Rational Solutions of x^2-x-193=y^2, x^2+193=z^2

[2012.10.28]x^2-x-193=y^2, x^2+193=z^2の有理数解

x^2-x-193=y^2,x^2+192=z^2の有理数解を計算するプログラム(pari/gp)

■問題
文献[1](p.49)の「(！)小さい式で定義される想像を絶する大きな値」で紹介されている２番目のDiophantus方程式系 S
     x²-x-193 = y², ----- (1)
     x²+193 = z² ----- (2)
の有理数解(x,y,z)を求める。

■この方程式系の有理数解を求めることは、ある楕円曲線の有理点を求めることに帰着できる。
２次曲線(2)は１パラメータ表示できる。そのためには、(2)より、
     (z+x)(z-x) = 193
を得るので、ある(0ではない)有理数uが存在して、
     z+x = 193u, ------ (3)
     z-x = 1/u ----- (4)
となる。(3),(4)から、x,yはuの有理式
     z = (193u+1/u)/2 = (193u²+1)/(2u), ------ (5)
     x = (19u-1/u)/2 = (193u²-1)/(2u) ----- (6)
で表現できる。
(6)を(1)に代入すると、
     y² = {(193u²-1)/(2u)}²-{(193u²-1)/(2u)}-193
         = (37249u⁴-386u³-1158u²+2u+1)/(4u²) ----- (7)
を得る。
     v = y*(2u) ------------- (8)
と置くと、
     C: v² = 37249u⁴-386u³-1158u²+2u+1 ------------- (9)
を得る。
よって、Sを満たす有理数解(x,y,z)を求めるには、楕円曲線Cの有理点(u,v)を求めれば良いことが分かる。

■楕円曲線CをWeierstrass標準形に変形する。
楕円曲線は有理点(0,1)を持つ。
(9)より、双有理変換φ:(u,v)→(X,Y)
    X = (2u + 2v + 2)/{u²},
    Y = (-2318u² + 4u + 4v + 4)/{u³}
により、Weierstrass標準形の楕円曲線
    E: Y²+2XY-772Y = X³ - 1159X² - 148996X + 172686364
に変換できる。
ここで、φの逆有理変換φ^-1:(X,Y)→(u,v)は
     u = (2X - 2318)/Y,
     v = (2X³ - 4636X² -2XY + 2686562X -Y² + 2318Y)/{Y²}
である。

■楕円曲線Eのねじれ点群E(Q)_torsを求める。
pari/gpで、E(Q)_torsを計算すると、Z/2Zであることが分かる。
    T = (386,0),
    E(Q)_tors = {T, O}

[pari/gpによる計算結果]

gp> ee=ellinit([2, -1159, -772, -148996, 172686364])
time = 1 ms.
%9 = [2, -1159, -772, -148996, 172686364, -4632, -299536, 691341440, -823003841344, 28644288, 0, 13600984846492749824, 1728, [1158.4998381876853183008526542204320979, 386.00000000000000000000000000000000000, -386.49983818768531830085265422043209785]~, 0.094339380858471040033762685397019258577, -0.094339380858471040033762685397019258577*I, 33.300967475108242771745648637589974879, 33.300967475108242771745648637589974880*I, 0.0088999187807596520664532281916298766976]
gp> elltors(ee,1)
time = 4 ms.
%10 = [2, [2], [[386, 0]]]

■楕円曲線Eの有理点群E(Q)/E(Q)_torsの基底を求める。
mwrankで、楕円曲線Eはrank 3であり、有理点群E(Q)/E(Q)_torsの基底は
    P1=(-386,1544),
    P2=(286,7760),
    P3=(1015897/36,-1008628621/216)
であることが分かる。
    E(Q) = Z³×(Z/2Z)

[mwrankによる計算結果]

Enter curve: [2, -1159, -772, -148996, 172686364]
Curve [2,-1159,-772,-148996,172686364] :        Working with minimal curve [0,0,0,-596756,0] via [u,r,s,t] = [1,386,-1,0]

1 points of order 2:
[0:0:1]

Using 2-isogenous curve [0,0,0,2387024,0] (minimal model [0,0,0,149189,0])
-------------------------------------------------------
First step, determining 1st descent Selmer groups
-------------------------------------------------------
After first local descent, rank bound = 3
rk(S^{phi}(E'))=        3
rk(S^{phi'}(E))=        2

-------------------------------------------------------
Second step, determining 2nd descent Selmer groups
-------------------------------------------------------
After second local descent, rank bound = 3
rk(phi'(S^{2}(E)))=     3
rk(phi(S^{2}(E')))=     2
rk(S^{2}(E))=   4
rk(S^{2}(E'))=  4

