Indag. Mathern.. N.S.. 19 (1 ). 65-72
March. 2008
On the number of solutions of Goormaghtigh equation for given
.\" and y
by 80 He a and Alain Togbe b
Department oj Mathematics. ABa Teacher s College. Wenchuan. Sichuan 623111111. P. R. China
Mathematics Department. Purdue University North Central. 14111 S. U.S. 421. Westville. IN 46391.
USA
a
b
Commumcated by Prof. R. TI]deman
ABSTRACT
In this note. we extend a result obtamed by Bugeaud and Shorey m Pacific J. Math. 207 (2002) 61-75.
In fact. we show that the Goormaghtlgh equation
.en _
1
;;11 - 1
x
~- y - l '
.>
1. y
.>
1. III
.>
1. n
.>
1. x
y. t,'r x. y. Ill. n E I T.
<-
has at most 1 solution (Ill. n ) t,'r given x and y.
1 INTRODUCTION
For any positive integer N with N
(x. Ill) of the equation
(1.1)
x lll -1
N = ---.
x-I
X.1Il
2, let s(N) denote the number of solutions
>
EN. x
~
2. III
>
2.11
>
2.
Ratat [13] in 1916 and Goormaghtigh [6] in 1917 observed that s(31) = 2 and
s (8191) = 2 respectively. We consider the equation
(1.2)
x lll -1
x-I
ylZ -1
y -1 .
x
>
1. y
>
1. III
>
1. 11
>
1. x
<
y. for
Ill.
11 . x. YEN.
It has been conjectured that equation (1.2) has only a finite number of solutions,
even that has only two solutions (x. y. Ill. 11 ) = (2.5.5.3). (2.90.13.3). Many auE-marls: hebo-one(a;hotmarl.com (B. He). [email protected] (A. TogM).
65
thors have proved that if two of the variables x, y, Ill, 11 are fixed then equation (1.2)
has a finite number of solutions. See for examples [1,2,4,5,8,9,11,15-20]. In
particular, Bugeaud and Shorey [2] considered x, y fixed and used Baker's theory
of linear forms in logarithms to prove the following result.
Theorem 1.1. Let y > x > 1 he integers. If gcd(x, y) > 1 or (f y ~ 10 11 , then
equation (1.2) has at most one solution. Furthel; (f y ~ 7, then (1.2) has at
most two solutions. FinallJ; the on()' solutions of (1.2) with y ~ 6 are given hy
(x,
y,
Ill, 11)
= (2, 5, 5, 3), (2, 6, 3, 2).
In fact, Theorem 1.1 extends a result obtained by Shorey [16]. In 2005, Yuan [21]
considered the case 11 = 3 and proved the following result.
Theorem 1.2. If 11 = 3, then equation (1.2) has no other solutions
odd than (2,5,5) and (2, 90, 13).
(x,
y,
Ill)
with
III
The aim of this note is also to extend Theorem 1.1. Here is our main result.
Theorem 1.3. Let y
one solution (Ill, 11 ).
>
x
>
1 he given integers. Then equation (1.2) has at most
The organization of this note is as follows. In Section 2, we recall or prove some
results useful for the proof of Theorem 1.3. We use Baker's theory oflinear forms in
logarithms to find an upper bound for an expression of III depending on y. The proof
of Theorem 1.3 is done in Section 3. We use some properties obtained by Bugeaud
and Shorey to find upper bounds for x, y, 11, Ill. Out of the remaining possibilities
we compute the solutions of equation (1.2) by the means of a program written in
PARI/OP [12].
2 SOME LEMMAS
The following result is contained in the proof of Theorem 1.1 by Bugeaud and
Shorey, see formulas (23) and (24) in [2]. We write these properties as a lemma.
Lemma 2.1. For given positive integers x, y and gcd(x, y) = 1, (f there exist pairs
ofpositive integers (Ill" 11,), i = 1. 2,3, with 1 ~ 1111 < 1Il2 < 1Il3 such that
(y -1)x1ll ,
then we have
and
66
-
(x _1)y"' = y -x,
Using an elementary method, Le [7] proved the same result on the equation
ax lll
-
by" =
C,
where a, b, C, x, yare positive integers, but with the conditions gcd(ax, by) = 1 and
1111, 111 > 1.
