manuscripta
math.
80,
373
 392
(1993)
manuscripta
mathematica
9 Spfinger~rlag 1993
Explicit c o m p l e t e solution in integers of a class
of equations
(ax 2  b)(ay
2  b)  z 2  c
Kenji K a s h i h a r a
D e d i c a t e d to Dr. Taro M o r i s h i m a
In this p a p e r we will s t u d y the equation for a r b i t r a r y integers a # 0, c and
b= •
• or •
When b = •
we suppose c is divisible by 4. T h e p a p e r will
provide one with a m e t h o d for finding algorithmically all integral nontrivial
solutions of the title equations, where an explicit unit of ~ ( ~ / a 2 n 2  ab) plays
an i m p o r t a n t role.
Introduction
In [2], L . J . M o r d e l l c o m m e n t e d on the quartic equation given by
2
a r s x r y ~ : d z 2,
(1)
r,S~O
where a's and d are integers. His c o m m e n t is t h a t when one integer solution
(x0, y0, z0) of (1) is known, an infinity can be found under certain conditions, and t h a t this leads to solutions (z0, Yl, zl), (xl, Yl, z2), ( z l , Y2, z3),
(z2, Y2, z4), ete . . . . , where from a Pellian equation, yl, z l , y2, x2, ... m a y
each have an infinity of values.
We will further consider this fact for the following special type :
( a z 2  b ) ( a y 2  b) = z 2  c,
(2)
where a , c 6 7/, a r 0, b = + 1 , •
•
When b = •
we suppose c =
0 ( m o d 4). In [3], we have investigated the equation for the case a = 1 and
b = 1. In this paper, we will show t h a t this equation can be dealt with generally
in the same m e t h o d .
If we fix z = n, equation (2) is written as
z 2  (a2n 2  a b ) y 2 "  a b n 2 + b ~ + c.
(3)
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Kashihara
We can solve equation (3) by the theory of binary quadratic forms as presented
in [4] or [5]. We will show a permutation group on all integral solutions of
equation (2), which will be denoted by G. And we will prove the possibility of
computing algorithmically a minimal finite set of integral solutions of the title
equation, such that the Gorbits of this set exhaust all integral solutions.
Here we introduce the notion of the t r w i a l solution. When ( a x 2  b)
( a y 2  b ) (  a b x 2 + b2 + c ) (  a b y 2 + b2 + c) = 0, the solution can be trivially
computed. If e = 0, then (x, =l=x, :t:(ax 2  b)) are trivially integral solutions.
Thus a trivial s o l u t i o n is defined as an integral solution such that
( a x 2  b)(ay 2  b ) (  a b x ~ + b~ + c ) (  a b y 2 + b 2 + c) = O,
or (only if c = 0) x 2 = y2.
The cases b = 1, 2 or 4, will be discussed in detail and for the other cases we
will state the results and give only the proofs different from the previous ones.
Up to the end of section 3 we suppose b = 1, 2, or 4. Since (  a x 2  b ) (  a y 2  b )
=
z 2  c is equivalent to ( a x ~ + b)(ay ~ + b) = z 2  c, we may suppose a > 0.
Notations.
F~c : the set of all real solutions of equation (2).
Sabr : the set of all integral solutions of equation (2).
C , : intersection of F~c and the plane x = n.
T~r : the set of all trivial solutions of equation (2).
C + y , C +~ and C + are the following branches of Cn:
c2~ := {(~, ~, z) e c . I y > 0},
c.+~:={(x,y,z)eC, lz > 0},
c . + := {(x, y, =) E c . ly > 0, z > 0}.
c~, v, Pl, P2 and P3 are the following permutations on Fbr or sbc:
~(x,
pl(~,
p~(~,
p3(x,
u,
y,
y,
y,
z):=
z):=
z):=
z):=
(y, ~, z),
(~, u, z),
( ~ ,  y , z),
(~, y, =).
G is the following permutation group and G1, G2, H and H1 are the following
subgroups of G:
G : = ~ 0", 7", ,O1, /)2, ,03 >,
G I "'"< (T, /91, P2, P3 >,
G2 : = < cr >,
H : = < T, P1, P2, P3 >,
H1 : = < pl, p2, P3 •
9
Kashihara
375
Let P and Q be points on F,br (or S~r I f Q = g P for some g E G, then P
and Q are called Gequivalent, otherwise Gindependent. These relations are
denoted by P , Q and P fi Q, respectively.
The following function is used:
~ ( z , y, z) := z 2 + y2.
1. T h e s t r u c t u r e o f G
As already noted, we assume b = 1, 2 or 4. If we fix z = n(> 0), equation (2) is
written as
Z 2  (a2n 2  ab)y 2 =  a b n 2 + b 2 + c.
(3)
or equivalently
N (z + y x / a ~ n ~  a b ) =  a b n ~ + b 2 § c,
(4)
where N denotes the norm from Q ( ~ / a 2 n 2  ab) to Q. Here we put
2 a n 2  b § 2 n x / a 2 n 2  ab
~n =
b
Since Ib} ~ {1, 2, 4}, it is straightforward to check that ~,~ is a unit in the ring
of integers of the above quadratic field with norm equal to +1; moreover, it is
useful to note that ~1 = v _ , . Let (Y0, z0) be one of the solutions of (4). Then
N { (zo + yo~/a2n 2  ab) on} =  a b n 2 + b2 + c.