Third step, determining E(Q)/phi(E'(Q)) and E'(Q)/phi'(E(Q))
-------------------------------------------------------
1. E(Q)/phi(E'(Q))
-------------------------------------------------------
(c,d)  =(0,-596756)
(c',d')=(0,2387024)
First stage (no second descent yet)...
(-1,0,0,0,596756):  (x:y:z) = (10:766:1)
        Curve E         Point [-100:-7660:1], height = 6.01302635531825
(193,0,0,0,-3092):  (x:y:z) = (9:689:4)
        Curve E         Point [62532:1196793:64], height = 7.31307551968803
After first global descent, this component of the rank = 3
-------------------------------------------------------
2. E'(Q)/phi'(E(Q))
-------------------------------------------------------
First stage (no second descent yet)...
(193,0,0,0,12368):  (x:y:z) = (1:1001:3)
        Curve E'        Point [579:193193:27], height = 6.91503269866986
        Curve E         Point [6012006:-1002616615:216], height = 13.8300653973397
After first global descent, this component of the rank = 2

-------------------------------------------------------
Summary of results:
-------------------------------------------------------
        rank(E) = 3
        #E(Q)/2E(Q) = 16

Information on III(E/Q):
        #III(E/Q)[phi']    = 1
        #III(E/Q)[2]       = 1

Information on III(E'/Q):
        #phi'(III(E/Q)[2]) = 1
        #III(E'/Q)[phi]    = 1
        #III(E'/Q)[2]      = 1

-------------------------------------------------------

List of points on E = [0,0,0,-596756,0]:

I.  Points on E mod phi(E')
Point [-100:-7660:1], height = 6.01302635531825
Point [62532:1196793:64], height = 7.31307551968803

II. Points on phi(E') mod 2E
Point [6012006:-1002616615:216], height = 13.8300653973397
-------------------------------------------------------
Computing full set of 8 coset representatives for
2E(Q) in E(Q) (modulo torsion), and sorting into height order....done.


Used descent via 2-isogeny with isogenous curve E' = [0,0,0,149189,0]
Rank = 3
Rank of S^2(E)  = 4
Rank of S^2(E') = 4
Rank of S^phi(E') = 3
Rank of S^phi'(E) = 2

Searching for points (bound = 8)...done:
  found points of rank 2
  and regulator 20.6710620571852
Processing points found during 2-descent...done:
2-descent increases rank to 3,   now regulator = 266.861830517483
Saturating (bound = 100)...done:
  points were already saturated.
Transferring points from minimal curve [0,0,0,-596756,0] back to original curve [2,-1159,-772,-148996,172686364]

Generator 1 is [-386:1544:1]; height 3.67153740401502
Generator 2 is [286:7760:1]; height 6.01302635531825
Generator 3 is [6095382:-1008628621:216]; height 13.8300653973397

Regulator = 266.861830517483

The rank and full Mordell-Weil basis have been determined unconditionally.
 (6.04183 seconds)

■楕円曲線Eの有理点をいくつか計算する。
pari/gpで計算すると、以下のようになる。

[pari/gpでの計算結果]

