Proof. Assume that (2.4) has solutions $(U,V)$ . Clearly, $2 \nmid U$ , $2 \nmid V$ , $D \equiv 7 \pmod 8$ and $(|U|,|V|)$ are positive integer solutions of (2.4). Since D is a nonsquare positive integer, $\sqrt {D}$ is irrational and (2.4) has a unique positive integer solution $(U_1,V_1)$ such that $U_1+V_1\sqrt {D}\le U+V\sqrt {D}$ , where $(U,V)$ runs through all positive integer solutions of (2.4). It follows that the fundamental solution $(U_1,V_1)$ of (2.4) exists.

Since $D \equiv 7\pmod 8$ , for any positive integer solution $(U,V)$ of (2.4), (2.1) has the positive integer solution

(2.5) $$ \begin{align} (u,v)=\bigg(\frac{U^2+DV^2}{2},UV\bigg) \end{align} $$

with $2\nmid v$ . By (2.5), these solutions are related by

(2.6) $$ \begin{align} \bigg(\frac{U+V\sqrt{D}}{\sqrt{2}}\bigg)^2=u+v\sqrt{D},\quad 2\nmid v. \end{align} $$

Let $(u_1,v_1)$ be the fundamental solution of (2.1). Recall that (2.1) has solutions $(u,v)$ with $2\nmid v$ . By Lemma 2.1, we have $2 \nmid v_1$ . Hence, from (2.6),

(2.7) $$ \begin{align} \bigg(\frac{U_1+V_1\sqrt{D}}{\sqrt{2}}\bigg)^2=(u_1+v_1\sqrt{D}\,)^{2s+1},\quad s \in {\mathbb{Z}}, s \ge 0. \end{align} $$

If $s \ge 1$ , let

(2.8) $$ \begin{align} U'+V'\sqrt{D}=(U_1+V_1\sqrt{D}\,)(u_1-v_1\sqrt{D}). \end{align} $$

From (2.7) and (2.8), $U'$ and $V'$ are positive integers satisfying $U^{\prime 2}-DV^{\prime 2}=2$ and

$$ \begin{align*} \bigg(\frac{U'+V'\sqrt{D}}{\sqrt{2}}\bigg)^2=(u_1+v_1\sqrt{D}\,)^{2s-1}\ge u_1+v_1\sqrt{D}. \end{align*} $$

It follows that $(U',V')$ with $U'+V'\sqrt {D}<U_1+V_1\sqrt {D}$ is a positive integer solution of (2.4), which contradicts the definition of the fundamental solution $(U_1,V_1)$ of (2.4). So we have $s=0$ and, by (2.7), proposition (ii) is proved.

For any positive integer r, let

(2.9) $$ \begin{align} u_r+v_r\sqrt{D}=(u_1+v_1\sqrt{D}\,)^r,\quad \frac{U_r+V_r\sqrt{D}}{\sqrt{2}}=\bigg(\frac{U_1+V_1\sqrt{D}}{\sqrt{2}}\bigg)^{2r-1}. \end{align} $$

Then,

(2.10) $$ \begin{align} u_r-v_r\sqrt{D}=(u_1-v_1\sqrt{D}\,)^r,\quad \frac{U_r-V_r\sqrt{D}}{\sqrt{2}}=\bigg(\frac{U_1-V_1\sqrt{D}}{\sqrt{2}}\bigg)^{2r-1}. \end{align} $$

By proposition (ii), from (2.9) and (2.10),

(2.11) $$ \begin{align} \frac{U_r+V_r\sqrt{D}}{\sqrt{2}} & =\bigg(\frac{U_{r-1}+V_{r-1}\sqrt{D}}{\sqrt{2}}\bigg)(u_1+v_1\sqrt{D}\,)=\cdots \notag\\ & =\bigg(\frac{U_1+V_1\sqrt{D}}{\sqrt{2}}\bigg)(u_{r-1}+v_{r-1}\sqrt{D}\,). \end{align} $$

From (2.11), using the inductive method, $(U,V)=(U_r,V_r)\ (r=1,2,\ldots )$ are positive integer solutions of (2.4) with

(2.12) $$ \begin{align} \bigg(\frac{U_r+V_r\sqrt{D}}{\sqrt{2}}\bigg)^2=u_{2r-1}+v_{2r-1}\sqrt{D},\quad r=1,2,\ldots. \end{align} $$