We recall the following result on linear forms in logarithms due to Matveev [10].
Lemma 2.2. Denote hy 0'1, ... , all algehraic nll1llhers, not 0 or 1, hy log 0'1, ... ,
log all determinations of their logarithms, hy D the degree over Q of the 111m/her field 1K = Q(a1, ... , all), and hy b1, ... , b ll rational integers. Define B =
max[lb11, .... Iblll), and A, = max[Dh(a,), Iloga,I,0.16) (l ~ i ~ II), where h(a)
denotes the ahsolute logarithmic Wei! height of a. Assume that the nll11lher A =
b1log 0'1 + ... + b ll log all does not vanish; then
where x = 1 if 1K c 1R and x = 2 otherwise and
Applying Lemma 2.2, we have the following result.
Lemma 2.3. Let y
we have
(2.3)
>
x
>
1 he integers and let
Ill-Ill
---- <
1 log III
+
(Ill, II)
he a solution of (1.2). Then
7
1.391·10 (log Y)-.
.
Proof. (Our proof is similar to that of Lemma 2 in [2].) Knowing that y
(this implies that II < Ill), we rewrite (1.2) as
>
x
>
1
Xlii
y"
1
1
0< - - - - . - = - - - - - ,
x-I
y-l x-I y-l
thus we get
o<
1 - y" X -111
(
-1)
X _
_
y-l
< X
-111 •
This inequality implies
Now we apply Lemma 2.2 with D =
1.11
= 3,0'1 = y, 0'2 = x, 0'3 =~::::~. Therefore
~~~
.
A 2 = logx,
A 3 = logy,
B=IIl.
67
So we have
100111100
y -1Il100b x
b
b.
>
+ 10 (XY -_
1)
1 1
0
b
-1.391.1011(logy)210gx(l+loglll).
Combining the above inequality with (2.4), we obtain the desired result (2.3).
D
3 PROOF OF THEOREM I 3
Suppose that equation (1.2) has two solutions (1Il2.112) and (1Il3.113) with 1 < 1Il2 <
> 1, the theorem is true (see [2]). We will assume gcd(x. y) = 1.
We rewrite (1.2) to
1Il3. If gcd(x. y)
(3.1)
(y -1)x1ll
(x -l)y" = y -x.
-
In [2], Bugeaud and Shorey proved that if equation (3.1) has three solutions
for 1 ~ i ~ 3, then
1Il,+1 -
Ill, ~
11,+1 - 11,
2.
~
2.
(Ill,. 11,)
i = 1. 2.
See [2], page 70. We apply Lemma 2.1 with 1Il1 = III = 1. Then we get 1Il3
y"r l + 1. Using this inequality and Lemma 2.3, we obtain
y"2- 1
--------;-1-1 +10g(y"2-
+ 1)
<
>
11
2
1.391·10 (logy).
.
Hence
Put 1(112. y) = 1.391 . 10 11 112(lOgy)3 - y"2- 1. If y is fixed then the function 1 is
decreasing. Since 112 ~ 3, from equation (3.2), we have
Thus we sharpened the bound for y from 1011 to 4.9 . 10 7 . A quick computation
helps to see that if y ~ 7, then we obtain 112 ~ 16. Moreover, for a fixed integer 112,
the upper bound for y is given in Table 1.
Table L Bounds t,'r y dependmg on 112
68
112
3
4
5
6
7
8
y
48054810
94200
4511
753
232
101
~
~
9
54
(i) Case 112 = 3: Yuan [21] solved the equation for the case 11 = 3 and 2 t Ill. So
we only consider 112 = 3 and 2111l2. Equation (1.2) can be rewritten to
(3.3)
x 1ll2 - 1
x-I
7
- - - = y-
.
+ .y + 1.
Using the bound for y in Table 1 and equation (3.3), we have
x 1ll2
2111r1 -1 ~
This implies that
1Il2
1
-
<
x-I
~
2.31.10 15 .
52 .
• If III 2 = 4, equation (1.2) gives
We use Magma [3] to see that the only rational point on the above elliptic curve
is (x. y) = (0.0). Therefore, there is no solution for 1Il2 = 4.
