Open Access

A note on the equation x y + y z = z x

Journal of Inequalities and Applications20142014:170

https://doi.org/10.1186/1029-242X-2014-170

Received: 11 March 2014

Accepted: 7 April 2014

Published: 7 May 2014

Abstract

In this paper, we shall use some simple inequalities and a deep result on the existence of primitive divisors of Lucas numbers to prove that the exponential Diophantine equation x y + y z = z x has no positive integer solution ( x , y , z ) with 2 y .

MSC:11D61.

Keywords

inequality exponential Diophantine equation primitive divisor of Lucas number existence

1 Introduction

Let , be the sets of all integers and positive integers, respectively. Recently, Zhang and Yuan [1] were interested in the equation
x y + y z = z x , x , y , z N .
(1.1)

Using the Gel’fond-Baker method, they proved that all solutions ( x , y , z ) of (1.1) satisfy max { x , y , z } < exp ( exp ( exp ( 5 ) ) ) . This upper bound is far beyond the computable scope at present. In this paper, we shall use some simple inequalities and a deep result on the existence of primitive divisors of Lucas numbers to prove the following result.

Theorem Equation (1.1) has no solution ( x , y , z ) with 2 y .

In addition, it is obvious that ( x , y , z ) = ( 1 , 1 , 2 ) is a solution of (1.1). Because one have not found the other solutions, we propose a conjecture as follows:

Conjecture Equation (1.1) has only the solution ( x , y , z ) = ( 1 , 1 , 2 ) .

Our theorem supports the above mentioned conjecture.

2 Preliminaries

Lemma 2.1 Let f ( X ) = X / log X , where X is a real number. Then f ( X ) is an increasing function for X > e .

Proof Since f ( X ) = ( log X 1 ) / ( log X ) 2 , we have f ( X ) > 0 for X > e . Thus, the lemma is proved. □

Lemma 2.2 Let g ( X ) = X 2 ( 2 + log ( 4 X ) ) / π , where X is a real number. Then we have g ( X ) > 0 for X 16 .

Proof Since g ( X ) = 1 / 2 X 2 / π X > 0 for X 16 , g(X) is an increasing function satisfying g ( X ) g ( 16 ) > 0 for X 16 . The lemma is proved. □

Lemma 2.3 ([2, 3])

The equation
X 2 + 2 m = Y n , X , Y , m , n N , gcd ( X , Y ) = 1 , n > 2
(2.1)

has only the solutions ( X , Y , m , n ) = ( 5 , 3 , 1 , 3 ) and ( 7 , 3 , 5 , 4 ) .

Lemma 2.4 ([[4], Theorem 8.4])

The equation
X 2 + Y m = 2 n , X , Y , m , n N , 2 Y , Y > 1 , m > 1
(2.2)

has only the solution ( X , Y , m , n ) = ( 13 , 7 , 3 , 9 ) .

Lemma 2.5 ([[4], Theorem 8.4])

The equation
X 2 Y m = 2 n , X , Y , m , n N , 2 Y , Y > 1 , m > 2 , n > 1
(2.3)

has only the solution ( X , Y , m , n ) = ( 71 , 17 , 3 , 7 ) .

Let D be a positive integer, and let h ( 4 D ) denote the class number of positive binary quadratic primitive forms of discriminant 4 D .

Lemma 2.6 h ( 4 D ) D .

