402
20114+
)
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ADVANCES IN MATHEMATICS
Vol.40, No.2
April, 2011
On the Positive Integer Solutions of
the Exponential Diophantine Equation
ax + (3a2 − 1)y = (4a2 − 1)z
HE bo1,∗ , Togbé Alain2,∗∗
(1. Department of Mathematics, Aba Teachers College, Wenchuan, Sichuan, 623000, P. R. China;
2. Department of Mathematics, Purdue University North Central, 1401, S.U.S. 421, Westville, IN 46391,
USA)
Abstract: In this paper, we use properties of the Jacobi symbol and lower bounds for
linear forms in two logarithms of algebraic numbers to prove that the diophantine equation
ax + (3a2 − 1)y = (4a2 − 1)z has only the positive integer solution (x, y, z) = (2, 1, 1) for the
integer a > 1.
Key words: exponential diophantine equations; Jacobi symbols; linear forms in the
logarithms
MR(2000) Subject Classification: 11D61 / CLC number: O156
Document code: A
Article ID: 1000-0917(2011)02-0227-08
0 Introduction
The Diophantine equation
ax + b y = cz ,
a, b, c, x, y, z ∈ N
(1)
has a very rich history. In 1956, Sierpiński[13] proved that (x, y, z) = (2, 2, 2) is the only positive
integral solution of the equation 3x + 4y = 5z . The same year, Jeśmanowicz[10] has conjectured
that if a, b, c are Pythagorean triples, i.e. positive integers satisfying a2 + b2 = c2 , then the
Diophantine equation ax + by = cz has only the positive integral solution (x, y, z) = (2, 2, 2).
Many special cases of this conjecture have been solved. Other conjectures related to equation (1)
were set and discussed. For example, the authors[7] completely solved the diophantine equation
nx + (n + 1)y = (n + 2)z recently. One extension of Jeśmanowicz’ conjecture is due to Terai[14, 15] .
In fact, Terai conjectured that if a, b, c, p, q, r ∈ N are fixed and ap + bq = cr , where p, q, r ≥ 2,
and gcd(a, b) = 1, then the Diophantine equation (1) has only the solution (x, y, z) = (p, q, r).
In 1999, Zhenfu Cao[3] pointed out that Terai’s conjecture, called also Terai-Jeśmanowicz con-
jecture, is clearly false. Moreover Cao added the condition max(a, b, c) > 7. However, the
Terai-Jeśmanowicz conjecture contained a mistake. Consequently, Cao and Dong[4] modified the
Terai-Jeśmanowicz conjecture and stated following
Received date: 2008-12-15. Revised date: 2009-04-29.
Foundation item: The first author is supported by the Applied Basic Research Foundation of Sichuan Provincial
Science and Technology Department(No. 2009JY0091).
E-mail: ∗ [email protected]; ∗∗ [email protected]
228
Conjecture
(
,
40
Equation (1) has at most one solution (x, y, z) with min(x, y, z) > 1.
Many authors have proved or disproved that the conjecture is true in some particular cases.
One can see for example [3, 5, 9, 11].
Recently, Yongzhong Hu[8] considered the Diophantine equation
a2x + (3a2 − 1)y = (4a2 − 1)z , x, y, z ∈ N
(2)
with positive integer a > 3. He used a deep theorem due to Bilu, Hanrot and Voutier[1] for the
primitive divisor of Lehmer numbers and showed that equation (2) has only the positive integer
solution (x, y, z) = (1, 1, 1).
In this paper, we consider the Diophantine equation
ax + (3a2 − 1)y = (4a2 − 1)z , x, y, z ∈ N
(3)
and we extend Hu’s work by proving the following result.
Theorem
If x is an odd integer, equation (3) has no positive integer solution (x, y, z).
Therefore, combining our theorem and Hu’s result in [8], we completely solve equation (3)
and we get
Corollary Equation (3) has only positive integer solution (x, y, z) = (2, 1, 1).
From the earlier work on equation (3), our result also holds when a > 1. One can refer to
[6] for 2x + 11y = 15z , and to [2] for 3x + 26y = 35z .
In this paper, we use properties of the Jacobi symbol and lower bounds for linear forms in
two logarithms of algebraic numbers[12] to obtain our result. In fact, we organize this paper as
follows. In Section 1, first we use properties of the Jacobi symbol to prove that a positive integer
solution (x, y, z) of equation (3) with 2 ∤ x satisfies x > 1, y = 1, 2 ∤ z and a ≡ 1(mod 8). Then
we recall a result due to Mignotte[12] that helps us to show that z ≤ 2835 log a. The proof of
Theorem is completely done in Section 2.
