索引
- 引理(Gauss):设 p p p是奇素数, a ∈ Z a\in \mathbb{Z} a∈Z, gcd ( a , p ) = 1 \gcd \left( a,p \right)=1 gcd(a,p)=1。 ∀ k ∈ Z & 1 ≤ k ≤ p − 1 2 \forall k\in \mathbb{Z}\And 1\le k\le \frac{p-1}{2} ∀k∈Z&1≤k≤2p−1,设 r k ∈ Z , 0 ≤ r k < p { {r}_{k}}\in \mathbb{Z},\text{ }0\le { {r}_{k}}<p rk∈Z, 0≤rk<p, a k ≡ r k m o d p ak\equiv { {r}_{k}}\text{ }\bmod p ak≡rk modp,则有 ( a p ) = ( − 1 ) m \left( \frac{a}{p} \right)={ {\left( -1 \right)}^{m}} (pa)=(−1)m,其中 m m m是大于 p 2 \frac{p}{2} 2p的 r k { {r}_{k}} rk的个数。
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- 推论1:设 p p p是奇素数, a ∈ Z a\in \mathbb{Z} a∈Z, gcd ( a , p ) = 1 \gcd \left( a,p \right)=1 gcd(a,p)=1。则有 ( a − 1 ) p 2 − 1 8 ≡ ( ∑ k = 1 p − 1 2 [ a k p ] ) + m m o d 2 \left( a-1 \right)\frac{ { {p}^{2}}-1}{8}\equiv \left( \sum\limits_{k=1}^{\frac{p-1}{2}}{\left[ \frac{ak}{p} \right]}\right)+m\text{ }\bmod 2 (a−1)8p2−1≡⎝⎛k=1∑2p−1[pak]⎠⎞+m mod2其中 m m m的意义同引理。
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- 推论1-1:设 p p p是奇素数,则有 ( 2 p ) = ( − 1 ) p 2 − 1 8 = { 1 , p ≡ ± 1 m o d 8 − 1 , p ≡ ± 3 m o d 8 \left( \frac{2}{p} \right)={ {\left( -1 \right)}^{\frac{ { {p}^{2}}-1}{8}}}=\left\{ \begin{aligned} & 1,\text{ }p\equiv \pm 1\text{ }\bmod 8 \\ & -1,\text{ }p\equiv \pm 3\text{ }\bmod 8 \\ \end{aligned} \right. (p2)=(−1)8p2−1={ 1, p≡±1 mod8−1, p≡±3 mod8
- 推论1-2:设 p p p是奇素数, a ∈ Z a\in \mathbb{Z} a∈Z是奇数, gcd ( a , p ) = 1 \gcd \left( a,p \right)=1 gcd(a,p)=1,则有 ( a p ) = ( − 1 ) ∑ k = 1 p − 1 2 [ a k p ] \left( \frac{a}{p} \right)={ {\left( -1 \right)}^{\sum\limits_{k=1}^{\frac{p-1}{2}}{\left[ \frac{ak}{p} \right]}}} (pa)=(−1)k=1∑2p−1[pak]
- 定理(二次反转定律):若 p p p和 q q q都是奇素数, gcd ( p , q ) = 1 \gcd \left( p,q \right)=1 gcd(p,q)=1,则有 ( q p ) = ( − 1 ) p − 1 2 ⋅ q − 1 2 ( p q ) \left( \frac{q}{p} \right)={ {\left( -1 \right)}^{\frac{p-1}{2}\centerdot \frac{q-1}{2}}}\left( \frac{p}{q} \right) (pq)=(−1)2p−1⋅2q−1(qp)
引理(Gauss):设 p p p是奇素数, a ∈ Z a\in \mathbb{Z} a∈Z, gcd ( a , p ) = 1 \gcd \left( a,p \right)=1 gcd(a,p)=1。 ∀ k ∈ Z & 1 ≤ k ≤ p − 1 2 \forall k\in \mathbb{Z}\And 1\le k\le \frac{p-1}{2} ∀k∈Z&1≤k≤2p−1,设 r k ∈ Z , 0 ≤ r k < p { {r}_{k}}\in \mathbb{Z},\text{ }0\le { {r}_{k}}<p rk∈Z, 0≤rk<p, a k ≡ r k m o d p ak\equiv { {r}_{k}}\text{ }\bmod p ak≡rk modp,则有 ( a p ) = ( − 1 ) m \left( \frac{a}{p} \right)={ {\left( -1 \right)}^{m}} (pa)=(−1)m,其中 m m m是大于 p 2 \frac{p}{2} 2p的 r k { {r}_{k}} rk的个数。
证明
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第一步,我们指出, ∀ k , a k ≡ 0 m o d p \forall k,\text{ }ak\cancel{\equiv }0\text{ }\bmod p ∀k, ak≡ 0 modp。
事实上有推理
p 是 素 数 , gcd ( a , p ) = 1 ⇒ p ∣ a ⇒ a ≡ 0 m o d p p 是 素 数 , 1 ≤ k ≤ p − 1 2 ⇒ gcd ( k , p ) = 1 ⇒ p ∣ k ⇒ k ≡ 0 m o d p \begin{aligned} & p是素数,\text{ }\gcd \left( a,p \right)=1\text{ }\Rightarrow \text{ }p\cancel{|}a\text{ }\Rightarrow \text{ }a\cancel{\equiv }0\text{ }\bmod p \\ & p是素数,\text{ }1\le k\le \frac{p-1}{2}\text{ }\Rightarrow \text{ }\gcd \left( k,p \right)=1\text{ }\Rightarrow \text{ }p\cancel{|}k\text{ }\Rightarrow \text{ }k\cancel{\equiv }0\text{ }\bmod p \\ \end{aligned} p是素数, gcd(a,p)=1 ⇒ p∣ a ⇒ a≡ 0 modpp是素数, 1≤k≤2p−1 ⇒ gcd(k,p)=1 ⇒ p∣ k ⇒ k≡ 0 modp
