# The Diophantine Equation8x+py=z2

Authors: Lan Qi; Xiaoxue Li
Journal: The Scientific World Journal
 (2015)

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## Abstract

Letp be a fixed odd prime. Using certain results of exponential Diophantine equations, we prove that (i) if p≡±3(mod8), then the equation 8x+py=z2 has no positive integer solutions (x,y,z); (ii) if p≡7(mod8), then the equation has only the solutions (p,x,y,z)=(2q-1,(1/3)(q+2),2,2q+1), where q is an odd prime with q≡1(mod3); (iii) if p≡1(mod8) and p≠17, then the equation has at most two positive integer solutions (x,y,z).


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## Body

## 1. Introduction

LetZ, N be the sets of all integers and positive integers, respectively. Let p be a fixed odd prime. Recently, the solutions (x,y,z) of the equation
(1)8x+py=z2,x,y,z∈N
were determined in the following cases:(1)
(Sroysang [1]) if p=19, then (1) has no solutions;(2)
(Sroysang [2]) if p=13, then (1) has no solutions;(3)
(Rabago [3]) if p=17, then (1) has only the solutions (x,y,z)=(1,1,5), (2,1,9), and (3,1,23).In this paper, using certain results of exponential Diophantine equations, we prove a general result as follows.Theorem 1.
Ifp≡±3(mod⁡8), then (1) has no solutions (x,y,z). If p≡7(mod⁡8), then (1) has only the solutions
(2)p,x,y,z=2q-1,q+23,2,2q+1,
where q is an odd prime with q≡1(mod⁡3).
Ifp≡1(mod⁡8) and p≠17, then (1) has at most two solutions (x,y,z).Obviously, the above theorem contains the results of [1, 2]. Finally, we propose the following conjecture.Conjecture 2.
Ifp≠17, then (1) has at most one solution (x,y,z).


## 2. Preliminaries

Lemma 3.
If2n-1 is a prime, where n is a positive integer, then n must be a prime.Proof.
See Theorem1.10.1 of [4].Lemma 4.
Ifp is an odd prime with p≡1(mod⁡4), then the equation
(3)u2-pv2=-1,u,v∈N
has solutions (u,v).Proof.
See Section 8.1 of [5].Lemma 5.
The equation(4)X2-2m=Yn,X,Y,m,n∈N,gcd⁡X,Y=1,Y>1,m>1,n>2
has only the solution (X,Y,m,n)=(71,17,7,3).Proof.
See Theorem 8.4 of [6].Lemma 6.
LetD be a fixed odd positive integer. If the equation
(5)u2-Dv2=-1,u,v∈N
has solutions (u,v), then the equation
(6)X2-D=2n,X,n∈N,n>2
has at most two solutions (X,n), except the following cases:(i)
D=22r-3·2r+1+1, (X,n)=(2r-3,3), (2r-1,r+2), (2r+1,r+3), and (3·2r-1,2r+3), where r is a positive integer with r≥3;(ii)
D=1/3(22r+1-17)2-32, (X,n)=1/3(22r+1-17),5, 1/3(22r+1+1,2r+3), and 1/3(17·22r+1-1),4r+7, where r is a positive integer with r≥3;(iii)
D=22r1+22r2-2r1+r2+1-2r1+1-2r2+1+1, (X,n)=(2r2-2r1-1,r1+2), (2r2-2r1+1,r2+2), and (2r2+2r1-1,r1+r2+2), where r1, r2 are positive integers with r2>r1+1>2.Proof.
See [7].Lemma 7.
IfD is an odd prime and D belongs to the exceptional case (i) of Lemma 6, then D=17.Proof.
We now assume thatD is an odd prime with D=22r-3·2r+1+1. Then we have
(7)2r-12-2r+2=D,(8)2r+12-2r+3=D.
If 2∣r, since r≥3, then r≥4, and by (7), we have
(9)2r-1+2r/2+1=D,2r-1-2r/2+1=1.
But, by the second equality of (9), we get 1≡(2r-1)-2r/2+1≡-1(mod⁡8), a contradiction.
If2∤r, then from (8) we get
(10)2r+1+2(r+3)/2=D,2r+1-2(r+3)/2=1.
Further, by the second equality of (10), we have 2r=2(r+3)/2, r=3, and D=17. Thus, the lemma is proved.Lemma 8.
IfD is an odd prime and D belongs to the exceptional case (iii) of Lemma 6, then D=17.Proof.
Using the same method as in the proof of Lemma7, we can obtain this lemma without any difficulty.Lemma 9.
IfD belongs to the exceptional case (ii), then (6) has at most one solution (X,n) with 3∣n.Proof.
Notice that, for any positive integerr, there exists at most one number of 5, 2r+3, and 4r+7 which is a multiple of 3. Thus, by Lemma 6, the lemma is proved.Lemma 10.
The equation(11)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).Proof.
See [8].


## 3. Proof of Theorem

We now assume that(x,y,z) is a solution of (1). Then we have gcd(2p,z)=1.If2∣y, since gcd(z+py/2,z-py/2)=2, then from (1) we get
(12)z+py/2=23x-1,z-py/2=2,
where we obtain
(13)z=23x-2+1,(14)py/2=23x-2-1.
Since p>1, applying Lemma 10 to (14), we get
(15)y=2,p=23x-2-1.
Further, by Lemma 3, we see from the second equality of (15) that
(16)p=2q-1,q=3x-2
is an odd prime with q≡1(mod⁡3).Therefore, by (13), (15), and (16), we obtain the solutions given in (2).Obviously, ifp satisfies (2), then p≡7(mod⁡8). Otherwise, since 2∤y, we see from (1) that p≡py≡z2-8x≡1(mod⁡8). It implies that if p≡±3(mod⁡8), then (1) has no solutions (x,y,z). If p≡7(mod⁡8), then (1) has only the solutions (2).Here and below, we consider the remaining cases thatp≡1(mod⁡8). By the above analysis, we have 2∤y. If y>1, then y≥3 and (4) has the solution (X,Y,m,n)=(z,p,3x,y) with 3∣m. But, by Lemma 5, it is impossible. Therefore, we have
(17)y=1.
Substituting (17) into (1), the equation
(18)X2-p=2n,X,n∈N,n>2
has the solution (X,n)=(z,3x) with 3∣n. Since p≡1(mod⁡8), by Lemma 4, (3) has solutions (u,v). Therefore, by Lemmas 6–9, (1) has at most two solutions (x,y,z). Thus, the theorem is proved.



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