索引
- 记号
- 引理1 设 n ∈ Z > 0 n\in { {\mathbb{Z}}_{>0}} n∈Z>0, x , y , z , w ∈ Z x,y,z,w\in \mathbb{Z} x,y,z,w∈Z, 成立恒等式 ( x 2 + n y 2 ) ( z 2 + n w 2 ) = ( x z + n y w ) 2 + n ( x w − y z ) 2 . \left( { {x}^{2}}+n{ {y}^{2}} \right)\left( { {z}^{2}}+n{ {w}^{2}} \right)={ {\left( xz+nyw \right)}^{2}}+n{ {\left( xw-yz \right)}^{2}}. (x2+ny2)(z2+nw2)=(xz+nyw)2+n(xw−yz)2.
- 推论1-1 设 n ∈ Z > 0 n\in { {\mathbb{Z}}_{>0}} n∈Z>0, x , y , z , w ∈ Z x,y,z,w\in \mathbb{Z} x,y,z,w∈Z, 成立恒等式 ( x 2 + n y 2 ) ( z 2 + n w 2 ) = ( x z − n y w ) 2 + n ( x w − y z ) 2 . \left( { {x}^{2}}+n{ {y}^{2}} \right)\left( { {z}^{2}}+n{ {w}^{2}} \right)={ {\left( xz-nyw \right)}^{2}}+n{ {\left( xw-yz \right)}^{2}}. (x2+ny2)(z2+nw2)=(xz−nyw)2+n(xw−yz)2.
- 引理2 设 n ∈ Z > 0 n\in { {\mathbb{Z}}_{>0}} n∈Z>0, N = a 2 + n b 2 N={ {a}^{2}}+n{ {b}^{2}} N=a2+nb2, a , b ∈ Z a,b\in \mathbb{Z} a,b∈Z. 设 q = x 2 + n y 2 q={ {x}^{2}}+n{ {y}^{2}} q=x2+ny2是素数, x , y ∈ Z x,y\in \mathbb{Z} x,y∈Z, gcd ( q , n ) = 1 \gcd \left( q,n \right)=1 gcd(q,n)=1, q ∣ N \left. q \right|N q∣N. 则存在 c , d ∈ Z c,d\in \mathbb{Z} c,d∈Z, 满足 N q = c 2 + n d 2 \frac{N}{q}={ {c}^{2}}+n{ {d}^{2}} qN=c2+nd2.
- 引理3 (下降) 设 a , b ∈ Z a,b\in \mathbb{Z} a,b∈Z, n ∈ { 1 , 2 , 3 } n\in \left\{ 1,2,3 \right\} n∈{ 1,2,3}, p p p是一奇素数, p ∣ N = a 2 + n b 2 \left. p \right|N={ {a}^{2}}+n{ {b}^{2}} p∣N=a2+nb2. 则 ∃ c , d ∈ Z \exists c,d\in \mathbb{Z} ∃c,d∈Z, 满足 p = c 2 + n d 2 p={ {c}^{2}}+n{ {d}^{2}} p=c2+nd2.
- 引理4 设 n ∈ Z n\in \mathbb{Z} n∈Z, p p p是素数, gcd ( p , n ) = 1 \gcd \left( p,n \right)=1 gcd(p,n)=1. 则 ( − n p ) = 1 ⇔ \left( \frac{-n}{p} \right)=1\text{ }\Leftrightarrow \text{ } (p−n)=1 ⇔ ∃ x , y ∈ Z \exists x,y\in \mathbb{Z} ∃x,y∈Z, gcd ( x , y ) = 1 \gcd \left( x,y \right)=1 gcd(x,y)=1, 满足 p ∣ ( x 2 + n y 2 ) \left. p \right|\left( { {x}^{2}}+n{ {y}^{2}} \right) p∣(x2+ny2).
- 定理5 (Fermat) 设 n ∈ { 1 , 2 , 3 } n\in \left\{ 1,2,3 \right\} n∈{ 1,2,3}, p p p是一奇素数, gcd ( p , n ) = 1 \gcd \left( p,n \right)=1 gcd(p,n)=1. 成立等价关系 p = c 2 + n d 2 , ∃ c , d ∈ Z ⇔ ( − n p ) = 1 ⇔ p ≡ { 1 m o d 4 , n = 1 , 1 , 3 m o d 8 , n = 2 , 1 m o d 3 , n = 3. p={ {c}^{2}}+n{ {d}^{2}},\text{ }\exists c,d\in \mathbb{Z}\text{ }\Leftrightarrow \text{ }\left( \frac{-n}{p} \right)=1\text{ }\Leftrightarrow \text{ }p\equiv \left\{ \begin{aligned} & 1\text{ }\bmod 4,\text{ }n=1, \\ & 1,3\text{ }\bmod 8,\text{ }n=2, \\ & 1\text{ }\bmod 3,\text{ }n=3. \\ \end{aligned} \right. p=c2+nd2, ∃c,d∈Z ⇔ (p−n)=1 ⇔ p≡⎩⎪⎨⎪⎧1 mod4, n=1,1,3 mod8, n=2,1 mod3, n=3.
- 推论6 不定方程 p = x 2 + y 2 p={ {x}^{2}}+{ {y}^{2}} p=x2+y2有(正)整数解 ⇔ \Leftrightarrow ⇔ p = 4 m + 1 p=4m+1 p=4m+1.
- 推论7 不定方程 p = x 2 + 2 y 2 p={ {x}^{2}}+2{ {y}^{2}} p=x2+2y2有(正)整数解 ⇔ \Leftrightarrow ⇔ ( − 2 p ) = 1 \left( \frac{-2}{p} \right)=1 (p−2)=1, 即 p = 8 m + 1 p=8m+1 p=8m+1或 p = 8 m + 3 p=8m+3 p=8m+3.
- 推论8 不定方程 p = x 2 + 3 y 2 p={ {x}^{2}}+3{ {y}^{2}} p=x2+3y2有(正)整数解 ⇔ \Leftrightarrow ⇔ ( − 3 p ) = 1 \left( \frac{-3}{p} \right)=1 (p−3)=1.
记号
(1) p i ∥ x ⇔ p i ∣ x ∧ p i + 1 ∣ x \left. {
{p}^{i}} \right\|x\text{ }\Leftrightarrow \text{ }\left. {
{p}^{i}} \right|x\text{ }\wedge \text{ }{
{p}^{i+1}}\cancel{|}x pi∥∥x ⇔ pi∣∣x ∧ pi+1∣
x.
