西南大学学报 (自然科学版)  2017, Vol. 39 Issue (8): 65-72.  DOI: 10.13718/j.cnki.xdzk.2017.08.009
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  • 关于Pell方程组x2-s2(s2-1)y2=1与y2-Dz2=4的解    [PDF全文]
    赵建红1, 杜先存2     
    1. 丽江师范高等专科学校 数学与计算机科学系,云南 丽江 674199;
    2. 红河学院 教师教育学院,云南 蒙自 661100
    摘要:设D=p1pj(1≤j≤3),p1,…,pj(1≤j≤3) 是互异的奇素数.利用初等方法讨论了Pell方程组x2-s2(s2-1)y2=1(sZ+s≥2) 与y2-Dz2=4的解的情况.
    关键词Pell方程    基本解    整数解    奇素数    递归序列    

    近年来,Pell方程x2-D1y2=k(D1$\mathbb{Z}$+k$\mathbb{Z}$)与y2-D2z2=m(D2$\mathbb{Z}$+m$\mathbb{Z}$)的公解问题一直备受人们的关注.当k=1,m=4,D1为偶数,D2为奇数时,已有如下结果:

    (ⅰ)文献[1-5]对D1=2的情况做了一些研究;

    (ⅱ)文献[6-7]对D1=6的情况做了一些研究.

    本文主要讨论当D1=s2(s2-1),(s$\mathbb{Z}$+s≥2) 时及D2为奇数时的情况,即证明了以下定理:

    定理1 若s$\mathbb{Z}$+s≥2,D=p1pj(1≤j≤3),其中p1,…,pj(1≤j≤3) 是互异的奇素数,则Pell方程组

    $\left\{ \begin{array}{l} {x^2} - {s^2}\left( {{s^2} - 1} \right){y^2} = 1\\ {y^2} - D{z^2} = 4 \end{array} \right.$ (1)

    D为16s4-16s2+3中不超过3个次数为奇次的素因子之积外均只有平凡解

    $\left( {x,y,z} \right) = \left( { \pm \left( {2{s^2} - 1} \right) \pm 2,0} \right)$
    1 关键性引理

    引理1[8] 不定方程x4-Dy2=1除当D=1 785,4×1 785,16×1 785时分别有两组正整数解(xy)=(13,4),(239,1 352);(xy)=(13,2),(239,676);(xy)=(13,1),(239,338) 外,最多只有一组正整数解(x1y1),且满足x12=x0或2x02-1,这里$\varepsilon = {x_0} + {y_0}\sqrt D $是Pell方程x2-Dy2=1的基本解.

    引理2[9] 当a>1且a是平方数时,方程ax4-by2=1至多有一组正整数解.

    引理3[10] 若D是一个非平方的正整数,方程x2-Dy4=1至多有两组正整数解,而且方程恰有两组正整数解的充要条件是D=1 785或D=28 560或2x1y1都是平方数,这里(x1y1)是方程x2-Dy2=1的基本解.

    引理4[11] 当A$\mathbb{N}$A>1,B$\mathbb{N}$AB不是平方数时,方程Ax2-By4=1至多只有1组正整数解.

    引理5 设Pell方程x2-s2(s2-1)y2=1的基本解为(x1y1),其全部整数解为(xnyn),n$\mathbb{Z}$,则对任意n$\mathbb{Z}$xnyn具有如下性质:

    (ⅰ)若2s2-1及8s2(s2-1)+1不为平方数,那么xn为平方数当且仅当n=0;若2s2-1为平方数,那么xn为平方数当且仅当n=1或n=0;若8s2(s2-1)+1为平方数,那么xn为平方数当且仅当n=2或n=0;

    (ⅱ) $\frac{{{x_n}}}{{2{s^2} - 1}}$为平方数当且仅当n=1;

    (ⅲ) $\frac{{{y_n}}}{2}$为平方数当且仅当n=0或n=1.

     设(x1y1)是Pell方程x2-s2(s2-1)y2=1的基本解,(xnyn) (n$\mathbb{Z}$)是Pell方程x2-s2(s2-1)y2=1的整数解.