Recall that $2 \nmid v_1$ . By Lemma 2.1, $(u,v)=(u_{2r-1},v_{2r-1})\ (r=1,2,\ldots )$ are all positive integer solutions of (2.1) with $2\nmid v$ . Thus, we see from (2.6) and (2.12) that $(U,V)=(U_r,V_r)\ (r=1,2,\ldots )$ are all positive integer solutions of (2.4). By (2.9), every positive integer solution $(U,V)$ of (2.4) can be expressed as

(2.13) $$ \begin{align} \frac{U+V\sqrt{D}}{\sqrt{2}}=\bigg(\frac{U_1+V_1\sqrt{D}}{\sqrt{2}}\bigg)^t,\quad t \in {\mathbb{N}},~2\nmid t. \end{align} $$

Furthermore, every solution $(U',V')$ of (2.4) can be expressed as

(2.14) $$ \begin{align} U'+V'\sqrt{D}=\lambda_1(U+\lambda_2V\sqrt{D}\,),\quad \lambda_1,\lambda_2\in \{1,-1\}, \end{align} $$

where $(U,V)$ is a positive integer solution of (2.4). The combination of (2.13) and (2.14) yields proposition (i). The lemma is proved.

Proof. By Lemma 2.1, (2.2) has positive integer solutions $(u,v)$ . Let $(u_1,v_1)$ be the fundamental solution of (2.2). Then,

(2.16) $$ \begin{align} u_1^2-pv_1^2=1,\quad u_1,v_1\in {\mathbb{N}}. \end{align} $$

If $2 \mid v_1$ , then from (2.16),

(2.17) $$ \begin{align} u_1+1=\begin{cases}2a^2,\\2pb^2,\end{cases} \ u_1-1=\begin{cases}2pb^2,\\2a^2,\end{cases}\ v_1=2ab,\quad a,b \in {\mathbb{N}}, \end{align} $$

whence we obtain either

(2.18) $$ \begin{align} a^2-pb^2=1 \end{align} $$

or

(2.19) $$ \begin{align} a^2-pb^2=-1. \end{align} $$

When (2.18) holds, we see from (2.17) that $(u,v)=(a,b)$ is a positive integer solution of (2.2) with $a+b\sqrt {p}<u_1+v_1\sqrt {p}$ , which is a contradiction. When (2.19) holds, we have $p \mid a^2+1$ and $p \not \equiv 7\pmod {8}$ , so we obtain $2 \nmid v_1$ . Then, by (2.16),

$$ \begin{align*} u_1+1=\begin{cases}a^2,\\pb^2,\end{cases}\ u_1-1=\begin{cases}pb^2,\\a^2,\end{cases}\ v_1=ab,\quad a,b,\in {\mathbb{N}},~2\nmid a,~2\nmid b, \end{align*} $$

whence

(2.20) $$ \begin{align} a^2-pb^2=\pm 2. \end{align} $$

Further, since $p \equiv 7 \pmod 8$ , we see from (2.20) that $a^2-pb^2=2$ and (2.15) has solutions $(U,V)$ . The lemma is proved.

Proof. From (1.5), $(2^{(q+1)/2}-1)^2-2(2^{(q-1)/2}-1)^2=2^q-1=p$ . Since $2^{(q+1)/2}-1$ and $2^{(q-1)/2}-1$ are coprime positive integers, (2.23) has a positive integer solution $(X,Y,Z)=(2^{(q+1)/2}-1,2^{(q-1)/2}-1,1)$ . Further, since

(2.26) $$ \begin{align} 1&<\frac{(2^{(q+1)/2}-1)+(2^{(q-1)/2}-1)\sqrt{2}}{(2^{(q+1)/2}-1)-(2^{(q-1)/2}-1)\sqrt{2}}=\bigg(\frac{2^{q/2}-1}{2^{q/2}+1}\bigg)\bigg(\frac{\sqrt{2}+1}{\sqrt{2}-1}\bigg)\notag\\ &<\frac{\sqrt{2}+1}{\sqrt{2}-1}=3+2\sqrt{2}, \end{align} $$

by Lemma 2.7, we find from (2.24) and (2.26) that (2.25) is the least solution of (2.23) for $p=2^q-1$ with $p \ge 7$ . The lemma is proved.

Proof. Assume that $(x,m,n)$ is a solution of (1.1) with $2\mid m$ . Since p is an odd prime, we see from (1.1) that $2\nmid x$ , $\gcd (x+2^{m/2},x-2^{m/2})=1$ and

$$ \begin{align*} x+2^{m/2}=p^n,\quad x-2^{m/2}=1. \end{align*} $$

It follows that

(2.28) $$ \begin{align} 2x=p^n+1 \end{align} $$

and

(2.29) $$ \begin{align} 2^{m/2+1}=p^n-1. \end{align} $$

When $n>1$ , by Lemma 2.9, we can only obtain $p=3$ , $m/2+1=3$ and $n=2$ from (2.29). Hence, by (2.28), we get (1.2).

