Elementary Number Theory Section 2.2 Solutions of Congruences
Definition 2.2.1 (a) Let r1, r2, . . . , rm denote a complete residue system modulo m. The number of solutions of f (x) ≡ 0(mod m) is the number of the ri such that f (ri) ≡ 0(mod m).
(b) Let f (x) = anxn + an−1xn−1 +· · · + a0 ∈ Z[x]. Let j be the largest integer such that aj 6≡ 0(mod m), then the degree of the congruence is j. If there is no such integer j, no degree is assigned to the congruence.
Lemma 2.2.2 If d|m, d > 0, and if u is a solution of f(x) ≡ 0(mod m), then u is a solution of f (x)≡ 0(mod d).
Proposition 2.2.3 Let a, b a and m > 0 be integers, and g = (a, m). The congruence ax ≡ b(mod m) has a solution if and only if g|b.
If this condition met, then the solution form an arithmetic progression with common difference mg , given g solutions (mod m).
Theorem 2.2.4 (The Chinese Remainder Theorem) Let m1, m2, . . . , mr denote r positive integers that relatively prime in pairs, and let a1, a2, . . . , ar denote any r integers. Then the congruence
x≡ a1(mod m)1, x≡ a2(mod m2),
· · · x≡ ar(mod mr),
has a common solution. If x0 is one such solution, then an integer x satisfies the congruence if and only if x = x0 + km1m2· · · mr for some k ∈ Z.
Example 2.2.5 Show that there is no x for which both x ≡ 29(mod 52) and x≡ 19(mod 72)
Example 2.2.6 Determine whether the system x≡ 3(mod 10), x ≡ 8(mod 15), x≡ 5(mod 84) has a solution, and find them all, if any exist.
Example 2.2.7 Exhibit the forgoing one-to-one correspondence explicitly, when m1 = 7, m2 = 9, m = 63.
Theorem 2.2.8 If m1, m2 ∈ N be relatively prime integers, then φ(m1)φ(m2) = φ(m1m2).
1
2
If m has the canonical factorization m = ∏
pα, then φ(m) = ∏
p|m(pα− pα−1) = m∏
p|m(1− 1p).
Theorem 2.2.9 Let f (x) ∈ Z[x] and for any m ∈ N, let N(m) denote the number of solutions of the congruence f (x) ≡ 0(mod m). If m = ∏
pα is the canonical factorization, then N (m) =∏
N (pα).
Example 2.2.10 Let f (x) = x2 + x + 27. Find all roots of the congruence f (x)≡ 0(mod 15).
Example 2.2.11 Let f (x) = x2 + x + 27. Find all roots of the congruence f (x)≡ 0(mod 189).
Remark 2.2.12 f (a + tpj)≡ f(a) + pjf0(a)(mod pj+1), j ≥ 1.
Theorem 2.2.13 (Hensel’s Lemma) Let f (x) ∈ Z[x]. If f(a) ≡ 0(mod pj) and f0(a) 6≡ 0(mod p), then there exists a unique t(mod p) such that f(a + tpj) ≡ 0(mod pj+1).
Definition 2.2.14 (a) Let j < k, f (a) ≡ 0(mod pj), f (b) ≡ 0(mod pk), and a≡ b(mod pj). We say that b lies above a, or a lifts to b.
(b) If f (a) ≡ 0(mod pj), then the root a is called nonsingular if f0(a) 6≡
0(mod pj).
Remark 2.2.15 If a1 is a nonsingular root of f (a) ≡ 0(mod p), then a1 can be lift to a unique roots a2(mod p2) by Hensel Lemma, and a2 can be lift to a unique roots a3(mod p3) by Hensel Lemma again. In general, a1 lifts to a unique root aj(mod pj) for all j.
Example 2.2.16 (a) Solve x2+ x + 47≡ 0(mod 73).
(b) Solve x2 + x + 7≡ 0(mod 81).
Theorem 2.2.17 Let F (x) ∈ Z[x]. Suppose that f(a) ≡ 0(mod pj), that pτkf0(a), and that j ≥ 2τ + 1. If b ≡ a(mod pj−τ), then f (b) ≡ f(a)(mod pj) and pτkf0(b).
Moreover, there is a unique t(mod p) such that f (a+tpj−τ) ≡ 0(mod pj+1).
Example 2.2.18 Discuss the solution of x2+ x + 223 ≡ 0(mod 3j).
Definition 2.2.19 Let f (x)∈ C[x] be a polynomial of degree n with leading co- efficient anand roots r1, . . . , rn. The discriminant of f is D(f ) = a2nn −2∏
1≤i<j≤n(ri− rj)2.
3
Lemma 2.2.20 Let f (z) = anzn+ an−1zn−1+ cdots + a1z + a0 be a polynomial of degree n, There is a homogeneous polynomial F ∈ Z[w0, w1, . . . , wn] of degree 2n− 2 such that D(f) = F (a0, . . . , an).
Moreover, if f (z) = (z− r)g(z), then D(f) = D(g)g(r)2.
Theorem 2.2.21 Let f (x)∈ Z[x] and suppose that pδkDf. If f(a) ≡ 0(mod pj), pτkf0(a), and j > δ, then j ≥ 2τ + 1.
Theorem 2.2.22 If the degree of f (x)≡ 0(mod p) is greater than or equal to p, then either every integer is a solution or there is a monic polynomial g(x)∈ Z[x]
such that g(x) ≡ 0(mod p) is of degree less p and the solution of g(x) ≡ 0(mod p) are precisely those of f (x)≡ 0(mod p).
Theorem 2.2.23 The congruence f (x) ≡ 0(mod p) of degree n has at most n solutions.
Corollary 2.2.24 If bnxn + bn−1xn−1+· · · + b0 ≡ 0(mod p) has more than n solutions, then all the coefficients bj are divisible by p.
Theorem 2.2.25 If F (x) is a function that maps residue classes (mod p) to residue classes (mod p), then there is a polynomial f (x) ∈ Z[x] and degree
≤ p − 1 such that F (x) ≡ f(x)(mod p) for all residue classes x(mod p).
Theorem 2.2.26 The congruence f (x) ≡ 0(mod p) of degree n, with leading coefficient an = 1, has n solutions if and only if f (x) is a factor of xp−x modulo p.
Corollary 2.2.27 If d|(p − 1), then xd ≡ 1(mod p) has d solutions.
Corollary 2.2.28 (a) (Wilson) (p− 1)! ≡ −1(mod p)
(b) (Wolstenholme’s congruence) Let (x− 1)(x − 2) · · · (x − p + 1) = xp−1− σ1xp−2 + σ2xp−3− · · · + σp−1. Then σp−2 ≡ 0(mod p2) if p≥ 5.