1 Some Lemmas
In this section, we recall and show some results that will be important for the proof of the
main theorem. Let ( ∗∗ ) denote the Jacobi symbol. First we show the following lemma.
Lemma 1 If equation (3) has a positive integer solution (x, y, z) with 2 ∤ x, then we have
x > 1, y = 1, 2 ∤ z and a ≡ 1(mod 8).
Proof Taking equation (3) modulo a, we get (−1)y ≡ (−1)z (mod a). Since a > 3, we have
y ≡ z(mod 2).
Suppose x = 1. Now if we take again equation (3) modulo a2 , then a ≡ 0(mod a2 ). So we obtain
1 ≡ 0(mod a). This is impossible. Hence, we consider x > 1.
Also we have
(−a2 )y ≡ −ax (mod (4a2 − 1))
(4)
Togbé Alain: Exponential Diophantine Equation a
2
x
+ (3a2 − 1)y = (4a2 − 1)z
229
by taking equation (3) modulo 4a2 − 1. From
(−a2 )y
(−1)y a2y
(−1)y
=
=
= (−1)y
4a2 − 1
4a2 − 1
4a2 − 1
and
we get
−ax
4a2 − 1
=
−a
4a2 − 1
a
4a2 − 1
=−
a
2
4a − 1
,
= −(−1)y .
(5)
Let us consider a = 2d a1 with 2 ∤ a1 . We have
a
2d
a1
=
.
4a2 − 1
22d+2 a21 − 1
22d+2 a21 − 1
(6)
Since
a1
22d+2 a21 − 1
= (−1)
a1 −1
2
22d+2 a21 − 1
a1
= (−1)
a1 −1
2
−1
a1
2
a1 −1
=1
= (−1) 2
and 22d+2 a21 − 1 ≡ −1(mod 8), then we obtain
2d
= 1.
22d+2 a21 − 1
These two results and equation (6) give us
a
2
4a − 1
= 1.
Combining this and equation (5) implies that y must be odd. Because y ≡ z(mod 2), z is also
odd.
As x > 1 is an odd integer, we have x ≥ 3. Then we take equation (3) modulo a3 . It results
that 3a2 y − 1 ≡ 4a2 z − 1(mod a3 ). Thus we get
3y ≡ 4z(mod a).
(7)
When a is even, then equation (7) implies that y is even. Therefore we get a contradiction.
Hence, a is an odd integer.
Now we take equation (3) modulo (3a2 − 1) to obtain ax ≡ a2z (mod (3a2 − 1)). As a is an
odd integer, then 3a2 − 1 ≡ 2(mod 8). We have
!
!
a
ax
= 3a2 −1 =
3a2 −1
2
It follows that
3a2 −1
2
2
a2z
3a2 −1
2
!
= 1.
≡ 1(mod 4) and we get
!
! 3a2 −1
a
6a2 − 2
−2
2
=
=
=
.
3a2 −1
a
a
a
2
230
(
,
40
Therefore, we have
a ≡ 1, 3(mod 8).
On the other hand, using equation (4) we get
a2y ≡ ax (mod (2a + 1)).
If a ≡ 3(mod 8), then 2a + 1 ≡ 7(mod 8). Thus we obtain the following contradiction
2y a
ax
a
2a + 1
1
1=
=
=
=−
=−
= −1.
2a + 1
2a + 1
2a + 1
a
a
Therefore,
a ≡ 1(mod 8).
have
(8)
Finally, we take equation (3) modulo 8 and we get 1x + 2y ≡ 3z (mod 8). Since z is odd, we
1 + 2y ≡ 3(mod 8).
(9)
If y ≥ 3, then equation (9) implies 1 ≡ 3(mod 8). This is impossible. Therefore, we have y = 1
and this completes the proof of Lemma 1.
Now we recall the following result due to Mignotte (see [12, Corollary of Theorem 2, page
110]) on linear forms in two logarithms. For any
non-zeroalgebraic number γ of degree d over
Qd
Q, whose minimal polynomial over Z is a j=1 X − γ (j) , we denote
h(γ) =
1
d
log |a| +
d
X
j=1
log max(1, |γ (j) |)
its absolute logarithmic height.
Lemma 2
Consider the linear form
Λ = b1 log α1 − b2 log α2 ,
where b1 and b2 are positive integers. Suppose that α1 , α2 are multiplicatively independent. Put
D = [Q(α1 , α2 ) : Q]/[R(α1 , α2 ) : R]
and let ρ, λ, a1 and a2 be positive real numbers with ρ ≥ 4, λ = log ρ,
ai ≥ max{1, (ρ − 1) log |αi | + 2Dh(αi )} (i = 1, 2)
and
a1 a2 ≥ max{20, 4λ2 }.