    (ⅰ)若xn=a2,将其代入原方程得

    ${a^4} - {s^2}\left( {{s^2} - 1} \right){y^2} = 1$ (2)

    又Pell方程x2-s2(s2-1)y2=1的基本解为(x1y1)=(2s2-1,2),则

    ${x_1} = 2{s^2} - 1\;\;\;\;2x_1^2 - 1 = 8{s^2}\left( {{s^2} - 1} \right) + 1$

    根据引理1得,方程(2) 至多有一组正整数解.当2s2-1及8s2(s2-1)+1均不为平方数时,根据引理1得,方程(2) 仅有平凡解(ay)=(±1,0),故xn=1,从而n=0;当2s2-1为平方数时,根据引理1得,方程(2) 有整数解$\left( {a,y} \right) = \left( {\sqrt {2{s^2} - 1} ,2} \right)$及平凡解(ay)=(±1,0),那么xn=2s2-1或xn=1,从而n=1或n=0;8s2(s2-1)+1为平方数时,根据引理1得,方程(2) 有整数解$\left( {a,y} \right) = \left( {\sqrt {8{s^2}\left( {{s^2} - 1} \right) + 1} ,4\left( {2{s^2} - 1} \right)} \right)$及平凡解(ay)=(±1,0),那么xn=8s2(s2-1)+1或xn=1,从而n=2或n=0.反之,显然.

    (ⅱ)若$\frac{{{x_n}}}{{2{s^2} - 1}} = {a^2}$,则xn=(2s2-1)a2,代入原方程得

    ${\left( {2{s^2} - 1} \right)^2}{a^4} - {s^2}\left( {{s^2} - 1} \right){y^2} = 1$ (3)

    根据引理2得,方程(3) 仅有整数解(ay)=(±1,±2),此时xn=2s2-1,从而n=1.反之,显然.

    (ⅲ)若$\frac{{{y_n}}}{2} = {b^2}$,则yn=2b2,代入原方程得

    ${x^2} - 4{s^2}\left( {{s^2} - 1} \right){b^4} = 1$ (4)

    根据引理3得,方程(4) 有整数解(xb)=(±(2s2-1),±1) 及平凡解(xb)=(±1,0),此时yn=2或0,从而n=1或n=0.反之,显然.

    2 定理1的证明

     设(x1y1)为Pell方程x2-s2(s2-1)y2=1(s$\mathbb{Z}$+s≥2) 的基本解,则有(x1y1)=(2s2-1,2),故Pell方程x2-s2(s2-1)y2=1(s$\mathbb{Z}$+s≥2) 的全部正整数解为:

    ${x_n} + {y_n}\sqrt {{s^2}\left( {{s^2} - 1} \right)} = {\left( {{x_1} + \sqrt {{s^2}\left( {{s^2} - 1} \right)} {y_1}} \right)^n} = {\left( {2{s^2} - 1 + 2\sqrt {{s^2}\left( {{s^2} - 1} \right)} } \right)^n},n \in {\mathbb{Z}^ + }$

    容易验证以下性质成立:

    性质a yn2-4=yn-1yn+1

    性质b y2n=2xnyn

    性质c x2n=2xn2-1;

    性质d gcd (xnyn)=1,gcd (xnxn+1)=1,gcd (ynyn+1)=2;

    性质e gcd (x2ny2n+1)=gcd (x2n+2y2n+1)=1,gcd (x2n+1y2n)=gcd (x2n+1y2n+2)=2s2-1;

    性质f xn≡1(mod 2),xn≡±1(mod s2),xn≡1(mod (s2-1)),x2n+1≡0(mod (2s2-1)),x2n≡±1(mod (2s2-1));

    性质g y2n+1≡2(mod 4),y2n≡0(mod 4),y2n+1≡±2(mod (2s2-1)),y2n≡0(mod (2s2-1)).

