西南大学学报 (自然科学版)  2017, Vol. 39 Issue (8): 83-88.  DOI: 10.13718/j.cnki.xdzk.2017.08.012
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  • 关于不定方程5x(x+1)(x+2)(x+3)=6y(y+1)(y+2)(y+3)    [PDF全文]
    李益孟, 罗明     
    西南大学 数学与统计学院,重庆 400715
    摘要:运用递归序列和平方剩余的方法,证明了不定方程5x(x+1)(x+2)(x+3)=6y(y+1)(y+2)(y+3) 仅有正整数解(xy)=(21,20).
    关键词不定方程    整数解    递归序列    平方剩余    

    当(pq)=1,pq$\mathbb{N} $时,形如

    $ px\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right) = qy\left( {y + 1} \right)\left( {y + 2} \right)\left( {y + 3} \right) $

    的不定方程已经有不少研究工作[1-10].在本文中,我们将运用递归数列方法证明当(pq)=(5,6) 时,不定方程

    $ 5x\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right) = 6y\left( {y + 1} \right)\left( {y + 2} \right)\left( {y + 3} \right) $ (1)

    仅有一组正整数解(xy)=(21,20).首先将方程(1) 化为

    $ {\left[{5\left( {{x^2} + 3x + 1} \right)} \right]^2} - 30{\left( {{y^2} + 3y + 1} \right)^2} = - 5 $ (2)

    容易知道方程x2-30y2=-5的全部整数解由一个结合类(歧类)给出

    $ {x_n} + {y_n}\sqrt {30} = \pm \left( {5 + \sqrt {30} } \right){\left( {11 + 2\sqrt {30} } \right)^n}\;\;\;\;\;\;\;\;n \in \mathbb{Z} $
    $ {{\bar x}_n} + {{\bar y}_n}\sqrt {30} = \pm \left( { - 5 + \sqrt {30} } \right){\left( {11 + 2\sqrt {30} } \right)^n}\;\;\;\;\;\;\;\;n \in \mathbb{Z} $

    其中$5+\sqrt{30}$x2-30y2=-5的最小正整数解,$11+2\sqrt{30}$是Pell方程u2-30v2=1的基本解.于是方程(2) 的解应满足

    $ {\left( {2y + 3} \right)^2} = 4{y_n} + 5 $
    $ {\left( {2y + 3} \right)^2} = 4{{\bar y}_n} + 5 $

    容易看出yn=y-n,于是方程(2) 的解应满足

    $ {\left( {2y + 3} \right)^2} = \pm 4{y_n} + 5\;\;\;\;\;\;n \in \mathbb{Z} $ (3)

    容易验证下列关系式成立

    $ {u_{n + 1}} = 22u - {u_{n - 1}}\;\;\;\;\;{u_0} = 1, {u_1} = 11 $ (4)
    $ {v_{n + 1}} = 22{v_n} - {v_{n - 1}}\;\;\;\;{v_0} = 0, {v_1} = 2 $ (5)
    $ {u_{2n}} = 2u_n^2 - 1\;\;\;\;\;{v_{2n}} = 2{u_n}{v_n} $ (6)
    $ {y_{n + 1}} = 22{y_n} - {y_{n - 1}}\;\;\;\;{y_0} = 1, {y_1} = 21 $ (7)
    $ {y_n} = {u_n} + 5{v_n} $ (8)
    $ {u_{n + 2km}} \equiv {\left( { - 1} \right)^k}{u_n}\left( {\bmod {u_m}} \right)\;\;\;\;\;\;{v_{n + 2km}} \equiv {\left( { - 1} \right)^k}{v_n}\left( {\bmod {u_m}} \right) $ (9)
    $ {y_{n + 2km}} \equiv {\left( { - 1} \right)^k}{y_n}\left( {\bmod {u_m}} \right) $ (10)

    下面将证明(3) 式仅当n等于0,2,-1,-3时成立,由此求得方程(2) 的全部整数解,进而求得(1) 的全部正整数解.

    1 (2y+3)2=4yn+5

    本节考察当n取何值时,4yn+5为完全平方数.在做此工作之前我们先介绍几个引理.

    引理1  设2|nn>0,则$\left( \frac{\pm 20{{v}_{2n}}+5}{{{u}_{2n}}} \right)=\left( \frac{{{u}_{n}}\pm 4{{v}_{n}}}{23} \right)$.

