数论 初步
0x01 整除
概念： 设 \(a, b \in \mathbb Z\) 且 \(a \neq 0\)，如果 \(\exists q \in \mathbb Z\)，使得 \(a \times q = b\)，则 \(b\) 能被 \(a\) 整除，记为 \(a \mid b\)。
性质：
1. 传递性：如果 \(a \mid b\) 且 \(b \mid c\)，则 \(a \mid c\)。
2. 若\(a \mid b\) 且 \(a \mid c\) 则对于 \(\forall a, b \in \mathbb Z\)，有 \(a \mid (bx+cy)\)。
3. 设 \(m\) 不为 \(0\)，则 \(a \mid b\) 等价于 \(ma \mid mb\)。
4. 设 \(x,y \in \mathbb Z\) 满足下式：\(ax+by=1\)，且 \(a \mid n\)，\(b \mid n\)，那么 \(ab \mid n\)
5. 若 \(\exists b, q, d, c \in \mathbb Z\) 且 \(b = q \times d + c\)，那么 \(d \mid b\) ，\(d \mid c\) 可以互推。
证明：
1. 记 \(b = ak_a(k_a \in \mathbb Z)\)，且 \(c = bk_b(k_b \in \mathbb Z)\)
则有 \(c = (ak_a)k_b\)
即 \(c = k_ak_ba\)
其中 \(k_a, k_b \in \mathbb Z\)
\(∴ a \mid c\)
2. 记 \(b = ak_b(k_b \in \mathbb Z)\)，且 \(c = ak_c(k_c \in \mathbb Z)\)
\(∴ bx + cy = (ak_b)x + (ak_c)y\)
即 \(bx + cy = a(k_bx + k_cy)\)
其中 \(k_b, k_c, x, y \in \mathbb Z\)
\(∴ (k_bx + k_cy) \in \mathbb Z\)
故 \(a \mid (bx+cy)\)
3. 记 \(b = ak_b(k_b \in \mathbb Z)\)
\(∴ mb = mak_b (k_b \in \mathbb Z)\)
且 \(m \in \mathbb Z, m \neq 0\)
\(∴ ma \mid mb\)
4. \(∵ ax + by = 1\)
\(∴ \frac x b + \frac y a = \frac 1 {ab}\)
\(∴ \frac n {ab} = \frac {xn} b + \frac {ny} a\)
又由题： \(ak_a = n, bk_b = n(k_a, k_b \in \mathbb Z)\)
\(∴ \frac n {ab} = k_ay + k_bx\)
即 \(n = ab(k_ay + k_bx)\)
\(∵ k_a, k_b, x, y \in \mathbb Z\)
\(∴ ab \mid n\)
5.1
记 \(b = d * k_b(k_b \in \mathbb Z)\)
\(∴ dk_b = qd + c\)
即 \(d(k_b - q) = c\)
\(∵ k_b, q \in \mathbb Z\)
\(∴ (k_b - q) \in \mathbb Z\)
故 \(d \mid c\)
5.2 记 \(c = d * k_c(k_c \in \mathbb Z)\)
\(∴ b = qd + qk_c\)
即 \(d(k_c + q) = b\)
\(∵ k_c, q \in \mathbb Z\)
\(∴ (k_c + q) \in \mathbb Z\)
故 \(d \mid b\)
0x02 mod
概念： 对于\(a,b \in \mathbb Z, b \neq 0\)，求 \(a\) 除以 \(b\) 的余数，称为 \(a\) 模 \(b\)，记为 \(a \bmod b\)。
性质： 暂且研究 \(a > 0, b \geq 0\)
0. 杂论：\((a + kb) \bmod b = a \bmod b\)
1. 分配率：模运算对加、减、乘具有分配率
1.1 \((a + b) \bmod c = (a \bmod c + b \bmod c) \bmod c\)
1.2 \((a - b) \bmod c = (a \bmod c - b \bmod c) \bmod c\)
1.3 \((a \times b) \bmod c = [(a \bmod c) \times (b \bmod c)] \bmod c\)
1.4 \((a^b) \bmod c = (a \bmod c)^b \bmod c\)
2. 放缩性:
2.1 如果 \(a \bmod b = c, d \neq 0\)，则有 \((a \times d) \bmod (b \times d) = c \times d\)
