初等数论 1.8 算数基本定理

定理:算数基本定理(fundamental theorem of arithmetic):设 1 < n ∈ Z 1<n\in\Z 1<nZ,有 n = p 1 p 2 ⋯ p s n=p_1p_2\cdots p_s n=p1p2ps.其中 p i ( 1 ≤ i ≤ s ) p_i(1\le i \le s) pi(1is)是素数.若不考虑素因数的次序,这种分解是唯一的.

证明:存在性:对 n ∈ Z n\in\Z nZ归纳.当 n = 2 n=2 n=2时,结论成立.
假设结论对小于 n n n的正整数均成立,现在考虑 n n n.若 n n n是素数,则结论显然成立.
n n n是合数,则 n n n有正的真因数 a , b a,b a,b,使得 n = a b , 1 < a , b < n n=ab,1<a,b<n n=ab1<a,b<n.由归纳假设, a , b a,b a,b均可分解为有限个素数之积.从而 n n n也可以分解为有限个素数之积.由数学归纳原理,对一切大于 1 1 1的正整数 n n n都能分解成 n = p 1 p 2 ⋯ p s n=p_1p_2\cdots p_s n=p1p2ps的形式.
唯一性:若 n = q 1 q 2 ⋯ q t , q j ( 1 ≤ j ≤ t ) n=q_1q_2\cdots q_t,q_j(1\le j \le t) n=q1q2qtqj(1jt)是素数,则有 p 1 p 2 ⋯ p s = q 1 q 2 ⋯ q t p_1p_2\cdots p_s=q_1q_2\cdots q_t p1p2ps=q1q2qt.从而 p 1 ∣ q 1 q 2 ⋯ q t p_1\mid q_1q_2\cdots q_t p1q1q2qt,则 p 1 ∣ q j ( 1 ≤ j ≤ t ) p_1\mid q_j\quad (1\le j \le t) p1qj(1jt).不妨 p 1 / m i d q 1 p_1/mid q_1 p1/midq1,又 p 1 , q 1 p_1,q_1 p1,q1都是素数,必有 p 1 = q 1 p_1=q_1 p1=q1.从而有 p 2 ⋯ p s = q 2 ⋯ q t p_2\cdots p_s=q_2\cdots q_t p2ps=q2qt.
s > t s>t s>t,则 p t + 1 ⋯ p s = 1 p_{t+1}\cdots p_s=1 pt+1ps=1 p t + 1 , ⋯   , p s p_{t+1},\cdots,p_s pt+1,,ps p t + 1 , ⋯   , p s p_{t+1},\cdots ,p_s pt+1,,ps为素数矛盾.于是 s = t s=t s=t.且适当调整顺序后, p i = q i , 1 ≤ i ≤ s p_i=q_i,1\le i \le s pi=qi1is.唯一性得证.