Therefore putting
zl 4 yl x[a2n 2  ab = (zo + yox[a2n 2  ab) ~n,
we have a new solution (Yl, zl). From this fact, if we define cr as above, cr is a
permutation on C,~, and we may replace Cn with F~c or S~r In the cases b = 1
or 2, r lies in the coefficient ring of the Z  m o d u l e {1, x / a 2 n 2  ab}. Consider
the case b = 4. Let (z, y, z) lies in S ~ , and put (z, r/, ~) = e ( z , y, z). Then
from (3) and c =_ 0 (rood 4),
(z + a x y ) ( z  a x y ) :  4 a y 2  4 a x 2 + 1 6 + c  0
(rood4).
And so z + a z y =_ z  a x y  0 ( m o d 2). Hence
77=
2 x ( a x y § z)  4 y
4
E T],
~ =
2 a x 2 ( a z y % z)  8 a x y  4z
4
E T/.
Therefore (x, r], ~) E S~4c, and so r is a permutation on $4c . From the symmetries of equation (2), we can obtain the other generators of G.
L e m m a 1. G is a p e r m u t a t i o n group on Fbar or Sbae.
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Kashihara
By easy calculation we have the following lemma.
Lemma
2. P e r m u t a t i o n s (r, r, Pl, P2, Pa satisfy the following relations,
PiPj = pjPi,
p~ = 1,
vpl = p2v,
O'pi = pi 0"1,
7P2 = Pl r,
=
72 = 1,
rp3 = p3r,
1,
,,xr = ,(T(r,) 1,
where i , j = 1,2,3.
Corollary
1. Let A = {or, (r t ,
r(rr, ( r ~ r )  l } ,
H = < r, Pl, P2, P3 >, then
AH = HA.
Corollary
2. A n y element of G has a representation in the form,
where a , b , c , d = 0 o r l a n d e i , f i
E 7/
Proof. Let g be an a r b i t r a r y element of G. Using Corollary 1 several times, g
takes the form h ( r " ' ( r ( r r ) / ' . . .(r'k(r(rr) lk, where h E H. By the relations p's
and v, h takes the form F.~a~b~c~d
[]
1 /'2/'3"
9
2. T h e p e r m u t a t i o n
o"
We continue to assume t h a t b = 1, 2 or 4. In this section, we fix z = n(>_ 0) and
regard (r as a p e r m u t a t i o n on Cn. Sometimes, for a point P = (n, y, z) E Cn,
we will simply write P = (y, z). T h e curve Cn varies as follows. In the case
an ~  b < O,  a b n 2 + b2 + c > O, Cn is an ellipse or a single point. (See Fig.
1.) In the case an 2  b =
O, c>_ 0, it degenerates to one or two lines. In the
case an 2  b > O,  a b n 2 + b2 + c > 0, it is a h y p e r b o l a with focuses on the z
axis. In t h e case an 2  b > O,  a b n 2 + b2 + c = 0, it degenerates to two lines.
A n d finally in the case an 2  b > O,  a b n 2 + b2 + c < 0, it is a h y p e r b o l a with
focuses on the y axis. (See Fig. 2.) We have the following lemma.
Lemma
3. Let n > O, except f o r the case ( i ) . For a point Po on Cn, put
P1 = (rPo, and let PoP1 be an arc of Cn, in which P1 is contained and Po is not.
( i ) I f n = O then (r = p2p3.
(ii) 1 f a n 2  b < 0 and  a b n ~ + b 2 + c > O, let Po = (Yo, zo) be a point on Cn
such that aPo = p~.Po, yo > O, zo > O. Then
Cn = U (ri PoP1,
i0
Kashihara
377
where r = 2, 3 or 5.
(iii) I f an 2  b > O,  a b n 24b 2 + c > 0 ,
let Po = (  y o , Zo) be a point o n e +z
such that aPo = p2Po, yo >_ O. Then
C+Z = U a' p o P , .
iEZ
(iv) I f an 2  b > 0 and  a b n ~ + b~ + c < O, let Po = (Yo,  Z o ) be a point on
C +y such lhat o'Po  P3Po, Zo > O. Then
= U
PoP,.
iEZ
The proof of case (i) is clear from the definition of r
Proof of case (ii). See Fig. 1. From an 2  b < 0 and n > 0, we have (a, b, n) =
(1,2, 1), (1, 4, 1), (2, 4, 1) or (3, 4, 1). or can be expressed by the following matrix
respectively:
1
10), A 2 : ~ (  1 3
Al:(21
1
_1),
1
A3:~(24
~),
1
A4:~(
_
13 11).
It is obvious that A14 = I, A 3 = I, A~ = I, A~ = I and that such Po exists.
Here we put Pi = ~iPo (i = 1,2,... ,6). Then it holds that P3 = Po, P4 = Po or
P6 = Po. By linearity ore, Pi+lPi+2 = ~rP~i+l. Therefore C,~ = U[=o P;Pi+] =
r
U~=0 ~ i P o P i , where r = 2, 3 or 5.
Proof of case (iii). First we note that the relation r
= P2Po is by the definition
of r and P2, equivalent to zo = nayo and now it is clear that such a point P0
exists on C,~. Next let (y, z) E C +z and put (~/, r = a(y, z). First we show
(~, ~) E C +Z. From the definition of or,
C=
2n(a2n2  ab)y + (2an 2  b)z
b
(5)
From (3) and the assumption  a b n ~ + b2 + c > 0, we have
z 2  (a2n 2  ab)y 2 > O.
(6)
Therefore
(2an 2  b)2 z 2 _ 4n2(a2n 2 _ ab)2y 2
> (a2n ~  ab)(2an ~  b)2y 2  4n2(a~n 2 _ ab)2y ~
= b2(a2n 2  ab)y 2 > O.
Hence
(2an 2  b)z > :t=2n(a2n 2  ab)y.