因此有 a k ≡ 0 m o d p ak\cancel{\equiv }0\text{ }\bmod p ak≡ 0 modp,也蕴含了 ∀ k , r k ≠ 0 \forall k,\text{ }{ {r}_{k}}\ne 0 ∀k, rk=0。 -
第二步,由于 p p p是奇数, p 2 ∉ Z \frac{p}{2}\notin \mathbb{Z} 2p∈/Z,因此不存在 r k = p 2 { {r}_{k}}=\frac{p}{2} rk=2p。又由上述可将 r k { {r}_{k}} rk的范围缩小至 0 < r k < p 0<{ {r}_{k}}<p 0<rk<p,因此可将全体 r k { {r}_{k}} rk划分为 A ⋃ B A\bigcup B A⋃B,其中
A = { r k : 0 < r k < p 2 } = { a 1 , ⋯ , a t } , B = { r k : p 2 < r k < p } = { b 1 , ⋯ , b m } , A ⋂ B = ∅ , \begin{aligned} & A=\left\{ { {r}_{k}}:0<{ {r}_{k}}<\frac{p}{2} \right\}=\left\{ { {a}_{1}},\cdots ,{ {a}_{t}} \right\}, \\ & B=\left\{ { {r}_{k}}:\frac{p}{2}<{ {r}_{k}}<p \right\}=\left\{ { {b}_{1}},\cdots ,{ {b}_{m}} \right\}, \\ & A\bigcap B=\varnothing , \\ \end{aligned} A={ rk:0<rk<2p}={ a1,⋯,at},B={ rk:2p<rk<p}={ b1,⋯,bm},A⋂B=∅, -
第三步,设 a k 1 ≡ r 1 m o d p , a k 2 ≡ r 2 m o d p a{ {k}_{1}}\equiv { {r}_{1}}\text{ }\bmod p,\text{ }a{ {k}_{2}}\equiv { {r}_{2}}\text{ }\bmod p ak1≡r1 modp, ak2≡r2 modp,我们指出有 r 1 + r 2 ≠ p { {r}_{1}}+{ {r}_{2}}\ne p r1+r2=p。
事实上,若 r 1 + r 2 = p { {r}_{1}}+{ {r}_{2}}=p r1+r2=p,则有 a k 1 + a k 2 ≡ r 1 + r 2 = p ≡ 0 m o d p a{ {k}_{1}}+a{ {k}_{2}}\equiv { {r}_{1}}+{ {r}_{2}}=p\equiv 0\text{ }\bmod p ak1+ak2≡r1+r2=p≡0 modp即
a ( k 1 + k 2 ) ≡ 0 m o d p a\left( { {k}_{1}}+{ {k}_{2}} \right)\equiv 0\text{ }\bmod p a(k1+k2)≡0 modp
p p p是素数, gcd ( a , p ) = 1 \gcd \left( a,p \right)=1 gcd(a,p)=1,因而 p ∣ a , a ≡ 0 m o d p p\cancel{|}a,\text{ }a\cancel{\equiv }0\text{ }\bmod p p∣ a, a≡ 0 modp,因此有
k 1 + k 2 ≡ 0 m o d p { {k}_{1}}+{ {k}_{2}}\equiv 0\text{ }\bmod p k1+k2≡0 modp
但这与 2 ≤ k 1 + k 2 ≤ p − 1 2\le { {k}_{1}}+{ {k}_{2}}\le p-1 2≤k1+k2≤p−1矛盾。
因此令 C = { c = p − b : b ∈ B } C=\left\{ c=p-b:b\in B \right\} C={ c=p−b:b∈B},则有 A ⋂ C = ∅ A\bigcap C=\varnothing A⋂C=∅,且 ∀ c ∈ C \forall c\in C ∀c∈C, 0 < c < p 2 0<c<\frac{p}{2} 0<c<2p。 -
第四步,我们也指出 ∀ k 1 ≠ k 2 , a k 1 ≡ a k 2 m o d p \forall { {k}_{1}}\ne { {k}_{2}},\text{ }a{ {k}_{1}}\cancel{\equiv }a{ {k}_{2}}\bmod p ∀k1=k2, ak1≡ ak2modp。
事实上有
{ k 1 ≠ k 2 ⇒ k 1 − k 2 ≠ 0 1 ≤ k 1 , k 2 ≤ p − 1 2 ⇒ 0 ≤ k 1 − k 2 ≤ p − 3 2 < p ⇒ k 1 − k 2 ≡ 0 m o d p \left\{ \begin{aligned} & { {k}_{1}}\ne { {k}_{2}}\Rightarrow { {k}_{1}}-{ {k}_{2}}\ne 0 \\ & 1\le { {k}_{1}},{ {k}_{2}}\le \frac{p-1}{2}\Rightarrow 0\le { {k}_{1}}-{ {k}_{2}}\le \frac{p-3}{2}<p \\ \end{aligned} \right.\text{ }\Rightarrow \text{ }{ {k}_{1}}-{ {k}_{2}}\cancel{\equiv }0\text{ }\bmod p ⎩⎨⎧k1=k2⇒k1−k2=01≤k1,k2≤2p−1⇒0≤k1−k2≤2p−3<p ⇒ k1−k2≡ 0 modp
gcd ( a , p ) = 1 \gcd \left( a,p \right)=1 gcd(a,p)=1, p p p是素数,因此 p ∣ a ⇒ a ≡ 0 m o d p p\cancel{|}a\text{ }\Rightarrow \text{ }a\cancel{\equiv }0\text{ }\bmod p p∣ a ⇒ a≡ 0 modp。
因此有
a ( k 1 − k 2 ) ≡ 0 m o d p ⇔ a k 1 ≡ a k 2 m o d p a\left( { {k}_{1}}-{ {k}_{2}} \right)\cancel{\equiv }0\text{ }\bmod p\text{ }\Leftrightarrow \text{ }a{ {k}_{1}}\cancel{\equiv }a{ {k}_{2}}\text{ }\bmod p a(k1−k2)≡ 0 modp ⇔ ak1≡ ak2 modp