推论:对于 ∀ 1 < n ∈ Z \forall 1<n\in\Z 1<nZ n n n可唯一表示为 p 1 α 1 p 2 α 2 ⋯ p k α k p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k} p1α1p2α2pkαk,其中 p 1 , p 2 , ⋯   , p k p_1,p_2,\cdots ,p_k p1,p2,,pk是素数,且 p 1 < p 2 < ⋯ < p k , α 1 , α 2 , ⋯   , α k ∈ Z + p_1<p_2<\cdots<p_k,\alpha_1,\alpha_2,\cdots,\alpha_k\in\Z^+ p1<p2<<pkα1,α2,,αkZ+.
定义:称满足以上条件, n = p 1 α 1 p 2 α 2 ⋯ p k α k n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k} n=p1α1p2α2pkαk n n n的标准分解式.
推论:设 n n n的标准分解式为 n = p 1 α 1 p 2 α 2 ⋯ p k α k n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k} n=p1α1p2α2pkαk,则
1. n n n的所有正因数 d = p 1 β 1 p 2 β 2 ⋯ p k β k ( β i ∈ N 且 0 ≤ β i ≤ α i , 1 ≤ i ≤ k ) \displaystyle d=p_1^{\beta_1}p_2^{\beta_2}\cdots p_k^{\beta_k}\quad (\beta_i\in\N且0\le \beta_i \le \alpha_i,1\le i \le k) d=p1β1p2β2pkβk(βiN0βiαi1ik).
2. n n n的所有正倍数 m = p 1 γ 1 p 2 γ 2 ⋯ p k γ k ( ∀ a ∈ N + , α i ≤ γ i ∈ N , 1 ≤ i ≤ k ) \displaystyle m=p_1^{\gamma_1}p_2^{\gamma_2}\cdots p_k^{\gamma_k}\quad (\forall a\in\N^+,\alpha_i\le \gamma_i\in\N,1\le i \le k) m=p1γ1p2γ2pkγk(aN+αiγiN1ik).
推论:若 a , b ∈ Z a,b\in\Z a,bZ a , b a,b a,b的标准分解式为 a = p 1 α 1 p 2 α 2 ⋯ p k α k b = p 1 β 1 p 2 β 2 ⋯ p k β k a=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k} \quad\quad b=p_1^{\beta_1}p_2^{\beta_2}\cdots p_k^{\beta_k} a=p1α1p2α2pkαkb=p1β1p2β2pkβk ( a , b ) = p 1 λ 1 p 2 λ 2 ⋯ p k λ k λ i = min ⁡ { α i , β i } 1 ≤ i ≤ k (a,b)=p_1^{\lambda_1}p_2^{\lambda_2}\cdots p_k^{\lambda_k}\quad \lambda_i=\min{\{\alpha_i,\beta_i\}}\quad 1\le i \le k (a,b)=p1λ1p2λ2pkλkλi=min{αi,βi}1ik [ a , b ] = p 1 μ 1 p 2 μ 2 ⋯ p k μ k μ i = max ⁡ { α i , β i } 1 ≤ i ≤ k [a,b]=p_1^{\mu_1}p_2^{\mu_2}\cdots p_k^{\mu_k}\quad \mu_i=\max{\{\alpha_i,\beta_i\}}\quad 1\le i \le k [a,b]=p1μ1p2μ2pkμkμi=max{αi,βi}1ik.
推论:若 n n n的标准分解式为 n = p 1 α 1 p 2 α 2 ⋯ p k α k n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k} n=p1α1p2α2pkαk
1. n n n的正因数个数 d ( n ) = ∏ i = 1 k ( α i + 1 ) \displaystyle d(n)=\prod_{i=1}^{k}{(\alpha_i+1)} d(n)=i=1k(αi+1).
2. n n n的正因数的和 S ( n ) = ∏ i = 1 k p i α i + 1 − 1 p i − 1 \displaystyle S(n)=\prod_{i=1}^{k}{\frac{p_{i}^{\alpha_i+1}-1}{p_{i}-1}} S(n)=i=1kpi1piαi+11.

特别地,若 ( a ⋅ b ) = 1 (a\cdot b)=1 (ab)=1,则有 d ( a ⋅ b ) = d ( a ) ⋅ d ( b ) , S ( a ⋅ b ) = S ( a ) ⋅ S ( b ) d(a\cdot b)=d(a)\cdot d(b),S(a\cdot b)=S(a)\cdot S(b) d(ab)=d(a)d(b)S(ab)=S(a)S(b).

定义:设 a , b ∈ Z a,b\in\Z a,bZ,若 a = b 2 a=b^2 a=b2,则称 a a a为平方数(square number).
定理: a a a是平方数当且仅当 a a a的标准分解式中所有素因数指数为偶数.
推论:若 a a a是平方数,则 d ( a ) d(a) d(a)是奇数.
定义:设 n ∈ Z + n\in\Z^+ nZ+,若 S ( n ) = 2 n S(n)=2n S(n)=2n,则称 n n n是完全数.
定理:1.平方数不是完全数.