• If III 2 ~ 6, using the bound for y in Table 1, we obtain
x6
-
1
-- ~
x-I
x 1ll2
Thus one can see that x
following:
(3.5)
112 = 3.
1
-
x-I
~
2~x~
7
= y-
+y +1<
2.31.10 15
1181. In fact, more precise upper bounds for x are the
1181
156
50
24
15
10
7
6
5
4
3
2
if III 2 =
if III 2 =
if III 2 =
if III 2 =
if III 2 =
if III 2 =
if III 2 =
if III 2 =
if III 2 =
if24 ~
if28 ~
if 34 ~
6,
8,
10,
12,
14,
16,
18,
20,
22,
26,
32,
~ 52.
1Il2
~
1Il2
~
1Il2
One can notice that equation (3.3) leads to
4·
x 1ll2 - 1
7
-3=(2y+1)-.
x-I
.
The problem is reduced to solving the above equation for each remaining triple
x. y. 1Il2. For example, if III 2 = 6 then we have to check if 4(x 6 - 1 )/(x - 1) - 3 is
a perfect square for x between 2 and 1181. We used PARI/OP [12] to write a short
program for the computations. We found that the left hand of the equation is not a
perfect square in these cases.
69
(ii) Case 112 = 4: Using the upper bound for y in Table 1, we obtain N = \~~11 <
835905761734201. As III 2 ~ 112 + 1 = 5 and x ~ 2, we obtain 1Il2 ~ 49 and x ~ 5376.
Here is the list of more precise bounds for x for each III 2 •
(3.6)
112 =
4.
2~x~
5376
964
306
135
73
45
30
22
17
13
11
if III 2 = 5,
if III 2 = 6,
iflll 2 = 7,
if III 2 = 8,
if III 2 = 9,
if III 2 = 10,
iflll 2 = 11,
if III 2 = 12,
if III 2 = 13,
if III 2 = 14,
if III 2 = 15,
if III 2 = 16,
if III 2 = 17,
if III 2 = 18,
iflll 2 = 19.20,
iflll 2 = 21. 22,
if23 ~ 1Il2 ~ 25,
if26 ~ 1Il2 ~ 31,
if32 ~ 1Il2 ~ 49.
9
8
7
6
5
4
3
2
We notice the following inequality
y3
<
x 1ll2 - 1
v4 - 1
= _._ _
x-I
y-l
This implies that if (Ill. 11) =
<
(1Il2.4)
(v
.
+ 1)3.
is a positive solution of equation (1.2), then y
must be equal to L 3t'~~11 J. Thus we need to determine the integer solutions of
in the range given by formula (3.6). We used PARI/OP [12] to write a short program
to see that there is no solution.
(iii) Case 5 ~ 112 ~ 16: The method is similar to that of the above cases. For
every 112, the bound for 1Il2 is lower than that in case 112 = 4. Moreover, when 1Il2
is fixed, the bound of x is lower than that in case 112 = 4 too. In order to find the
solutions of equation (1.2), we consider the following inequality
V" 2 - 1 <
.
70
x 1ll2 - 1
V" 2 - 1
= -'- x-I
y-l
<
(v
.
+ 1)/r1 .
Therefore if (Ill,
11) = (1Il2, 112)
is a positive solution of (1.2), then y must be equal
to L 1Z2-1t'~~11 j. Thus we must determine the integer solutions of
2
1 = (lIZ2-1 x
,_\,1
- 11
x-I
1ll2
-1
x-I
J 2 -1)/(lIZ2-1 x x-I 1 J-1).
Il
1ll2
-
The program was developed in PARI/OP [12] and executed on a 2.4 OHz Pentium-4
computer running under Windows XP. It took 3 minutes and 3 seconds to run the
program. We found that there are no solutions (x, y, 1Il2, 112). This completes the
proof of Theorem 1.3.
ACKNOWLEDGEMENTS
The authors express their gratitude to the anonymous referee for constructive
suggestions on an earlier draft of this paper, particularly for comments leading to
the change of Lemma 2.2. The first author is supported by the Natural Science
Foundation of Education Department of Sichuan Province (No. 2006C057). The
second author is partially supported by Purdue University North Central.
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'v-I'
(Received November 2(07)
72