Combining (5) with (7), we have r > 0. And so (T/, ~) e C +Z.
(7)
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Kashihara
Next we show ~ > y. From the definition of a, we have
:
u
2
2  b)y + nz}.
+ ~{(~n
(8)
From (6),
n2z 2 _ (an 2  b)2y 2 > n2(a2n 2 _ ab)y 2  (an 2  b)2y 2
= b(an 2  b)y 2 ~ O,
which implies
nz > +(an 2  b)y.
(9)
Combining (8) with (9), we have ~ > y. Now we put Pi+l = crPi, Pi1 =
( r  l P i and Pi = (Yi, z~) for all i G 77. Then, by (8) and (9) Yi+l >_ Yi + 2/b for
all i E 7]. Therefore Yi
> qc<) as i
> + o c . So we have
C + " = U PiPi+I.
iEZ
By linearity of ~, e,~',+, =
~PZlg, = ~'eo~P~. The
res,lt follows.
Case (iv) is proved similarly.
[]
R e m a r k 1. Sometimes we suppose an arc PoP1 contains b o t h P0 and P1. T h e n
Lemraa 3 still holds.
Lemma
4. Let n > 0 and let ( a , b , n ) 7s (1,4, 1).
I
(i)
The case
 a b n 2 + b2 + c > 0 : Define

Then
C,,
(ii) The case
Eba~
Then
G2E~'~
"
an 2  b
V
ab
i f an 2   b < O , and
C +~
"=
/ a b n 2  52  c
:
"" E
~'T2
"
"~
E ab"
otherwise.
~2
r
> O and  a b n 2 + b2 + c < 0 : D e f i n e
= ! (n, u, z) e c:~ V  J ~ : ~
C +u
a~
/ a b n 2  52  c
<Y<nv a~:~
J"
b"
ac'
II I f P is any point in E~nr and o'P, o '  l P do not belong to H 1 P , then
9~(~P) > ~ ( P ) ,
~(alp)
> 9~(e),
respectively. Moreover, if an 2  b > 0 and P is any point in Ebb, while
Q  (n, y, z) any point not belonging to Eba'~, then, unless Q E H I P
~(q) > ~(P).
Kashihara
Proof of case (i).
379
From c'P0 = P2Po, we see that
Yo =
 ( 2 a n ~  b)yo + 2nzo
b
hence z0 = a n y o and now, since Po E Ca, n 2 + Y~o = ( b2 + c)/ab. Therefore,
P0 = (n, Y0, zo), P1 = r
= (n, Y0, zo), with Yo = ~ /  n 2 + (b 2 + c)/ab and if
we choose the arc PoP1 on Ca, which lies in the half plane z > 0 then, obviously,
PoP1 = E a cb n. Since, by the previous lemma, Ca(resp. C +*) is a union of arcs
r
with i E 7], we may conclude that Ca(resp. C +z) is equal to G2Eba~.
Proof of case (ii). From ~P0 = p3Po we see that
Yo =
(2an 2  b)yo  2nzo
b
'
hence nzo = (an ~  b)yo and now, since Po E C + y , y ~ = n2(abn 2  b~  c)
/(abn 2  b2). Thus, P0 = (n, y o ,  z o ) , P1 = aPo = (n,yo,zo), with Y0 =
nx/(abn~  b~  c)/(abn ~  b2) and the projection of the arc PoP1 on the yaxis is the interval
/abn
b 2c
/abn 2  b 2  c
V 7 z z z , "V 7z:
]
]
As y runs through the values of this interval, the point (n, y, z) runs through
Eabcn , therefore PoP1 = E a cba
. By the previous lemma, C +~ = ~Jiez ~ipoP~
= G 2 E aba
c"
Proof of part II. In the proof of part I, we saw that E ~ = PoP1 ;hence P E E ~
means, in case an 2  b > 0, that P is a point on the arc PoP1 of one of the
hyperbolas in Fig.2. Then, ~ P E P'd'P2 and ~ r  l p E PoP1, from which it is
clear that, unless P = P0 or P1, the ycoordinate of P is strictly less than
the y coordinate of ~P(resp. of a  l p ) . Thus in view of the definition of 9,
unless crp, crlp E H i P , we have T ( P ) < T ( t r P ) , ~o(alP). In the case
an ~  b < 0 we are in one of the four cases explicitly stated at the begining of
the proof of the previous lemma and we check every case separately. Consider
for example, the case (a, b, n) = (3, 4, 1); then, for P = (1, y, z) E C1 we have
crP = (1, (y + z ) / 2 , (  3 y + z)/2) and the relation ~o(P) < ~o(aP) is equivalent
to y~ < ( y + z ) 2 / 4 and this, in turn, means  1 / 3 < y / z < 1. The last relation
is seen to be true as follows. By (1, y, z) E E4r1 it follows that y2 < (4 + c)/12
380
Kashihara
and since ( 1 , y , z ) is a solution to the title equation, 2 = 4 + c  3y s >_
(12 + 3 c ) / 4 , hence (y/z) s <_ 1/9; consequently  1 / 3 _< y/z < 1 and it is easy
to see that we can have equality only if lYl = ~/(4 + c)/12, z = V/(12 + 3c)/4,
in which case #P E H1P. We deal with the other cases analogously. The proof
of the last statement is obvious from Fig. 2.
[]
In the case of (a, b) = (1,4), part II of this lemma does not hold, because the
order of As is equal to 3; however instead of this lemma we have the following.
L e m m a 5.
I Define
E~ =
{
I
(1,y,z) e C +~ 0 < _ y < v
12
j
Then C 1 = G 1 E l41
c.