证明:设 n n n是平方数,且 n = p 1 2 α 1 p 2 2 α 2 ⋯ p s 2 α s α i ≥ 1 p i ( 1 ≤ i ≤ s ) n=p_1^{2\alpha_1}p_2^{2\alpha_2}\cdots p_s^{2\alpha_s}\quad \alpha_i\ge 1 \quad p_i(1\le i \le s) n=p12α1p22α2ps2αsαi1pi(1is)为素数, p 1 < p 2 < ⋯ < p s p_1<p_2<\cdots<p_s p1<p2<<ps,则 S ( n ) = ∏ i = 1 s ( p 1 2 α 1 + p 2 2 α 1 − 1 + ⋯ + p 1 + 1 ) S(n)=\prod_{i=1}^{s}{(p_1^{2\alpha1}+p_2^{2\alpha_1-1}+\cdots+p_1+1)} S(n)=i=1s(p12α1+p22α11++p1+1) S ( n ) S(n) S(n)是奇数,所以 n n n不是完全数.

2.若 n n n是无平方因子的完全数,则 n = 6 n=6 n=6.

证明:设 n = p 1 ⋯ p t p i ( 1 ≤ i ≤ t ) n=p_1\cdots p_t\quad p_i(1\le i \le t) n=p1ptpi(1it)为素数, p 1 < ⋯ < p t p_1<\cdots<p_t p1<<pt,则 S ( n ) = ∏ i = 1 t ( p i + 1 ) S(n)=\prod_{i=1}^{t}{(p_i+1)} S(n)=i=1t(pi+1).
S ( n ) = 2 n S(n)=2n S(n)=2n,若 t = 1 t=1 t=1 S ( n ) = p 1 + 1 = 2 p 1 , p 1 = 1 S(n)=p_1+1=2p_1,p_1=1 S(n)=p1+1=2p1p1=1,矛盾.
t ≥ 2 t\ge 2 t2,当 n n n为奇数时, p i ( ∀ 1 ≤ i ≤ t ) p_i(\forall 1\le i \le t) pi(1it)为奇数,则 4 ∣ S ( n ) 4\mid S(n) 4S(n),即 4 ∣ 2 n 4\mid 2n 42n,与 n n n为奇数矛盾.所以 n n n有因子 2 2 2.
t = 2 t=2 t=2时, n = 6 n=6 n=6. t = 3 t=3 t=3时, S ( n ) = 3 ( p 2 + 1 ) ( p 3 + 1 ) = 2 ⋅ 2 p 2 p 3 S(n)=3(p_2+1)(p_3+1)=2\cdot2p_2p_3 S(n)=3(p2+1)(p3+1)=22p2p3,无解.
t ≥ 3 t\ge 3 t3时有 8 / m i d S ( n ) 8/mid S(n) 8/midS(n),从而 4 ∣ n 4\mid n 4n,与 n n n无平方因子矛盾.
综上, n = 6 n=6 n=6.


练习:1.设 n ∈ Z n\in\Z nZ p p p为素数,若当 p ∣ n p\mid n pn时, p 2 ∣ n p^2\mid n p2n.称 n n n为重幂的(powerful).证明:每个重幂数可写为完全平方数和完全立方数的乘积.
2.设 p p p为素数, n ∈ Z + n\in\Z^+ nZ+,若 p a ∣ n p^a\mid n pan p a + 1 ∤ n p^{a+1} \nmid n pa+1n,称 p a p^a pa恰整除(exactly divides) n n n.记为 p a ∥ n p^a\parallel n pan.证明:若 p a ∣ m , p b ∣ n p^a\mid m,p^b\mid n pampbn,且 a ≠ b a\neq b a̸=b,则 p m i n { a , b } ∥ ( m + n ) p^{min{\{a,b\}}}\parallel (m+n) pmin{a,b}(m+n).
3.设 m , n ∈ Z m,n\in\Z m,nZ,求方程 m n = n m m^n=n^m mn=nm的所有解.

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