1. 8.3 If a is even, then ...
MATH10111 - SOLUTIONS TO EXERCISES 8
8.1 (i), (ii), (iv) and (v) are true, (iii) and (vi) are false. 8.2 (i) 9; (ii) 9; (iii) 0; (iv) 12; (v) 4; (vi) 1. 8.3 If a is even, then a = 2k for some k ∈ Z. Hence a2 = 4k 2 ≡ 0 mod 4. If a is odd, then a = 2k + 1 for some k ∈ Z. So a2 = (2k + 1)2 = 4k 2 + 4k + 1 = 4k(k + 1) + 1 ≡ 1 mod 4. Let b, c ∈ Z and suppose b2 ≡ e mod 4 and c2 ≡ f mod 4. Then b2 + c2 ≡ e + f mod 4. Hence b2 + c2 ≡ 0, 1 or 2 mod 4. So as a ≡ 3 mod 4, this means that x2 + y 2 = a cannot have an integer solution. 8.4 Since a is odd, a = 2k + 1 for some k ∈ Z. So a2 = (2k + 1)2 = 4k 2 + 4k + 1 = 4k(k + 1) + 1. Since one of k and k + 1 is even, 2|k(k + 1). Hence 8|4k(k + 1), and so a2 ≡ 1 mod 8. 8.5(i) 23x ≡ 16 mod 107. 107 = 4.23 + 15 23 = 1.15 + 8 15 = 1.8 + 7 8 = 1.7 + 1 7 = 7.1 So gcd(107, 23) = 1. So there is a unique solution modulo 107. Now 1 = 8 − 17 = 8 − (15 − 1.8) = 2.8 − 1.15 = 2(23 − 15) − 15 = 2.23 − 3.15 = 2.23 − 3(107 − 4.23) = −3.107 + 14.23 Hence 16 = 16.(−3).107 + 16.14.23, so 16.14 = 224 ≡ 10 mod 107 is the unique solution modulo 107 of 23x ≡ 16 mod 107. The solution set is {10 + 107k : k ∈ Z}. (ii) gcd(366, 234) = 6. Since 6 does not divide 20, there are no solutions. (iii) gcd(366, 234) = 6, so there will be 6 solutions modulo 366. 6 = 16.366 − 25.234, so 1 = 16.61 − 25.39. Hence x = −25 is a solution to 39x ≡ 1 mod 61. Note that −25 ≡ 36 mod 61. So the six solutions modulo 366 of 234x ≡ 6 mod 366 are 36+61w, where w = 0, 1, 2, 3, 4, 5, i.e., 36, 97, 158, 219, 280, 341. 1
