By I.G. Macdonald

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1, 2, 5, . 6. 1 Congruences Modulo n 23 We use the notation Z/nZ because Z/nZ is the quotient of the ring Z by the “ideal” nZ of multiples of n. Because Z/nZ is the quotient of a ring by an ideal, the ring structure on Z induces a ring structure on Z/nZ. We often let a or a (mod n) denote the equivalence class a + nZ of a. 7 (Field). A field K is a ring such that for every nonzero element a ∈ K there is an element b ∈ K such that ab = 1. 12). 8 (Reduction Map and Lift). We call the natural reduction map Z → Z/nZ, which sends a to a + nZ, reduction modulo n.

B) The prime number theorem implies π(x) is asymptotic to How close is π(y) to y/ log(y), where y is as in (a)? x log(x) . 8 Let a, b, c, n be integers. Prove that (a) if a | n and b | n with gcd(a, b) = 1, then ab | n. (b) if a | bc and gcd(a, b) = 1, then a | c. 9 Let a, b, c, d, and m be integers. Prove that (a) if a | b and b | c then a | c. (b) if a | b and c | d then ac | bd. (c) if m = 0, then a | b if and only if ma | mb. (d) if d | a and a = 0, then |d| ≤ |a|. 10 In each of the following, apply the division algorithm to find q and r such that a = bq + r and 0 ≤ r < |b|: a = 300, b = 17, a = 729, b = 31, a = 300, b = −17, a = 389, b = 4.

When n = 5, R = {0, 1, −1, 2, −2} is a complete set of residues. 12. If R is a complete set of residues modulo n and a ∈ Z with gcd(a, n) = 1, then aR = {ax : x ∈ R} is also a complete set of residues modulo n. Proof. 10 implies that x ≡ x (mod n). Because R is a complete set of residues, this implies that x = x . Thus the elements of aR have distinct reductions modulo n. It follows, since #aR = n, that aR is a complete set of residues modulo n. 13 (Units). If gcd(a, n) = 1, then the equation ax ≡ b (mod n) has a solution, and that solution is unique modulo n.