Research Article The Exponential Diophantine Equation...
Transcript of Research Article The Exponential Diophantine Equation...
Research ArticleThe Exponential Diophantine Equation 2π₯ + ππ¦ = ππ§
Yahui Yu1 and Xiaoxue Li2
1 Luoyang Institute of Science and Technology, Luoyang, Henan 471023, China2 School of Mathematics, Northwest University, Xiβan, Shaanxi 710127, China
Correspondence should be addressed to Xiaoxue Li; [email protected]
Received 1 April 2014; Accepted 6 May 2014; Published 18 May 2014
Academic Editor: Abdelghani Bellouquid
Copyright Β© 2014 Y. Yu and X. Li. This is an open access article distributed under the Creative Commons Attribution License,which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Let π and π be fixed coprime odd positive integers with min{π, π} > 1. In this paper, a classification of all positive integer solutions(π₯, π¦, π§) of the equation 2
π₯
+ ππ¦
= ππ§ is given. Further, by an elementary approach, we prove that if π = π + 2, then the equation
has only the positive integer solution (π₯, π¦, π§) = (1, 1, 1), except for (π, π₯, π¦, π§) = (89, 13, 1, 2) and (2π
β 1, π + 2, 2, 2), where π is apositive integer with π β₯ 2.
1. Introduction
Let N be the set of all positive integers. Let π, π, π be fixedcoprime positive integers with min{π₯, π¦, π§} > 1. In recentyears, the solutions (π₯, π¦, π§) of the equation
ππ₯
+ ππ¦
= ππ§
, π₯, π¦, π§ β N (1)
have been investigated in many papers (see [1β3] and itsreferences). In this paper we deal with (1) for the case thatπ = 2. Then (1) can be rewritten as
2π₯
+ ππ¦
= ππ§
, π₯, π¦, π§ β N, (2)
where π and π are fixed coprime odd positive integers withmin{π, π} > 1. We will give a classification of all solutions(π₯, π¦, π§) of (2) as follows.
Theorem 1. Every solution (π₯, π¦, π§) of (2) satisfies one of thefollowing types:
(i) (π, π, π₯, π¦, π§) = (7, 3, 5, 2, 4);(ii) (π, π, π₯, π¦, π§) = (2
π
β 1, 2π
+ 1, π + 2, 2, 2), where π is apositive integer with π β₯ 2;
(iii) (π, π, π₯, π¦, π§) = (5, 3, 1, 2, 3);(iv) (π, π, π₯, π¦, π§) = (11, 5, 2, 2, 3);(v) 2 | π¦ and π§ = 1;(vi) (π, π, π₯, π¦, π§) = (17, 71, 7, 3, 2);
(vii) π₯ = 1, π¦ > 1, 2 β€ π¦ and 2 | π§;(viii) π₯ > 1, π¦ = 1, 2 | π§ and 2
π₯
< π50/13;
(ix) 2 β€ π¦π§.
Recently, Miyazaki and Togbe [4] showed that if π β₯ 5 andπ = π + 2, then (2) has only the solution (π₯, π¦, π§) = (1, 1, 1),except for (π, π₯, π¦, π§) = (89, 13, 1, 2). However, there are someexceptional cases missing from the result of [4]. In this paper,by an elementary approach, we prove the following result.
Corollary 2. If π = π + 2, then (2) has only the solution(π₯, π¦, π§) = (1, 1, 1), except for (π, π₯, π¦, π§) = (89, 13, 1, 2) and(2π
β 1, π + 2, 2, 2), where π is a positive integer with π β₯ 2.
2. Preliminaries
Lemma 3 (see [5, Formula 1.76]). For any positive integer πand any complex numbers πΌ and π½, one has
πΌπ
+ π½π
=
[π/2]
β
π=0
[
π
π] (πΌ + π½)
πβ2π
(βπΌπ½)π
, (3)
where [π/2] is the integer part of π/2;
[
π
π] =
(π β π β 1)!π
(π β 2π)!π!
, π = 0, . . . , [
π
2
] (4)
are positive integers.