II If P is any point in E~le and o'P close not coincide with PaP, then
~(~P) > ~(p),
Proof.
~(#~P) > ~(P).
As we saw, in this case, g is expressed by the matrix
As = ~
3
1
'
the order of which is equal to 3. Let P/'s be the same points that are defined in the proof of the previous lemma. We consider a point Q0 E C +* such
that aQo = p3Qo and put Qi = #iQ0 (i = 1,2). Next we consider a point
R0 = (1,0, zr) E C +z and put R/ = #iR0 (i = 1,2). (See Fig.l.) Then from
both the ycoordinate of P2 and the zcoordinate of Q2 are equal to 0. From
Fig.i, it is obvious that QoP1 = ~  I Q I p 2 = #lpaQoRo and PoRo = 02RoPt.
2
By case (ii) of Lemma 3, 6'i = Ui=0 #iPoP1, therefore C1 = alRoQ0 = a l E ~ .
Next we consider a point P E P~Q0 pr0m Fig.1 we can see that o'P E R1Q1
and a  l P E RsQ2. This proves part II of the lemma.
[]
Kashihara
38"1
p_~
z
P2
P2
Qe
P~
P1
P1
Pe
Y
y
PB
R
1
I
Fig. 1. Ca & E ~
Fig. 2. Cn & E ~
3. T h e M a i n R e s u l t s in t h e Cases b = 1, 2~ 4
In section 2, we have investigated c~ as a permutation on Cn. Now we consider
the permutation group G on S~c.
G~
o', 7", P l , P2, P3 > .
1. Lel b = 1,2,4, (a,b) 7s (1,4) and let Tbac be the set of integral
trivial solutions of equation (2). Define
Theorem
Rl = { ( x , y , z ) E S ~ l O < x < y, x2 +y~ <_ b2ab
+c , z>O } ,
and if c < O,
{
R~= (~,y,z)~S]o
V~
/abx2b2c }
<~o
.
( i ) Put Rbar = Rz u Tbar if c > O and Rbac = Rt U R2 u Tbae if c < O. Then the
set of all integral solutions of (2) coincides with G.R~c.
(ii) If (z, y, z) E Rz then
~
, /~_ + c
}~b , z < y <
~f~ + c
ab
If (x, y, z) e R2(c < O) then, either x • y, in which case
V~ab <
x <
 c + x/c 2 + 16ab 3
4ab
/abx 2  b2  c
, x < y < x V .~~ __b~ ,
or x = y aria V47~ < x < ~/(b  ~)/a. In pa,'tic,,Za,', R1, n~ are at most
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382
finite and algorithmically computable.
Proof. Let P = (x, y, z) be a nontrivial solution of (2), so that  a b x 2 + b 2 + c #
0. Since in the Hiorbit of P there is a point with all its three coordinates nonnegative, it suffices to consider only the case in which z, y and z are nonnegative.
Consider a point P0 = (x0, y0,z0) in the Gorbit of P, such that !o(P0) is
b Indeed,
minimal and x0, Y0, z0 are nonnegative. We will show that P E GRar
if P0 E Tabc then P0 is already in Rbc, so we may suppose that P0 is not a trivial
solution. Suppose first that  a b x ~ + b2 + c > 0 or  a b y o2 + b2 + c > 0. If the first
of these inequalities holds and Xo = 0, then, by (3) (with n = x0), P0 is already
in R1, therefore we may suppose that xo > 0. Similarly, if the second inequality
holds, then we may suppose yo > 0. By case (i) of L e m m a ~ , Po or rPo,
respectively, belongs to the G2orbit of some point P~ = (n, Yl, z~) E E ~ , ( w h e r e
b2 + c
ab ' by the definition of E]~ in this
n = z0 or Y0, respectively) and ~o(P1) <
case. If n < Yl, the point P1, is already in R1, otherwise rPI,p~P1 or "rp2P1
belongs to R1. Thus, P is in the Gorbit of P1 E R1.
Next, let abz2o + b2 + c < 0 and  a b y o 2 + b2 + c < 0, so that by (3) (with
n = xo) ax2o  b > 0 and ayo~  b > 0. Suppose that P0 ~ acb:b~~By case (ii) of
L e m m a ~,Po is in the G2orbit of some point P1 E Eab~~ and then, by part II
of the same lemma, ~o(P0) > ~o(P1). But P1 is also in the Gorbit of P, hence,
the last inequality contradicts the minimality of ~o(P0). Thus, P0 E E ~ ~ Since
!o(TP0) = ~(P0), we can prove, in exactly the same way, that rP0 E acF.bY~If
c > 0 we will be led to a contradiction. Indeed, by the definition of the sets F) ~~
a n d ]~byo we must have

ac
/ a b x ~  b2  c
/abyo2  b2  c
V

0<
0 0. Multiplication and squaring of the last two relations
gives(as we previously saw, nominators and denominators are positive)
(abx~o  b2  c)(abyo2  b2  c) > (abx~  b2)(abyo2  b2),
(11)
i.e. c k ab(xo~ + yo2)  2b 2. However, if we add the relations  a b x ~ + b 2 + c < 0
and  a b y o 2 + b2 + c < 0 we get ab(xo2 + yo2)  2b 2 > 2c > c, arriving at a
contradiction. If c < 0 we put PI = P0 if x0 _< Yo and P1 = rP0 otherwise. If we
put P1 = ( x l , y l z l ) , t h e n 0 < x l < y l zl > 0 and P1 E F,b~ hence, by the
definition of this set in the case under c o n d i t i o n (  a b x 2 + b ~ + c < O, a x ~  b > 0),
we see that P1 E R2 and P is in GR2.