    情形1 n为正偶数,由(1) 式得

    ${s^2}\left( {{s^2} - 1} \right)D{z^2} = x_n^2 - {\left( {2{s^2} - 1} \right)^2} = \left( {{x_n} + 2{s^2} - 1} \right)\left( {{x_n} - 2{s^2} + 1} \right)$ (5)

    n=2m(m$\mathbb{Z}$+),则(5) 式成为

    ${s^2}\left( {{s^2} - 1} \right)D{z^2} = \left( {{x_{2m}} + 2{s^2} - 1} \right)\left( {{x_{2m}} - 2{s^2} - 1} \right)$ (6)

    由性质c得,(6) 式可化为

    ${s^2}\left( {{s^2} - 1} \right)D{z^2} = \left( {2x_m^2 + 2{s^2} - 2} \right)\left( {2x_m^2 - 2{s^2}} \right) = 4\left( {x_m^2 + {s^2} - 1} \right)\left( {x_m^2 - {s^2}} \right)$

    ${s^2}\left( {{s^2} - 1} \right)D{z^2} = 4\left( {x_m^2 + {s^2} - 1} \right)\left( {x_m^2 - {s^2}} \right)$ (7)

    由性质f知

    $x_m^2 \equiv 0,1\left( {\bmod 2{s^2} - 1} \right)$

    $x_m^2 - {s^2} \equiv \pm {s^2}\left( {\bmod 2{s^2} - 1} \right)$

    $\gcd \left( {x_m^2 + {s^2} - 1,x_m^2 - {s^2}} \right) = \gcd \left( {2{s^2} - 1, \pm {s^2}} \right) = \gcd \left( {1,{s^2}} \right) = 1$

    由性质f知,

    $x_m^2 \equiv 1\left( {\bmod {s^2}} \right)\;\;\;\;\;\;x_m^2 \equiv 1\left( {\bmod \left( {{s^2} - 1} \right)} \right)$

    所以(7) 式可分解为下面2种情形:

    情形Ⅰ 4(xm2+s2-1)=D1s2z12xm2-s2=D2(s2-1)z22D=D1D2z=z1z2

    情形Ⅱ xm2+s2-1=D1s2z12,4(xm2-s2)=D2(s2-1)z22D=D1D2z=z1z2

    其中gcd (D1D2)=1,gcd (z1z2)=1.

    先讨论情形Ⅰ:

    $4\left( {x_m^2 + {s^2} - 1} \right) = {D_1}{s^2}z_1^2\;及\;x_m^2 - {s^2}\left( {{s^2} - 1} \right)y_m^2 = 1$

    ${D_1}z_1^2 = 4\left[ {x_m^2 - {{\left( {{s^2} - 1} \right)}^2}y_m^2} \right] = 4\left[ {{x_m} + \left( {{s^2} - 1} \right){y_m}} \right]\left[ {{x_m} - \left( {{s^2} - 1} \right){y_m}} \right]$

    ${D_1}z_1^2 = 4\left[ {{x_m} + \left( {{s^2} - 1} \right){y_m}} \right]\left[ {{x_m} - \left( {{s^2} - 1} \right){y_m}} \right]$ (8)

    $\begin{array}{l} \gcd \left( {{x_m} + \left( {{s^2} - 1} \right){y_m},{x_m} - \left( {{s^2} - 1} \right){y_m}} \right) = \gcd \left( {2\left( {{s^2} - 1} \right){y_m},{x_m} - \left( {{s^2} - 1} \right){y_m}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\gcd \left( {{s^2} - 1,{x_m}} \right) = 1 \end{array}$

    则(8) 式可分解为下式:

    ${x_m} + \left( {{s^2} - 1} \right){y_m} = {D_3}z_3^2,{x_m} - \left( {{s^2} - 1} \right){y_m} = {D_4}z_4^2,{D_1} = {D_3}{D_4},{z_1} = 2{z_3}{z_4}$ (9)

    其中

    $\gcd \left( {{D_3},{D_4}} \right) = 1\;\;\;\;\;\;\gcd \left( {{z_3},{z_4}} \right) = 1$

    $x_m^2 - {s^2} = {D_2}\left( {{s^2} - 1} \right)z_2^2\;及\;x_m^2 - {s^2}\left( {{s^2} - 1} \right)y_m^2 = 1$