Further suppose h is a real number with
b1
b2
h ≥ max 3.5, 1.5λ, D log
+
+ log λ + 1.377 + 0.023 ,
a2
a1
Togbé Alain: Exponential Diophantine Equation a
2
x
+ (3a2 − 1)y = (4a2 − 1)z
231
χ = hλ , υ = 4χ + 4 + χ1 . Then we have the lower bound
log |Λ| ≥ −(C0 + 0.06)(λ + h)2 a1 a2 ,
(10)
where


2
s
√
3

2
1
2(1
+
χ)
1 
1
1
4λ
1
1
32
 +

C0 = 3
2+
+
+
+
.
√

λ 
2χ(χ + 1)
3
9 3v a1
a2
3v 2 a1 a2
Now we use Lemma 2 to prove the following result.
Lemma 3 If equation (3) has a positive integer solution (x, y, z), then we have
z ≤ 2835 log a.
(11)
Proof Let us consider the linear form of two logarithms
Λ = z log(4a2 − 1) − x log a.
(12)
From Lemma 1, we get 1 < a ≡ 1(mod 8). It implies that a ≥ 9 and then 3a2 − 1 < a2.5 . As
y = 1, we have
(4a2 − 1)z
3a2 − 1
0 < Λ < eΛ − 1 =
−
1
=
< a2.5−x .
(13)
ax
ax
It follows that
log Λ < −(x − 2.5) log a.
(14)
Now, we apply Lemma 2 with
D = 1, α1 = 4a2 − 1, α2 = a, b1 = z, b2 = x,
and
a1 = (ρ + 1) log(4a2 − 1),
a2 = (ρ + 1) log a.
We choose ρ = 5. It satisfies a1 a2 ≥ max{20, 4λ2}. From inequalities (13), we deduce
z
x
2z
+
<
.
log a log(4a2 − 1)
log a
Let us consider
h = max 10.52, log
2z
log a
+ 1.88 .
(15)
If h = 10.52, then equation (15) gives
z≤
If h = log
2z
log a
1 10.52−1.88
e
log a < 2827 log a.
2
+ 1.84 > 10.52, since a ≥ 9 and a1 = 6 log(4a2 − 1) > 34.6, a2 = 6 log a >
13.1, one can see that C0 < 0.475. By (10) we obtain
2
2z
log Λ > −19.26 3.49 + log
log a log(4a2 − 1).
log a
(16)
232
(
,
40
Combining equations (14) and (16), we get
2
2z
x − 2.5 < 19.26 3.49 + log
log(4a2 − 1).
log a
It follows
One can notice that
2
x
2z
<
19.26
3.49
+
log
+ 0.44.
log(4a2 − 1)
log a
z
x
1
x
<
+ 0.5
<
+ 0.09.
2
2
log a
log(4a − 1) a (log a)(log(4a − 1))
log(4a2 − 1)
Thus we have
2
2z
z
< 19.26 3.49 + log
+ 0.55.
log a
log a
From equation (17), we get
z
log a
(17)
< 2835. This completes the proof of Lemma 3.
2 The Proof of Our Theorem
Let (x, y, z) be a positive integer solution of equation (3) with x odd. From Lemma 1, we
have x ≥ 3, y = 1, 2 ∤ z and a ≡ 1(mod 8). We rewrite equation (3) into
ax + 3a2 − 1 = (4a2 − 1)z , x, z ∈ N.
(18)
Since a > 3, we have
1
1
(4a2 − 1) 3 < a < (4a2 − 1) 2 .
(19)
First, we want to determine a lower bound for x. If equation (18) has a positive integer
solution (x, z) with x = 3, then one can use equations (18) and (19) to obtain
3
4a2 − 1 < a3 < a3 + 3a2 − 1 < (4a2 − 1) 2 + 3a2 − 1 < (4a2 − 1)2 .
It follows that 1 < z < 2. Thus, (3, z) is not a solution of equation (18). If (5, z) is a solution
(x, z) of equation (18), then
5
5
(4a2 − 1) 3 < a5 < a5 + 3a2 − 1 < (4a2 − 1) 2 + 3a2 − 1 < (4a2 − 1)3 .
But 2 ∤ z, so we don’t have any odd integer z verifying
Again here, if x = 7, then
7
5
3
< z < 3. Hence, we deduce that x 6= 5.