      当2|n时,u2n≡1(mod 8),vn≡0(mod 2),un≡1mod 8,于是

    $ \begin{array}{l} \left( {\frac{{ \pm 20{v_{2n}} + 5}}{{{u_{2n}}}}} \right) = \left( {\frac{{ \pm 40{u_n}{v_n} + 10u_n^2}}{{{u_{2n}}}}} \right) = \left( {\frac{2}{{{u_{2n}}}}} \right)\left( {\frac{{{u_n}}}{{{u_{2n}}}}} \right)\left( {\frac{{ \pm 20{v_n} + 5{u_n}}}{{{u_{2n}}}}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {\frac{{{u_{2n}}}}{{{u_n}}}} \right)\left( {\frac{{{u_{2n}}}}{{ \pm 20{v_n} + 5{u_n}}}} \right) = \left( {\frac{{ - 1}}{{{u_n}}}} \right)\left( {\frac{{30v_n^2 + u_n^2}}{{ \pm 20{v_n} + 5{u_n}}}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left( {\frac{{46}}{{ \pm 40{v_n} + {u_n}}}} \right) = \left( {\frac{{23}}{{ \pm 40{v_n} + {u_n}}}} \right) = \left( {\frac{{{u_n} \pm 4{v_n}}}{{23}}} \right) \end{array} $

    证毕.

    引理2  若4yn+5是平方数,则n≡0,2,-1,-3(mod 2×52).

      在此次证明中我们采用对序列{4yn+5}取模的方法证明,分两步

    第一步:

    mod 101,排除n≡1,3(mod 5),此时4yn+5≡89,89(mod 101),余n≡0,2,4(mod 5),即n≡0,2,4,5,7,9,10,12,14,15,17,19,20,22,24(mod 25). mod 101是对{4yn+5}取的,mod 5指出所得剩余序列周期为5. 89为mod 101的平方非剩余,为节省篇幅,我们以下均按这种方式叙述.

    mod 701,排除n≡4,7,17,20(mod 25),此时4yn+5≡642,540,540,642(mod 701).余n≡0,2,5,9,10,12,14,15,19,22,24(mod 25).

    mod 14401,排除n≡5,12,19(mod 25),此时4yn+5≡14255,6834,14255(mod 14401),余n≡0,2,9,10,14,15,22,24(mod 25).

    mod 515401,排除n≡9,10,14,15(mod 25),此时4yn+5≡502844,226595,226595,502844mod 515401,余n≡0,2,22,24(mod 25)

    n≡0,2,22,24,25,27,47,49,50,52,72,74,75,77,97,99(mod 100).

    第二步:

    mod 231841,排除n≡3,4,5,7,12,14,15,16(mod 20),排除n≡24,25,27,47,52,72,74,75(mod 100).

    综上即余n≡0,2,22,47,49,50,77,97,99(mod 100),即n≡0,2,22,27,47,49(mod 50),mod 83401排除n≡22,27(mod 50),此时4yn+5≡81562,81562(mod 50).

    综上n≡0,2,-1,-3(mod 50).

    引理3  设n≡0(mod 2×52),则仅当n=0时,4yn+5为平方数.

      如果n≡0(mod 2×52)且n≠0则可令

    $ n = 2 \times k \times {5^2} \times {2^t}\;\;\;\;\;\;t \ge 0.2\;ᚾ\;k $

    对{un±4vn}取模23,所得的两个剩余序列周期均为3,而对{2t}模3的剩余序列具有周期2.下面对k分两种情况讨论.

    情况Ⅰ  k≡1(mod 4) 时,令

    $ m\left\{ \begin{array}{l} {2^t}\;\;\;\;\;\;\;\;\;\;\;t \equiv 0\left( {\bmod 2} \right)\\ 5 \cdot {2^t}\;\;\;\;\;\;\;\;t \equiv 1\left( {\bmod 2} \right) \end{array} \right. $

    则有表 1.

    表 1 k≡1(mod 4) 数据情况表

    对表中所有m,均有

    $ \left( {\frac{{{u_m} + 4{v_m}}}{{23}}} \right) = - 1 $

    于是,由(9),(10) 及引理1,有

    $ 4{y_n} + 5 \equiv 4{y_{2m}} + 5 \equiv 20{v_{2m}} + 5\left( {\bmod {u_{2m}}} \right) $

    进而得

    $ \left( {\frac{{4{y_n} + 5}}{{{u_{2m}}}}} \right) = \left( {\frac{{20{v_{2m}} + 5}}{{{u_{2m}}}}} \right) = \left( {\frac{{{u_m} + 4{v_m}}}{{23}}} \right) = - 1 $

    从而4yn+5是非平方数.

    情况Ⅱ  k≡-1(mod 4) 时,令

    $ m\left\{ \begin{array}{l} 5 \cdot {2^t}\;\;\;\;\;\;\;\;t \equiv 0\left( {\bmod 2} \right)\\ {2^t}\;\;\;\;\;\;\;\;\;\;\;t \equiv 1\left( {\bmod 2} \right) \end{array} \right. $

    则有表 2.