引理1 设 n ∈ Z > 0 n\in { {\mathbb{Z}}_{>0}} n∈Z>0, x , y , z , w ∈ Z x,y,z,w\in \mathbb{Z} x,y,z,w∈Z, 成立恒等式 ( x 2 + n y 2 ) ( z 2 + n w 2 ) = ( x z + n y w ) 2 + n ( x w − y z ) 2 . \left( { {x}^{2}}+n{ {y}^{2}} \right)\left( { {z}^{2}}+n{ {w}^{2}} \right)={ {\left( xz+nyw \right)}^{2}}+n{ {\left( xw-yz \right)}^{2}}. (x2+ny2)(z2+nw2)=(xz+nyw)2+n(xw−yz)2.
证明
( x 2 + n y 2 ) ( z 2 + n w 2 ) = x 2 z 2 + n y 2 z 2 + n x 2 w 2 + n 2 y 2 w 2 = [ ( x z ) 2 + 2 n x y z w + ( n y w ) 2 ] + n [ ( x w ) 2 − 2 x y z w + ( y z ) 2 ] = ( x z + n y w ) 2 + n ( x w − y z ) 2 . \begin{aligned} & \left( {
{x}^{2}}+n{
{y}^{2}} \right)\left( {
{z}^{2}}+n{
{w}^{2}} \right) \\ & ={
{x}^{2}}{
{z}^{2}}+n{
{y}^{2}}{
{z}^{2}}+n{
{x}^{2}}{
{w}^{2}}+{
{n}^{2}}{
{y}^{2}}{
{w}^{2}} \\ & =\left[ {
{\left( xz \right)}^{2}}+2nxyzw+{
{\left( nyw \right)}^{2}} \right]+n\left[ {
{\left( xw \right)}^{2}}-2xyzw+{
{\left( yz \right)}^{2}} \right] \\ & ={
{\left( xz+nyw \right)}^{2}}+n{
{\left( xw-yz \right)}^{2}}. \\ \end{aligned} (x2+ny2)(z2+nw2)=x2z2+ny2z2+nx2w2+n2y2w2=[(xz)2+2nxyzw+(nyw)2]+n[(xw)2−2xyzw+(yz)2]=(xz+nyw)2+n(xw−yz)2.
推论1-1 设 n ∈ Z > 0 n\in { {\mathbb{Z}}_{>0}} n∈Z>0, x , y , z , w ∈ Z x,y,z,w\in \mathbb{Z} x,y,z,w∈Z, 成立恒等式 ( x 2 + n y 2 ) ( z 2 + n w 2 ) = ( x z − n y w ) 2 + n ( x w − y z ) 2 . \left( { {x}^{2}}+n{ {y}^{2}} \right)\left( { {z}^{2}}+n{ {w}^{2}} \right)={ {\left( xz-nyw \right)}^{2}}+n{ {\left( xw-yz \right)}^{2}}. (x2+ny2)(z2+nw2)=(xz−nyw)2+n(xw−yz)2.
证明 将引理1中的 y y y替换为 − y -y −y即可得到该结论.
引理2 设 n ∈ Z > 0 n\in { {\mathbb{Z}}_{>0}} n∈Z>0, N = a 2 + n b 2 N={ {a}^{2}}+n{ {b}^{2}} N=a2+nb2, a , b ∈ Z a,b\in \mathbb{Z} a,b∈Z. 设 q = x 2 + n y 2 q={ {x}^{2}}+n{ {y}^{2}} q=x2+ny2是素数, x , y ∈ Z x,y\in \mathbb{Z} x,y∈Z, gcd ( q , n ) = 1 \gcd \left( q,n \right)=1 gcd(q,n)=1, q ∣ N \left. q \right|N q∣N. 则存在 c , d ∈ Z c,d\in \mathbb{Z} c,d∈Z, 满足 N q = c 2 + n d 2 \frac{N}{q}={ {c}^{2}}+n{ {d}^{2}} qN=c2+nd2.
证明
第一步, 我们指出成立
q ∣ n ( x b − a y ) ( x b + a y ) . (2.1) \left. q \right|n\left( xb-ay \right)\left( xb+ay \right). \tag{2.1} q∣n(xb−ay)(xb+ay).(2.1)
事实上, 由于 q ∣ N \left. q \right|N q∣N, 因此有 q ∣ ( x 2 N − a 2 q ) \left. q \right|\left( {
{x}^{2}}N-{
{a}^{2}}q \right) q∣(x2N−a2q), 即有
q ∣ x 2 ( a 2 + n b 2 ) − a 2 ( x 2 + n y 2 ) = n ( x 2 b 2 − a 2 y 2 ) = n ( x b − a y ) ( x b + a y ) . \left. q \right|{
{x}^{2}}\left( {
{a}^{2}}+n{
{b}^{2}} \right)-{
{a}^{2}}\left( {
{x}^{2}}+n{
{y}^{2}} \right)=n\left( {
{x}^{2}}{
{b}^{2}}-{
{a}^{2}}{
{y}^{2}} \right)=n\left( xb-ay \right)\left( xb+ay \right). q∣x2(a2+nb2)−a2(x2+ny2)=n(x2b2−a2y2)=n(xb−ay)(xb+ay).