    ${D_2}z_2^2 = {s^4}y_m^2 - x_m^2 = \left( {{s^2}{y_m} + {x_m}} \right)\left( {{s^2}{y_m} - {x_m}} \right)$

    ${D_2}z_2^2 = \left( {{s^2}{y_m} + {x_m}} \right)\left( {{s^2}{y_m} - {x_m}} \right)$ (10)

    $\gcd \left( {{s^2}{y_m} + {x_m},{s^2}{y_m} - {x_m}} \right) = \gcd \left( {2{x_m},{s^2}{y_m} - {x_m}} \right) = \gcd \left( {{x_m},{s^2}} \right)\gcd \left( {1,{s^2}} \right) = 1$

    则(10) 可分解为下式:

    ${s^2}{y_m} + {x_m} = {D_5}z_5^2,{s^2}{y_m} - {x_m} = {D_6}z_6^2,{D_2} = {D_5}{D_6},{z_2} = {z_5}{z_6}$ (11)

    其中

    $\gcd \left( {{D_5},{D_6}} \right) = 1\;\;\;\;\;\;\gcd \left( {{z_5},{z_6}} \right) = 1$

    由(9),(11) 式得D=D1D2=D3D4D5D6,且D3D4D5D6两两互素,而D=p1psp1,…,ps(1≤s≤3) 是互异的奇素数,则D3D4D5D6中至少有一个为1.

    D3=1时,(9) 式的第一式为

    ${x_m} + \left( {{s^2} - 1} \right){y_m} = z_3^2$

    ${x_m} = z_3^2 - \left( {{s^2} - 1} \right){y_m}$

    代入xm2-s2(s2-1)ym2=1,整理得

    ${s^2}z_3^4 - \left( {{s^2} - 1} \right){\left( {{y_m} + z_3^2} \right)^2} = 1$

    由引理2得

    ${s^2}z_3^4 - \left( {{s^2} - 1} \right){\left( {{y_m} + z_3^2} \right)^2} = 1$

    仅有一组正整数解(z3ym+z32)=(1,1),则有

    $\left( {{z_3},{y_m}} \right) = \left( {1,0} \right)$

    此时方程(1) 无正整数解.

    D4=1时,(9) 式的第二式为

    ${x_m} - \left( {{s^2} - 1} \right){y_m} = z_4^2$

    则有

    ${x_m} = \left( {{s^2} - 1} \right){y_m} + z_4^2$

    代入

    $x_m^2 - {s^2}\left( {{s^2} - 1} \right)y_m^2 = 1$

    整理得

    ${s^2}z_4^4 - \left( {{s^2} - 1} \right){\left( {{y_m} - z_4^2} \right)^2} = 1$

    由引理2得

    ${s^2}z_4^4 - \left( {{s^2} - 1} \right){\left( {{y_m} - z_4^2} \right)^2} = 1$

    仅有一组正整数解

    $\left( {{z_4},\left| {{y_m} - z_4^2} \right|} \right) = \left( {1,1} \right)$

    则有(z4ym)=(1,0) 或(1,2),从而m=1,则n=2,此时有

    $D{z^2} = y_2^2 - 4 = {\left( {8{s^2} - 4} \right)^2} - 4 = {2^2} \times \left( {16{s^4} - 16{s^2} + 3} \right)$

    则若16s4-16s2+3中次数为奇次的素因子的个数不超过3个,则此时方程(1) 有正整数解;若16s4-16s2+3中次数为奇次的素因子的个数超过3个,则此时方程(1) 仅有平凡解.

    D5=1时,(11) 式的第一式为s2ym+xm=z52,则有

    ${x_m} = z_5^2 - {s^2}{y_m}$

    代入

    $x_m^2 - {s^2}\left( {{s^2} - 1} \right)y_m^2 = 1$

    整理得

    ${s^2}{\left( {{y_m} - z_5^2} \right)^2} - \left( {{s^2} - 1} \right)z_5^4 = 1$

    由引理4得

    ${s^2}{\left( {{y_m} - z_5^2} \right)^2} - \left( {{s^2} - 1} \right)z_5^4 = 1$

    仅有1组正整数解

    $\left( {\left| {{y_m} - z_5^2} \right|,{z_5}} \right) = \left( {1,1} \right)$

    则有(z5ym)=(1,0) 或(1,2),从而m=1,则n=2,m=2,此时方程(1) 的解的情况与D4=1的情形相同.