Hindawi Publishing Corporatione Scientific World JournalVolume 2014, Article ID 401816, 3 pageshttp://dx.doi.org/10.1155/2014/401816
2 The Scientific World Journal
Lemma 4 (see [6]). Let π΄ and π΅ be coprime odd positiveintegers withmin{π΄, π΅} > 1. If the equation
π΄π’2
β π΅V2 = 2, π’, V β N (5)
has solutions (π’, V), then it has a unique solution (π’1, V1) such
that π’1βπ΄ + V
1βπ΅ β€ π’βπ΄ + Vβπ΅, where (π’, V) through all
solutions of (5). The solution (π’1, V1) is called the least solution
of (5). Every solution (π’, V) of (5) can be expressed as
π’βπ΄ + Vβπ΅
β2
= (
π’1βπ΄ + V
1βπ΅
β2
)
π
, π β N, 2 β€ π. (6)
Further, by (6), we have π’1| π’ and V
1| V.
Lemma 5. Equation (5) has no solutions (π’, V) such that π’ >
π’1, V > V
1, and every prime divisor of π’/π’
1and V/V
1divides π΄
and π΅, respectively.
Proof. Wenow assume that (π’, V) is a solution of (5) satisfyingthe hypothesis. Since π’ > π’
1, by Lemma 4, the (π’
π, Vπ) is all
solutions of (5). Let
πΌ =
π’1βπ΄ + V
1βπ΅
β2
, π½ =
π’1βπ΄ β V
1βπ΅
β2
. (7)
We get
π’π
π’1
=
πΌπ
β (βπ½)π
πΌ β (βπ½)
,
Vπ
V1
=
πΌπ
β (π½)π
πΌ β (π½)
, (8)
where π is odd. Numbers πΌ and π½ are such that (πΌ, βπ½) satisfyπ₯2
β β2π΅V21π₯ + 1 = 0 and (πΌ, π½) satisfy π₯
2
β β2π΄π’2
1π₯ +
1 = 0. Thus, {π’π/π’1}πβ₯1
and {Vπ/V1}πβ₯1
are the odd indexedsubsequences of the two Lehmer sequences of roots (πΌ, βπ½)and (πΌ, π½). Their discriminants are (πΌ + π½)
2
= 2π΄π’2
1and
(πΌ β π½)2
= 2π΅V21, respectively. Saying that all prime factors
of π’π/π’1divide π΄ implies that all primes of the πth term of a
Lehmer sequence divide its discriminant.The same is true forVπ/V1. Hence,π’
π/π’1and Vπ/V1are terms of a Lehmer sequence
of real roots lacking primitive divisors. By Table 2 in [7], thisis possible only for π = 3, 5. Even more, in the present case,
(πΌ2
)
π
β (π½2
)
π
πΌ2β π½2
=
π’πVπ
π’1V1
(9)
is the πth term of the Lucas sequence of positive real roots(πΌ2
, π½2
) whose all prime factors divide its discriminant (πΌ2 βπ½2
)2
= 4π΄π΅π’2
1V22, and by Table 1 in [7] this is possible for π
odd only if π = 3 or π = 5. Furthermore, when π = 5, wemust have πΌ2 = (1 + β5)/2, but this is not possible since πΌ =
β(1 + β5)/2 is not of the form (π’1βπ΄ + V
1βπ΅)/β2 for some
positive integers π΄ > 1, π΅ > 1, π’1and V
1. So, only π = 3
is possible. Now by some simple numerical computation forπ’3and V3, we see that it is not possible that all prime factors
of π’3and all prime factors of V
3divide B. Thus, Lemma 5 is
proved.
Lemma 6 (see [8]). The equation
π2
+ 7 = 2π+2
, π, π β N (10)
has only the solutions (π, π) = (1, 1), (3, 2), (5, 3), (11, 5), and(181, 13).
Lemma 7 (see [9]). Letπ· be an odd positive integer withπ· >
1. If (π, π) is a solution of the equation
π2
β π· = 2π
, π, π β N, (11)
then 2π
< π·50/13.
Lemma 8 (see [10, 11]). The equation
π2
+ 2π
= ππ
, π, π,π, π β N, gcd (π, π) = 1, π β₯ 3
(12)
has only the solutions (π, π,π, π) = (5, 3, 1, 3), (7, 3, 5, 4), and(11, 5, 2, 3).
Lemma 9 (see [12]). The equation
π2
β 2π
= ππ
, π, π,π, π β N, gcd (π, π) = 1,
π > 1, π > 1, π β₯ 3
(13)
has only the solution (π, π,π, π) = (71, 17, 7, 3).
Lemma 10 (see [13]). The equation
ππ
β ππ
= 1, π, π,π, π β N, min {π, π,π, π} > 1 (14)
has only the solution (π, π,π, π) = (3, 2, 2, 3).