第二步, 我们指出成立
q ∣ x b − a y 或 q ∣ x b + a y . (2.2) \left. q \right|xb-ay\text{ }或\text{ }\left. q \right|xb+ay. \tag{2.2} q∣xb−ay 或 q∣xb+ay.(2.2)
事实上, 由 gcd ( q , n ) = 1 \gcd \left( q,n \right)=1 gcd(q,n)=1和式(2.1), 有 q ∣ ( x b − y a ) ( x b + y a ) \left. q \right|\left( xb-ya \right)\left( xb+ya \right) q∣(xb−ya)(xb+ya). 又由于 q q q是素数, 因此式(2.2)成立. 基于此, 不妨设 q ∣ x b − a y \left. q \right|xb-ay q∣xb−ay, 即 ∃ d ∈ Z \exists d\in \mathbb{Z} ∃d∈Z, 使得
x b − a y = d q . (2.3) xb-ay=dq. \tag{2.3} xb−ay=dq.(2.3)
第三步, 由式(2.3)和 q = x 2 + n y 2 q={
{x}^{2}}+n{
{y}^{2}} q=x2+ny2, 成立等式
( a + n d y ) y = a y + n d y 2 = ( x b − d q ) + n d y 2 = x b − d ( q − n y 2 ) = x b − d x 2 = ( b − d x ) x . (2.4) \begin{aligned} & \left( a+ndy \right)y=ay+nd{
{y}^{2}} \\ & =\left( xb-dq \right)+nd{
{y}^{2}} \\ & =xb-d\left( q-n{
{y}^{2}} \right) \\ & =xb-d{
{x}^{2}} \\ & =\left( b-dx \right)x. \\ \end{aligned} \tag{2.4} (a+ndy)y=ay+ndy2=(xb−dq)+ndy2=xb−d(q−ny2)=xb−dx2=(b−dx)x.(2.4)
因此成立
x ∣ ( a + n d y ) y . (2.5) \left. x \right|\left( a+ndy \right)y. \tag{2.5} x∣(a+ndy)y.(2.5)
第四步, 我们指出成立
gcd ( x , y ) = 1. (2.6) \gcd \left( x,y \right)=1. \tag{2.6} gcd(x,y)=1.(2.6)
事实上, 有
{ gcd ( x , y ) ∣ x gcd ( x , y ) ∣ y ⇒ { gcd ( x , y ) 2 ∣ x 2 gcd ( x , y ) 2 ∣ n y 2 ⇒ gcd ( x , y ) 2 ∣ ( x 2 + n y 2 ) = q . \left\{ \begin{aligned} & \left. \gcd \left( x,y \right) \right|x \\ & \left. \gcd \left( x,y \right) \right|y \\ \end{aligned} \right.\text{ }\Rightarrow \text{ }\left\{ \begin{aligned} & \left. \gcd {
{\left( x,y \right)}^{2}} \right|{
{x}^{2}} \\ & \left. \gcd {
{\left( x,y \right)}^{2}} \right|n{
{y}^{2}} \\ \end{aligned} \right.\text{ }\Rightarrow \text{ }\left. \gcd {
{\left( x,y \right)}^{2}} \right|\left( {
{x}^{2}}+n{
{y}^{2}} \right)=q. {
gcd(x,y)∣xgcd(x,y)∣y ⇒ ⎩⎪⎨⎪⎧gcd(x,y)2∣∣∣x2gcd(x,y)2∣∣∣ny2 ⇒ gcd(x,y)2∣∣∣(x2+ny2)=q.
而 q q q是素数, 因此只能是
gcd ( x , y ) 2 = 1 或 q ⇒ gcd ( x , y ) = 1 或 q ∉ Z ( 舍 ) . \gcd {
{\left( x,y \right)}^{2}}=1或q\text{ }\Rightarrow \text{ }\gcd \left( x,y \right)=1\text{ }或\text{ }\sqrt[{}]{q}\notin \mathbb{Z}\left( 舍 \right). gcd(x,y)2=1或q ⇒ gcd(x,y)=1 或 q∈/Z(舍).
第五步, 由式(2.5), 式(2.6), 有 x ∣ ( a + n d y ) \left. x \right|\left( a+ndy \right) x∣(a+ndy), ∃ c ∈ Z \exists c\in \mathbb{Z} ∃c∈Z, 使得
a + n d y = c x . (2.7) a+ndy=cx. \tag{2.7} a+ndy=cx.(2.7)
由式(2.7), 式(2.4), 成立
c x y = ( a + n d y ) y = ( b − d x ) x . (2.8) cxy=\left( a+ndy \right)y=\left( b-dx \right)x. \tag{2.8} cxy=(a+ndy)y=(b−dx)x.(2.8)
第六步, 我们指出一定成立
x ≠ 0. (2.9) x\ne 0. \tag{2.9} x=0.(2.9)
事实上, 若 x = 0 x=0 x=0, 则 q = 0 2 + n y 2 = n y 2 q={
{0}^{2}}+n{
{y}^{2}}=n{
{y}^{2}} q=02+ny2=ny2. 此时有
1 = gcd ( q , n ) = gcd ( n y 2 , n ) = n . 1=\gcd \left( q,n \right)=\gcd \left( n{
{y}^{2}},n \right)=n. 1=gcd(q,n)=gcd(ny2,n)=n.
因此有 q = y 2 q={
{y}^{2}} q=y2是素数, 而这是不可能的.
第七步, 由式(2.8), 式(2.9), 成立
c y = b − d x . (2.10) cy=b-dx. \tag{2.10} cy=b−dx.(2.10)
至此, 由式(2.7), 式(2.10), q = x 2 + n y 2 q={
{x}^{2}}+n{
{y}^{2}} q=x2+ny2和推论1-1, 成立
N = a 2 + n b 2 = ( c x − n d y ) 2 + n ( d x + c y ) 2 = ( c 2 + n d 2 ) ( x 2 + n y 2 ) = ( c 2 + n d 2 ) q . \begin{aligned} & N={
{a}^{2}}+n{
{b}^{2}} \\ & ={
{\left( cx-ndy \right)}^{2}}+n{
{\left( dx+cy \right)}^{2}} \\ & =\left( {
{c}^{2}}+n{
{d}^{2}} \right)\left( {
{x}^{2}}+n{
{y}^{2}} \right) \\ & =\left( {
{c}^{2}}+n{
{d}^{2}} \right)q. \\ \end{aligned} N=a2+nb2=(cx−ndy)2+n(dx+cy)2=(c2+nd2)(x2+ny2)=(c2+nd2)q.
即有
N q = c 2 + n d 2 . \frac{N}{q}={
{c}^{2}}+n{
{d}^{2}}. qN=c2+nd2.
引理3 (下降) 设 a , b ∈ Z a,b\in \mathbb{Z} a,b∈Z, n ∈ { 1 , 2 , 3 } n\in \left\{ 1,2,3 \right\} n∈{ 1,2,3}, p p p是一奇素数, p ∣ N = a 2 + n b 2 \left. p \right|N={ {a}^{2}}+n{ {b}^{2}} p∣N=a2+nb2. 则 ∃ c , d ∈ Z \exists c,d\in \mathbb{Z} ∃c,d∈Z, 满足 p = c 2 + n d 2 p={ {c}^{2}}+n{ {d}^{2}} p=c2+nd2.