When $n=1$ , by (2.29),

(2.30) $$ \begin{align} p=2^{m/2+1}+1. \end{align} $$

It follows that p is a Fermat prime. Therefore, since $p=2^{2^l}+1$ , where $l \in {\mathbb {N}}$ , by (2.28) and (2.30), we get (2.27). The lemma is proved.

Proof. Assume that $(x,m,n)$ is a solution of (1.1) with $2 \mid n$ . Since $2 \nmid x$ , by (1.1), we have $\gcd (x+p^{n/2},x-p^{n/2})=2$ and

$$ \begin{align*} x+p^{n/2}=2^{m-1},\quad x-p^{n/2}=2, \end{align*} $$

giving

(2.31) $$ \begin{align} x=2^{m-2}+1 \end{align} $$

and

(2.32) $$ \begin{align} p^{n/2}=2^{m-2}-1. \end{align} $$

Applying Lemma 2.9 to (2.32), we obtain $n/2=1$ and

(2.33) $$ \begin{align} p=2^{m-2}-1. \end{align} $$

From (2.33), p is a Mersenne prime, so p can be expressed as (1.5). By (1.5), (2.31) and (2.33), we obtain the solutions (1.6). The lemma is proved.

Proof. Assume that $(x,n)$ is a solution of (4.2). Then, $2 \nmid x$ and $p^n\equiv x^2-2\equiv 1-2\equiv 7\pmod 8$ . It follows that $2\nmid n$ and $p\equiv 7\pmod 8$ . Hence, we see from (4.2) that (2.15) has a positive integer solution

(4.4) $$ \begin{align} (U,V)=(x,p^{(n-1)/2}). \end{align} $$

Applying Lemma 2.4(i) to (4.4),

(4.5) $$ \begin{align} \frac{x+p^{(n-1)/2}\sqrt{p}}{\sqrt{2}}=\bigg(\frac{U_1+V_1\sqrt{p}}{\sqrt{2}}\bigg)^t,\quad t \in {\mathbb{N}},~2\nmid t, \end{align} $$

where $(U_1,V_1)$ is the fundamental solution of (2.15). By (4.5),

(4.6) $$ \begin{align} 2^{(t-1)/2}p^{(n-1)/2}=V_1\sum_{i=0}^{(t-1)/2}\binom{t}{2i+1}U_1^{t-2i-1}(pV_1^2)^i. \end{align} $$

If $t>1$ , since $t \ge 3$ , $U_1 \ge 3$ and

$$ \begin{align*} \sum_{i=0}^{(t-1)/2}\binom{t}{2i+1}U_1^{t-2i-1}(pV_1^2)^i>tU_1^{t-1}>2^{(t-1)/2}, \end{align*} $$

it follows from (4.6) that $n>1$ and $p \mid tV_1$ . Let

(4.7) $$ \begin{align} p^{\alpha}\,\|\, tV_1,\quad p^{\beta_i}\,\|\, 2i+1,\quad i \in {\mathbb{N}}. \end{align} $$

Since $p^{\beta _i}\le 2i+1$ and

(4.8) $$ \begin{align} \beta_i \le \frac{\log(2i+1)}{\log p}<i,\quad i \in {\mathbb{N}}, \end{align} $$

by (4.7) and (4.8),

(4.9) $$ \begin{align} V_1\binom{t}{2i+1}U_1^{t-2i-1}(pV_1^2)^i \equiv tV_1\binom{t-1}{2i}U_1^{t-2i-1}\frac{(pV_1^2)^i}{2i+1} \equiv 0 \pmod{p^{\alpha+1}},\quad i \in {\mathbb{N}}. \end{align} $$

Hence, from (4.6), (4.7) and (4.9), $\alpha =(n-1)/2$ and $tV_1\ge p^{(n-1)/2}$ . So $2^{(t-1)/2}p^{(n-1)/2}>tU_1^{t-1}V_1\ge U_1^{t-1}p^{(n-1)/2}$ and $2^{(t-1)/2}>U_1^{t-1}\ge 3^{t-1}$ , which is a contradiction. It follows that $t=1$ and by (4.5), we get (4.3). The lemma is proved.