gp> rpE(ee,1,1,1)
[-1]P1+[-1]P2+[-1]P3=[187488113/155236, 441095833713/61162984]
[-1]P1+[-1]P2+[-1]P3+T=[-224855678/660969, 4205149664648/537367797]
[-1]P1+[-1]P2+[0]P3=[21809/16, 1134261/64]
[-1]P1+[-1]P2+[0]P3+T=[-18206/81, 8966800/729]
[-1]P1+[-1]P2+[1]P3=[2603819689697/1541504644, 2214056175778511061/60522555332728]
[-1]P1+[-1]P2+[1]P3+T=[-748734565022/10408284441, 14623192139811139064/1061863586955261]
[-1]P1+[0]P2+[-1]P3=[-19491842/58081, -93594159576/13997521]
[-1]P1+[0]P2+[-1]P3+T=[65852167/54289, -117854913176/12649337]
[-1]P1+[0]P2+[0]P3=[-386, 0]
[-1]P1+[0]P2+[0]P3+T=[1159, -1546]
[-1]P1+[0]P2+[1]P3=[-6447355298/18258529, 534152378606328/78018694417]
[-1]P1+[0]P2+[1]P3+T=[20861416663/17480761, 428593208255212/73087061741]
[-1]P1+[1]P2+[-1]P3=[942088338697/524776464, -548833367203892561/12021579237312]
[-1]P1+[1]P2+[-1]P3+T=[-143559579502/3831733801, -3053077653203514096/237188154015701]
[-1]P1+[1]P2+[0]P3=[2496841/1764, -1693745033/74088]
[-1]P1+[1]P2+[0]P3+T=[-1822414/9409, -10691053800/912673]
[-1]P1+[1]P2+[1]P3=[6796546407673/5521084416, -4430911654812961997/410238656446464]
[-1]P1+[1]P2+[1]P3+T=[-7740377368078/24173097529, -28616757709261448448/3758360684516333]
[0]P1+[-1]P2+[-1]P3=[56266030/157609, -258309533952/62570773]
[0]P1+[-1]P2+[-1]P3+T=[23954634847/1142761, -3628187173375954/1221611509]
[0]P1+[-1]P2+[0]P3=[286, -7560]
[0]P1+[-1]P2+[0]P3+T=[158839/25, -57885332/125]
[0]P1+[-1]P2+[1]P3=[1121045230/6416089, -172094479699248/16251953437]
[0]P1+[-1]P2+[1]P3+T=[1088021936287/338891281, -918421573117644694/6238649591929]
[0]P1+[0]P2+[-1]P3=[1015897/36, 996604609/216]
[0]P1+[0]P2+[-1]P3+T=[365289170/1002001, 3607823264856/1003003001]
[0]P1+[0]P2+[0]P3+T=[386, 0]
[0]P1+[0]P2+[1]P3=[1015897/36, -1008628621/216]
[0]P1+[0]P2+[1]P3+T=[365289170/1002001, -3564813866424/1003003001]
[0]P1+[1]P2+[-1]P3=[1121045230/6416089, 178961772617432/16251953437]
[0]P1+[1]P2+[-1]P3+T=[1088021936287/338891281, 883179018952399116/6238649591929]
[0]P1+[1]P2+[0]P3=[286, 7760]
[0]P1+[1]P2+[0]P3+T=[158839/25, 56393442/125]
[0]P1+[1]P2+[1]P3=[56266030/157609, 261938942888/62570773]
[0]P1+[1]P2+[1]P3+T=[23954634847/1142761, 3577915248158016/1221611509]
[1]P1+[-1]P2+[-1]P3=[6796546407673/5521084416, 3737594729038163021/410238656446464]
[1]P1+[-1]P2+[-1]P3+T=[-7740377368078/24173097529, 33925113461821383936/3758360684516333]
[1]P1+[-1]P2+[0]P3=[2496841/1764, 1541206325/74088]
[1]P1+[-1]P2+[0]P3+T=[-1822414/9409, 11749185672/912673]
[1]P1+[-1]P2+[1]P3=[942088338697/524776464, 514951307049355673/12021579237312]
[1]P1+[-1]P2+[1]P3+T=[-143559579502/3831733801, 3253959871165141872/237188154015701]
[1]P1+[0]P2+[-1]P3=[-6447355298/18258529, -418822848139696/78018694417]
[1]P1+[0]P2+[-1]P3+T=[20861416663/17480761, -546613162727166/73087061741]
[1]P1+[0]P2+[0]P3=[-386, 1544]
[1]P1+[0]P2+[0]P3+T=[1159, 0]
[1]P1+[0]P2+[1]P3=[-19491842/58081, 113795313632/13997521]
[1]P1+[0]P2+[1]P3+T=[65852167/54289, 96933091518/12649337]
[1]P1+[1]P2+[-1]P3=[2603819689697/1541504644, -2371795100375412273/60522555332728]
[1]P1+[1]P2+[-1]P3+T=[-748734565022/10408284441, -13650660152565458648/1061863586955261]
[1]P1+[1]P2+[0]P3=[21809/16, -1259325/64]
[1]P1+[1]P2+[0]P3+T=[-18206/81, -8076304/729]
[1]P1+[1]P2+[1]P3=[187488113/155236, -541618643109/61162984]
[1]P1+[1]P2+[1]P3+T=[-224855678/660969, -3424686392936/537367797]
time = -5 ms.