Proof Notice that h ( 4 ) = h ( 8 ) = h ( 28 ) = 1 , h ( 12 ) = h ( 16 ) = h ( 20 ) = h ( 24 ) = h ( 32 ) = h ( 36 ) = h ( 40 ) = h ( 52 ) = h ( 60 ) = 2 , h ( 44 ) = 3 , h ( 48 ) = h ( 56 ) = 4 . The lemma holds for D 15 . By Theorems 11.4.3, 12.10.1, and 12.14.3 of [5], if D 1 , then
h ( 4 D ) < 2 D π ( 2 + log ( 4 D ) ) .
(2.4)
Therefore, if h ( 4 D ) > D , then from (2.4) we get
D < 2 π ( 2 + log ( 4 D ) ) .
(2.5)

But, by Lemma 2.2, (2.5) is impossible for D 16 . Thus, the lemma is proved. □

Lemma 2.7 Let k be a positive integer with gcd ( k , 2 D ) = 1 . Every solution ( X , Y , Z ) of the equation
X 2 + D Y 2 = k Z , X , Y , Z Z , gcd ( X , Y ) = 1 , Z > 0 ,
(2.6)
can be expressed as
Z = Z 1 t , t N , X + Y D = λ 1 ( X 1 + λ 2 Y 1 D ) t , λ 1 , λ 2 { ± 1 } ,
where X 1 , Y 1 , Z 1 are positive integers satisfying
X 1 2 + D Y 1 2 = k Z 1 , gcd ( X 1 , Y 1 ) = 1 , Z 1 h ( 4 D ) .

Proof This lemma is the special case of [[6], Theorems 1 and 2] for D 1 = 1 and D 2 < 0 .

Let α, β be algebraic integers. If α + β and αβ are nonzero coprime integers and α / β is not a root of unity, then ( α , β ) is called a Lucas pair. Further, let a = α + β and c = α β . Then we have
α = 1 2 ( a + λ b ) , β = 1 2 ( a λ b ) , λ { ± 1 } ,
where b = a 2 4 c . We call ( a , b ) the parameters of the Lucas pair ( α , β ) . Two Lucas pairs ( α 1 , β 1 ) and ( α 2 , β 2 ) are equivalent if α 1 / α 2 = β 1 / β 2 = ± 1 . Given a Lucas pair ( α , β ) , one defines the corresponding sequence of Lucas numbers by
L n ( α , β ) = α n β n α β , n = 0 , 1 , 2 , .

For equivalent Lucas pairs ( α 1 , β 1 ) and ( α 2 , β 2 ) , we have L n ( α 1 , β 1 ) = ± L n ( α 2 , β 2 ) for any n 0 . A prime p is called a primitive divisor of L n ( α , β ) ( n > 1 ) if p L n ( α , β ) and p b L 1 ( α , β ) L n 1 ( α , β ) . A Lucas pair ( α , β ) such that L n ( α , β ) has no primitive divisor will be called an n-defective Lucas pair. Further, a positive integer n is called totally non-defective if no Lucas pair is n-defective. □

Lemma 2.8 ([7])

Let n satisfy 4 < n 30 and n 6 . Then, up to equivalence, all parameters of n-defective Lucas pairs are given as follows:
  1. (i)

    n = 5 , ( a , b ) = ( 1 , 5 ) , ( 1 , 7 ) , ( 2 , 40 ) , ( 1 , 11 ) , ( 1 , 15 ) , ( 12 , 76 ) , ( 12 , 1 , 364 ) .

     
  2. (ii)

    n = 7 , ( a , b ) = ( 1 , 7 ) , ( 1 , 19 ) .

     
  3. (iii)

    n = 8 , ( a , b ) = ( 2 , 24 ) , ( 1 , 7 ) .

     
  4. (iv)

    n = 10 , ( a , b ) = ( 2 , 8 ) , ( 5 , 3 ) , ( 5 , 47 ) .

     
  5. (v)

    n = 12 , ( a , b ) = ( 1 , 5 ) , ( 1 , 7 ) , ( 1 , 11 ) , ( 2 , 56 ) , ( 1 , 15 ) , ( 1 , 19 ) .

     
  6. (vi)

    n { 13 , 18 , 30 } , ( a , b ) = ( 1 , 7 ) .

     

Lemma 2.9 ([8])

If n > 30 , then n is totally non-defective.

