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

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 2y.

MSC:11D61.

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,zN.
(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 2y.

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/logX, where X is a real number. Then f(X) is an increasing function for X>e.

Proof Since f (X)=(logX1)/ ( 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(4X))/π, where X is a real number. Then we have g(X)>0 for X16.

Proof Since g (X)=1/2 X 2/πX>0 for X16, g(X) is an increasing function satisfying g(X)g(16)>0 for X16. The lemma is proved. □

Lemma 2.3 ([2, 3])

The equation

X 2 + 2 m = Y n ,X,Y,m,nN,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,nN,2Y,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,nN,2Y,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(4D) denote the class number of positive binary quadratic primitive forms of discriminant 4D.

Lemma 2.6 h(4D)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 D15. By Theorems 11.4.3, 12.10.1, and 12.14.3 of [5], if D1, then

h(4D)< 2 D π ( 2 + log ( 4 D ) ) .
(2.4)

Therefore, if h(4D)>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 D16. Thus, the lemma is proved. □

Lemma 2.7 Let k be a positive integer with gcd(k,2D)=1. Every solution (X,Y,Z) of the equation

X 2 +D Y 2 = k Z ,X,Y,ZZ,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(4D).

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=αβ. Then we have

α= 1 2 (a+λ b ),β= 1 2 (aλ b ),λ{±1},

where b= a 2 4c. 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 n0. A prime p is called a primitive divisor of L n (α,β) (n>1) if p L n (α,β) and pb 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<n30 and n6. 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 ,2xz,min{x,z}3
(3.2)

or

x y + y 2 = 2 x ,2xy,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 3x<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 3z<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 2y. Since (x,y,z)(1,1,2), by Lemmas 3.1, 3.2 and 3.3, we have 2xz, 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,ZZ,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,tN,
(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(4y).
(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 αβ= 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 ,4y 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 t30. Further, since 2x and 2t 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 d3 and d{1,3}. Further, since tx, we get 3x, 3y and d3 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 2y and z3, we get from (4.11) that 2 X 1 and 0 y z 3 X 1 2 1310(mod8), a contradiction.

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

xh(4y).
(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 2y. The theorem is proved.

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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.

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Correspondence to Xiaoxue Li.

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Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

YL obtained the theorems and completed the proof. XL corrected and improved the final version. Both authors read and approved the final manuscript.

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Keywords

  • inequality
  • exponential Diophantine equation
  • primitive divisor of Lucas number
  • existence