因此有 ∣ A ∣ + ∣ C ∣ = ∣ A ∣ + ∣ B ∣ = t + m = p − 1 2 \left| A \right|+\left| C \right|=\left| A \right|+\left| B \right|=t+m=\frac{p-1}{2} ∣A∣+∣C∣=∣A∣+∣B∣=t+m=2p−1。 -
基于上述几步的结论,有以下推理
{ A ⋂ C = ∅ ∣ A ∣ + ∣ C ∣ = p − 1 2 ∀ a ∈ A , ∀ c ∈ C , 0 < a , c ≤ p − 1 2 ⇒ A ⋃ C = { 1 , 2 , ⋯ , p − 1 2 } \left\{ \begin{aligned} & A\bigcap C=\varnothing \\ & \left| A \right|+\left| C \right|=\frac{p-1}{2} \\ & \forall a\in A,\text{ }\forall c\in C,\text{ }0<a,c\le \frac{p-1}{2} \\ \end{aligned} \right.\Rightarrow A\bigcup C=\left\{ 1,2,\cdots ,\frac{p-1}{2} \right\} ⎩⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎧A⋂C=∅∣A∣+∣C∣=2p−1∀a∈A, ∀c∈C, 0<a,c≤2p−1⇒A⋃C={ 1,2,⋯,2p−1}
因此有
( p − 1 2 ) ! = ( ∏ i = 1 t a i ) ( ∏ j = 1 m p − b j ) ≡ ( ∏ i = 1 t a i ) ( ∏ j = 1 m ( − b j ) ) m o d p = ( − 1 ) m ( ∏ i = 1 t a i ) ( ∏ j = 1 m b j ) = ( − 1 ) m ∏ k = 1 p − 1 2 r k ≡ ( − 1 ) m ∏ k = 1 p − 1 2 ( a k ) m o d p = ( − 1 ) m a p − 1 2 ( p − 1 2 ) ! \begin{aligned} & \left( \frac{p-1}{2} \right)!=\left( \prod\limits_{i=1}^{t}{ { {a}_{i}}} \right)\left( \prod\limits_{j=1}^{m}{p-{ {b}_{j}}} \right) \\ & \equiv \left( \prod\limits_{i=1}^{t}{ { {a}_{i}}} \right)\left( \prod\limits_{j=1}^{m}{\left( -{ {b}_{j}} \right)} \right)\text{ }\bmod p \\ & ={ {\left( -1 \right)}^{m}}\left( \prod\limits_{i=1}^{t}{ { {a}_{i}}} \right)\left( \prod\limits_{j=1}^{m}{ { {b}_{j}}} \right) \\ & ={ {\left( -1 \right)}^{m}}\prod\limits_{k=1}^{\frac{p-1}{2}}{ { {r}_{k}}} \\ & \equiv { {\left( -1 \right)}^{m}}\prod\limits_{k=1}^{\frac{p-1}{2}}{\left( ak \right)}\text{ }\bmod p \\ & ={ {\left( -1 \right)}^{m}}{ {a}^{\frac{p-1}{2}}}\left( \frac{p-1}{2} \right)! \\ \end{aligned} (2p−1)!=(i=1∏tai)(j=1∏mp−bj)≡(i=1∏tai)(j=1∏m(−bj)) modp=(−1)m(i=1∏tai)(j=1∏mbj)=(−1)mk=1∏2p−1rk≡(−1)mk=1∏2p−1(ak) modp=(−1)ma2p−1(2p−1)!
即有
( p − 1 2 ) [ ( − 1 ) m a p − 1 2 − 1 ] ≡ 0 m o d p \left( \frac{p-1}{2} \right)\left[ { {\left( -1 \right)}^{m}}{ {a}^{\frac{p-1}{2}}}-1 \right]\equiv 0\text{ }\bmod p (2p−1)[(−1)ma2p−1−1]≡0 modp
由于 p p p是素数,因此 ∀ 1 ≤ k ≤ p − 1 2 \forall 1\le k\le \frac{p-1}{2} ∀1≤k≤2p−1, gcd ( k , p ) = 1 \gcd \left( k,p \right)=1 gcd(k,p)=1,因此有 gcd ( ( p − 1 2 ) ! , p ) = 1 \gcd \left( \left( \frac{p-1}{2} \right)!,p \right)=1 gcd((2p−1)!,p)=1,因此有
( − 1 ) m a p − 1 2 − 1 ≡ 0 m o d p ⇒ ( a p ) ≡ a p − 1 2 ≡ ( − 1 ) m m o d p \begin{aligned} & { {\left( -1 \right)}^{m}}{ {a}^{\frac{p-1}{2}}}-1\equiv 0\text{ }\bmod p \\ & \Rightarrow \left( \frac{a}{p} \right)\equiv { {a}^{\frac{p-1}{2}}}\equiv { {\left( -1 \right)}^{m}}\text{ }\bmod p \\ \end{aligned} (−1)ma2p−1−1≡0 modp⇒(pa)≡a2p−1≡(−1)m modp
又 gcd ( a , p ) = 1 \gcd \left( a,p \right)=1 gcd(a,p)=1的情况下 ( a p ) = ± 1 \left( \frac{a}{p} \right)=\pm 1 (pa)=±1,因此只能有
( a p ) = ( − 1 ) m \left( \frac{a}{p} \right)={ {\left( -1 \right)}^{m}} (pa)=(−1)m