3 Further lemmas on the solutions of (1.1)

Throughout this section, we assume that ( x , y , z ) is a solution of (1.1) with ( x , y , z ) ( 1 , 1 , 2 ) .

Lemma 3.1 ([1])

x, y and z are coprime.

Lemma 3.2 min { x , y , z } 3 .

Proof Since z x = x y + y z > 1 , we have z > 1 . If x = 1 , since ( x , y , z ) ( 1 , 1 , 2 ) , then y > 1 and z = 1 + y z 1 + 2 z z + 3 , a contradiction. Similarly, if y = 1 , then x > 1 and x + 1 = z x 2 x x + 2 , a contradiction. Therefore, we have min { x , y , z } 2 .

If x = 2 , then
2 y + y z = z 2 .
(3.1)

Further, by Lemma 3.1, y and z are odd integers with min { y , z } 3 . Hence, we see from (3.1) that (2.3) has the solution ( X , Y , m , n ) = ( z , y , z , y ) . But, by Lemma 2.5, it is impossible.

Similarly, if y = 2 or z = 2 , then we have
x 2 + 2 z = z x , 2 x z , min { x , z } 3
(3.2)
or
x y + y 2 = 2 x , 2 x y , min { x , y } 3 .
(3.3)

But, by Lemmas 2.3 and 2.4, (3.2) and (3.3) are impossible. Thus, we get min { x , y , z } 3 . The lemma is proved. □

Lemma 3.3 y < x .

Proof By (1.1), we have z x > x y and z x > y z . Hence,
x log x > y log z
(3.4)
and
x log y > z log z .
(3.5)

In addition, by Lemmas 3.1 and 3.2, x, y and z are distinct.

If x < y < z , by Lemma 3.2, then 3 x < y < z . Hence, by Lemma 2.1, we get
z log z > x log x > x log y ,
(3.6)

which contradicts (3.5). Similarly, we can remove the case that x < z < y .

If z < x < y , then 3 z < x < y and
y log z > y log y > x log x ,
(3.7)

which contradicts (3.4). Thus, we get y < x . The lemma is proved. □

4 Proof of theorem

We now assume that ( x , y , z ) is a solution of (1.1) with 2 y . Since ( x , y , z ) ( 1 , 1 , 2 ) , by Lemmas 3.1, 3.2 and 3.3, we have 2 x z , gcd ( y , z ) = 1 , min { x , y , z } 3 and x > y .

We see from (1.1) that the equation
X 2 + y Y 2 = z Z , X , Y , Z Z , gcd ( X , Y ) = 1 , Z > 0
(4.1)
has the solution
( X , Y , Z ) = ( x y / 2 , y ( z 1 ) / 2 , x ) .
(4.2)
Applying Lemma 2.7 to (4.1) and (4.2), we have
x = Z 1 t , t N ,
(4.3)
x y / 2 + y ( z 1 ) / 2 y = λ 1 ( X 1 + λ 2 Y 1 y ) t , λ 1 , λ 2 { ± 1 } ,
(4.4)
where X 1 , Y 1 , Z 1 are positive integers satisfying
X 1 2 + y Y 1 2 = z Z 1 , gcd ( X 1 , Y 1 ) = 1
(4.5)
and
Z 1 h ( 4 y ) .
(4.6)
Let
α = X 1 + Y 1 y , β = X 1 Y 1 y .
(4.7)
We see from (4.5) and (4.7) that α + β = 2 X 1 and α β = z Z 1 are coprime nonzero integers, α / β = ( ( X 1 2 y Y 1 2 ) + 2 X 1 Y 1 y ) / z Z 1 is not a root of unity. Hence, ( α , β ) is a Lucas pair with parameters ( 2 X 1 , 4 y Y 1 2 ) . Further, Let L n ( α , β ) ( n = 0 , 1 , 2 , ) denote the corresponding Lucas numbers. By (4.4) and (4.7), we have
y ( z 1 ) / 2 = | L t ( α , β ) | .
(4.8)

We find from (4.7) and (4.8) that the Lucas number L t ( α , β ) has no primitive divisor. Therefore, by Lemma 2.9, we have t 30 . Further, since 2 x and 2 t by (4.3), it is easy to remove all cases in Lemma 2.8 and conclude that t { 1 , 3 } .