■Diophantine方程式系Sの有理数解(x,y,z)で、y > 0, z > 0であるものをいくつか計算する。

pari/gpで計算すると、以下のようになる。
楕円曲線Eのrankは3であるので、Sの有理数解(x,y,z)は無数に存在することが分かる。

[pari/gpでの計算結果]

gp> rpC(ee,1,1,1)
[-19396119395969/543769658184, 18157345034243/543769658184, 20815300568737/543769658184]
[-14033/720, 10343/720, 17233/720]
[-10799022573088415201/747114858526457976, 4118026307905926347/747114858526457976, 14978242440869286913/747114858526457976]
[56688420193/1349019672, 52781354221/1349019672, 59706019969/1349019672]
[-17809239006817/429198681912, 17007641657749/429198681912, 18780889626241/429198681912]
[4033487130169746241/264490585687822704, 1304159484782056727/264490585687822704, 5456224475158629313/264490585687822704]
[1639537/81480, 1128347/81480, 1992337/81480]
[710073141797244769/19611307490616576, 645019341841261343/19611307490616576, 760547381383745377/19611307490616576]
[710073141797244769/19611307490616576, 645019341841261343/19611307490616576, 760547381383745377/19611307490616576]
[1639537/81480, 1128347/81480, 1992337/81480]
[4033487130169746241/264490585687822704, 1304159484782056727/264490585687822704, 5456224475158629313/264490585687822704]
[-17809239006817/429198681912, 17007641657749/429198681912, 18780889626241/429198681912]
[56688420193/1349019672, 52781354221/1349019672, 59706019969/1349019672]
[-10799022573088415201/747114858526457976, 4118026307905926347/747114858526457976, 14978242440869286913/747114858526457976]
[-14033/720, 10343/720, 17233/720]
[-19396119395969/543769658184, 18157345034243/543769658184, 20815300568737/543769658184]
[-60283851291848018086035733633/2002665897865292285565289728, 54596810300113698576254419199/2002665897865292285565289728, 66394293213971765574330268801/2002665897865292285565289728]
[-434537739066001/24275139951720, 292643134389539/24275139951720, 550049592009649/24275139951720]
[-3363833832533175476966497/240866369746305160834032, 963538861768162807691591/240866369746305160834032, 4744742712869080651900129/240866369746305160834032]
[13663121426129679178721/255498479856391347504, 13061050571120671532713/255498479856391347504, 14116650645053407764577/255498479856391347504]
[-2383937/12360, 2383933/12360, 2390113/12360]
[-420059707455329/12310460823264, 390349368365137/12310460823264, 453540312258913/12310460823264]
[73856394810921853671913217/4654024141095773012441688, 30506868960994666777426451/4654024141095773012441688, 98158726615140532397771809/4654024141095773012441688]
[301393649/13634640, 225498071/13634640, 355974001/13634640]
[1298935495786913/29296072259112, 1218004746730139/29296072259112, 1361204456806081/29296072259112]
time = 1 ms.

■参考
文献[1](p.49)の「(！)小さい式で定義される想像を絶する大きな値」で紹介されている１番目のDiophantine方程式系
    x²-157 = y²,
    x²+157 = z²
の有理数解(x,y,z)は、合同数157に関連する有理数であるので、楕円曲線
    E₁₅₇: Y² = X³-157²X
の自明でない(つまりY!=0である)有理点(X,Y)を求めると、それから容易に構成できる。

この有理点(X,Y)の2倍点のX-座標は、
    (X²+157²)²/(4*Y²) = {(X²+157²)/(2*Y)}²
であり、有理数の平方数になるので、
    x=|(X²+157²)/(2*Y)|
とする。このとき、x²-157およびx²+157はどちらも有理数の平方数となるので、y,zは、それぞれ
    y=|(X²-2*157*X-157²)/(2*Y)|
    z=|(X²+2*157*X-157²)/(2*Y)|
とすれば良い。

例えば、E₁₅₇の有理点群E₁₅₇(Q)の基底
    (-166136231668185267540804/2825630694251145858025, -167661624456834335404812111469782006/150201095200135518108761470235125)
から、
    (x,y,z) = (224403517704336969924557513090674863160948472041/17824664537857719176051070357934327140032961660,
          21796977171070247104112455266586147721935979809/17824664537857719176051070357934327140032961660,
          316605068345983991287469841722668300352741098609/17824664537857719176051070357934327140032961660)
を得る。実際に、この(x,y,z)は、x^2-157=y^2, x^2+157=z^2を満たす。

[参考文献]

[1]Clifford A. Pickover(著), 糸川洋(訳), "数学のおもちゃ箱　下 代数、幾何、確率の話編", 日経BP社, 2011(1刷), ISBN978-4-8222-8456-5, {2000円}.
[2]Joseph H.Silverman, John Tate(著), 足立恒雄, 木田雅成, 小松啓一, 田谷久雄(訳), "楕円曲線論入門", シュプリンガー・フェアラーク東京, 1995, ISBN4-431-70683-6, {3900円}.
[3]Joseph H. Silverman, "The Arithmetic of Elliptic Curves", GTM 106, Springer-Verlag New York Inc., 1986, ISBN0-387-96203-4.

Last Update: 2013.07.20

H.Nakao