Now we prove part (ii). If ( x , y , z ) e R1 then both x 2 and y2 must be _<
,

1
,
 
ac
,
b2+ c
b2ab
+ c and the minimum of these two, i.e. x 2 cannot exceed ~ .
If (x, y, z) E
R2 and x = y,then ( a x e  b ) 2 = z ~  c >  c hence (az 2  b + z ) ( a x 2  b  z )
= c
and it follows that a z 2  b <  c , as claimed. Finally, if (z, y, z) E R2 and x < y,
then, from the inequalities in the definition of R2, it follows that
Kashihara
xv
383
/ abx 2  b2  c
7x ~ b5
x > I ,
(12)
from which,
 c x 2 > (2x + 1)(abx ~  b2) > 2x(abx 2  b2)
(13)
and, consequently, 2abx 2 + cx  2b2 < 0, hence x is strictly less than the larger
root of the lefthand side. This completes the proof.
[]
R e m a r k 2. Note that the Gorbit of a trivial solution contains both trivial and
nontrivial solutions. Example: (a, b, c) = (1, 1, 9); a trivial solution is (1,2, 3)
and ~rr~r(1, 2, 3) = (8,175, 1389) which is not trivial.
For the case (a, b) = (1,4) we have the following analogous theorem.
T h e o r e m 2. Let (a, b) = (1, 4) and let T14c be the set of integral trivial solutions
of equation (2). Define
= {(x,u,z)e $14r
R]' =
16+c
~ , z4 > O } ,
xr162
y2 < 1 2 + c
~ ( 1 , y , z ) e $4c
k

z>0


12
'

)
'
and if c < O,
/ 4 x 2  16  c
R2= ( ,v,z)eS2c 2O
}
= I~I U R ", H T#r if c _> O and a L = Ri U Rf U
U
Then the set of all integral solutions of (2) coincides with GR4c.
if c < O.
Put R ~
(ii) I f (x, y, z) E R] then
O 0,
so that P1 E R]~. Therefore P is in the Gorbit of P1 E R]~. This proves part (i)
of the theorem. Part (ii) is proved in exactly the same way as in Theorem 1. []
Theorem 1 and 2 give us a criterion for solvability of equation (2) with
b  1,2 or 4. It is solvable if and only if R~c is not empty. Moreover we can
derive effectively all solutions by GRit. Next we will show that R~c \ T~ is
a minimal set of integral solutions of equation (2) such that the Gorbits of
this set exhaust all nontrivial integral solutions except for that derived from a
trivial solution.
P r o p o s i t i o n 1. get b = 1, 2, 4. For any points A, B belonging to R~r \ T~r
it holds that if A "~ B then A = B.
Proof. Let B = gA , g E G then by Corollary 2 of Lemma 2, we have a
representation
g = p~ p~ p~ , d ~ e , ( v ~ v ) f , . . . ~ e ~ ( , ~ v ) f ~ ,
where k is some nonnegative integer, el, fi E 7/ (i=1,2.....k), a, b, c, d = 0 or 1.
We can show g E H by reduction to absurdity. Suppose g ~ H and
A=P0
a, )P1 a2 P2
''"
{r
'P~I
a. p,~ h 'P,~+l = B ,
k
+
By Corollary
1 of Lemma 2 we may suppose P i r HP~ (i < j < n). Consider a point Pm in
these Pi (i=l, 2..... n+l), such that ~a(P,,) is maximal.
First we show m r 0, n, n + 1. By Lemma d and 5,
where
C
h C H, n =
~(~rPo) > ~(Po),
~(alpo) > to(Po),
where we may replace cr with r a y . So we have m r 0. Similarly we have also
m r n, n + 1. Now we may suppose
~(Pm1) < ~(Pm),
~o(Pm+l)< ~(Pm).
(14)
Let Qm = (~, 77, ~) be a p o i n t such that Qm E HPm, ~, r], ~ > O, 7? > ~.
By Corollary 1 of Lemma 2,
Pm+ e
u
U
U
By the definition of r and a,
2~(a2~ 2  ab)~+ (2a~ 2  b)r
b
)
cr_lQ m _ (~, (2a~ 2  5)7/ 2~i 2~(a2~ 2  ab)~ + (2a~ 2  b ) ~
b
)
b
[ (2aT/2 b)~ + 2z?~
2r/(a2~?2  ab)~ T (2a7/2 b)~'~
To'TQm
b
]
b
' ~'
2r/(a2~ 2  ab)~ + (2a~ 2  b)~'~
r~r_lrQr n = ( ( 2 a t / 2  b)~  2~I
b
]
\
b
, 7,
oQm
(~, (2a~ 2  b)W + 2 ~
b
Kashihara
385
First we consider the case:
~ ( ( r  l Q m ) < ~(Qm),
~(r~
~_ ~(Qm).
(15)
From the first inequality, we obtain b~7 < 2 a ~ r /  b~7  2 ~ < b~/ , hence
0 < ~ ( a ~  ~) < ~ .
(16)
0 < , ( a ~ ,  ~ ) < b~.
(17)
Similarly from the second,
In view of (16), 77 > 0, hence multiplication of (16) and (17) gives
0 < 5 ~ ( a 5 r /  r < b25r/, from which
0 < (a~
~)~ < b2.
(18)
In the case b = 1, this is a contradiction. Consider the case b  2. From (18)
a ~ 7 7  ~   1 . Here we put S = a  I Q m ,
T = r c r  l r Q m . Then
S  (~, ~  r], a ~ ( r ]  ~) § 1),
T   (  ~  F 77, r/, a r / ( ~  r]) q 1).