2.2 如果 \(a \bmod b = c, d \neq 0\)，则有 \(\frac a d \bmod \frac b d = \frac c d\)
证明：
0.1 \(a < b\)
记 \(x = a + kb\)
根据 \(\bmod\) 的定义，\(x \bmod b = a\)
所以 \((a + kb) \bmod b = a\)
0.2 \(a \geqslant b\)
\(∴(a + kb) \bmod b = [(a - \lfloor \frac a b \rfloor \times b) + (k + \lfloor \frac a b \rfloor)b] \bmod b\)
由定义得：\(a \bmod b = a - \lfloor \frac a b \rfloor \times b\)
\(∴[(a - \lfloor \frac a b \rfloor \times b) + (k + \lfloor \frac a b \rfloor)b] \bmod b = [a \bmod b + (k + \lfloor \frac a b \rfloor)b] \bmod b\)
\(∵ a \bmod b < b\)
然后用 0.1 解决即可。
1.1 记 \(a = c * q_a + r_a, b = c * q_b + r_b (q_a, q_b, r_a, r_b \in \mathbb Z, r_a,rb < c)\)
\(∴ (a + b) \bmod c = (c * q_a + r_a + c * q_b + r_b) \bmod c\)
即 \((a + b) \bmod c = [r_a + r_b + (q_a + q_b)c] \bmod c\)
由 0 得：\((a + b) \bmod c = (r_a + r_b) \bmod c)\)
\(∵r_a = a \bmod c, r_b = b \bmod c\)
\(∴ (a + b) \bmod c = (a \bmod c + b \bmod c) \bmod c\)
1.2 记 \(a = c * q_a + r_a, b = c * q_b + r_b (q_a, q_b, r_a, r_b \in \mathbb Z, r_a,rb < c)\)
\(∴ (a - b) \bmod c = (c * q_a + r_a - c * q_b - r_b) \bmod c\)
即 \((a - b) \bmod c = [r_a - r_b + (q_a - q_b)c] \bmod c\)
由 0 得：\((a - b) \bmod c = (r_a - r_b) \bmod c)\)
\(∵r_a = a \bmod c, r_b = b \bmod c\)
\(∴ (a - b) \bmod c = (a \bmod c - b \bmod c) \bmod c\)
1.3 记 \(a = c * q_a + r_a, b = c * q_b + r_b (q_a, q_b, r_a, r_b \in \mathbb Z, r_a,rb < c)\)
\(∴ (a * b) \bmod c = [(c * q_a + r_a) * (c * q_b + r_b)] \bmod c\)
\(∴ (a * b) \bmod c = (q_aq_bc^2 + q_ar_bc + q_br_ac + r_ar_b) \bmod c\)
即 $ (a * b) \bmod c = [r_a * r_b + c(q_aq_bc + q_ar_b + q_br_a)] \bmod c$
由 0 得：\((a * b) \bmod c = (r_a * r_b) \bmod c)\)
\(∵r_a = a \bmod c, r_b = b \bmod c\)
\(∴ (a \times b) \bmod c = [(a \bmod c) \times (b \bmod c)] \bmod c\)
1.4
由 1.3 得：
\((a^b) \bmod c = (a \times a^{b - 1}) \bmod c = [(a \bmod c) \times (a ^ {b - 1} \bmod c)] \bmod c\)
\((a^b) \bmod c = [(a \bmod c) \times (a \bmod c) \times (a ^ {b - 2} \bmod c)] \bmod c\)
\(...\)
\((a^b) \bmod c = [(a \bmod c) \times (a \bmod c) ...\times (a \bmod c)] \bmod c\)
2.1 记 \(a = b \times q_a + c\)
$∴ (a \times d) \bmod (b \times d) = (cd + bdq_a) \bmod bd $
\(∵ a \bmod b = c\)
\(∴ c < b\)
又 \(∵ d > 0\)
\(∴ cd < bd\)
