Research on the Number of Solutions to a Special Type of Diophantine Equation (a^x−1)(b^y−1) = 2z²

Abstract

Let $b$ be an odd number. By using elementary methods, we prove that: (1) When $x$ is an odd number and $y$ is an even number, the Diophantine equation $(2^x-1)(b^y-1)=$$2z^2$ has no positive integer solution except when $b$ is two special types of odd number. (2) When $x$ is an odd number and $b\equiv \pm 3(\mathrm{mod}8)$, the Diophantine equation $(2^x-$$1)(b^y-1)=2z^2$ has no positive integer solution except where $b=3$ and is another special type of the odd number.

1. Preface

Regarding the exponential Diophantine equation $(a^x-1)(b^y-1)=c{z}^2$, many scholars have conducted extensive research on the cases of $c=1$ and $c=2$. In Section 2, we describe the main results of the equation $(a^x-1)(b^y-1)=c{z}^2$ in three parts. In the first part, we introduce the situation where $c=1$ [1,2,3,4,5,6,7,8,9], which is the main result of the equation $(a^x-1)(b^y-1)=z^2$. In the second part, we introduce the situation where $c=2$ [10], which is the main result of the equation $(a^x-1)(b^y-1)=2z^2$. In the third part, we introduce the main results of this paper, which are the main results of equation $(2^x-1)(b^y-1)=2z^2$. In Section 3, we describe the basic concepts such as the fundamental solution and the solution of the equations to be used in this paper. In Section 4, we describe the lemmas that are used in the theorem proving process. In Section 5, we describe the process of proving theorems using elementary methods such as the recursive sequence method and quadratic residue method. In Section 6, we describe the conclusion of this article and propose some unresolved issues and conjectures.