    表 2 k≡-1(mod 4) 数据情况表

    对表中所有m,均有

    $ \left( {\frac{{{u_m} - 4{v_m}}}{{23}}} \right) = - 1 $

    于是,由(9),(10) 式及引理1,有

    $ 4{y_n} + 5 \equiv - 4{y_{2m}} + 5 \equiv - 20{v_{2m}} + 5\left( {\bmod {u_{2m}}} \right) $

    进而得

    $ \left( {\frac{{4{y_n} + 5}}{{{u_{2m}}}}} \right) = \left( {\frac{{ - 20{v_{2m}} + 5}}{{{u_{2m}}}}} \right) = \left( {\frac{{{u_m} - 4{v_m}}}{{23}}} \right) = - 1 $

    从而4yn+5是非平方数.

    n=0时,4yn+5=32.证毕

    引理4  设n≡-1(mod 2×52),则仅当n=-1时,4yn+5为完全平方数.

      如果n≡-1(mod 2×52)且n≠-1,则可令

    $ n = - 1 + 2 \times k \times {5^2} \times {2^t}\;\;\;\;\;\;t \ge 0, 2\;ᚾ\;k $

    若取m为2t,5×2t之一,则由(10) 知:

    $ \begin{array}{l} {y_n} \equiv {y_{ - 1 \pm 2m}} \equiv {u_{ - 1 \pm 2m}} + 5{v_{ - 1 \pm 2m}} \equiv \\ \;\;\;\;\;\;\;{u_{ - 1}}{u_{ \pm 2m}} + 30{v_{ - 1}}{v_{ \pm 2m}} + 5\left( {{v_{ - 1}}{u_{ \pm 2m}} + {u_{ - 1}}{v_{ \pm 2m}}} \right) \equiv \\ \;\;\;\;\;\;\;30{v_{ - 1}}{v_{ \pm 2m}} + 5{u_{ - 1}}{v_{ \pm 2m}} \equiv - 5{v_{ \pm 2m}}\left( {\bmod {u_{2m}}} \right) \end{array} $

    于是

    $ 4{y_n} + 5 \equiv - 20{v_{ \pm 2m}} + 5 \equiv \mp 20{v_{2m}} + 5\left( {\bmod {u_{2m}}} \right) $

    由引理1可知

    $ \left( {\frac{{ \pm 20{v_{2m}} + 5}}{{{u_{2m}}}}} \right) = \left( {\frac{{{u_m} \pm 4{v_m}}}{{23}}} \right) $

    只要完全按照引理3证明过程中的方式取m,可得

    $ \left( {\frac{{4{y_n} + 5}}{{{u_{2m}}}}} \right) = \left( {\frac{{ \pm 20{v_{2m}} + 5}}{{{u_{2m}}}}} \right) = \left( {\frac{{{u_m} \pm 4{v_m}}}{{23}}} \right) = - 1 $

    从而4yn+5不为平方数,假设不成立.

    n=-1时,4yn+5=32,证毕.

    引理5  设n≡2(mod 2×52),则仅当n=2时,4yn+5为完全平方数.

      如果n≡2(mod 2×52)且n≠2,则可令

    $ n = 2 + 2 \times k \times {5^2} \times {2^t}\;\;\;\;\;t \ge 0, 2\;ᚾ\;k $

    若取m为2t,5×2t之一,则由(10) 知,

    $ 4{y_n} + 5 \equiv - 4{y_2} + 5 \equiv - 1839\left( {\bmod {u_m}} \right) $

    由于2∣m时,um≡1(mod 8),um≡1(mod 3).故

    $ \left( {\frac{{4{y_n} + 5}}{{{u_m}}}} \right) = \left( {\frac{{ - 1839}}{{{u_m}}}} \right) = \left( {\frac{{ - 1}}{{{u_m}}}} \right)\left( {\frac{3}{{{u_m}}}} \right)\left( {\frac{{613}}{{{u_m}}}} \right) = \left( {\frac{{{u_m}}}{3}} \right)\left( {\frac{{{u_m}}}{{613}}} \right) = \left( {\frac{{{u_m}}}{{613}}} \right) $

    {um}对mod 613的周期为612,而{2t}对mod 612的周期为24.

