索引
- 定义
- 意义
- 若干性质
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- 1. ( a p ) ≡ a p − 1 2 m o d p \left( \frac{a}{p} \right)\equiv { {a}^{\frac{p-1}{2}}}\text{ }\bmod p (pa)≡a2p−1 modp
- 2. ( 1 p ) = 1 \left( \frac{1}{p} \right)=1 (p1)=1
- 3. ( − 1 p ) = ( − 1 ) p − 1 2 = { 1 , p ≡ 1 m o d 4 − 1 , p ≡ 3 m o d 4 \left( \frac{-1}{p} \right)={ {\left( -1 \right)}^{\frac{p-1}{2}}}=\left\{ \begin{aligned} & 1,\text{ }p\equiv 1\text{ }\bmod 4 \\ & -1,\text{ }p\equiv 3\text{ }\bmod 4 \\ \end{aligned} \right. (p−1)=(−1)2p−1={ 1, p≡1 mod4−1, p≡3 mod4
- 4. a ≡ b m o d p ⇒ ( a p ) = ( b p ) a\equiv b\text{ }\bmod p\text{ }\Rightarrow \text{ }\left( \frac{a}{p} \right)=\left( \frac{b}{p} \right) a≡b modp ⇒ (pa)=(pb)
- 5-1. ( a b p ) = ( a p ) ( b p ) \left( \frac{ab}{p} \right)=\left( \frac{a}{p} \right)\left( \frac{b}{p} \right) (pab)=(pa)(pb)
- 5-2. ( a 1 a 2 ⋯ a n p ) = ( a 1 p ) ( a 2 p ) ⋯ ( a n p ) \left( \frac{ { {a}_{1}}{ {a}_{2}}\cdots { {a}_{n}}}{p} \right)=\left( \frac{ { {a}_{1}}}{p} \right)\left( \frac{ { {a}_{2}}}{p} \right)\cdots \left( \frac{ { {a}_{n}}}{p} \right) (pa1a2⋯an)=(pa1)(pa2)⋯(pan)
- 6. p ∣ b ⇒ ( a b 2 p ) = ( a p ) p\cancel{|}b\text{ }\Rightarrow \text{ }\left( \frac{a{ {b}^{2}}}{p} \right)=\left( \frac{a}{p} \right) p∣ b ⇒ (pab2)=(pa)
- 7. ( 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
- 8. 若 gcd ( a , p ) = 1 , 2 ∣ a \gcd \left( a,p \right)=1,\text{ }2\cancel{|}a gcd(a,p)=1, 2∣ a，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]
- 9. 若 p , q p,q p,q均为奇素数， gcd ( p , q ) = 1 \gcd \left( p,q \right)=1 gcd(p,q)=1（即奇素数 p ≠ q p \ne q p=q），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)
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- 推论：设 p ≠ q p\ne q p=q都是奇素数，则有 ( q p ) = { ( p q ) , p ≡ 1 m o d 4 或 q ≡ 1 m o d 4 − ( p q ) , p ≡ 3 m o d 4 且 q ≡ 3 m o d 4 \left( \frac{q}{p} \right)=\left\{ \begin{aligned} & \left( \frac{p}{q} \right),p\equiv 1\text{ }\bmod 4\text{ }或\text{ }q\equiv 1\text{ }\bmod 4 \\ & -\left( \frac{p}{q} \right),p\equiv 3\text{ }\bmod 4\text{ }且\text{ }q\equiv 3\text{ }\bmod 4 \\ \end{aligned} \right. (pq)=⎩⎪⎪⎪⎨⎪⎪⎪⎧(qp),p≡1 mod4 或 q≡1 mod4−(qp),p≡3 mod4 且 q≡3 mod4
- 命题: 存在无穷多个素数 p p p，满足 p ≡ 1 m o d 4 p\equiv 1\text{ }\bmod 4 p≡1 mod4。
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