推论1:设 p p p是奇素数, a ∈ Z a\in \mathbb{Z} a∈Z, gcd ( a , p ) = 1 \gcd \left( a,p \right)=1 gcd(a,p)=1。则有 ( a − 1 ) p 2 − 1 8 ≡ ( ∑ k = 1 p − 1 2 [ a k p ] ) + m m o d 2 \left( a-1 \right)\frac{ { {p}^{2}}-1}{8}\equiv \left( \sum\limits_{k=1}^{\frac{p-1}{2}}{\left[ \frac{ak}{p} \right]}\right)+m\text{ }\bmod 2 (a−1)8p2−1≡⎝⎛k=1∑2p−1[pak]⎠⎞+m mod2其中 m m m的意义同引理。
证明
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由带余除法, ∀ 1 ≤ k ≤ p − 1 2 \forall 1\le k\le \frac{p-1}{2} ∀1≤k≤2p−1,有
a k = p [ a k p ] + r k ak=p\left[ \frac{ak}{p} \right]+{ {r}_{k}} ak=p[pak]+rk
r k { {r}_{k}} rk的意义同引理。因此有
∑ k = 1 p − 1 2 ( a k ) = a ∑ k = 1 p − 1 2 k = a ( 1 + p − 1 2 ) × p − 1 2 2 = a p 2 − 1 8 \sum\limits_{k=1}^{\frac{p-1}{2}}{\left( ak \right)}=a\sum\limits_{k=1}^{\frac{p-1}{2}}{k}=a\frac{\left( 1+\frac{p-1}{2} \right)\times \frac{p-1}{2}}{2}=a\frac{ { {p}^{2}}-1}{8} k=1∑2p−1(ak)=ak=1∑2p−1k=a2(1+2p−1)×2p−1=a8p2−1 -
另一方面,
∑ k = 1 p − 1 2 ( a k ) = ∑ k = 1 p − 1 2 ( p [ a k p ] + r k ) = p ∑ k = 1 p − 1 2 [ a k p ] + ∑ k = 1 p − 1 2 r k ≡ ∑ k = 1 p − 1 2 [ a k p ] + ∑ k = 1 p − 1 2 r k m o d 2 ( p ≡ 1 m o d 2 ) \begin{aligned} & \sum\limits_{k=1}^{\frac{p-1}{2}}{\left( ak \right)}=\sum\limits_{k=1}^{\frac{p-1}{2}}{\left( p\left[ \frac{ak}{p} \right]+{ {r}_{k}} \right)} \\ & =p\sum\limits_{k=1}^{\frac{p-1}{2}}{\left[ \frac{ak}{p} \right]}+\sum\limits_{k=1}^{\frac{p-1}{2}}{ { {r}_{k}}} \\ & \equiv \sum\limits_{k=1}^{\frac{p-1}{2}}{\left[ \frac{ak}{p} \right]}+\sum\limits_{k=1}^{\frac{p-1}{2}}{ { {r}_{k}}}\text{ }\bmod 2\text{ }\left( p\equiv 1\text{ }\bmod 2 \right) \\ \end{aligned} k=1∑2p−1(ak)=k=1∑2p−1(p[pak]+rk)=pk=1∑2p−1[pak]+k=1∑2p−1rk≡k=1∑2p−1[pak]+k=1∑2p−1rk mod2 (p≡1 mod2)
而对于 ∑ k = 1 p − 1 2 r k \sum\limits_{k=1}^{\frac{p-1}{2}}{ { {r}_{k}}} k=1∑2p−1rk,还有
∑ k = 1 p − 1 2 r k = ∑ i = 1 t a i + ∑ j = 1 m b j = ( ∑ i = 1 t a i + ∑ j = 1 m ( p − b j ) ) − ∑ j = 1 m ( p − b j ) + ∑ j = 1 m b j = ∑ k = 1 p − 1 2 k − ∑ j = 1 m p + 2 ∑ j = 1 m b j ≡ ( 1 + p − 1 2 ) × p − 1 2 2 − p m m o d 2 ≡ p 2 − 1 8 − m m o d 2 ( p ≡ 1 m o d 2 ) \begin{aligned} & \sum\limits_{k=1}^{\frac{p-1}{2}}{ { {r}_{k}}}=\sum\limits_{i=1}^{t}{ { {a}_{i}}}+\sum\limits_{j=1}^{m}{ { {b}_{j}}} \\ & =\left( \sum\limits_{i=1}^{t}{ { {a}_{i}}}+\sum\limits_{j=1}^{m}{\left( p-{ {b}_{j}} \right)} \right)-\sum\limits_{j=1}^{m}{\left( p-{ {b}_{j}} \right)}+\sum\limits_{j=1}^{m}{ { {b}_{j}}} \\ & =\sum\limits_{k=1}^{\frac{p-1}{2}}{k}-\sum\limits_{j=1}^{m}{p}+2\sum\limits_{j=1}^{m}{ { {b}_{j}}} \\ & \equiv \frac{\left( 1+\frac{p-1}{2} \right)\times \frac{p-1}{2}}{2}-pm\text{ }\bmod 2 \\ & \equiv \frac{ { {p}^{2}}-1}{8}-m\text{ }\bmod 2\text{ }\left( p\equiv 1\text{ }\bmod 2 \right) \\ \end{aligned} k=1∑2p−1rk=i=1∑tai+j=1∑mbj=(i=1∑tai+j=1∑m(p−bj))−j=1∑m(p−bj)+j=1∑mbj=k=1∑2p−1k−j=1∑mp+2j=1∑mbj≡2(1+2p−1)×2p−1−pm mod2≡8p2−1−m mod2 (p≡1 mod2)
其中 a i , b j a_i, b_j ai,bj的意义同引理。 -
因此最终有
a p 2 − 1 8 ≡ ∑ k = 1 p − 1 2 [ a k p ] + p 2 − 1 8 − m m o d 2 a\frac{ { {p}^{2}}-1}{8}\equiv \sum\limits_{k=1}^{\frac{p-1}{2}}{\left[ \frac{ak}{p} \right]}+\frac{ { {p}^{2}}-1}{8}-m\text{ }\bmod 2 a8p2−1≡k=1∑2p−1[pak]+8p2−1−m mod2