2. Introduction to Main Results

2.1. The Equation $(a^x-1)(b^y-1)=z^2$

In 2000, Walsh [1] proved that: the equation $(2^x-1)(3^y-1)=z^2$ has no positive integer solution. Szalay [2] proved that: (I) the equation $(2^x-1)(3^x-1)=z^2$ has no positive integer solution; (II) the equation $(2^x-1)(5^x-1)=z^2$ has only the positive integer solution $(x,z)=(1,2)$; (III) the equation $(2^x-1)({(2^k)}^x-1)=z^2$ has only the positive integer solution $(k,x,z)=(2,3,21)$ in positive integers $k>1$. Thus, it can be seen that Walsh generalized the results (I) of Szalay. In 2002, Cohn [3] proved that: (I) if $a^l=b^k,$ then the equation $(a^x-1)(b^x-1)=z^2$ has only the following solutions: $(a,b,x,z)=(c^2$$-1,a^2,1,c(c^2-2))$ with any $c\ge 2$, $(2,4,3,$ $21)$, $(3,243,1,22)$, and $(7,2401,1,$$120)$; (II) if $4|x$, then the equation $(a^x-1)(b^x-1)=z^2$ has no solution except for $(a,b,x,z)=$$(13,239,4,9653280)$, etc. In 2011, Tang [4] proved that: (I) if $a\equiv 0(\mathrm{mod}2),b\equiv 15(\mathrm{mod}20),$ then the equation $(a^x-1)(b^x-1)=$ $z^2$ has no positive integer solution; (II) if $a\equiv 2(\mathrm{mod}6),b\equiv 3(\mathrm{mod}12)$, then the equation $(a^x-1)(b^y-1)=z^2$ has no positive integer solution. He [5] proved that: (I) when $x<y,$ the equation $(a^x-1)(b^y-1)=z^2$ has only the positive integer solutions $(a,b,x,y,z)=$$(2,2,3,6,21)$, $(3,3,1,5,22)$, $(3,3,2,5,$$44)$, $(7,7,1,4,120)$, and $(k^2-1,{(k^2-1)}^2,$$1,2,k(k^2$$-2))$, wherein any $k>1$; (II) when $2|a,$$2\cancel{|}b$ and $b-1=2^{s}c$ with $2\cancel{|}sc$, $p\equiv \pm 3(\mathrm{mod}8)$ is an odd prime, and $a\equiv -1(\mathrm{mod}p),b\equiv$ $0(\mathrm{mod}p),$ then the equation $(a^x-1)(b^y-1)=z^2$ has no positive integer solution. Clearly, He generalized some of the results of Cohn. In 2012, Yuan and Zhang [6] proved that: When $x$$>2$ and $a,b$ are positive integers, and supposing that one of the following properties is satisfied: (i) $a\equiv 2(\mathrm{mod}3)$, $b\equiv 0(\mathrm{mod}3)$; (ii) $a\equiv 3(\mathrm{mod}4)$, $b\equiv 0(\mathrm{mod}2)$; (iii) $a\equiv -1(\mathrm{mod}5)$, $b\equiv 0(\mathrm{mod}5)$; then, the equation $(a^x-1)(b^x-1)$ $=z^2$ has no positive integer solution. In 2013, Guo [7] proved that: if $v(a-1)$ and $v(b-1)$ have opposite parity, then the equation $(a^x-1)(b^x-1)$$=z^2$ has only the following solutions: (i) $(a,b,x,z)=$$(13,239,4,$$9653280)$ or $(239,13,4,$$9653280)$; (ii) $(a,b,x,z)=$$(u_r,u_s,2,d{v}_r{v}_s)$, where $d=c^2-1,$ $c$ is a positive integer, $c>1,c\equiv$ $0,2,3(\mathrm{mod}4),$ $v(c+1)$ is even, $(u_r,v_r)$ and $(u_s,v_s)$ satisfy $u_k+v_k\sqrt{d}={(c+\sqrt{d})}^k,$ with $k,r,s$ positive integers, and $(r,s)=1,2|rs$. In 2019, Noubissie and Togbé [8] proved that: (I) Let $a,b$ be positive integers and $a,b>1,$ and suppose that one of the following conditions is satisfied: (i) $a\equiv 0(\mathrm{mod}2),$$b\equiv 3(\mathrm{mod}12)$; (ii) $a$ is even, $v(b-1)=1$ and $5|b$; then, the equation $(a^x-1)(b^x-1)$$=z^2$ has no positive integer solution; (II) Let $a,b$ be positive integers such that $a,b>1,$ and suppose that $a\equiv 4(\mathrm{mod}10),$ $b\equiv 0(\mathrm{mod}5)$; then, the equation $(a^x-1)(b^x-1)$$=z^2$ has no positive integer solution. Let $d=(a^x-1,b^x-1),$$(u_1,v_1)$ be the fundamental solution of the equation $u^2-d{v}^2=1,$ and let $(u_k,v_k)$ satisfy $u_k+v_k\sqrt{d}={(c+\sqrt{d})}^k,$ where $k$ is a positive integer. Noubissie and Togbé [8] proved that: (III) Suppose that $a\equiv 4($$\mathrm{mod}5),$ $b\equiv 0(\mathrm{mod}5)$, then the equation $(a^x-1)(b^x-1)=$$z^2$ has a positive integer solution $(x,z)$ if and only if $(a,b)=$$(u_r,u_s)$ with nonsquare $d\equiv \pm 1(\mathrm{mod}5)$ satisfying $u_1\equiv 0(\mathrm{mod}5),r\equiv$$2(\mathrm{mod}4)$ and $s$ is odd. In this case, the solution is $(x,z)$$=(2,d{v}_r{v}_s)$. (IV) Suppose that $a\equiv 3(\mathrm{mod}4)$ and $b\equiv 0(\mathrm{mod}2)$, then the equation $(a^x-1)(b^x-1)=$$z^2$ has a positive integer solution $(x,z)$ if and only if $(a,b)=$$(u_r,u_s)$ with nonsquare $d\equiv 3(\mathrm{mod}4)$ satisfying $u_1\equiv$$0(\mathrm{mod}2),r\equiv 2($ $\mathrm{mod}4)$ and $s$ is odd. In this case, the solution is $(x,z)$$=(2,d{v}_r{v}_s)$. Keskin [9] proved that: (I) Let $(a,b)=1$; if the equation $(a^x-1)(b^x-1)=z^2$ for some integers $z$ with $2|x$ and $4\cancel{|}x$, then $x=2$. (II) Let $v_2(a)\ne v_2(b)$ and $(a,b)>1,$ then the equation $(a^x-1)(b^x-1)=$$z^2$ has no integer solution $(x,z)$ with $2|x$. (III) Let $a\cancel{|}b$ and $b\cancel{|}a$ with $g=(a,b)>1$; if $g^2>a$ or $g^2>b$, then the equation $(a^x-1)(b^x-1)=$$z^2$ has no integer solution $(x,z)$ with $2|x$. If $a|b$ and $a^2>b$, then the same is true. (IV) Let $a,b$ be odd and $g=(a,b)>1$. If $\frac{a}{g}\equiv 3(\mathrm{mod}4)$ or $\frac{b}{g}\equiv 3(\mathrm{mod}4)$, then the equation $(a^x-1)(b^x-1)$$=z^2$ has no integer solution $(x,z)$ with $2|x$ and $4\cancel{|}x$. (V) The equation $(2^x-1)({50}^x-1)=$$z^2$ has only the solution $(x,z)=(1,7)$. (VI) Let $b$ be even. Then, the equation $(a^x-1)(b^{2x}{a}^x-1)$$=z^2$ has no integer solution.