证明
第一步, ∃ a ′ , b ′ ∈ Z \exists a',\text{ }b'\in \mathbb{Z} ∃a′, b′∈Z, 满足 ∣ a ′ ∣ ≤ p 2 , ∣ b ′ ∣ ≤ p 2 \left| a' \right|\le \frac{p}{2},\text{ }\left| b' \right|\le \frac{p}{2} ∣a′∣≤2p, ∣b′∣≤2p, 成立
a = p s + a ′ , b = p t + b ′ . (3.1) a=ps+a',\text{ }b=pt+b'. \tag{3.1} a=ps+a′, b=pt+b′.(3.1)
且由于 p p p是奇数, 因此实际上成立
∣ a ′ ∣ < p 2 , ∣ b ′ ∣ < p 2 . (3.2) \left| a' \right|<\frac{p}{2},\text{ }\left| b' \right|<\frac{p}{2}. \tag{3.2} ∣a′∣<2p, ∣b′∣<2p.(3.2)
此时有
N = a 2 + n b 2 = ( p s + a ′ ) 2 + n ( p t + b ′ ) 2 = ( a ′ 2 + n b ′ 2 ) + ( p 2 s 2 + 2 p s a ′ + n ( p 2 t 2 + 2 p t b ′ ) ) = ( a ′ 2 + n b ′ 2 ) + p ( p s 2 + 2 s a ′ + n p t 2 + 2 n t b ′ ) . \begin{aligned} & N={
{a}^{2}}+n{
{b}^{2}}={
{\left( ps+a' \right)}^{2}}+n{
{\left( pt+b' \right)}^{2}} \\ & =\left( a{
{'}^{2}}+nb{
{'}^{2}} \right)+\left( {
{p}^{2}}{
{s}^{2}}+2psa'+n\left( {
{p}^{2}}{
{t}^{2}}+2ptb' \right) \right) \\ & =\left( a{
{'}^{2}}+nb{
{'}^{2}} \right)+p\left( p{
{s}^{2}}+2sa'+np{
{t}^{2}}+2ntb' \right). \\ \end{aligned} N=a2+nb2=(ps+a′)2+n(pt+b′)2=(a′2+nb′2)+(p2s2+2psa′+n(p2t2+2ptb′))=(a′2+nb′2)+p(ps2+2sa′+npt2+2ntb′).
又由于 p ∣ N \left. p \right|N p∣N, 于是有
p ∣ ( a ′ 2 + n b ′ 2 ) = : N ′ . (3.3) \left. p \right|\left( a{
{'}^{2}}+nb{
{'}^{2}} \right)=:N'. \tag{3.3} p∣(a′2+nb′2)=:N′.(3.3)
且成立
0 ≤ N ′ = a ′ 2 + n b ′ 2 < ( p 2 ) 2 + n ( p 2 ) 2 = ( 1 + n ) p 2 4 ≤ p 2 . (3.4) 0\le N'=a{
{'}^{2}}+nb{
{'}^{2}}<{
{\left( \frac{p}{2} \right)}^{2}}+n{
{\left( \frac{p}{2} \right)}^{2}}=\frac{\left( 1+n \right){
{p}^{2}}}{4}\le {
{p}^{2}}. \tag{3.4} 0≤N′=a′2+nb′2<(2p)2+n(2p)2=4(1+n)p2≤p2.(3.4)
第二步, 设 q q q是素数, q ∣ N \left. q \right|N q∣N, 则成立推理
若 q ≠ p ⇒ q < p . (3.5) 若q\ne p\text{ }\Rightarrow \text{ }q<p. \tag{3.5} 若q=p ⇒ q<p.(3.5)
事实上, 由于 p , q p,q p,q均为素数, p ≠ q p\ne q p=q, 且 p ∣ N , q ∣ N \left. p \right|N,\text{ }\left. q \right|N p∣N, q∣N, 因此有
q ∣ N p ⇒ q ≤ N p < p 2 p = p . \left. q \right|\frac{N}{p}\text{ }\Rightarrow \text{ }q\le \frac{N}{p}<\frac{
{
{p}^{2}}}{p}=p. q∣pN ⇒ q≤pN<pp2=p.
第三步, 我们指出对于任意素数 q q q, 成立推理
若 q 具 有 形 式 x 2 + n y 2 , n ∈ { 1 , 2 , 3 } ⇒ gcd ( q , n ) = 1. (3.6) 若q具有形式{
{x}^{2}}+n{
{y}^{2}},\text{ }n\in \left\{ 1,2,3 \right\}\text{ }\Rightarrow \text{ }\gcd \left( q,n \right)=1. \tag{3.6} 若q具有形式x2+ny2, n∈{
1,2,3} ⇒ gcd(q,n)=1.(3.6)
若 q = 2 q=2 q=2, 则有 q = 1 2 + 1 × 1 2 = 2 q={
{1}^{2}}+1\times {
{1}^{2}}=2 q=12+1×12=2, gcd ( 2 , 1 ) = 1 \gcd \left( 2,1 \right)=1 gcd(2,1)=1.
若 q = 3 q=3 q=3, 则有 q = 1 2 + 2 × 1 2 = 3 q={
{1}^{2}}+2\times {
{1}^{2}}=3 q=12+2×12=3, gcd ( 3 , 2 ) = 1 \gcd \left( 3,2 \right)=1 gcd(3,2)=1.
若 q ≥ 5 q\ge 5 q≥5, 则由于 n ∈ { 1 , 2 , 3 } , n < q n\in \left\{ 1,2,3 \right\},\text{ }n<q n∈{
1,2,3}, n<q, q q q是素数, 因此一定有 gcd ( q , n ) = 1 \gcd \left( q,n \right)=1 gcd(q,n)=1.
综上, 推理(3.6)成立.
第四步, 我们指出若 p p p非 x 2 + n y 2 {
{x}^{2}}+n{
{y}^{2}} x2+ny2形式, 则存在素数 q < p q<p q<p非 x 2 + n y 2 {
{x}^{2}}+n{
{y}^{2}} x2+ny2形式.
事实上, 由式(3.4), 有
p 2 ∣ N ′ . (3.7) {
{p}^{2}}\cancel{|}N'. \tag{3.7} p2∣
N′.(3.7)
由式(3.3), (3.7), 有
p ∥ N ′ . (3.8) \left. p \right\|N'. \tag{3.8} p∥N′.(3.8)
N ′ N' N′存在不等于 p p p的其他素因子(否则由式(3.8), N ′ = p N'=p N′=p非 x 2 + n y 2 {
{x}^{2}}+n{
{y}^{2}} x2+ny2形式, 与 N ′ = a ′ 2 + n b ′ 2 N'=a{
{'}^{2}}+nb{
{'}^{2}} N′=a′2+nb′2矛盾). 基于此, 可将 N ′ N' N′进行素因子分解得到
N ′ = p ( p 1 α 1 p 2 α 2 ⋯ p s α s ) , N'=p\left( {
{p}_{1}}^{
{
{\alpha }_{1}}}{
{p}_{2}}^{
{
{\alpha }_{2}}}\cdots {
{p}_{s}}^{
{
{\alpha }_{s}}} \right), N′=p(p1α1p2α2⋯psαs),
其中 p 1 , p 2 , ⋯ , p s {
{p}_{1}},{
{p}_{2}},\cdots ,{
{p}_{s}} p1,p2,⋯,ps是不等于 p p p的素数.