By easy calculation, we have
S = pip2p3r~T
or equivalently
T = pip2p3r(r~r)S.
(19)
It follows that P,~+i = h'g* P,~_ i with some h* E H, g* E {or, z  i, rzr, rcr lr}.
By Corollary 1 of Lemma 2, we have a representation
h I~I
g,h. = h,
i=m+2
Ii
g:,
i=m+2
with some h I E H, g~ E {cr, ~r1, rc~r, r a  l r } (i=m+~,...,,~). So we obtain a new
sequence of points from A to B, where the number of Pi's decreases by one.
In the case b = 4, from (18) and ~2 _ a2~2~2 = 0 (rood 4), we have a~,j  r = 2,
hence we obtain the relation (19) and the same result.
In the case
~(~Q,~) < ~(Qm),
~(~r
_< ~(Qm),
we have ~,77 > 0, a~ ~  b < 0, at/2  b < 0, hence ~ = q = 1. Therefore
oQm = p3r (r~rQ,,~),
which contradicts the assumption Pm+l ~ HPm1.
Next we consider the case
~(~XQm ) < ~(Qm),
~ ( r ~ r Q m ) < ~(Q~).
(20)
From these inequalities we have
0 0, a~ ~  b < 0. After consideration of
Pm+l ~ HPm1, Prol, Pm+l ~ HPm, only one case remains, that is (a, b,~) =
(1,4, 1). In this case, the order ofo" as a permutation on C1 is 3, hence o'lQm =
o'(o'Qm). Therefore we come to the same result. Similarly
<
_<
leads also to the same result.
Now the assumption leads us to a contradiction or a new sequence of points
from A to B, where the number of Pi's decreases by one. And if n = 1, either
~ ( B ) > ~(A)
or
~(A) > ~(B),
which contradicts A, B E Rbar Consequently g E H, which implies A = B.
[]
R e m a r k 3. In the case b2+c < 0, the proof becomes simple. We consider a point
Qm as in the proof. From (3) with ab~ 2 + b2 + c < 0, we have a~ 2  b > 0,
similarly we have a~?2  b > 0. From the first we have ( 2 a ~ 
b)g + 2 ~ _
b
2(a~ 2  b)q + 2 ~
(2a~ 2  b)q + 2~(
> 0, consequently
> ~, from which
r]=
b
b
we have ~(~Qm) > ~(Qm). Likewise, from the second we have ~(r~rQm) >
~(Q,~). From these relations it is clear that it suffices to consider only the case
<
_<
Kashihara
387
4. The Cases bl,  2 or  4
In this section, we examine the cases b =  1 ,  2 or  4 . We state the results
and give the proof only at the points in which the proof differs essentially from
the previous one. We preserve the notations for the case b > 0, except for the
p e r m u t a t i o n cr and Tabc.
In the case n = 0, we can solve equation (3) by the theory of binary quadratic
forms (see [4] or [5]), the integral solutions of which are algorithmically computable. T h u s we define a trivial solution of equation (2) as an integral solution
such t h a t x y (  a b x 2 + b2 + c ) (  a b y 2 + b2 + c) = 0 or (only if c = 0) x 2 = y2.
(Note that it always holds t h a t (ax 2  b)(ay 2  b) r 0 in this case.) 1
We define cr as
(
o'(x, y, z ) : =
x,
(2ax2b)y+2xz
b
'
2x(a2x2ab)y+(2ax2b)z)
b
"
Since
2an 2  b + 2n~/a2n 2  ab
~n ~
b
is a unit with n o r m + 1 in the ring of integers of ff~(x/a2n 2  ab), ~ is a permutation on Sabc , F~c or C~ and satisfies lemmas and corollaries in Sec.1. We can
prove the following in the same way as in L e m m a 3.
L e m m a 6. Let n > O. For a point Po on C~, put P1 = ~P0, and let PoP1 be an
arc of Cn, in which P1 is contained and Po is not .
( i ) I f  a b n 2 + b2 + c < O, let Po = (Yo,  Z o ) be a point on C +v such that
rrP0 = P3P0, Zo > O. Then
C+Y =
U ~iP~
i~z
(ii) I f  a b n 2 + b2 + c > O, let Po = (  Y o , Zo) be a point on C +z such that
crPo = P2Po, Yo >_ O. Then
cZ" = U
iEZ
Also we can prove the following in the same way as in L e m m a 4.
Lemma
7. Let n > O.
I
( i ) I f  a b n 2 + b2 + c < 0 ,
define
1 Finding all trivial solutions with x = 0 or y = 0 amounts to solving a Pell equation
and, in that sense, these solutions are not trivial in a strict sense.
Kashihara
388
+c
E2~ : { ( . , y , z ) e c . +~ V~l/abn2a~n2 b2ab c < Y < v~/b2ao'7
Then
C +y
=
n2 } .
G 2 E ab"c '
(ii) If  a b n 2 + b2 + c > 0, define Eban as
dg~?~ : ( p ) ,
~ ( ~  l p ) > ~(p),
respectively. Moreover, if P is any point in E~'~, while Q = (n,y, z) any
point not belonging to E~'~, then, unless Q 6 HI P
:(Q) > :(P).
T h e o r e m 3. Let b =  1 ,  2 ,  4
solutions of equation (2). Define
y,z) ES~e
and let Tbac be the set of integral trivial
OO ,
and if c > O,
R2=
l
/  a b x 2 + b2 + c
(=,y,z)~SL 0O
}
(i) Put Rbac = R2 U Tbae if c > 0 and Rbae = R1 U Tbac if C < O. Then the set of
all integral solutions of (2) coincides with GRbae.