整理得
( a − 1 ) p 2 − 1 8 ≡ ( ∑ k = 1 p − 1 2 [ a k p ] ) + m m o d 2 \left( a-1 \right)\frac{ { {p}^{2}}-1}{8}\equiv \left( \sum\limits_{k=1}^{\frac{p-1}{2}}{\left[ \frac{ak}{p} \right]}\right)+m\text{ }\bmod 2 (a−1)8p2−1≡⎝⎛k=1∑2p−1[pak]⎠⎞+m mod2
推论1-1:设 p p p是奇素数,则有 ( 2 p ) = ( − 1 ) p 2 − 1 8 = { 1 , p ≡ ± 1 m o d 8 − 1 , p ≡ ± 3 m o d 8 \left( \frac{2}{p} \right)={ {\left( -1 \right)}^{\frac{ { {p}^{2}}-1}{8}}}=\left\{ \begin{aligned} & 1,\text{ }p\equiv \pm 1\text{ }\bmod 8 \\ & -1,\text{ }p\equiv \pm 3\text{ }\bmod 8 \\ \end{aligned} \right. (p2)=(−1)8p2−1={ 1, p≡±1 mod8−1, p≡±3 mod8
证明
在引理的推论1中,令 a = 2 a=2 a=2,则有
p 2 − 1 8 ≡ ( ∑ k = 1 p − 1 2 [ 2 k p ] ) + m m o d 2 \frac{
{
{p}^{2}}-1}{8}\equiv \left( \sum\limits_{k=1}^{\frac{p-1}{2}}{\left[ \frac{2k}{p} \right]} \right)+m\text{ }\bmod 2 8p2−1≡⎝⎛k=1∑2p−1[p2k]⎠⎞+m mod2
由于 1 ≤ k ≤ p − 1 2 1\le k\le \frac{p-1}{2} 1≤k≤2p−1,因此 2 ≤ 2 k ≤ p − 1 < p 2\le 2k\le p-1<p 2≤2k≤p−1<p, 0 < 2 k p < 1 0<\frac{2k}{p}<1 0<p2k<1,因此 ∀ 1 ≤ k ≤ p − 1 2 \forall 1\le k\le \frac{p-1}{2} ∀1≤k≤2p−1, [ 2 k p ] = 0 \left[ \frac{2k}{p} \right]=0 [p2k]=0,也就有
∑ k = 1 p − 1 2 [ 2 k p ] = 0 \sum\limits_{k=1}^{\frac{p-1}{2}}{\left[ \frac{2k}{p} \right]}=0 k=1∑2p−1[p2k]=0
因此有
p 2 − 1 8 ≡ m m o d 2 \frac{
{
{p}^{2}}-1}{8}\equiv m\text{ }\bmod 2 8p2−1≡m mod2
即 ∃ k ∈ Z , s.t. m = 2 k + p 2 − 1 8 \exists k\in \mathbb{Z},\text{ s}\text{.t}\text{. }m=2k+\frac{
{
{p}^{2}}-1}{8} ∃k∈Z, s.t. m=2k+8p2−1。又 gcd ( 2 , p ) = 1 \gcd \left( 2,p \right)=1 gcd(2,p)=1,因此由引理有
( 2 p ) = ( − 1 ) m = ( − 1 ) 2 k + p 2 − 1 8 = ( − 1 ) p 2 − 1 8 \left( \frac{2}{p} \right)={
{\left( -1 \right)}^{m}}={
{\left( -1 \right)}^{2k+\frac{
{
{p}^{2}}-1}{8}}}={
{\left( -1 \right)}^{\frac{
{
{p}^{2}}-1}{8}}} (p2)=(−1)m=(−1)2k+8p2−1=(−1)8p2−1
由于 p p p是奇数,因此 p p p模8的结果只可能是下列情况
p ≡ ± 1 , ± 3 m o d 8 p\equiv \pm 1,\pm 3\text{ }\bmod 8 p≡±1,±3 mod8
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当 p ≡ ± 1 m o d 8 p\equiv \pm 1\text{ }\bmod 8 p≡±1 mod8时, ∃ t ∈ Z , s.t. p = 8 t ± 1 \exists t\in \mathbb{Z},\text{ s}\text{.t}\text{. }p=8t\pm 1 ∃t∈Z, s.t. p=8t±1,于是有
( 2 p ) = ( − 1 ) ( 8 t ± 1 ) 2 − 1 8 = ( − 1 ) 64 t 2 ± 16 t 8 = ( − 1 ) 8 t 2 ± 2 t = 1 \left( \frac{2}{p} \right)={ {\left( -1 \right)}^{\frac{ { {\left( 8t\pm 1 \right)}^{2}}-1}{8}}}={ {\left( -1 \right)}^{\frac{64{ {t}^{2}}\pm 16t}{8}}}={ {\left( -1 \right)}^{8{ {t}^{2}}\pm 2t}}=1 (p2)=(−1)8(8t±1)2−1=(−1)864t2±16t=(−1)8t2±2t=1 -
当 p ≡ ± 3 m o d 8 p\equiv \pm 3\text{ }\bmod 8 p≡±3 mod8时, ∃ t ∈ Z , s.t. p = 8 t ± 3 \exists t\in \mathbb{Z},\text{ s}\text{.t}\text{. }p=8t\pm 3 ∃t∈Z, s.t. p=8t±3,于是有
( 2 p ) = ( − 1 ) ( 8 t ± 3 ) 2 − 1 8 = ( − 1 ) 64 t 2 ± 48 t + 8 8 = ( − 1 ) 8 t 2 ± 6 t + 1 = − 1 \left( \frac{2}{p} \right)={ {\left( -1 \right)}^{\frac{ { {\left( 8t\pm 3 \right)}^{2}}-1}{8}}}={ {\left( -1 \right)}^{\frac{64{ {t}^{2}}\pm 48t+8}{8}}}={ {\left( -1 \right)}^{8{ {t}^{2}}\pm 6t+1}}=-1 (p2)=(−1)8(8t±3)2−1=(−1)864t2±48t+8=(−1)8t2±6t+1=−1