若 p 1 , p 2 , ⋯ , p s {
{p}_{1}},{
{p}_{2}},\cdots ,{
{p}_{s}} p1,p2,⋯,ps均具有 x 2 + n y 2 {
{x}^{2}}+n{
{y}^{2}} x2+ny2的形式, 其中 n ∈ { 1 , 2 , 3 } n\in \left\{ 1,2,3 \right\} n∈{
1,2,3}, 由推理(3.6), 成立
gcd ( p j , n ) = 1 , j = 1 , 2 , ⋯ , s . \gcd \left( {
{p}_{j}},n \right)=1,\text{ }j=1,2,\cdots ,s. gcd(pj,n)=1, j=1,2,⋯,s.
由素数 p 1 ∣ N ′ = a ′ 2 + n b ′ 2 \left. {
{p}_{1}} \right|N'=a{
{'}^{2}}+nb{
{'}^{2}} p1∣N′=a′2+nb′2, p 1 {
{p}_{1}} p1具有形式 x 2 + n y 2 {
{x}^{2}}+n{
{y}^{2}} x2+ny2, gcd ( p 1 , n ) = 1 \gcd \left( {
{p}_{1}},n \right)=1 gcd(p1,n)=1, 根据引理2, ∃ a 1 ′ , b 1 ′ ∈ Z \exists {
{a}_{1}}',\text{ }{
{b}_{1}}'\in \mathbb{Z} ∃a1′, b1′∈Z使得 N ′ p 1 = ( a 1 ′ ) 2 + n ( b 1 ′ ) 2 . \frac{N'}{
{
{p}_{1}}}={
{\left( {
{a}_{1}}' \right)}^{2}}+n{
{\left( {
{b}_{1}}' \right)}^{2}}. p1N′=(a1′)2+n(b1′)2.
由素数 p 1 ∣ N ′ p 1 = ( a 1 ′ ) 2 + n ( b 1 ′ ) 2 \left. {
{p}_{1}} \right|\frac{N'}{
{
{p}_{1}}}={
{\left( {
{a}_{1}}' \right)}^{2}}+n{
{\left( {
{b}_{1}}' \right)}^{2}} p1∣p1N′=(a1′)2+n(b1′)2, p 1 {
{p}_{1}} p1具有形式 x 2 + n y 2 {
{x}^{2}}+n{
{y}^{2}} x2+ny2, gcd ( p 1 , n ) = 1 \gcd \left( {
{p}_{1}},n \right)=1 gcd(p1,n)=1, 根据引理2, ∃ a 2 ′ , b 2 ′ ∈ Z \exists {
{a}_{2}}',\text{ }{
{b}_{2}}'\in \mathbb{Z} ∃a2′, b2′∈Z使得 N ′ p 1 2 = ( a 2 ′ ) 2 + n ( b 2 ′ ) 2 \frac{N'}{
{
{p}_{1}}^{2}}={
{\left( {
{a}_{2}}' \right)}^{2}}+n{
{\left( {
{b}_{2}}' \right)}^{2}} p12N′=(a2′)2+n(b2′)2.
继续下去, 可得到 ∃ a α 1 ′ , b α 1 ′ ∈ Z \exists {
{a}_{
{
{\alpha }_{1}}}}',\text{ }{
{b}_{
{
{\alpha }_{1}}}}'\in \mathbb{Z} ∃aα1′, bα1′∈Z使得 N ′ p 1 α 1 = ( a α 1 ′ ) + n ( b α 1 ′ ) 2 \frac{N'}{
{
{p}_{1}}^{
{
{\alpha }_{1}}}}=\left( {
{a}_{
{
{\alpha }_{1}}}}' \right)+n{
{\left( {
{b}_{
{
{\alpha }_{1}}}}' \right)}^{2}} p1α1N′=(aα1′)+n(bα1′)2, 继而最终可以得到 ∃ a ∑ j = 1 s α j ′ , b ∑ j = 1 s α j ′ ∈ Z \exists {
{a}_{\sum\limits_{j=1}^{s}{
{
{\alpha }_{j}}}}}',\text{ }{
{b}_{\sum\limits_{j=1}^{s}{
{
{\alpha }_{j}}}}}'\in \mathbb{Z} ∃aj=1∑sαj′, bj=1∑sαj′∈Z满足
p = N ′ ∏ j = 1 s p j α j = ( a ∑ j = 1 s α j ′ ) 2 + n ( b ∑ j = 1 s α j ′ ) 2 . p=\frac{N'}{\prod\limits_{j=1}^{s}{
{
{p}_{j}}^{
{
{\alpha }_{j}}}}}={
{\left( {
{a}_{\sum\limits_{j=1}^{s}{
{
{\alpha }_{j}}}}}' \right)}^{2}}+n{
{\left( {
{b}_{\sum\limits_{j=1}^{s}{
{
{\alpha }_{j}}}}}' \right)}^{2}}. p=j=1∏spjαjN′=⎝⎛aj=1∑sαj′⎠⎞2+n⎝⎛bj=1∑sαj′⎠⎞2.
这与 p p p非 x 2 + n y 2 {
{x}^{2}}+n{
{y}^{2}} x2+ny2形式矛盾. 由反证法, ∃ q ∈ { p 1 , p 2 , ⋯ , p s } \exists q\in \left\{ {
{p}_{1}},{
{p}_{2}},\cdots ,{
{p}_{s}} \right\} ∃q∈{
p1,p2,⋯,ps}非 x 2 + n y 2 {
{x}^{2}}+n{
{y}^{2}} x2+ny2形式且 q ≠ p q\ne p q=p, 由推理(3.5), q < p q<p q<p(且由第三步讨论的第一种情况, 一定有 q ≠ 2 q\ne 2 q=2, 即 q q q是奇素数).
第五步, 第四步中得到了非 x 2 + n y 2 {
{x}^{2}}+n{
{y}^{2}} x2+ny2形式的奇素数 q < p q \lt p q<p, q ∣ N ′ = a ′ 2 + n b ′ 2 \left. q \right|N'=a{
{'}^{2}}+nb{
{'}^{2}} q∣N′=a′2+nb′2. 将上述步骤中的 p p p替换为 q q q, 重新经历步骤一, 步骤四, 又可以得到非 x 2 + n y 2 {
{x}^{2}}+n{
{y}^{2}} x2+ny2形式的奇素数 q 1 , q 2 , ⋯ {
{q}_{1}},\text{ }{
{q}_{2}},\cdots q1, q2,⋯满足 q 1 > q 2 > ⋯ {
{q}_{1}}>{
{q}_{2}}>\cdots q1>q2>⋯.