2.2. The Equation $(a^x-1)(b^y-1)=2z^2$

In 2021, Tong [10] studied the case of $a=2,b=p$, with $p$ being an odd prime, and proved that: (I) Let $p$ be an odd prime; when $2\cancel{|}x,$ $p\equiv$$\pm 3(\mathrm{mod}8),$ the equation $(2^x-1)($$p^y-1)$$=2z^2$ has no positive integer solution except that when $p=3$, it has only positive integer solutions $(x,y,z)=(1,5,$$11)$, $(1,2,2)$, and $(1,1,1)$, and when $p=2a_0{}^2+1$, it has only the positive integer solution $(x,y,z)=(1,1,a_0)$ with $2\cancel{|}{a}_0>1$. (II) Let $p$ be an odd prime; when $2\cancel{|}x$, $2|y,$ and $y\ne 2,4$, the equation $(2^x-1)(p^y-1)=$ $2z^2$ has no positive integer solution.

In addition, Osipov and Tikhomirov [11] studied the equation $c^2-2a^2$ $=n.$ Given a positive integer $n,$ denote by $P_n$ the number of pairs $(c,a)\in$Z${}^2$ with $c^2-2a^2$$=n,0\le 2a<c.$ Let $Q(n)$ be the maximal number of pairwise nonassociated solutions $\alpha =c+a\sqrt{2}\in$Z$[\sqrt{2}]$ of the norm equation $N(\alpha )=\left|c^2-2a^2\right|=n$ in the ring of integers of the quadratic field Q$(\sqrt{2}).$ Let $m=c-2a,m\ge 1,a\ge 0,$$(m,a)\in$ Z${}^2.$ Osipov and Tikhomirov [11] proved that: if $n=m^2,$ then $P_n=\frac{Q(n)+1}{2};$ if $n=2m^2,$ then $P_n=\frac{Q(n)-1}{2};$ if $n\ne m^2$ and $n\ne 2m^2,$ then $P_n=\frac{Q(n)}{2}.$ The content of the paper provides enlightenment for my reseach work.

In this paper, we generalize the results of Tong and study the case of $a=2$ with $b$ being an odd number. Therefore, it is meaningful. The main achievements of this article are as follows.

2.3. The Theorems in This Paper

We study the following equation:

$$(2^x-1)(b^y-1)=2z^2,b>0,b\mathrm{is}\mathrm{an}\mathrm{odd}\mathrm{number}$$

(1)

and prove the following theorems:

Theorem 1. 

Let $2\cancel{|}x,$$2|y$; the Diophantine equation (Equation (1)) has no positive integer solution except that when $b=2^{2a+3}-2^{a+3}+1,$$2\cancel{|}a$, it has only the positive integer solution $x=a,$$y=2,z=$$2^{\frac{a+3}{2}}(2^a-1)(2^{a+1}-1)$, and when $b=2^{a+1}-1,$$2\cancel{|}a$, it has only the positive integer solution$x=a,y=2,z=2^{\frac{a+1}{2}}(2^a-1)$.

Theorem 2. 

When $2\cancel{|}x$, $b\equiv \pm 3(\mathrm{mod}8),$ the Diophantine equation (Equation(1)) has no positive integer solution except that when$b=3$, it has only positive integer solutions$x=1,y=5,z=11$; $x=1,y=2,z=2$;$x=1,y=1,z=1$; and when $b=2a_0{}^2+1$, it has only the positive integer solution $x=1,y=1,z=a_0$ with $2\cancel{|}{a}_0>1$.

3. The Definitions Used in the Paper

Before proving the theorems in this paper, it is necessary to state some definitions involved in the proof.