推论1-2:设 p p p是奇素数, a ∈ Z a\in \mathbb{Z} a∈Z是奇数, gcd ( a , p ) = 1 \gcd \left( a,p \right)=1 gcd(a,p)=1,则有 ( a p ) = ( − 1 ) ∑ k = 1 p − 1 2 [ a k p ] \left( \frac{a}{p} \right)={ {\left( -1 \right)}^{\sum\limits_{k=1}^{\frac{p-1}{2}}{\left[ \frac{ak}{p} \right]}}} (pa)=(−1)k=1∑2p−1[pak]
证明
由引理的推论1有
( a − 1 ) p 2 − 1 8 ≡ ( ∑ k = 1 p − 1 2 [ a k p ] ) + m m o d 2 \left( a-1 \right)\frac{
{
{p}^{2}}-1}{8}\equiv \left( \sum\limits_{k=1}^{\frac{p-1}{2}}{\left[ \frac{ak}{p} \right]} \right)+m\text{ }\bmod 2 (a−1)8p2−1≡⎝⎛k=1∑2p−1[pak]⎠⎞+m mod2
a a a是奇数, a − 1 a-1 a−1是偶数, a − 1 ≡ 0 mod2 a-1\equiv 0\text{ mod2} a−1≡0 mod2,因此有
( ∑ k = 1 p − 1 2 [ a k p ] ) + m ≡ 0 m o d 2 ⇒ ( ∑ k = 1 p − 1 2 [ a k p ] ) ≡ − m ≡ − m + 2 m = m m o d 2 \begin{aligned} & \left( \sum\limits_{k=1}^{\frac{p-1}{2}}{\left[ \frac{ak}{p} \right]} \right)+m\equiv 0\text{ }\bmod 2 \\ & \Rightarrow \left( \sum\limits_{k=1}^{\frac{p-1}{2}}{\left[ \frac{ak}{p} \right]} \right)\equiv -m\equiv -m+2m=m\text{ }\bmod 2 \\ \end{aligned} ⎝⎛k=1∑2p−1[pak]⎠⎞+m≡0 mod2⇒⎝⎛k=1∑2p−1[pak]⎠⎞≡−m≡−m+2m=m mod2
即 ∃ t ∈ Z , s.t. m = 2 t + ∑ k = 1 p − 1 2 [ a k p ] \exists t\in \mathbb{Z},\text{ s}\text{.t}\text{. }m=2t+\sum\limits_{k=1}^{\frac{p-1}{2}}{\left[ \frac{ak}{p} \right]} ∃t∈Z, s.t. m=2t+k=1∑2p−1[pak]。又 gcd ( a , p ) = 1 \gcd \left( a,p \right)=1 gcd(a,p)=1,因此由引理有
( a p ) = ( − 1 ) m = ( − 1 ) 2 t + ∑ k = 1 p − 1 2 [ a k p ] = ( − 1 ) ∑ k = 1 p − 1 2 [ a k p ] \left( \frac{a}{p} \right)={
{\left( -1 \right)}^{m}}={
{\left( -1 \right)}^{2t+\sum\limits_{k=1}^{\frac{p-1}{2}}{\left[ \frac{ak}{p} \right]}}}={
{\left( -1 \right)}^{\sum\limits_{k=1}^{\frac{p-1}{2}}{\left[ \frac{ak}{p} \right]}}} (pa)=(−1)m=(−1)2t+k=1∑2p−1[pak]=(−1)k=1∑2p−1[pak]
定理(二次反转定律):若 p p p和 q q q都是奇素数, gcd ( p , q ) = 1 \gcd \left( p,q \right)=1 gcd(p,q)=1,则有 ( q p ) = ( − 1 ) p − 1 2 ⋅ q − 1 2 ( p q ) \left( \frac{q}{p} \right)={ {\left( -1 \right)}^{\frac{p-1}{2}\centerdot \frac{q-1}{2}}}\left( \frac{p}{q} \right) (pq)=(−1)2p−1⋅2q−1(qp)
证明
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第一步,由于 q q q是素数, gcd ( p , q ) = 1 \gcd \left( p,q \right)=1 gcd(p,q)=1,因此 q ∣ p q\cancel{|}p q∣ p。再由Legendre符号的定义,有 ( p q ) ≠ 0 \left( \frac{p}{q} \right)\ne 0 (qp)=0, ( p q ) = 1 \left( \frac{p}{q} \right)=1 (qp)=1或 ( p q ) = − 1 \left( \frac{p}{q} \right)=-1 (qp)=−1,即有
( p q ) 2 = 1 { {\left( \frac{p}{q} \right)}^{2}}=1 (qp)2=1
基于此,可将原问题转化为等价形式
( q p ) = ( − 1 ) p − 1 2 ⋅ q − 1 2 ( p q ) ⇔ ( q p ) ( p q ) = ( − 1 ) p − 1 2 ⋅ q − 1 2 ( p q ) 2 = ( − 1 ) p − 1 2 ⋅ q − 1 2 \begin{aligned} & \left( \frac{q}{p} \right)={ {\left( -1 \right)}^{\frac{p-1}{2}\centerdot \frac{q-1}{2}}}\left( \frac{p}{q} \right) \\ & \Leftrightarrow \left( \frac{q}{p} \right)\left( \frac{p}{q} \right)={ {\left( -1 \right)}^{\frac{p-1}{2}\centerdot \frac{q-1}{2}}}{ {\left( \frac{p}{q} \right)}^{2}}={ {\left( -1 \right)}^{\frac{p-1}{2}\centerdot \frac{q-1}{2}}} \\ \end{aligned} (pq)=(−1)2p−1⋅2q−1(qp)⇔(pq)(qp)=(−1)2p−1⋅2q−1(qp)2=(−1)2p−1⋅2q−1 -