总结起来即: 若 p p p非 x 2 + n y 2 {
{x}^{2}}+n{
{y}^{2}} x2+ny2形式, 则可以产生一个无穷递降的, 非 x 2 + n y 2 {
{x}^{2}}+n{
{y}^{2}} x2+ny2形式的奇素数列
{ p , q , q 1 , q 2 , ⋯ } . \left\{ p,q,{
{q}_{1}},{
{q}_{2}},\cdots \right\}. {
p,q,q1,q2,⋯}.
而这是不可能的(不存在无穷递降的正整数列). 由反证法, p p p具有 x 2 + n y 2 {
{x}^{2}}+n{
{y}^{2}} x2+ny2的形式, 即 ∃ c , d ∈ Z \exists c,d\in \mathbb{Z} ∃c,d∈Z, 使得 p = c 2 + n d 2 p={
{c}^{2}}+n{
{d}^{2}} p=c2+nd2.
引理4 设 n ∈ Z n\in \mathbb{Z} n∈Z, p p p是素数, gcd ( p , n ) = 1 \gcd \left( p,n \right)=1 gcd(p,n)=1. 则 ( − n p ) = 1 ⇔ \left( \frac{-n}{p} \right)=1\text{ }\Leftrightarrow \text{ } (p−n)=1 ⇔ ∃ x , y ∈ Z \exists x,y\in \mathbb{Z} ∃x,y∈Z, gcd ( x , y ) = 1 \gcd \left( x,y \right)=1 gcd(x,y)=1, 满足 p ∣ ( x 2 + n y 2 ) \left. p \right|\left( { {x}^{2}}+n{ {y}^{2}} \right) p∣(x2+ny2).
证明
( ⇒ ) \left( \Rightarrow \right) (⇒) 设Legendre符号 ( − n p ) = 1 \left( \frac{-n}{p} \right)=1 (p−n)=1, 则同余式 x 2 ≡ − n m o d p {
{x}^{2}}\equiv -n\text{ }\bmod p x2≡−n modp有解 x ≡ x 0 m o d p x\equiv {
{x}_{0}}\text{ }\bmod p x≡x0 modp, 此时有
x 0 2 ≡ − n m o d p ⇒ x 0 2 + n ≡ 0 m o d p ⇒ p ∣ ( x 0 2 + n ) = ( x 0 2 + n × 1 ) , gcd ( x 0 , 1 ) = 1. \begin{aligned} & {
{x}_{0}}^{2}\equiv -n\text{ }\bmod p\text{ }\Rightarrow \text{ }{
{x}_{0}}^{2}+n\equiv 0\text{ }\bmod p\text{ } \\ & \Rightarrow \text{ }\left. p \right|\left( {
{x}_{0}}^{2}+n \right)=\left( {
{x}_{0}}^{2}+n\times 1 \right),\text{ }\gcd \left( {
{x}_{0}},1 \right)=1. \\ \end{aligned} x02≡−n modp ⇒ x02+n≡0 modp ⇒ p∣(x02+n)=(x02+n×1), gcd(x0,1)=1.
( ⇐ ) \left( \Leftarrow \right) (⇐) 设 ∃ x 0 , y 0 ∈ Z \exists {
{x}_{0}},\text{ }{
{y}_{0}}\in \mathbb{Z} ∃x0, y0∈Z, gcd ( x 0 , y 0 ) = 1 \gcd \left( {
{x}_{0}},{
{y}_{0}} \right)=1 gcd(x0,y0)=1, 满足 p ∣ ( x 0 2 + n y 0 2 ) \left. p \right|\left( {
{x}_{0}}^{2}+n{
{y}_{0}}^{2} \right) p∣(x02+ny02), 即有
x 0 2 + n y 0 2 ≡ 0 m o d p . (4.1) {
{x}_{0}}^{2}+n{
{y}_{0}}^{2}\equiv 0\text{ }\bmod p. \tag{4.1} x02+ny02≡0 modp.(4.1)
第一步, 我们指出成立
gcd ( p , y 0 ) = 1. (4.2) \gcd \left( p,{
{y}_{0}} \right)=1. \tag{4.2} gcd(p,y0)=1.(4.2)
事实上, 若 p ∣ y 0 \left. p \right|{
{y}_{0}} p∣y0, 则有
{ p ∣ ( x 0 2 + n y 0 2 ) p ∣ y 0 ⇒ p ∣ n y 0 2 ⇒ p ∣ ( x 0 2 + n y 0 2 ) − n y 0 2 = x 0 2 ⇒ p ∣ x 0 ( ∵ p 是 素 数 ) . \left\{ \begin{aligned} & \left. p \right|\left( {
{x}_{0}}^{2}+n{
{y}_{0}}^{2} \right) \\ & \left. p \right|{
{y}_{0}}\Rightarrow \left. p \right|n{
{y}_{0}}^{2} \\ \end{aligned} \right.\text{ }\Rightarrow \text{ }\left. p \right|\left( {
{x}_{0}}^{2}+n{
{y}_{0}}^{2} \right)-n{
{y}_{0}}^{2}={
{x}_{0}}^{2}\text{ }\Rightarrow \text{ }\left. p \right|{
{x}_{0}}\text{ }\left( \because p是素数 \right). {
p∣(x02+ny02)p∣y0⇒p∣ny02 ⇒ p∣(x02+ny02)−ny02=x02 ⇒ p∣x0 (∵p是素数).
而 { p ∣ x 0 p ∣ y 0 ⇒ p ∣ gcd ( x 0 , y 0 ) \left\{ \begin{aligned} & \left. p \right|{
{x}_{0}} \\ & \left. p \right|{
{y}_{0}} \\ \end{aligned} \right.\Rightarrow \left. p \right|\gcd \left( {
{x}_{0}},{
{y}_{0}} \right) {
p∣x0p∣y0⇒p∣gcd(x0,y0), 与 gcd ( x 0 , y 0 ) = 1 \gcd \left( {
{x}_{0}},{
{y}_{0}} \right)=1 gcd(x0,y0)=1矛盾, 因此 p ∣ y 0 p\cancel{|}{
{y}_{0}} p∣
y0, 式(4.2)成立.