7
(4a2 − 1) 3 < a7 < a7 + 3a2 − 1 < (4a2 − 1) 2 + 3a2 − 1 < (4a2 − 1)4 .
This gives us
7
3
< z < 4. We have z = 3. Let us consider the function
f (a) = a7 + 3a2 − 1 − (4a2 − 1)3 = a2 (a5 − 64a4 + 48a2 − 9).
One can easily see that the only integer solution of f (a) = 0 is a = 0. So, x = 7 cannot satisfy
equation (18). Therefore we deduce that x ≥ 9.
2
Togbé Alain: Exponential Diophantine Equation a
x
+ (3a2 − 1)y = (4a2 − 1)z
233
Second, we want to find a lower bound for z by taking (18) modulo ak for even values of
k ≤ 9. So we take equation (18) modulo a4 , we have 3a2 − 1 ≡ 4za2 − 1(mod a4 ). This implies
4z ≡ 3(mod a2 ).
There exists a positive integer r such that
z=
ra2 + 3
.
4
(20)
Now taking equation (18) modulo a6 , we obtain
3a2 − 1 ≡ −
z(z − 1)
(4a2 )2 + z(4a2 ) − 1(mod a6 ).
2
We combine this and (20), then we simplify to get
2r ≡ −3(mod a2 ).
So there exists a positive integer s satisfying
r=
sa2 − 3
.
2
(21)
Finally, we take equation (18) modulo a8 . It results
3a2 − 1 ≡
z(z − 1)(z − 2)
z(z − 1)
(4a2 )3 −
(4a2 )2 + z(4a2 ) − 1(mod a8 ).
6
2
Using this, (20) and (21), we have
s ≡ −8(mod a2 ).
There exists a positive integer t such that
s = ta2 − 8.
(22)
8z = 2ra2 + 6 > 2ra2 = (sa2 − 3)a2 = ((ta2 − 8)a2 − 3)a2 ≥ a6 − 8a4 − 3a2 .
(23)
From equations (20)–(22), we obtain
Now, combining (23) and Lemma 3, we deduce
a6 − 8a4 − 3a2 < 8 · 2835 log a.
(24)
After a straightforward computation, we find that a < 7. But, in Lemma 1 we have a ≥ 9.
Therefore we get a contradiction. This completes the proof of Theorem.
References
[1] Bilu, Y., Hanrot, G. and Voutier, P., (with an appendix by Mignotte, M.), Existence of primitive divisors
of Lucas and Lehmer numbers, J. Reine Angew. Math., 2001, 539: 75-122.
234
(
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40
[2] Cao Z.F., On the Diophantine equation ax + by = cz , I, Chinese Sci. Bull., 1987, 32: 1519-1521; II, ibid.,
1988, 33: 237(in Chinese).
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253-265.
[5] Cao Z.F. and Dong X.L., An application of a lower bound for linear forms in two logarithms to the
Terai-Jeśmanowicz conjecture, Acta Arith., 2003, 110(2): 153-164.
[6] Hadano, T., On the Diophantine equation ax = by + cz , Math. J. Okayama Univ., 1976/77, 19: 25-29.
[7] He B. and Togbé, A., The Diophantine equation nx + (n + 1)y = (n + 2)z revisited, Glasgow Math. J.,
2009, 51(3): 659-667.
[8] Hu Y.Z., On the Exponential Diophantine equation a2x +(3a2 −1)y = (4a2 −1)z , Advances in Mathematics
(China), 2007, 36: 429-434.
[9] Hu Y.Z. and Yuan P.Z., On the exponential diophantine equation ax + by = cz , Acta Mathematica Sinica,
Chinese Series, 2005, 48(6): 1175-1178.
[10] Jeśmanowicz, L., Some remarks on Pythagorean numbers, Wiadom. Mat., 1956, 1: 196-202.
[11] Le M.H., On the Terai’s conjecture concerning the exponential diophantine equation ax + by = cz , Acta
Mathematica Sinica, Chinese Series, 2003, 46(2): 245-250.
[12] Mignotte, M., A corollary to a theorem of Laurent-Mignotte-Nesterenko, Acta Arith. 1998, 86: 101-111.
[13] Sierpiński, W., On the equation 3x + 4y = 5z , Wiadom. Mat., 1956, 1: 194-195.
[14] Terai, N., The Diophantine equation x2 + q m = pn , Acta Arith., 1993, 63: 351-358.
[15] Terai, N., The Diophantine equation ax + by = cz , Proc. Japan Acad. A Math. Sci., 1994, 70: 22-26.
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