第二步,由于 q q q是奇素数, p p p是奇数, gcd ( p , q ) = 1 \gcd \left( p,q \right)=1 gcd(p,q)=1,因此由上面的推论1-2有
( p q ) = ( − 1 ) ∑ k = 1 q − 1 2 [ p k q ] \left( \frac{p}{q} \right)={ {\left( -1 \right)}^{\sum\limits_{k=1}^{\frac{q-1}{2}}{\left[ \frac{pk}{q} \right]}}} (qp)=(−1)k=1∑2q−1[qpk]
同理有
( q p ) = ( − 1 ) ∑ h = 1 p − 1 2 [ q h p ] \left( \frac{q}{p} \right)={ {\left( -1 \right)}^{\sum\limits_{h=1}^{\frac{p-1}{2}}{\left[ \frac{qh}{p} \right]}}} (pq)=(−1)h=1∑2p−1[pqh]
因此原问题又可以转化为等价形式
( − 1 ) p − 1 2 ⋅ q − 1 2 = ( q p ) ( p q ) = ( − 1 ) ( ∑ h = 1 p − 1 2 [ q h p ] ) + ( ∑ k = 1 q − 1 2 [ p k q ] ) { {\left( -1 \right)}^{\frac{p-1}{2}\centerdot \frac{q-1}{2}}}=\left( \frac{q}{p} \right)\left( \frac{p}{q} \right)={ {\left( -1 \right)}^{\left( \sum\limits_{h=1}^{\frac{p-1}{2}}{\left[ \frac{qh}{p} \right]} \right)+\left( \sum\limits_{k=1}^{\frac{q-1}{2}}{\left[ \frac{pk}{q} \right]} \right)}} (−1)2p−1⋅2q−1=(pq)(qp)=(−1)(h=1∑2p−1[pqh])+(k=1∑2q−1[qpk]) -
第三步,我们指出有 ∑ h = 1 p − 1 2 [ q h p ] + ∑ k = 1 q − 1 2 [ p k q ] = p − 1 2 ⋅ q − 1 2 \sum\limits_{h=1}^{\frac{p-1}{2}}{\left[ \frac{qh}{p} \right]}+\sum\limits_{k=1}^{\frac{q-1}{2}}{\left[ \frac{pk}{q} \right]}=\frac{p-1}{2}\centerdot \frac{q-1}{2} h=1∑2p−1[pqh]+k=1∑2q−1[qpk]=2p−1⋅2q−1
不妨假定 p > q p>q p>q,用几何的形式来证明该结论。
上图中 p 1 = p − 1 2 , q 1 = q − 1 2 { {p}_{1}}=\frac{p-1}{2},{ {q}_{1}}=\frac{q-1}{2} p1=2p−1,q1=2q−1。称坐标都是整数的点为格子点。-
一方面,观察到矩形 O A B C OABC OABC除去边 O A , O C OA,OC OA,OC后,刚好有 p − 1 2 ⋅ q − 1 2 \frac{p-1}{2}\centerdot \frac{q-1}{2} 2p−1⋅2q−1个格子点。
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另一方面,直线 O K OK OK的方程是 k = q p h k=\frac{q}{p}h k=pqh,由于 p , q p,q p,q互素, p p p是素数,因此对于直线 O K OK OK,在 h ∈ [ 0 , p − 1 2 ] h\in \left[ 0,\frac{p-1}{2} \right] h∈[0,2p−1]的范围中,只有在 h = 0 h=0 h=0处 k = q p h k=\frac{q}{p}h k=pqh是整数即只有 O O O点是格子点。因此,矩形 O A B C OABC OABC除去边 O A , O C OA,OC OA,OC后的格子点不可能在直线 O K OK OK上,只可能在 O K OK OK上方和 O K OK OK下方。
现在我们证明矩形 O A B C OABC OABC除去边 O A , O C OA,OC OA,OC后在 O K OK OK下方的格子点的个数恰好就是 ∑ h = 1 p − 1 2 [ q h p ] \sum\limits_{h=1}^{\frac{p-1}{2}}{\left[ \frac{qh}{p} \right]} h=1∑2p−1[pqh]个。- 记 O K OK OK与直线 h = p − 1 2 h=\frac{p-1}{2} h=2p−1的交点为 M M M。
显然地,矩形 O A B C OABC OABC除去 O A , O C OA,OC OA,OC后在直线 O K OK OK下方的格子点都在除去边 O A OA OA后的三角形 O A M OAM OAM中。而三角形 O A M OAM OAM中除去边 O A OA OA后的格子点都在直线 h = a h=a h=a上, a = 1 , 2 , ⋯ , p − 1 2 a=1,2,\cdots ,\frac{p-1}{2} a=1,2,⋯,2p−1,且对于某一条直线 h = a h=a h=a,除去边 O A OA OA后的三角形 O A M OAM OAM在其上的格子点个数恰好为 [ q p a ] \left[ \frac{q}{p}a \right] [pqa]个。于是除去边 O A OA OA后的三角形 O A M OAM OAM中的格子点个数为
∑ h = 1 p − 1 2 [ q h p ] \sum\limits_{h=1}^{\frac{p-1}{2}}{\left[ \frac{qh}{p} \right]} h=1∑2p−1[pqh] - 同时,我们指出除去边 O A OA OA后的三角形 O A M OAM OAM中的所有格子点均为除去 O A , O C OA,OC OA,OC后的矩形 O A B C OABC OABC里的格子点。事实上,