(ii) If (=, y, z) E R1 then
0 o) then, eitheT z # y, in which case
o<~<
c + x/c 2 + 16ab 3
/  a b x 2 + b2 + c
4ab
' = 0 or abyo~ + b2 + c > 0" in t h e p r o o f
of Theorem 1 w i t h "abx~ + b~ + c < 0 or abyo2 + b2 + c < 0" a n d t h e case
"  a b z ~ + b 2 + c < 0 a n d  a b y 0 ~ + b 2 + c < 0" with "  a 0z 02 + b2 + c > 0 a n d
abyo2 + b 2 + c > 0", and using Lemma 7 i n s t e a d of Lemma ~, we can prove
this in t h e s a m e way as in Theorem 1.
Proposition
2. Letb =  1 ,  2 ,  4 . For any poinls A, B belonging to R~c\T~r
it holds that if
A ..~ B
then
A = B.
Proof. T h e p r o o f is similar to t h a t of Proposition 1 a n d we give only a sketch
of it. We preserve the definitions a n d n o t a t i o n s in t h e p r o o f of Proposition 1. For
a p o i n t P = (x, y, z) e R~c \ Tabc with x, y > O, z > O, unless crP, vt77"P E H P ,
we have
W(trP) > #(P),
~ ( v a r P ) > ~(P).
C o n s e q u e n t l y if P,~ is a p o i n t such t h a t ~ ( P m ) is m a x i m a l in t h e sequence of
p o i n t s from A to B, t h e n we m a y a s s u m e
~(~lPm) < ~(Pm),
~(r~'*Pm) < ~(P~).
F r o m t h e s e i n e q u a l i t i e s we o b t a i n
0 < ;  a~7 <  b .
(23)
If b =  1 this is a c o n t r a d i c t i o n . If b =  2 we have (  a ( y = 1, hence
S = (~, 77  ~, a~(~  7) + 1), T = (~  7, q, a 7 ( 7  ( ) + 1), f r o m which we have
S = p2vcrT
or equivalently
T = plv(r~v)S,
(24)
hence, t h e r e is a new s e q u e n c e of p o i n t s f r o m A to B where t h e n u m b e r of P ' s
decreases by one. In t h e case b =  4 , from (23) a n d (2 _ a2~272  O(mod 4) we
have (  a ~ q = 2, hence we o b t a i n t h e r e l a t i o n (24) a n d come to the s a m e result.
A n d if n = 1 we are led to a c o n t r a d i c t i o n in the s a m e way as in t h e p r o o f of
Proposition 1. T h u s if we s u p p o s e g ~ H t h e n we are led to a c o n t r a d i c t i o n .
C o n s e q u e n t l y g E H , hence we a r r i v e a t A = B .
[3
5. E x a m p l e s
In Table I and 2, we show R ~ for t h e cases a = 2, b = t1,  8 5 < c < 85, e ~ 0,
which we have o b t a i n e d by using UBASIC.
Theorem
5. Let a > 2, b = 1, c =  1 . Then equation (2) has only one solution
(0, O, 0).
Proof. It is obvious that T~_ 1 = { ( 0 , 0 , 0 ) } and R1 = {(0,0,0)}, hence we
have Rla_l : {(0, O, 0)}. Therefore by Theorem 1 we obtain sin,1 = GRin,1  ~ {(o,o,o)}.
[]
390
Kashihara
T a b l e l . R~r of (2x 2  1 ) ( 2 y 2  1 )   z 2  c ,
c
85
83
81
80
78
75
73
72
71
70
67
64
63
62
61
56
55
54
51
48
49
46
45
43
40
38
35
33
x y z
4 646
2 3 6
1 7 4
1 9 9
1 8 7
4 538
2 4 12
1 7 5
I 6 0
1 6 1
2 3 7
1 6 2
3 3 15
1 8 8
1 6 3
1 7 6
4 430
2 621
1 6 4
238
2 5 17
3 634
3 528
I 5 1
i 7 7
1 5 0
2 2 0
2 2 1
2 4 13
1 6 5
1 5 2
2 2 2
3 422
1 5 3
223
2 3 9
1 6 6
1 5 4
2 2 4
c
33
31
30
27
24
22
21
19
17
16
15
13
8
7
6
3
1
1
2
3
5
7
8
9
10
11
15
16
17
18
x y z
3 3 16
1 4 0
1 4 1
1 4 2
1 5 5
2 2 5
1 4 3
2 4 14
2 3 I0
1 3 0
1 3 1
I 4 4
1 3 2
2 2 6
1 3 3
1 2 0
1 2 1
1 2 2
0 0 0
1 i0
0 1 0
0 1 1
0 0 2
I 1 2
0 1 2
0 2 0
0 0 3
0 2 1
1 1 3
1 2 4
0 1 3
0 2 2
0 0 4
1 1 4
228
0 2 3
0 1 4
0 3 0
0 3 1
for  8 5 < c <
c
19
21
23
24
25
26
29
31
32
33
35
40
42
37
39
43
47
48
49
50
51
x y z
1 2 5
1 3 6
0 3 2
0 2 4
0 0 5
1 1 5
2 312
0 1 5
0 3 3
1 2 6
0 4 0
0 2 5
0 4 1
1 3 7
2 2 9
0 3 4
1 4 8
0 0 6
0 4 2
116
3 318
0 4 3
0 3 5
1 2 7
0 1 6
2 416
0 2 6
0 4 4
1 3 8
0 0 7
1 1 7
0 5 0
3 424
0 1 7
051
1 4 9
2 313
1 5 10
2 210
85, c r
c
53