    D6=1时,(11) 式的第二式为s2ym-xm=z62,则有

    ${x_m} = {s^2}{y_m} - z_6^2$

    代入

    $x_m^2 = {s^2}\left( {{s^2} - 1} \right)y_m^2 = 1$

    整理得

    ${s^2}{\left( {{y_m} - z_6^2} \right)^2} - \left( {{s^2} - 1} \right)z_6^4 = 1$

    由引理4得

    ${s^2}{\left( {{y_m} - z_6^2} \right)^2} - \left( {{s^2} - 1} \right)z_6^4 = 1$

    仅有1组正整数解

    $\left( {\left| {{y_m} - z_6^2} \right|,{z_6}} \right) = \left( {1,1} \right)$

    则有(z6ym)=(1,0) 或(1,2),从而m=1,则n=2,m=2,此时方程(1) 的解的情况与D4=1的情形相同.

    综上知情形Ⅰ不成立.

    下面讨论情形Ⅱ:

    $\left( {x_m^2 + {s^2} - 1} \right) = {D_1}{s^2}z_1^2\;及\;x_m^2 - {s^2}\left( {{s^2} - 1} \right)y_m^2 = 1$

    ${D_1}z_1^2 = x_m^2 - {\left( {{s^2} - 1} \right)^2}y_m^2 = \left[ {{x_m} + \left( {{s^2} - 1} \right){y_m}} \right]\left[ {{x_m} - \left( {{s^2} - 1} \right){y_m}} \right]$

    ${D_1}z_1^2 = \left[ {{x_m} + \left( {{s^2} - 1} \right){y_m}} \right]\left[ {{x_m} - \left( {{s^2} - 1} \right){y_m}} \right]$ (12)

    $\gcd \left( {x_m^2 + \left( {{s^2} - 1} \right){y_m}} \right.\;\;\;\;\;\left. {{x_m} - \left( {{s^2} - 1} \right){y_m}} \right) = 1$

    则(12) 式可分解为下式:

    ${x_m} + \left( {{s^2} - 1} \right){y_m} = {D_7}z_7^2,{x_m} - \left( {{s^2} - 1} \right){y_m} = {D_8}z_8^2,{D_1} = {D_7}{D_8},{z_1} = {z_7}{z_8}$ (13)

    其中

    $\gcd \left( {{D_7},{D_8}} \right) = 1\;\;\;\;\;\gcd \left( {{z_7},{z_8}} \right) = 1$

    $4\left( {x_m^2 - {s^2}} \right) = {D_2}\left( {{s^2} - 1} \right)z_2^2\;及\;x_m^2 - {s^2}\left( {{s^2} - 1} \right)y_m^2 = 1$

    ${D_2}z_2^2 = 4\left( {{s^4}y_m^2 - x_m^2} \right) = 4\left( {{s^2}{y_m} + {x_m}} \right)\left( {{s^2}{y_m} - {x_m}} \right)$

    ${D_2}z_2^2 = 4\left( {{s^2}{y_m} + {x_m}} \right)\left( {{s^2}{y_m} - {x_m}} \right)$ (14)

    $\gcd \left( {{s^2}{y_m} + {x_m},{s^2}{y_m} - {x_m}} \right) = 1$

    则(14) 式可分解为下式:

    ${s^2}{y_m} + {x_m} = {D_9}z_9^2,{s^2}{y_m} - {x_m} = {D_{10}}z_{10}^2,{D_2} = {D_9}{D_{10}},{z_2} = 2{z_9}{z_{10}}$ (15)

    其中

    $\gcd \left( {{D_9},{D_{10}}} \right) = 1\;\;\;\;\;\gcd \left( {{z_9},{z_{10}}} \right) = 1$

    由(13),(15) 式得D=D1D2=D7D8D9D10D7D8D9D10两两互素,而D=p1psp1,…,ps(1≤s≤3) 是互异的奇素数,则D7D8D9D10中至少有一个为1.