Definition 1. 

Let $D$be a positive interger that is not a perfect square. Then, the equation $x^2-D{y}^2=1$has an infinity of positive interger solutions. If $(x,y)=(x_0,y_0)$, where $x_0>0,y_0>0$ is the solution with the least positive $x$, then $x_0+y_0\sqrt{D}$ is called a fundamental solution of the equation.

Definition 2. 

A set of equations that combines several equations so that the unknowns satisfy each equation simultaneously is called a system of equations or simultaneous equations. The value of the unknown number that can simultaneously satisfy each equation in the equations is called the solution of the equations or the solution of the simultaneous equations.

Definition 3. 

If $y$ is divisible by every prime factor of $x$, then $y$ is divisible by asterisk $x$, which is denoted as $x|*y$.

4. Preliminary Lemmas

In order to prove the main results in this paper, we first state the following lemmas.

Lemma 1. 

[12]. Pell equation

$$x^2-D{y}^2=1,(x,y)=1,D>0 \mathrm{non}-\mathrm{square} \mathrm{number}$$

(2)

Let $x>0,y>0$ be a solution of Equation (2). $\epsilon =x_0+$$y_0\sqrt{D}$ is the fundamental solution of Equation (2). If $x{|*x}_0$, then $x+y\sqrt{D}=\epsilon$.

Lemma 2. 

[13]. The Diophantine equation $x^4-2y^2=1$ has only positive integer solutions $x=\pm 1,y=0$.

Lemma 3. 

[13]. Let $p$ be an odd prime. If the Diophantine equation $x^p+1=2y^2$ has a positive integer solution, then $2p|y$ is true except that it has the positive integer solution $(x,y)=$ $(1,1)$, and when $p=3$, it has only positive integer solutions $(x,y)=(1,1),(23,78)$.

Lemma 4. 

[10]. The Diophantine equation

$$a^y-1=2z^2,y\ne 2$$

has only positive integer solutions $a=3,y=5,z=11$ and $a=2a_0^2+1,y=1,z=a_0$.

5. Proof of the Theorems

Proof of the Theorem 1. 

Assume $(2^x-1,b^y-1)=d$, then Equation (1) is converted into

$$\{\begin{array}{l}{2}^x-1=d{s}^2\\ b^y-1=2d{t}^2\\ z=dst\end{array}$$

(3)

There, $(s,t)=1$, $2\cancel{|}s,$ $2\cancel{|}d$.

Since $2\cancel{|}x$, $2|y$, from the first equation in (3), we obtain:

$${\left(2{\left(2^{\frac{x-1}{2}}\right)}^2+d{s}^2\right)}^2-2d{\left(2^{\frac{x+1}{2}}s\right)}^2=1$$

(4)

Let the Pell equation be:

$$X^2-2d{Y}^2=1,(X,Y)=1$$

We assume that ${\epsilon }_{2d}=T+U\sqrt{2d}$ is the fundamental solution of the above equation. Then, all positive integer solutions of the above equation are $X_k+Y_k\sqrt{2d}=$${(T+U\sqrt{2d})}^k$, $k\ge 0$. Therefore, from the second equation in (3) and Equation (4), we obtain simultaneous equations:

$$\{\begin{array}{l}2{\left(2^{\frac{x-1}{2}}\right)}^2+d{s}^2+\left(2^{\frac{x+1}{2}}s\right)\sqrt{2d}=X_r+Y_r\sqrt{2d}\\ b^{\frac{y}{2}}+t\sqrt{2d}=X_l+Y_l\sqrt{2d}\end{array}$$

(5)

Next, we find the solution of the system of Equation (5) by the recursive sequence method and discuss the second equation in (5) in two cases of $l=2h$ and $l=2h+1$.

• The case of $l=2h$:

From the second equation in (5), since $l=2h$, $b^{\frac{y}{2}}=X_{2h}=2X_h^2-1$, that is:

$$b^{\frac{y}{2}}=2X_h^2-1$$

(6)

We distinguish two cases to discuss Equation (6).

(i)

The case of $b\equiv \pm 1(\mathrm{mod}8)$:

Since $2|y$, let $y=2^w{y}_1,$$w\ge 1,2\cancel{|}{y}_1>0$. If $y_1>1$, then there exists an odd prime $q$ such that $q|y_1$; when $q=3$, Equation (6) is converted into ${\left(b^{\frac{y}{6}}\right)}^3+1$$=2X_h^2$; and from Lemma 3, it has only positive integer solutions $(b^{\frac{y}{6}},X_h)=(1,1),(23,78)$.