If t = 3 , then from (4.4) we get
y ( z 1 ) / 2 = λ 1 λ 2 Y 1 ( 3 X 1 2 y Y 1 2 ) .
(4.9)
Let d = gcd ( Y 1 , 3 X 1 2 y Y 1 2 ) . Since gcd ( X 1 , Y 1 ) = 1 , we have d 3 and d { 1 , 3 } . Further, since t x , we get 3 x , 3 y and d 3 by (4.9). Therefore, we have d = 1 and, by (4.9), gcd ( y , 3 X 1 2 y Y 1 2 ) = 1 and
Y 1 = y ( z 1 ) / 2 , 3 X 1 2 y Y 1 2 = ± 1 .
(4.10)
It implies that
3 X 1 2 1 = y z .
(4.11)

But, since 2 y and z 3 , we get from (4.11) that 2 X 1 and 0 y z 3 X 1 2 1 3 1 0 ( mod 8 ) , a contradiction.

If t = 1 , then from (4.3) and (4.6) that x = Z 1 , x h ( 4 y ) and
x h ( 4 y ) .
(4.12)

But recall that x > y , by Lemma 2.6, (4.12) is impossible. Thus, (1.1) has no solution ( x , y , z ) with 2 y . The theorem is proved.

Declarations

Acknowledgements

The authors would like to thank the referee for his very helpful and detailed comments, which have significantly improved the presentation of this paper. This work is supported by the P. S. F. (2013JZ001) and N.S.F. (11371291) of P.R. China.

Authors’ Affiliations

(1)
College of Mathematics and Information Science, Weinan Normal University
(2)
School of Mathematics, Northwest University

References

  1. Zhang Z-F, Yuan P-Z:On the diophantine equation a x y + b y z + c z x = 0 . Int. J. Number Theory 2012,8(3):813-821. 10.1142/S1793042112500467MathSciNetView ArticleMATHGoogle Scholar
  2. Cohn JHE:The diophantine equation x 2 + 2 k = y n . Arch. Math. Basel 1992,59(3):341-343.MathSciNetView ArticleMATHGoogle Scholar
  3. Le M-H:On Cohn’s conjecture concerning the diophantine equation x 2 + 2 m = y n . Arch. Math. Basel 2002,78(1):26-35. 10.1007/s00013-002-8213-5MathSciNetView ArticleMATHGoogle Scholar
  4. Bennett MA, Skinner CM: Ternary diophantine equations via Galois representations and modular forms. Can. J. Math. 2004,56(1):23-54. 10.4153/CJM-2004-002-2MathSciNetView ArticleMATHGoogle Scholar
  5. Hua L-K: Introduction to Number Theory. Springer, Berlin; 1982.Google Scholar
  6. Le M-H:Some exponential diophantine equations I: the equation D 1 x 2 D 2 y 2 = λ k z . J. Number Theory 1995,55(2):209-221. 10.1006/jnth.1995.1138MathSciNetView ArticleGoogle Scholar
  7. Voutier PM: Primitive divisors of Lucas and Lehmer sequences. Math. Comput. 1995, 64: 869-888. 10.1090/S0025-5718-1995-1284673-6MathSciNetView ArticleMATHGoogle Scholar
  8. Bilu Y, Hanrot G, Voutier PM: Existence of primitive divisors of Lucas and Lehmer numbers. J. Reine Angew. Math. 2001, 539: 75-122. (with an appendix by M Mignotte)MathSciNetMATHGoogle Scholar

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© Lu and Li; licensee Springer. 2014

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