3. Proof of Theorem
Let (π₯, π¦, π§) be a solution of (2). If 2 | π¦ and 2 | π§, then wehave π₯ β₯ 3, ππ§/2 + π
π¦/2
= 2π₯β1, and π
π§/2
β ππ¦/2
= 2. It followsthat
ππ§/2
= 2π₯β2
+ 1, ππ¦/2
= 2π₯β2
β 1. (15)
Applying Lemma 10 to (15), we can only obtain the solutionsof types (i) and (ii).
If 2 | π¦ and 2 β€ π§, then we have
(ππ¦/2
)
2
+ 2π₯
= ππ§
, 2 β€ π§. (16)
Applying Lemma 8 to (16), we can only get the solutions oftypes (iii), (iv), and (v).
Similarly, if 2 β€ π¦ and 2 | π§, using Lemmas 7 and 9, thenwe can only obtain the solutions of types (vi), (vii), and (viii).Finally, if 2 β€ π¦π§, then the solutions are of type (ix). Thus, thetheorem is proved.
4. Proof of Corollary
Since π = π + 2, (2) can be rewritten as
2π₯
+ ππ¦
= (π + 2)π§
, π₯, π¦, π§ β N. (17)
The Scientific World Journal 3
Let (π₯, π¦, π§) be a solution of (17). By the theorem, (17) has onlythe solutions
(π, π₯, π¦, π§) = (2π
β 1, π + 2, 2, 2) , π β N, π β₯ 2 (18)
satisfying 2 | π¦ and 2 | π§.If π₯ = 1, 2 β€ π¦ and 2 | π§, then from (17) we get
2 + ππ¦
= 2 + ((π + 1) β 1)π¦
= 1 + (π + 1)
π¦
β
π=1
(β1)πβ1
(
π¦
π) (π + 1)
πβ1
= 1 + (π + 1)
π§
β
π=1
(
π§
π) (π + 1)
πβ1
= ((π + 1) + 1)π§
= (π + 2)π§
,
(19)
whence we obtain
π¦ β‘ π§ (mod (π + 1)) . (20)
But, since 2 | π+1 and 2 β€ π¦βπ§, congruence (20) is impossible.If π₯ > 1, 2 β€ π¦ and 2 | π§, by the theorem, then we have
π¦ = 1 and 2π₯
< π50/13. Hence, by (17), we get
π2
< (π + 2)2
β€ (π + 2)π§
= 2π₯
+ π < π50/13
+ π. (21)
Since π β₯ 3 and 2 | π§, we see from (21) that π§ = 2. Substitutingit into (17), we have π2 + 3π β 4(2
π₯β2
β 1) = 0 and
π =
1
2
(β3 + β2π₯+2
β 7) . (22)
By (22), we get
π =
1
2
(π β 3) , π₯ = π, (23)
where (π, π) is a solution of (10). Since 2 β€ π and π > 1, byLemma 6, we can only have (π, π) = (181, 13) and
(π, π₯, π¦, π§) = (89, 13, 1, 2) , (24)
by (23).If 2 β€ π¦π§, then π
π¦β1
β‘ (π + 2)π§β1
β‘ 1(mod 8). Hence, by(17), we get 2π₯ β‘ (π + 2)
π§
β ππ¦
β‘ (π + 2) β π β‘ 2(mod 8) andπ₯ = 1. It implies that the equation
(π + 2) π’2
β πV2 = 2, π’, V β N (25)
has the solution
(π’, V) = ((π + 2)(π§β1)/2
, π(π¦β1)/2
) . (26)
Notice that the least solution of (25) is (π’1, V1) = (1, 1); π¦
and π§ satisfy either π¦ = π§ = 1 or min{π¦, π§} > 1. ApplyingLemma 5 to (26), we only obtain that (π’, V) = (1, 1) and
(π₯, π¦, π§) = (1, 1, 1) . (27)
Thus, (17) has only the solutions (18), (24), and (27). Thecorollary is proved.
Conflict of Interests
The authors declare that there is no conflict of interestsregarding the publication of this paper.
Acknowledgments
Theauthorswould like to thank the referee for his very helpfuland detailed comments, which have significantly improvedthe presentation of this paper.This work is supported by N. S.F. (11371291) and P. N. S. F. (2013ZJ001) of China and G. I. C.F. (YZZ13075) of NWU.
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