Since $2\cancel{|}{X}_h^{}$, we have $b=1$, and from Equation (1), we know $z=0$, which contradicts $z>0$. When $q>3$, Equation (6) is converted into ${\left(b^{\frac{y}{2q}}\right)}^q+1=2X_h^2$, since $(b^{\frac{y}{2q}},X_h)\ne$$(1,1)$. From Lemma 3, we know $2q{|X}_h$, which contradicts $2\cancel{|}{X}_h^{}$, hence $y_1=1$. Therefore, $y=2^w.$ If $w\ge 3$, then Equation (6) is converted into ${\left(b^{2^{w-3}}\right)}^4-2X_h^2=-1$, and from Lemma 2, it has only the positive integer solution $(b^{2^{w-3}},X_h)=$$(1,1),$ which contradicts $b\ne 1$. Hence, $w\le 2$, which yields $y=2^w=2,4$. When $y=2$, Equation (6) is converted into $b=2X_h^2-1$, and from the first equation in (5), we have $X_h=2(2^{\frac{x-1}{2}}{)}^2+$ $d{s}^2=2^{x+1}-1$ and deduce that $b=2^{x+3}(2^x-1)+1$. From Equation (1), we obtain:

$$2z^2=(2^x-1)(b^2-1)=2^{x+4}{(2^x-1)}^2(2^{x+2}(2^x-1)+1)$$

Therefore, $z=2^{\frac{x+3}{2}}(2^x-1)(2^{x+1}-1),$ that is, $b=2^{2a+3}-2^{a+3}+1$$,2\cancel{|}a$. Equation (1) has only the positive integer solution $(x,y,z)=(a,2,2^{\frac{a+3}{2}}(2^a-1)(2^{a+1}-1))$.

When $y=4$, Equation (6) is converted into $b^2=2X_h^2-1$. Since all positive integer solutions of the Pell equation $H^2-2K^2=-1$ are $(H_k,K_k)=$$({\lambda }_{2k-1},{\mu }_{2k-1})$, $k=1,2,3,\cdots$, ${\lambda }_n+{\mu }_n\sqrt{2}$$={(1+\sqrt{2})}^n,$$n=1,2,3,\cdots .$

Since ${\mu }_{n+1}={\lambda }_n+{\mu }_n,{\lambda }_{n+1}={\mu }_n+{\mu }_{n+1}$, ${\mu }_{2k-1}\equiv 1(\mathrm{mod}4)$,$k=1,2,3,\cdots$.

Hence, $X_h=K=K_k={\mu }_{2k-1}\equiv 1(\mathrm{mod}4)$, when $x\ge 1$, since $X_h=2^{x+1}-1$$\equiv -1(\mathrm{mod}4),$—a contradiction.

(ii)

The case of $b\equiv \pm 3(\mathrm{mod}8)$:

We use the quadratic residual method and obtain $1=\left(\frac{1}{b}\right)=\left(\frac{2X_h^2-b^{\frac{y}{2}}}{b}\right)=$$\left(\frac{2X_h^2}{b}\right)=-1$ from Equation (6)—a contradiction.

2.

The case of $l=2h+1$:

Let $l=2h+1$, then $b^{\frac{y}{2}}=X_{2h+1}$. Since $X_1|X_{2h+1}$, we obtain $T|b^{\frac{y}{2}}$. If $X_1=T=1$, then $b^{\frac{y}{2}}=X_1=1$, which contradicts $z>0$. Hence, $b^{\frac{y}{2}}\ge T$$>1$, and let $T=p_1^{{\alpha }_1}{p}_2^{{\alpha }_2}\cdots p_u^{{\alpha }_u},b=q_1^{{\beta }_1}{q}_2^{{\beta }_2}\cdots q_v^{{\beta }_v}$, where $u,v,{\alpha }_i(i=1,2,\cdots ,u),{\beta }_j(j=1,$ $2,\cdots ,v)$ are positive integers, $p_i(i=1,2,\cdots ,u),q_j(j=1,2,\cdots ,v)$ are odd primes, and $p_1<p_2<\cdots <p_u,q_1<q_2<\cdots <q_v$. When $h=0,$ from $b^{\frac{y}{2}}=X_1=T$, we have $b|T$. Hence, $q_j^{{\beta }_j}|T,j=1,2,\cdots ,v$. Therefore, there exists $p_i(1\le i\le u)$ such that $q_j=p_i$; hence, $v\le u$. From $T|b^{\frac{y}{2}}$, we deduce $p_i|b^{\frac{y}{2}}(i=1,2,\cdots ,u)$. Therefore, there exists $q_j(1\le j\le v)$ such that $p_i=q_j$; hence, $u\le v$, namely $u=v$. It follows that $T=p_1^{{\alpha }_1}{p}_2^{{\alpha }_2}\cdots p_u^{{\alpha }_u},b=p_1^{{\beta }_1}{p}_2^{{\beta }_2}\cdots p_u^{{\beta }_u},{\beta }_i\le {\alpha }_i(i=1,2,\cdots ,u)$. From $T|b^{\frac{y}{2}}$, we know that there exists $T_0$ such that $T=b^{T_0},{\alpha }_i=T_0{\beta }_i(i=1,2,\cdots ,u),T_0\ge 1$. From Lemma 1, we know that the second equation in (5) has only the positive integer solution $(b^{\frac{y}{2}},t)=$$(b^{T_0},U)$, that is, the second equation in (5) has only the positive integer solution $(y,t)=(2T_0,U)$.

From the first equation in (5), $2{\left(2^{\frac{x-1}{2}}\right)}^2+d{s}^2=X_r$ is an odd number; if $r=2h$, then $X_r=2X_h^2-1$ and $2^{\frac{x+1}{2}}s=Y_r=2X_h{Y}_h$; since $2\cancel{|}{X}_h$,$(X_h,Y_h)=1$, we obtain $X_h=b_1,Y_h=2^{\frac{x-1}{2}}{b}_2,(b_1,b_2)=1$,$2\cancel{|}{b}_1$. So:

$$2b_1^2-1=X_r=2{\left(2^{\frac{x-1}{2}}\right)}^2+d{s}^2=2\cdot 2^x-1$$

that is, $b_1^2=2^x$, which contradicts the fact. Therefore, $2\cancel{|}r$.

Since $2\cancel{|}d$, if $2\cancel{|}Y$, then $X^2=2d{Y}^2+1\equiv 2d+1\equiv 3(\mathrm{mod}4)$, which contradicts the fact. Hence, $2|Y$. Therefore, $U=2^{U_0}{U}_1$,$2\cancel{|}{U}_1$, and $U_0\ge 1$; that is, $Y_1=U=2^{U_0}{U}_1$ and $Y_2=2X_1{Y}_1=b^{T_0}\cdot 2^{U_0+1}{U}_1$. Let $r=2h+1$; since $Y_1|Y_r$, we deduce

$$Y_{2h+1}=2X_1{Y}_{2h}-Y_{2h-1}=2X_1{Y}_1\frac{Y_{2h}}{Y_1}-Y_{2h-1}=b^{T_0}\cdot 2^{U_0+1}{U}_1\frac{Y_{2h}}{Y_1}-Y_{2h-1}\equiv -Y_{2h-1}(\mathrm{mod}{2}^{U_0+1})$$

$$Y_{2h+1}\equiv Y_{2h-1}\equiv Y_{2h-3}\equiv \cdots \cdots \equiv Y_3\equiv Y_1\equiv 2^{U_0}{U}_1\equiv 2^{U_0}(\mathrm{mod}{2}^{U_0+1})$$

that is, $2^{\frac{x+1}{2}}s=Y_{2h+1}\equiv 2^{U_0}(\mathrm{mod}{2}^{U_0+1})$, when $\frac{x+1}{2}>U_0$ and $0\equiv 2^{U_0}(\mathrm{mod}{2}^{U_0+1})$—a contradiction. Therefore, $\frac{x+1}{2}\le U_0$, that is, $x\le 2U_0-1$. Here, $X_{2h+1}$$=2^x+d{s}^2=2^{x+1}-1$, that is

$$X_{2h+1}=2^{x+1}-1$$

(7)

We discuss Equation (7) in two cases.