56
56
57
58
63
64
65
66
67
69
71
72
73
74
75
77
79
80
81
82
83
85
x y z
0 3 6
0 5 2
0 2 7
0 4 5
1 2 8
2 520
0 5 3
0 0 8
i 1 8
4 432
1 3 9
0 1 8
0 5 4
0 3 7
0 4 6
3 530
1 4 i0
0 2 8
0 6 0
061
1 5 ii
3 3 19
1 6 12
0 5 5
2 211
2 4 17
12
9
0 6 2
2 3 14
2 624
0 0 9
0 4 7
0 6 3
1 1 9
038
4 540
0 1 9
1 3 10
0 5 6
Kashihara
Table 2. R~~ of (2x 2+1)(2y ~ + l ) = z
c
85
82
81
80
77
75
74
73
72
71
69
65
64
63
59
57
56
53
51
50
x y z
0 7 4
1 4 4
4 540
0 9 9
2 624
2 2 0
0 8 7
2 2 1
2 2 2
1 6 12
0 7 5
14 5
0 6 0
0 6 1
1 5 9
2 415
223
3 317
2 3 10
0 6 2
3 530
0 8 8
2 2 4
4 432
0 6 3
0 7 6
1 4 6
2 520
0 6 4
1 3 0
1 3 1
2 2 5
1 5 10
1 3 2
0 5 0
3 424
0 7 7
c
50
48
47
45
42
41
37
35
33
32
29
27
26
24
23
21
19
18
17
15
11
10
9
8
x y z
0 5 1
1 4 7
2 311
0 6 5
1 3 3
0 5 2
2 2 6
0 5 3
1 3 4
2 4 16
0 6 6
3 318
0 5 4
1 4 8
0 4 0
0 4 1
135
2 2 7
0 4 2
1 2 0
2 312
0 5 5
1 2 1
0 4 3
1 2 2
1 3 6
0 3 0
0 3 1
1 2 3
0 4 4
2 2 8
0 3 2
1 2 4
0 3 3
0 2 0
1 1 0
0 2 1
2c,
391
for 85 < c < 8 5 ,
c
8
5
3
2
1
1
3
6
7
8
9
13
15
16
17
19
22
24
25
27
30
31
33
35
37
39
40
43
x y z
1 1 1
0 2 2
1 1 2
0 1 0
0 1 1
0 0 0
0 1 2
0 0 2
0 1 3
0 2 4
1 1 4
0 0 3
1 2 6
0 1 4
0 0 4
0 2 5
115
0 3 6
2 2 10
0 1 5
1 2 7
0 0 5
1 3 9
2 3 14
0 2 6
2 4 18
1 1 6
0 3 7
0 4 8
0 1 6
0 0 6
1 2 8
3 320
0 2 7
1 1 7
2 2 11
1 3 10
c~0
c
45
46
48
49
54
55
57
61
62
63
64
67
70
71
72
73
78
80
81
85
82
x y z
0 3 8
1 4 12
0 1 7
0 0 7
0 4 9
0 5 10
3 426
1 2 9
2 3 15
0 2 8
3 532
1 18
3 638
0 1 8
0 3 9
0 0 8
2212
1 3 11
2 4 19
0 6 10
4 434
0 5 11
1 413
2 523
0 6 12
0 2 9
1 5 15
2 627
1 1 9
1 2 10
0 1 9
0 0 9
3 321
0 3 10
4 542
2 3 16
0 1 9
N o t e . T h e solution (0, 7, ~) such that ~ + T/x/~ = (Zo + yox/2)(3 + 2 v ~ ) k,
k >_ 0, k E 7] is expressed by (0, yo, z0).
392
Kashihara
B y Theorem 3 we can p r o v e the following analogously.
6. Let b =  1 , c =  1 , and let a be not a square. Then equation (2)
has only one solution (0, O, 0).
Theorem
Acknowledgement. The author would like to thank the referee for giving much advice.
He made the author a number of proposals for the purpose of making the paper more
easily readable. In paticular, Theorem I and its proof as they finally appeared are owed
to him. The author also would like to thank Dr. Shinichi Katayama for suggesting
this problem and for giving the idea of permutations on the set of all real or integral
solutions. Finally the author would like to thank John Plant for checking the English
in the paper.
References
1. Kashihara, K.: The diophantain equation x 2  1 = (y2 _ 1 ) ( 2  1). Res. Rep.
Anan College Tech. 26, 119130 (in Japanese) 1990
2. Mordell, L. J.: Diophantain equations. London & New York, Academic Press, 1969
3. Katayama. S. & Kashihara, K.: On the structure of the integer solutions of z 2 =
(z 2  1)(y a  1) + a. J. Math. Tokushima Univ. 24, 111 1990
4. Takagi, T.: Elementary theory of numbers, 2nd ed. Tokyo, Kyoritsu, (in Japanese)
1971
5. Borevich, Z.I. & Shafarevich, I.R.: Number theory. London & New York, Academic
Press, 1971
6. Katayama,S.: On Products of Consecutive Integers. Proc. Japan Acad. 66, Ser.A,
No.10, 305306 1990
7. Kida, Y.: UBASIC 86. Tokyo, Nihonhyoronsha, 1990
Kenji Kashihara
Department of Mathematics
Anan College of Technology
265 Aoki Minobayashicho
Ananshi Tokushima 774 Japan
This article was processed by the author
using the SpringerVerlag TEX m a m a t h macro package 1990.
(Received October 23, 1992;
in revised form May 27, 1993)