Solutions - Problem Set 2

Problem Set 2 2Since 100 = 64 + 32 + 4,3 100 ≡≡ 3 64 3 32 3 4 ≡ 81 · 41 · 81 ≡ 1 (mod 100).You could also observe that 3 20 ≡ 3 16 · 3 4 ≡ 21 · 81 ≡ 1 (mod 100).Problem 32, page 58: Show that 2, 4, 6, · · · , 2m is a complete residue system modulo m if m isodd.Solution: It suffices to show that no two of the numbers in the list are congruent modulo m.If two such numbers are congruent, then, without loss of generality, we may assume 2i ≡ 2j(mod m), where 1 ≤ i < j ≤ m. Hence 2(j − i) ≡ 0 (mod m). Since (2, m) = 1 we have that(j − i) ≡ 0 (mod m). This is not possible since 0 < j − i < m.Problem 38, page 58: Prove that there are infinitely primes of the form 4n + 1. (Hint provided)Solution: Suppose, by way of contradiction, that p 1 , p 2 , ·p n is the list of all primes of theform 4n + 1. (Note that there is at least one such prime.) Let N = 4(p 1 p 2 · · · p n ) 2 + 1. If q|N, then(2p 1 p 2 · · · p n ) 2 ≡ −1 (mod q). This means that q = 2 or q ≡ 1 (mod 4), by Theorem 2.12. ButN has been constructed to be relatively prime to 2 and to all prime numbers of the form 4n + 1.This is a contradiction to our assumption that there are a finite number of primes of the form4n + 1.Problem 43, page 58: If p is an odd prime, prove that:1 2 · 3 2 · 5 2 · · · (p − 2) 2 ≡ (−1) (p+1)/2 (mod p)and2 2 · 4 2 · 6 2 · · · (p − 1) 2 ≡ (−1) (p+1)/2 (mod p)Solution:Now,1 2 · 3 2 · 5 2 · · · (p − 2) 2 ≡ 1(p − 1)2(p − 2) · (p − 2)(p − (p − 2)(−1) (p−1)/2 (mod p),Since there are (p − 1)/2 numbers in the list 1, 3, · · · , p − 2. Now1(p − 1)2(p − 2) · (p − 2)(p − (p − 2)(−1) (p−1)/2 ≡ (p − 1)!(−1) (p−1)/2 (mod p).Using Wilson’s Theorem,(p − 1)!(−1) (p−1)/2 ≡ (−1)(−1) (p−1)/2 ≡ (−1) (p+1)/2 (mod p).The second part of the problem is done similarly.Problem 5a, page 63: Find all solutions to the congruence 20x ≡ 4 (mod 30),Solution: We want to find x and y such that 20x+30y = 4. Since 10 = (20, 30), the existenceof such an x and y would imply that 10|4, we have a contradiction. Hence no solutions exist.Problem 5c, page 63: Find all solutions to the congruence 353x ≡ 254 (mod 400),