# The Exponential Diophantine Equation2x+by=cz

Authors: Yahui Yu; Xiaoxue Li
Journal: The Scientific World Journal
 (2014)

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## Abstract

Letb and c be fixed coprime odd positive integers with min{b,c}>1. In this paper, a classification of all positive integer solutions (x,y,z) of the equation 2x+by=cz is given. Further, by an elementary approach, we prove that if c=b+2, then the equation has only the positive integer solution (x,y,z)=(1,1,1), except for (b,x,y,z)=(89,13,1,2) and (2r-1,r+2,2,2), where r is a positive integer with r≥2.


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## Body

## 1. Introduction

LetN be the set of all positive integers. Let a,b,c be fixed coprime positive integers with min⁡{x,y,z}>1. In recent years, the solutions (x,y,z) of the equation(1)ax+by=cz,x,y,z∈N
have been investigated in many papers (see [1–3] and its references). In this paper we deal with (1) for the case that a=2. Then (1) can be rewritten as
(2)2x+by=cz,x,y,z∈N,
where b and c are fixed coprime odd positive integers with min⁡{b,c}>1. We will give a classification of all solutions (x,y,z) of (2) as follows.Theorem 1.
Every solution(x,y,z) of (2) satisfies one of the following types:(i)
(b,c,x,y,z)=(7,3,5,2,4);(ii)
(b,c,x,y,z)=(2r-1,2r+1,r+2,2,2), where r is a positive integer with r≥2;(iii)
(b,c,x,y,z)=(5,3,1,2,3);(iv)
(b,c,x,y,z)=(11,5,2,2,3);(v)
2∣y and z=1;(vi)
(b,c,x,y,z)=(17,71,7,3,2);(vii)
x=1, y>1, 2∤y and 2∣z;(viii)
x>1, y=1, 2∣z and 2x<b50/13;(ix)
2∤yz.
Recently, Miyazaki and Togbé [4] showed that if  b≥5 and c=b+2, then (2) has only the solution (x,y,z)=(1,1,1), except for (b,x,y,z)=(89,13,1,2). However, there are some exceptional cases missing from the result of [4]. In this paper, by an elementary approach, we prove the following result.Corollary 2.
Ifc=b+2, then (2) has only the solution (x,y,z)=(1,1,1), except for (b,x,y,z)=(89,13,1,2) and (2r-1,r+2,2,2), where r is a positive integer with r≥2.


## 2. Preliminaries

Lemma 3 (see [5, Formula 1.76]).
For any positive integern and any complex numbers α and β, one has
(3)αn+βn=∑i=0[n/2]‍[ni](α+β)n-2i(-αβ)i,
where [n/2] is the integer part of n/2;
(4)[ni]=(n-i-1)!n(n-2i)!i!,i=0,…,[n2]
are positive integers.Lemma 4 (see [6]).
LetA and B be coprime odd positive integers with min⁡{A,B}>1. If the equation
(5)Au2-Bv2=2,u,v∈N
has solutions (u,v), then it has a unique solution (u1,v1) such that u1A+v1B≤uA+vB, where (u,v) through all solutions of (5). The solution (u1,v1) is called the least solution of (5). Every solution (u,v) of (5) can be expressed as
(6)uA+vB2=(u1A+v1B2)n,n∈N,2∤n.
Further, by (6), we have u1∣u and v1∣v.Lemma 5.
Equation (5) has no solutions (u,v) such that u>u1, v>v1, and every prime divisor of u/u1 and v/v1 divides A and B, respectively.Proof.
We now assume that(u,v) is a solution of (5) satisfying the hypothesis. Since u>u1, by Lemma 4, the (un,vn) is all solutions of (5). Let
(7)α=u1A+v1B2,β=u1A-v1B2.
We get(8)unu1=αn-(-β)nα-(-β),vnv1=αn-(β)nα-(β),
where n is odd. Numbers α and β are such that (α,-β) satisfy x2-2Bv12x+1=0 and (α,β) satisfy x2-2Au12x+1=0. Thus, {un/u1}n≥1 and {vn/v1}n≥1 are the odd indexed subsequences of the two Lehmer sequences of roots (α,-β) and (α,β). Their discriminants are (α+β)2=2Au12 and (α-β)2=2Bv12, respectively. Saying that all prime factors of un/u1 divide A implies that all primes of the nth term of a Lehmer sequence divide its discriminant. The same is true for vn/v1. Hence, un/u1 and vn/v1 are terms of a Lehmer sequence of real roots lacking primitive divisors. By Table 2 in [7], this is possible only for n=3,5. Even more, in the present case,
(9)(α2)n-(β2)nα2-β2=unvnu1v1
is the nth term of the Lucas sequence of positive real roots (α2,β2) whose all prime factors divide its discriminant (α2-β2)2=4ABu12v22, and by Table 1 in [7] this is possible for n odd only if n=3 or n=5. Furthermore, when n=5, we must have α2=(1+5)/2, but this is not possible since α=(1+5)/2 is not of the form (u1A+v1B)/2 for some positive integers A>1, B>1, u1 and v1. So, only n=3 is possible. Now by some simple numerical computation for u3 and v3, we see that it is not possible that all prime factors of u3 and all prime factors of v3 divideB. Thus, Lemma 5 is proved.Lemma 6 (see [8]).
The equation(10)X2+7=2n+2,X,n∈N
has only the solutions (X,n)=(1,1),(3,2),(5,3),(11,5), and (181,13).Lemma 7 (see [9]).
LetD be an odd positive integer with D>1. If (X,n) is a solution of the equation
(11)X2-D=2n,X,n∈N,
then 2n<D50/13.Lemma 8 (see [10, 11]).
The equation(12)X2+2m=Yn,X,Y,m,n∈N,gcd(X,Y)=1,n≥3
has only the solutions (X,Y,m,n)=(5,3,1,3),(7,3,5,4), and (11,5,2,3).Lemma 9 (see [12]).
The equation(13)X2-2m=Yn,X,Y,m,n∈N,gcd(X,Y)=1,hhhhhhhjjjjjjjjjjjjjjjjjhhY>1,m>1,n≥3
has only the solution (X,Y,m,n)=(71,17,7,3).Lemma 10 (see [13]).
The equation(14)Xm-Yn=1,X,Y,m,n∈N,min⁡{X,Y,m,n}>1
has only the solution (X,Y,m,n)=(3,2,2,3).