O K OK OK与直线 h = p − 1 2 h=\frac{p-1}{2} h=2p−1的交点 M M M坐标为 ( h M , k M ) = ( p − 1 2 , q ( p − 1 ) 2 p ) \left( { {h}_{M}},{ {k}_{M}} \right)=\left( \frac{p-1}{2},\frac{q\left( p-1 \right)}{2p} \right) (hM,kM)=(2p−1,2pq(p−1)),点 B B B坐标为 ( h B , k B ) = ( p − 1 2 , q − 1 2 ) \left( { {h}_{B}},{ {k}_{B}} \right)=\left( \frac{p-1}{2},\frac{q-1}{2} \right) (hB,kB)=(2p−1,2q−1),其中有
k M = q ( p − 1 ) 2 p > p ( q − 1 ) 2 p = q − 1 2 = k B { {k}_{M}}=\frac{q\left( p-1 \right)}{2p}>\frac{p\left( q-1 \right)}{2p}=\frac{q-1}{2}={ {k}_{B}} kM=2pq(p−1)>2pp(q−1)=2q−1=kB
因此点 M M M在点 B B B上方。
因此三角形 O A M OAM OAM中只有三角形区域 B L M BLM BLM(不含边 B L BL BL)不在矩形 O A B C OABC OABC中。所以只需要证明不含边 B L BL BL的三角形 B L M BLM BLM中没有格子点即可。
O K OK OK与直线 k = q − 1 2 k=\frac{q-1}{2} k=2q−1的交点 L L L坐标为 ( h L , k L ) = ( p ( q − 1 ) 2 q , q − 1 2 ) \left( { {h}_{L}},{ {k}_{L}} \right)=\left( \frac{p\left( q-1 \right)}{2q},\frac{q-1}{2} \right) (hL,kL)=(2qp(q−1),2q−1)。
除去边 B L BL BL后的三角形 B L M BLM BLM的所有格子点都在直线 h = a h=a h=a上, a ∈ Z a\in \mathbb{Z} a∈Z且 h L < a ≤ p − 1 2 { {h}_{L}}<a\le \frac{p-1}{2} hL<a≤2p−1。
∣ B M ∣ = k M − k B = q ( p − 1 ) 2 p − q − 1 2 = p − q 2 p < p 2 p = 1 2 < 1 \left| BM \right|={ {k}_{M}}-{ {k}_{B}}=\frac{q\left( p-1 \right)}{2p}-\frac{q-1}{2}=\frac{p-q}{2p}<\frac{p}{2p}=\frac{1}{2}<1 ∣BM∣=kM−kB=2pq(p−1)−2q−1=2pp−q<2pp=21<1
边 B L BL BL上的点的纵坐标均为整数 q − 1 2 \frac{q-1}{2} 2q−1,而三角形 B L M BLM BLM中任意平行于边 B M BM BM的线段长度均 ≤ ∣ B M ∣ < 1 \le \left| BM \right|<1 ≤∣BM∣<1,因此 ∀ a ∈ Z & h L < a ≤ p − 1 2 \forall a\in \mathbb{Z}\text{ }\And \text{ }{ {h}_{L}}<a\le \frac{p-1}{2} ∀a∈Z & hL<a≤2p−1,直线 h = a h=a h=a在除去边 B L BL BL后的三角形 B L M BLM BLM中的部分均没有格子点,因此不含边 B L BL BL的三角形 B L M BLM BLM中没有格子点。 - 因此除去边 O A , O C OA,OC OA,OC后的矩形 O A B C OABC OABC在直线 O K OK OK下方的格子点总共有 ∑ h = 1 p − 1 2 [ q h p ] \sum\limits_{h=1}^{\frac{p-1}{2}}{\left[ \frac{qh}{p} \right]} h=1∑2p−1[pqh]个。同理,除去边 O A , O C OA,OC OA,OC后的矩形 O A B C OABC OABC在直线 O K OK OK上方的格子点总共有 ∑ k = 1 q − 1 2 [ p k q ] \sum\limits_{k=1}^{\frac{q-1}{2}}{\left[ \frac{pk}{q} \right]} k=1∑2q−1[qpk]个。
因此除去边 O A , O C OA,OC OA,OC后的矩形 O A B C OABC OABC的格子点总个数为
∑ h = 1 p − 1 2 [ q h p ] + ∑ k = 1 q − 1 2 [ p k q ] \sum\limits_{h=1}^{\frac{p-1}{2}}{\left[ \frac{qh}{p} \right]}+\sum\limits_{k=1}^{\frac{q-1}{2}}{\left[ \frac{pk}{q} \right]} h=1∑2p−1[pqh]+k=1∑2q−1[qpk]
- 记 O K OK OK与直线 h = p − 1 2 h=\frac{p-1}{2} h=2p−1的交点为 M M M。
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因此证得了
∑ h = 1 p − 1 2 [ q h p ] + ∑ k = 1 q − 1 2 [ p k q ] = p − 1 2 ⋅ q − 1 2 \sum\limits_{h=1}^{\frac{p-1}{2}}{\left[ \frac{qh}{p} \right]}+\sum\limits_{k=1}^{\frac{q-1}{2}}{\left[ \frac{pk}{q} \right]}=\frac{p-1}{2}\centerdot \frac{q-1}{2} h=1∑2p−1[pqh]+k=1∑2q−1[qpk]=2p−1⋅2q−1
因此也就有
( − 1 ) p − 1 2 ⋅ q − 1 2 = ( − 1 ) ∑ h = 1 p − 1 2 [ q h p ] + ∑ k = 1 q − 1 2 [ p k q ] { {\left( -1 \right)}^{\frac{p-1}{2}\centerdot \frac{q-1}{2}}}={ {\left( -1 \right)}^{\sum\limits_{h=1}^{\frac{p-1}{2}}{\left[ \frac{qh}{p} \right]}+\sum\limits_{k=1}^{\frac{q-1}{2}}{\left[ \frac{pk}{q} \right]}}} (−1)2p−1⋅2q−1=(−1)h=1∑2p−1[pqh]+k=1∑2q−1[qpk]
从而原结论得证。
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