第二步, 由式(4.2), 关于 y ′ y' y′的同余式 y 0 y ′ ≡ 1 m o d p {
{y}_{0}}y'\equiv 1\text{ }\bmod p y0y′≡1 modp有解 y ′ ≡ y 0 ′ m o d p y'\equiv {
{y}_{0}}'\text{ }\bmod p y′≡y0′ modp, 成立
y 0 y 0 ′ ≡ 1 m o d p . (4.3) {
{y}_{0}}{
{y}_{0}}'\equiv 1\text{ }\bmod p. \tag{4.3} y0y0′≡1 modp.(4.3)
由式(4.2), 式(4.3), 得
x 0 2 y 0 ′ 2 y 0 2 = x 0 2 ( y 0 y 0 ′ ) 2 ≡ − n y 0 2 × 1 = − n y 0 2 m o d p . {
{x}_{0}}^{2}{
{y}_{0}}{
{'}^{2}}{
{y}_{0}}^{2}={
{x}_{0}}^{2}{
{\left( {
{y}_{0}}{
{y}_{0}}' \right)}^{2}}\equiv -n{
{y}_{0}}^{2}\times 1=-n{
{y}_{0}}^{2}\text{ }\bmod p. x02y0′2y02=x02(y0y0′)2≡−ny02×1=−ny02 modp.
由式(4.2), 进一步可得
( x 0 y 0 ′ ) 2 = x 0 2 y 0 ′ 2 ≡ − n m o d p . (4.4) {
{\left( {
{x}_{0}}{
{y}_{0}}' \right)}^{2}}={
{x}_{0}}^{2}{
{y}_{0}}{
{'}^{2}}\equiv -n\text{ }\bmod p. \tag{4.4} (x0y0′)2=x02y0′2≡−n modp.(4.4)
即同余式 z 2 ≡ − n m o d p {
{z}^{2}}\equiv -n\text{ }\bmod p z2≡−n modp有解 z ≡ x 0 y 0 ′ m o d p . z\equiv {
{x}_{0}}{
{y}_{0}}'\text{ }\bmod p. z≡x0y0′ modp. 又由于 gcd ( p , n ) = 1 \gcd \left( p,n \right)=1 gcd(p,n)=1, 因此有
( − n p ) = 1. \left( \frac{-n}{p} \right)=1. (p−n)=1.
定理5 (Fermat) 设 n ∈ { 1 , 2 , 3 } n\in \left\{ 1,2,3 \right\} n∈{ 1,2,3}, p p p是一奇素数, gcd ( p , n ) = 1 \gcd \left( p,n \right)=1 gcd(p,n)=1. 成立等价关系 p = c 2 + n d 2 , ∃ c , d ∈ Z ⇔ ( − n p ) = 1 ⇔ p ≡ { 1 m o d 4 , n = 1 , 1 , 3 m o d 8 , n = 2 , 1 m o d 3 , n = 3. p={ {c}^{2}}+n{ {d}^{2}},\text{ }\exists c,d\in \mathbb{Z}\text{ }\Leftrightarrow \text{ }\left( \frac{-n}{p} \right)=1\text{ }\Leftrightarrow \text{ }p\equiv \left\{ \begin{aligned} & 1\text{ }\bmod 4,\text{ }n=1, \\ & 1,3\text{ }\bmod 8,\text{ }n=2, \\ & 1\text{ }\bmod 3,\text{ }n=3. \\ \end{aligned} \right. p=c2+nd2, ∃c,d∈Z ⇔ (p−n)=1 ⇔ p≡⎩⎪⎨⎪⎧1 mod4, n=1,1,3 mod8, n=2,1 mod3, n=3.
证明
先证明成立等价关系
p = c 2 + n d 2 , ∃ c , d ∈ Z ⇔ ( − n p ) = 1. p={
{c}^{2}}+n{
{d}^{2}},\text{ }\exists c,d\in \mathbb{Z}\text{ }\Leftrightarrow \text{ }\left( \frac{-n}{p} \right)=1. p=c2+nd2, ∃c,d∈Z ⇔ (p−n)=1.
( ⇒ ) \left( \Rightarrow \right) (⇒) 设 ∃ c , d ∈ Z \exists c,d\in \mathbb{Z} ∃c,d∈Z, 满足 p = c 2 + n d 2 p={
{c}^{2}}+n{
{d}^{2}} p=c2+nd2, 当然也有
p ∣ p = ( c 2 + n d 2 ) . (5.1) \left. p \right|p=\left( {
{c}^{2}}+n{
{d}^{2}} \right). \tag{5.1} p∣p=(c2+nd2).(5.1)
{ gcd ( c , d ) ∣ c gcd ( c , d ) ∣ d ⇒ { gcd ( c , d ) 2 ∣ c 2 gcd ( c , d ) 2 ∣ n d 2 ⇒ gcd ( c , d ) 2 ∣ ( c 2 + n d 2 ) = p . \left\{ \begin{aligned} & \left. \gcd \left( c,d \right) \right|c \\ & \left. \gcd \left( c,d \right) \right|d \\ \end{aligned} \right.\text{ }\Rightarrow \text{ }\left\{ \begin{aligned} & \left. \gcd {
{\left( c,d \right)}^{2}} \right|{
{c}^{2}} \\ & \left. \gcd {
{\left( c,d \right)}^{2}} \right|n{
{d}^{2}} \\ \end{aligned} \right.\text{ }\Rightarrow \text{ }\left. \gcd {
{\left( c,d \right)}^{2}} \right|\left( {
{c}^{2}}+n{
{d}^{2}} \right)=p. {
gcd(c,d)∣cgcd(c,d)∣d ⇒ ⎩⎪⎨⎪⎧gcd(c,d)2∣∣∣c2gcd(c,d)2∣∣∣nd2 ⇒ gcd(c,d)2∣∣∣(c2+nd2)=p.
由于 p p p是素数, 因此有 gcd ( c , d ) 2 = 1 \gcd {
{\left( c,d \right)}^{2}}=1 gcd(c,d)2=1或 gcd ( c , d ) 2 = p \gcd {
{\left( c,d \right)}^{2}}=p gcd(c,d)2=p, 因此只能是
gcd ( c , d ) = 1. (5.2) \gcd \left( c,d \right)=1. \tag{5.2} gcd(c,d)=1.(5.2)
由式(5.1), 式(5.2), gcd ( p , n ) = 1 \gcd \left( p,n \right)=1 gcd(p,n)=1和引理4, 即得到 ( − n p ) = 1 \left( \frac{-n}{p} \right)=1 (p−n)=1.