(i)

The case of $b\equiv \pm 1(\mathrm{mod}8)$:

From (7), when $h=0$, then $2^{x+1}-1$$=X_1=b^{T_0}$. If $2|T_0$ or $b\equiv 1(\mathrm{mod}8)$, then $-1\equiv 2^{x+1}-1$$=b^{T_0}\equiv 1(\mathrm{mod}4)$—a contradiction. Hence, $2\cancel{|}{T}_0$,$b\equiv -1(\mathrm{mod}8)$.

Since $2|(x+1)$, we obtain $(2^{\frac{x+1}{2}}+1)$$(2^{\frac{x+1}{2}}-1)$$=b^{T_0}$. Since $(2^{\frac{x+1}{2}}+1$,$2^{\frac{x+1}{2}}-1)$$=1$, we have:

$$\{\begin{array}{l}{2}^{\frac{x+1}{2}}+1=b_1^{T_0}\\ 2^{\frac{x+1}{2}}-1=b_2^{T_0}\\ \\ b=b_1{b}_2\end{array}(b_1,b_2)=1,2\cancel{|}{b}_1{b}_2$$

(8)

We add up the first two equations in (8) and obtain $b_1^{T_0}+b_2^{T_0}=2^{\frac{x+3}{2}}$, that is, $\frac{b_1^{T_0}+b_2^{T_0}}{b_1+b_2}(b_1+b_2)=2^{\frac{x+3}{2}}$; since $2\cancel{|}$$\frac{b_1^{T_0}+b_2^{T_0}}{b_1+b_2}$, we have

$$\{\begin{array}{l}\frac{b_1^{T_0}+b_2^{T_0}}{b_1+b_2}=1\\ b_1+b_2=2^{\frac{x+3}{2}}\end{array}$$

If $b_2$$=b_1=1$, then $1=b^{\frac{y}{2}}\ge T>1$, which contradicts the fact. Therefore, without loss of generality, we assume $b_1>1$. If $T_0>1$, then $b_1^{}+b_2^{}=$$b_1^{T_0}+b_2^{T_0}>$$b_1+b_2^{T_0}\ge$$b_1^{}+b_2^{}$—a contradiction. Hence, $T_0=1$. Therefore, $b=2^{x+1}-1,$ and $y=2T_0=2,$ and plugging them into Equation (1), we obtain

$$2z^2=(2^x-1)(b^2-1)=(2^x-1)\left({(2^{x+1}-1)}^2-1\right)$$

that is, $z=2^{\frac{x+1}{2}}(2^x-1)$. So, when $b=2^{a+1}-1,$$2\cancel{|}a>1$, Equation (1) has only the positive integer solution $(x,y,z)=(a,2,2^{\frac{a+1}{2}}(2^a-1))$. From (7), when $h>0$, since $x+1\le$$2U_0$, we deduce

$$2^{x+1}-1=X_{2h+1}={\displaystyle \sum _{t=0}^h{C}_{2h+1}^{2t}}{\left(b^{T_0}\right)}^{2h+1-2t}{2}^{2t{U}_0}{U}_1^{2t}{\left(2d\right)}^t\equiv {\left(b^{2h+1}\right)}^{T_0}\left(\mathrm{mod}{2}^{2U_0+1}\right)$$

(9)

When $b\equiv 1(\mathrm{mod}8)$, Equation (9) obviously does not hold. When $b\equiv$$-1($$\mathrm{mod}8)$, that is, $b=2^{u}v-1,u\ge 3,$$2\cancel{|}v>0$, then $b^2={\left(2^{u}v-1\right)}^2\equiv 1(\mathrm{mod}{2}^{u+1})$. From Equation (9), we deduce that $-1\equiv {\left(b^{2h+1}\right)}^{T_0}\left(\mathrm{mod}{2}^{x+1}\right)$ holds, and when $x\ge u$, we have $-1\equiv {\left(b^{2h+1}\right)}^{T_0}\equiv b^{T_0}\equiv 1,b\equiv 1,2^u-1\left(\mathrm{mod}{2}^{u+1}\right)$. This is a contradiction. Hence, $x\le$$u-1$, and from Equation (9), we obtain

$$2^u-1\ge 2^{x+1}-1={\displaystyle \sum _{t=0}^h{C}_{2h+1}^{2t}}{\left(b^{T_0}\right)}^{2h+1-2t}{2}^{2t{U}_0}{U}_1^{2t}{\left(2d\right)}^t$$

$$>{\left(b^{2h+1}\right)}^{T_0}>b^2=2^{2u}{v}^2-2^{u+1}v+1>2^{u+1}+1$$

which contradicts the fact.