    D7=1时仿D3=1的证明可知此时情形Ⅱ不成立;D8=1时仿D4=1的证明可知此时情形Ⅱ不成立;D9=1时仿D5=1的证明可知此时情形Ⅱ不成立;D10=1时仿D6=1的证明可知此时情形Ⅱ不成立.综上可知情形Ⅱ不成立.

    情形2 n为正奇数,令n=2m-1,m$\mathbb{Z}$+.则由性质a得

    $D{z^2} = y_{2m - 1}^2 - 4 = {y_{2\left( {m - 1} \right)}}{y_{2m}}$ (16)

    由性质b知,(16) 式可化为

    $D{z^2} = 4{x_{m - 1}}{y_{m - 1}}{x_m}{y_m}$ (17)

    由性质d

    $\gcd \left( {{x_{m - 1}},{y_{m - 1}}} \right) = \gcd \left( {{x_m},{y_m}} \right) = 1\;\;\;\;\gcd \left( {{x_m},{x_{m - 1}}} \right) = 1\;\;\;\;\;\gcd \left( {{y_m},{y_{m - 1}}} \right) = 2$

    $\gcd \left( {\frac{{{y_m}}}{2},\frac{{{y_{m - 1}}}}{2}} \right) = 1$

    m为正偶数时,由性质e知

    $\gcd \left( {{x_m},{y_{m - 1}}} \right) = 1\;\;\;\;\;\gcd \left( {{x_{m - 1}},{y_m}} \right) = 2{s^2} - 1$

    $\gcd \left( {\frac{{{x_{m - 1}}}}{{2{s^2} - 1}},\frac{{{y_m}}}{{2{s^2} - 1}}} \right) = 1$

    则有xm$\frac{{{y_{m - 1}}}}{2},\frac{{{x_{m - 1}}}}{{2{s^2} - 1}},\frac{{{y_m}}}{{2\left( {2{s^2} - 1} \right)}}$两两互素.

    由性质f知xm$\frac{{{x_{m - 1}}}}{{2{s^2} - 1}}$均为奇数,由性质g知m为偶数时,$\frac{{{y_{m - 1}}}}{2}$为奇数;由引理5(ⅰ)知仅当m=0时,xm为平方数;由引理5(ⅱ)知仅当m=2时,xm-12s2-1为平方数;由引理5(ⅲ)知仅当m=1或m=2时,$\frac{{{y_{m - 1}}}}{2}$为平方数.故有xm$\frac{{{y_{m - 1}}}}{2},\frac{{{x_{m - 1}}}}{{2{s^2} - 1}},\frac{{{y_m}}}{{2\left( {2{s^2} - 1} \right)}}$两两互素,因此m>2为正偶数时,xm$\frac{{{y_{m - 1}}}}{2},\frac{{{x_{m - 1}}}}{{2{s^2} - 1}}$均为奇数且均不为平方数,故xm$\frac{{{y_{m - 1}}}}{2},\frac{{{x_{m - 1}}}}{{2{s^2} - 1}}$至少为D提供3个互异的奇素数.

    m>2为正偶数时,因为D=p1ps(1≤s≤3),p1,…,ps(1≤s≤3) 是互异的奇素数,所以$\frac{{{y_m}}}{{2\left( {2{s^2} - 1} \right)}}$为平方数.令m=2kk$\mathbb{Z}$+,且k≥2,则有

    $\frac{{{y_m}}}{{2\left( {2{s^2} - 1} \right)}} = \frac{{{y_{2k}}}}{{2\left( {2{s^2} - 1} \right)}} = \frac{{2{x_k}{x_k}}}{{2\left( {2{s^2} - 1} \right)}} = \frac{{{x_k}{y_k}}}{{2{s^2} - 1}}$