( ⇐ ) \left( \Leftarrow \right) (⇐) 设 ( − n p ) = 1 \left( \frac{-n}{p} \right)=1 (p−n)=1. 由于 gcd ( p , n ) = 1 \gcd \left( p,n \right)=1 gcd(p,n)=1, 由引理4, ∃ x , y ∈ Z \exists x,y\in \mathbb{Z} ∃x,y∈Z满足 gcd ( x , y ) = 1 \gcd \left( x,y \right)=1 gcd(x,y)=1, 使得 p ∣ ( x 2 + n y 2 ) \left. p \right|\left( { {x}^{2}}+n{ {y}^{2}} \right) p∣(x2+ny2). 又由于 n ∈ Z ≤ 3 n\in { {\mathbb{Z}}_{\le 3}} n∈Z≤3, 根据引理3 (下降), 即得 ∃ c , d ∈ Z \exists c,d\in \mathbb{Z} ∃c,d∈Z, 使得 p = c 2 + n d 2 p={ {c}^{2}}+n{ {d}^{2}} p=c2+nd2.
再证明成立等价关系
( − n p ) = 1 ⇔ p ≡ { 1 m o d 4 , n = 1 , 1 , 3 m o d 8 , n = 2 , 1 m o d 3 , n = 3. \left( \frac{-n}{p} \right)=1\text{ }\Leftrightarrow \text{ }p\equiv \left\{ \begin{aligned} & 1\text{ }\bmod 4,\text{ }n=1, \\ & 1,3\text{ }\bmod 8,\text{ }n=2, \\ & 1\text{ }\bmod 3,\text{ }n=3. \\ \end{aligned} \right. (p−n)=1 ⇔ p≡⎩⎪⎨⎪⎧1 mod4, n=1,1,3 mod8, n=2,1 mod3, n=3.
由博文《Legendre符号的定义和基本性质》, 有以下讨论.
n = 1 n=1 n=1时, 直接有 ( − 1 p ) ≡ 1 ⇔ p ≡ 1 m o d 4. \left( \frac{-1}{p} \right)\equiv 1\text{ }\Leftrightarrow \text{ }p\equiv 1\text{ }\bmod 4. (p−1)≡1 ⇔ p≡1 mod4.
n = 2 n=2 n=2时, 我们指出成立 ( − 2 p ) = 1 ⇔ p ≡ 1 , 3 m o d 8. (5.3) \left( \frac{-2}{p} \right)=1\text{ }\Leftrightarrow \text{ }p\equiv 1,3\text{ }\bmod 8. \tag{5.3} (p−2)=1 ⇔ p≡1,3 mod8.(5.3)
事实上, ( − 2 p ) = ( − 1 p ) ( 2 p ) \left( \frac{-2}{p} \right)=\left( \frac{-1}{p} \right)\left( \frac{2}{p} \right) (p−2)=(p−1)(p2), 欲成立 ( − 2 p ) = 1 \left( \frac{-2}{p} \right)=1 (p−2)=1, 只有两种情况:
(1) ( − 1 p ) = ( 2 p ) = 1 ⇔ { p ≡ 1 m o d 4 p ≡ ± 1 m o d 8 ⇔ p ≡ 1 m o d 8. \left( \frac{-1}{p} \right)=\left( \frac{2}{p} \right)=1\text{ }\Leftrightarrow \text{ }\left\{ \begin{aligned} & p\equiv 1\text{ }\bmod 4 \\ & p\equiv \pm 1\text{ }\bmod 8 \\ \end{aligned} \right.\text{ }\Leftrightarrow \text{ }p\equiv 1\text{ }\bmod 8. (p−1)=(p2)=1 ⇔ {
p≡1 mod4p≡±1 mod8 ⇔ p≡1 mod8.
(2) ( − 1 p ) = ( 2 p ) = − 1 ⇔ { p ≡ 3 m o d 4 p ≡ ± 3 m o d 8 ⇔ p ≡ 3 m o d 8. \left( \frac{-1}{p} \right)=\left( \frac{2}{p} \right)=-1\text{ }\Leftrightarrow \text{ }\left\{ \begin{aligned} & p\equiv 3\text{ }\bmod 4 \\ & p\equiv \pm 3\text{ }\bmod 8 \\ \end{aligned} \right.\text{ }\Leftrightarrow \text{ }p\equiv 3\text{ }\bmod 8. (p−1)=(p2)=−1 ⇔ {
p≡3 mod4p≡±3 mod8 ⇔ p≡3 mod8.
于是也就成立等价关系(5.3).
n = 3 n=3 n=3时, 我们指出成立 ( − 3 p ) = 1 ⇔ p ≡ 1 m o d 3. (5.4) \left( \frac{-3}{p} \right)=1\text{ }\Leftrightarrow \text{ }p\equiv 1\text{ }\bmod 3. \tag{5.4} (p−3)=1 ⇔ p≡1 mod3.(5.4)
事实上, 由二次互反律, 有
( p 3 ) = ( − 1 ) p − 1 2 3 − 1 2 ( 3 p ) = ( − 1 ) p − 1 2 ( 3 p ) = ( − 1 p ) ( 3 p ) = ( − 3 p ) . \left( \frac{p}{3} \right)={
{\left( -1 \right)}^{\frac{p-1}{2}\frac{3-1}{2}}}\left( \frac{3}{p} \right)={
{\left( -1 \right)}^{\frac{p-1}{2}}}\left( \frac{3}{p} \right)=\left( \frac{-1}{p} \right)\left( \frac{3}{p} \right)=\left( \frac{-3}{p} \right). (3p)=(−1)2p−123−1(p3)=(−1)2p−1(p3)=(p−1)(p3)=(p−3).
而
{ p ≡ 0 m o d 3 ⇔ ( p 3 ) = 0 p ≡ 1 m o d 3 ⇔ ( p 3 ) = 1 p ≡ 2 m o d 3 ⇔ ( p 3 ) = − 1 \left\{ \begin{aligned} & p\equiv 0\text{ }\bmod 3\text{ }\Leftrightarrow \text{ }\left( \frac{p}{3} \right)=0 \\ & p\equiv 1\text{ }\bmod 3\text{ }\Leftrightarrow \text{ }\left( \frac{p}{3} \right)=1 \\ & p\equiv 2\text{ }\bmod 3\text{ }\Leftrightarrow \text{ }\left( \frac{p}{3} \right)=-1 \\ \end{aligned} \right. ⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧p≡0 mod3 ⇔ (3p)=0p≡1 mod3 ⇔ (3p)=1p≡2 mod3 ⇔ (3p)=−1
于是也就成立等价关系(5.4).