\(∴\) 由 0.1 得：\((a \times d) \bmod (b \times d) = (cd + bdq_a) \bmod bd = cd\)
2.1 记 \(a = b \times q_a + c\)
$∴ \frac a d \bmod \frac b d = (\frac c d + \frac {bq_a} d) \bmod \frac b d $
\(∵ a \bmod b = c\)
\(∴ c < b\)
又 \(∵ d > 0\)
\(∴ \frac c d < \frac b d\)
\(∴\) 由 0.1 得：\(\frac a d \bmod \frac b d = (\frac c d + \frac {bq_a} d) \bmod \frac b d = \frac c d\)
0x03 同余
概念： 设 \(m\) 为给定正整数，若满足 \(m \mid (a - b), a, b \in \mathbb Z\)，则称 \(a\) 与 \(b\) 对 \(m\) 同余。记作 \(a \equiv b \pmod m\)
性质：
0. 一些基本性质：
0.1 反身性：\(a \equiv a \pmod m\)
0.2 对称性：\(a \equiv b \pmod m\)，则 \(b \equiv a \pmod m\)
0.3 传递性：\(a \equiv b \pmod m\)，则 \(b \equiv c \pmod m\)，则 \(a \equiv c \pmod m\)
1. 同加性：若 \(a \equiv b \pmod m\)，则 \(a + c \equiv b + c \pmod m\)
2. 同减性：若 \(a \equiv b \pmod m\)，则 \(a - c \equiv b - c \pmod m\)
3. 同乘性：若 \(a \equiv b \pmod m\)，则 \(a \times c \equiv b \times c \pmod m\)
4. 同除性：若 \(a \equiv b \pmod m, c \mid a, c \mid b, (c, m) = 1\)，则 \(\frac a c \equiv \frac b c \pmod m\)
5. 同幂性：若 \(a \equiv b \pmod m\)，则 \(a^c \equiv b^c \pmod m\)
6. 若 \(a \bmod p = x, a \bmod q = x, (p, q) = 1\)，则 \(a \bmod (p \times q) = x\)
1. 由题：\(m \mid (a - b)\)
\(∴ m \mid [(a + c) - (b + c)]\)
\(∴ a + c \equiv b + c \pmod m\)
2. 由题：\(m \mid (a - b)\)
\(∴ m \mid [(a - c) - (b - c)]\)
\(∴ a - c \equiv b - c \pmod m\)
3. 由题：\(m \mid (a - b)\)
\(∴ m \mid [ac - bc]\)
\(∴ a \times c \equiv b \times c \pmod m\)
4. 记 \(a = k_ac, b = k_bc\)
\(∴ p \mid (k_a - k_b)c\)
\(∵ (c, p) = 1\)
\(∴ p \mid (k_a - k_b)\)
\(∴ pc \mid (k_ac - k_bc)\)
即 \(pc \mid (a - b)\)
\(∴ p \mid \frac {a - b} c\)
即 \(p \mid (\frac a c - \frac b c)\)
\(∴\frac a c \equiv \frac b c \pmod m\)
5. 由题：\(m \mid (a - b)\)
\(∴ a^c - b^c = (a - b)(a^{c - 1} + a^{c - 2}b + ... + ab^{c - 2} + b^{c - 1})\)
\(∵a, b \in \mathbb Z\)
\(∴ (a^{c - 1} + a^{c - 2}b + ... + ab^{c - 2} + b^{c - 1}) \in \mathbb Z\)
\(∴ (a - b) \mid (a^c - b^c)\)
\(∴ m \mid (a^c - b^c)\)
即 \(a^c \equiv b^c \pmod m\)
6. \(∵ a \bmod p = x, a \bmod q = x\)
\(∴p \mid (a - x), q \mid (a - x)\)
\(∴ a - x = pk_p, a - x = qk_q(k_p, k_q \in \mathbb Z)\)
即 \(pk_p = qk_q\)
\(∵(p, q) = 1\)
\(∴ k_p = kq(k \in \mathbb Z)\)
\(∴ a - x = pqk\)
故 \(pq \mid a - x\)
即 \(a \bmod (p \times q) = x\)