    $\frac{{{y_m}}}{{2\left( {2{s^2} - 1} \right)}} = \frac{{{x_k}{y_k}}}{{2{s^2} - 1}}$ (18)

    k≥2为正偶数,则由性质b知

    $\frac{{{y_m}}}{{2\left( {2{s^2} - 1} \right)}} = {x_k} \cdot \frac{{{y_k}}}{{2{s^2} - 1}}$

    由性质d知

    $\gcd \left( {{x_k},\frac{{{y_k}}}{{2{s^2} - 1}}} \right) = 1$

    $\frac{{{y_m}}}{{2\left( {2{s^2} - 1} \right)}}$为平方数需xk$\frac{{{y_k}}}{{2{s^2} - 1}}$同时为平方数,由引理5的(ⅰ)知k≥2时xk不为平方数,则$\frac{{{y_m}}}{{2\left( {2{s^2} - 1} \right)}}$不为平方数,故此时方程(1) 无正整数解.

    k≥3为正奇数,由性质b知

    $\frac{{{y_m}}}{{2\left( {2{s^2} - 1} \right)}} = {y_k} \cdot \frac{{{x_k}}}{{2{s^2} - 1}}$

    由性质d知

    $\gcd \left( {{y_k},\frac{{{x_k}}}{{2{s^2} - 1}}} \right) = 1$

    $\frac{{{y_m}}}{{2\left( {2{s^2} - 1} \right)}}$为平方数,需yk$\frac{{{x_k}}}{{2{s^2} - 1}}$同时为平方数,由引理6(ⅰ)知k≥3时,xk不为平方数,则$\frac{{{y_m}}}{{2\left( {2{s^2} - 1} \right)}}$不为平方数,故此时方程(1) 无正整数解.

    m=0时,(17) 式为

    $D{z^2} = 4{x_{ - 1}}{y_{ - 1}}{x_0}{y_0} = 0$

    从而得方程(1) 的平凡解

    $\left( {x,y,z} \right) = \left( { \pm \left( {2{s^2} - 1} \right), \pm 2,0} \right)$

    m=2时,(17) 式为

    $\begin{array}{l} D{z^2} = 4{x_1}{y_1}{x_2}{y_2} = 4 \times \left( {2{s^2} - 1} \right) \times 2 \times \left( {8{s^4} - 8{s^2} + 1} \right) \times 4\left( {2{s^2} - 1} \right) = \\ \;\;\;\;\;\;\;\;\;\;{2^5}{\left( {2{s^2} - 1} \right)^2}\left( {8{s^4} - 8{s^2} + 1} \right) \end{array}$

    $D{z^2} = {2^5}{\left( {2{s^2} - 1} \right)^2}\left( {8{s^4} - 8{s^2} + 1} \right)$ (19)

    D为奇数,故(19) 式左边2的次数为偶数次,而(19) 式右边2的次数为5次,矛盾,故此时方程(1) 无正整数解.

    m为正奇数时,仿情形1的证明可知此时方程(1) 亦无正整数解.

    3 相关推论

    推论1 Pell方程组x2-12y2=1与y2-195z2=4有解(xyz)=(±97,±28,±2),(±7;±2,0);

    推论2 Pell方程组x2-72y2=1与y2-1155z2=4有解(xyz)=(±577,±68,±2),(±17;±2,0);

    推论3 Pell方程组x2-240y2=1与y2-427z2=4有解(xyz)=(±1921,±124,±6),(±31;±2,0).

    参考文献
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    On the Solutions of System of Pell Equations x2-s2(s2-1)y2=1 and y2-Dz2=4
    ZHAO Jian-hong1, DU Xian-cun2     
    1. Department of Mathematics and Computer Science, Lijiang Teachers College, Lijiang Yunnan 674199, China;
    2. College of Teachers Education, Honghe University, Mengzi Yunnan 661100, China
    Abstract: Let D be not a perfect square positive integer which has at most three distinct prime factors. The integer solutions of the system of Pell equations in title are discussed with the help of the elementary method.
    Key words: Pell equation    fundamental solution    integer solution    odd prime    recursive sequence    
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