## 3. Proof of Theorem

Let(x,y,z) be a solution of (2). If 2∣y and 2∣z, then we have x≥3, cz/2+by/2=2x-1, and cz/2-by/2=2. It follows that
(15)cz/2=2x-2+1,by/2=2x-2-1.
Applying Lemma 10 to (15), we can only obtain the solutions of types (i) and (ii).If2∣y and 2∤z, then we have
(16)(by/2)2+2x=cz,2∤z.
Applying Lemma 8 to (16), we can only get the solutions of types (iii), (iv), and (v).Similarly, if2∤y and 2∣z, using Lemmas 7 and 9, then we can only obtain the solutions of types (vi), (vii), and (viii). Finally, if 2∤yz, then the solutions are of type (ix). Thus, the theorem is proved.


## 4. Proof of Corollary

Sincec=b+2, (2) can be rewritten as
(17)2x+by=(b+2)z,x,y,z∈N.
Let (x,y,z) be a solution of (17). By the theorem, (17) has only the solutions
(18)(b,x,y,z)=(2r-1,r+2,2,2),r∈N,r≥2
satisfying 2∣y and 2∣z.Ifx=1, 2∤y and 2∣z, then from (17) we get
(19)2+by=2+((b+1)-1)yhhh=1+(b+1)∑i=1y‍(-1)i-1(yi)(b+1)i-1hhhhh=1+(b+1)∑j=1z‍(zj)(b+1)j-1hhhhh=((b+1)+1)z=(b+2)z,
whence we obtain
(20)y≡z(mod⁡(b+1)).
But, since 2∣b+1 and 2∤y-z, congruence (20) is impossible.Ifx>1, 2∤y and 2∣z, by the theorem, then we have y=1 and 2x<b50/13. Hence, by (17), we get
(21)b2<(b+2)2≤(b+2)z=2x+b<b50/13+b.
Since b≥3 and 2∣z, we see from (21) that z=2. Substituting it into (17), we have b2+3b-4(2x-2-1)=0 and
(22)b=12(-3+2x+2-7).
By (22), we get
(23)b=12(X-3),x=n,
where (X,n) is a solution of (10). Since 2∤b and b>1, by Lemma 6, we can only have (X,n)=(181,13) and
(24)(b,x,y,z)=(89,13,1,2),
by (23).If2∤yz, then by-1≡(b+2)z-1≡1(mod⁡8). Hence, by (17), we get 2x≡(b+2)z-by≡(b+2)-b≡2(mod⁡8) and x=1. It implies that the equation
(25)(b+2)u2-bv2=2,u,v∈N
has the solution
(26)(u,v)=((b+2)(z-1)/2,b(y-1)/2).
Notice that the least solution of (25) is (u1,v1)=(1,1); y and z satisfy either y=z=1 or min⁡{y,z}>1. Applying Lemma 5 to (26), we only obtain that (u,v)=(1,1) and
(27)(x,y,z)=(1,1,1).
Thus, (17) has only the solutions (18), (24), and (27). The corollary is proved.



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