    $ m = \left\{ \begin{array}{l} {2^t}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;t \equiv 0, 1, 7, 8, 9, 10, 11, 13, 14, 15, 18, 19, 22, 23\left( {\bmod 24} \right)\\ 5 \cdot {2^t}\;\;\;\;\;\;\;\;\;\;\;\;\;\;t \equiv 3, 6, 16, 17\left( {\bmod 24} \right)\\ {5^2} \cdot {2^t}\;\;\;\;\;\;\;\;\;\;\;\;\;t \equiv 2, 4, 12\left( {\bmod 24} \right)\\ {5^3} \cdot {2^t}\;\;\;\;\;\;\;\;\;\;\;\;\;t \equiv 20\left( {\bmod 24} \right)\\ {5^4} \cdot {2^t}\;\;\;\;\;\;\;\;\;\;\;\;\;t \equiv 5, 21\left( {\bmod 24} \right) \end{array} \right. $

    对所有m,均有$\left( \frac{{{u}_{m}}}{613} \right)=-1$.于是由上可知

    $ \left( {\frac{{4{y_n} + 5}}{{{u_m}}}} \right) = - 1 $

    从而4yn+5不为平方数,假设不成立.

    n=2时,4yn+5=432,证毕.

    引理6  设n≡-3(mod 2×52),则仅当n=-3时,4yn+5为完全平方数.

      如果n≡-3(mod 2×52)且n≠-3,则可令

    $ n = - 3 + 2 \times k \times {5^2} \times {2^t}\;\;\;\;\;t \ge 0, 2\;ᚾ\;k $

    若取m为2t,5×2t之一,则由(10) 知,

    $ 4{y_n} + 5 \equiv - 4{y_{ - 3}} + 5 \equiv - 1839\left( {\bmod {u_m}} \right) $

    由于2∣m时,um≡1(mod 8),um≡1(mod 3).故

    $ \left( {\frac{{4{y_n} + 5}}{{{u_m}}}} \right) = \left( {\frac{{ - 1839}}{{{u_m}}}} \right) = \left( {\frac{{ - 1}}{{{u_m}}}} \right)\left( {\frac{3}{{{u_m}}}} \right)\left( {\frac{{613}}{{{u_m}}}} \right) = \left( {\frac{{{u_m}}}{3}} \right)\left( {\frac{{{u_m}}}{{613}}} \right) = \left( {\frac{{{u_m}}}{{613}}} \right) $

    只要完全按照引理5证明过程中的方式取m,可得

    $ \left( {\frac{{4{y_n} + 5}}{{{u_m}}}} \right) = - 1 $

    从而4yn+5不为平方数,假设不成立.

    n=-3时,4yn+5=432,证毕.

    2 关于(2y+3)2=4yn+5的证明

    引理7  仅当n=0时,4yn+5是平方数.

      要使4yn+5≥0成立,即yn≤1,当且仅当n=0时成立,此时4yn+5=1.证毕.

    3 结论

    根据前面的讨论,现给出本文的主要结论.

    定理  不定方程5x(x+1)(x+2)(x+3)=6y(y+1)(y+2)(y+3) 仅有一组正整数解(xy)=(21,20).

      由引理7知

    $ {\left( {2y + 3} \right)^2} = 4{{\bar y}_0} + 5 = 1 $

    因此y=-1,-2;

    由引理3知

    $ {\left( {2y + 3} \right)^2} = 4{y_0} + 5 = 9 $

    因此y=0,-3;

    由引理4知

    $ {\left( {2y + 3} \right)^2} = 4{y_{ - 1}} + 5 = 9 $

    因此y=0,-3;

    由引理5知

    $ {\left( {2y + 3} \right)^2} = 4{y_2} + 5 = {43^2} $

    因此y=20,-23;

    由引理6知

    $ {\left( {2y + 3} \right)^2} = 4{y_{ - 3}} + 5 = {43^2} $

    因此y=20,-23.

    由此,容易知道方程(1) 共有20组整数解,其中有16组平凡解使得(1) 式两端都?#8838;悖?即(0,-1),(-3,-1),(-2,-1),(-1,-1),(0,-2),(-3,-2),(-2,-2),(-1,-2),(0,0),(-3,0),(-2,0),(-1,0),(0,-3),(-3,-3),(-2,-3),(-1,-3);另外4组非平凡解,它们是(-24,20),(21,20),(-24,-23),(21,-23).因此,不定方程5x(x+1)(x+2)(x+3)=6y(y+1)(y+2)(y+3) 仅有正整数解(xy)=(21,20).证毕.

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    On the Diophantine Equation 5x(x+1)(x+2)(x+3)=6y(y+1)(y+2)(y+3)
    LI Yi-meng, LUO Ming     
    School of Mathematics and Statistics, Southwest University, Chongqing 400715, China
    Abstract: In this paper, with the primary methods of recurrence sequences and quadratic remainders, the authors show that the diophantine equation 5x(x+1)(x+2)(x+3)=6y(y+1)(y+2)(y+3) has a unique positive integer (x, y)=(21, 20).
    Key words: diophantine equation    integer solution    recurrence sequence    quadratic remainder    
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