(ii)

The case of $b\equiv \pm 3(\mathrm{mod}8)$

From Equation (7), when $h=0$ and $2^{x+1}-1$$=X_{2h+1}=X_1=b^{T_0}$, if $x>1,$ then $-1\equiv 2^{x+1}-1$$=b^{T_0}\equiv {\left(\pm 3\right)}^{T_0}$$\equiv \pm 3,1(\mathrm{mod}8)$. This is a contradiction. Therefore, $x=1$, which yields $b^{T_0}=3$, that is, $b=3,T_0=1$. Here, $y=2T_0=2$, and from Equation (3), we obtain $s=d$$=1,t=2,z=dst=2$, so Equation (1) has the positive integer solution $b=3,x=1,$$y=2,z=2$. We incorporate this integer solution into $b=2^{a+1}-1,$ $2\cancel{|}a\ge 1$, $(x,y,z)$$=(a,2,2^{\frac{a+1}{2}}(2^a-1))$.

When $h>0$, from Equation (9), we obtain $-1\equiv {\left(b^{2h+1}\right)}^{T_0}\left(\mathrm{mod}{2}^{x+1}\right)$, and when $x>1$,$-1\equiv {\left(b^{2h+1}\right)}^{T_0}\equiv$${\left(\pm 3\right)}^{T_0}\equiv \pm 3,1\left(\mathrm{mod}8\right)$, which contradicts the fact. Hence, $x=1$.

Since $b\ge 3$, we obtain:

$$3=2^{x+1}-1=X_{2h+1}={\displaystyle \sum _{t=0}^h{C}_{2h+1}^{2t}}{\left(b^{T_0}\right)}^{2h+1-2t}{2}^{2t{U}_0}{U}_1^{2t}{\left(2d\right)}^t>3^{T_0}+C_3^2\ge 6$$

which contradicts the fact. □

Proof of Theorem 2. 

Assuming $2\cancel{|}x$, $b\equiv \pm 3(\mathrm{mod}8)$, when $2|y$, from Theorem 1, and when $b=3$, we know Equation (1) has only the positive integer solution $x=1,y=2,$$z=2$. When $2\cancel{|}y$, if $b\equiv -3(\mathrm{mod}8),$ from Equation (1), we obtain $2z^2\equiv$$(2^x-1)(b-$$1)\equiv 4(2^x-1)(\mathrm{mod}8)$, that is, $z^2\equiv$$2(2^x-1)(\mathrm{mod}4)$, hence $z^2\equiv$$2(\mathrm{mod}4)$, which is impossible. If $b\equiv 3(\mathrm{mod}8),$ from Equation (1), we obtain $2z^2\equiv$$(2^x-1)(b-1)\equiv$$2(2^x$$-1)(\mathrm{mod}8)$, that is, $z^2\equiv$ $(2^x-1)(\mathrm{mod}4)$.

If $x\ge 2$, then $z^2\equiv$$2^x-1\equiv$$-1($$\mathrm{mod}4)$,which contradicts the fact. Hence, $x=1$. Here, Equation (1) is converted into

$$b^y-1=2z^2,b\equiv 3(\mathrm{mod}8)$$

From Lemma 4, we deduce that the above equation has only the positive integer solution $b=3,y=5,z=11$ and $b=2a_0{}^2+1,y=1,z=a_0$,$2\cancel{|}{a}_0$, so Equation (1) has only the positive integer solution $b=3,x=1,y=5,$$z=11,$ and $b=2a_0{}^2+1,x=1,y=1,z=a_0$, $2\cancel{|}{a}_0$. This completes the proof. □

6. Conclusions

In this paper, we generalized Tong’s results on the Diophantine equation $(2^x-1)(p^y-1)=2z^2$. Let $b$ be an odd number where $2\cancel{|}x,2|y$ or $2\cancel{|}x,b\equiv \pm 3(\mathrm{mod}8)$. We completely solved the Diophantine equation $(2^x-$$1)(b^y-1)=2z^2$. However, when $2|x,2|y$, we considered it difficult to solve the equation $(2^x-$$1)(b^y-1)=2z^2$. We conjectured that when $2|x,2|y$, the equation $(2^x-$$1)(3^y-1)=2z^2$ has no positive integer solution.