Proof. Assume that $(x,n)$ is a solution of (4.2) with $n>1$ . By Lemma 4.4, $p \mid V_1$ , where $(U_1,V_1)$ is the fundamental solution of (2.15). Further, by Lemma 2.4(ii), $v_1=U_1V_1$ , where $(u_1,v_1)$ is the fundamental solution of (2.2). So $p \mid v_1$ . Therefore, by Lemma 2.2, $p>1.5\times 10^{12}$ and by (1.5),

(4.10) $$ \begin{align} q>40. \end{align} $$

According to well-known properties of Mersenne primes, by (4.10), we just need to consider the case

(4.11) $$ \begin{align} q \ge 61,\quad p \ge 2^{61}-1>2\times10^{18}. \end{align} $$

We see from (4.2) that (2.23) has a solution

(4.12) $$ \begin{align} (X,Y,Z)=(x,1,n). \end{align} $$

Applying Lemmas 2.7 and 2.8 to (4.12),

(4.13) $$ \begin{align} x+\sqrt{2}=(X_1+\lambda Y_1\sqrt{2}\,)^n(u-\lambda v\sqrt{2}\,),\quad \lambda \in \{1,-1\}, \end{align} $$

where $X_1$ , $Y_1$ satisfy (2.25) and $(u,v)$ is a positive integer solution of (2.22). Since the fundamental solution of (2.22) is $(u_1,v_1)=(3,2)$ , by Lemma 2.1,

(4.14) $$ \begin{align} u+v\sqrt{2}=(3+2\sqrt{2}\,)^t,\quad t \in {\mathbb{N}}. \end{align} $$

Hence, by (4.13) and (4.14), $x-\sqrt {2}=(X_1-\lambda Y_1\sqrt {2}\,)^n(u+\lambda v\sqrt {2}\,)$ and

(4.15) $$ \begin{align} \frac{x+\sqrt{2}}{x-\sqrt{2}}=\begin{cases}\bigg(\dfrac{X_1+Y_1\sqrt{2}}{X_1-Y_1\sqrt{2}}\bigg)^n(3+2\sqrt{2}\,)^{-2t}&\text{if }\lambda=1,\\[8pt] \bigg(\dfrac{X_1-Y_1\sqrt{2}}{X_1+Y_1\sqrt{2}}\bigg)^n(3+2\sqrt{2}\,)^{2t}&\text{if }\lambda=-1. \end{cases} \end{align} $$

Here, $n>1$ , $2 \nmid n$ , $x>p^{n/2}\ge p^{3/2}>10^{27}$ , $1.01>(x+\sqrt {2}\,)/(x-\sqrt {2}\,)>1$ and $X_1$ , $Y_1$ satisfy (2.24) by Lemmas 2.7 and 2.8. If $\lambda =1$ , then from (2.24) and (4.15),

$$ \begin{align*} (3+2\sqrt{2}\,)^{2t}<\bigg(\frac{x+\sqrt{2}}{x-\sqrt{2}}\bigg)(3+2\sqrt{2}\,)^{2t}=\bigg(\frac{X_1+Y_1\sqrt{2}}{X_1-Y_1\sqrt{2}}\bigg)^n<(3+2\sqrt{2}\,)^n, \end{align*} $$

whence we obtain $2t<n$ and

(4.16) $$ \begin{align} n \ge 2t+1. \end{align} $$

Similarly, if $\lambda =-1$ , then

$$ \begin{align*} (3+2\sqrt{2}\,)^{2t}=\bigg(\frac{x+\sqrt{2}}{x-\sqrt{2}}\bigg)\bigg(\frac{X_1+Y_1\sqrt{2}}{X_1-Y_1\sqrt{2}}\bigg)^n<1.01(3+2\sqrt{2}\,)^n<(3+2\sqrt{2}\,)^{n+1}, \end{align*} $$

whence $n+1>2t$ and (4.16) is still true, since $2 \nmid n$ .

Further, by (2.26) and (4.15),

(4.17) $$ \begin{align} \frac{x+\sqrt{2}}{x-\sqrt{2}}=\begin{cases}(3+2\sqrt{2}\,)^{n-2t}\theta^{-n}&\text{if }\lambda=1,\\ (3+2\sqrt{2}\,)^{-(n-2t)}\theta^{n}&\text{if }\lambda=-1, \end{cases} \end{align} $$

where

(4.18) $$ \begin{align} \theta=\frac{2^{q/2}+1}{2^{q/2}-1}. \end{align} $$

If $\lambda =1$ , then from (4.16) and (4.17),

(4.19) $$ \begin{align} 1.01\theta^n>\bigg(\frac{x+\sqrt{2}}{x-\sqrt{2}}\bigg)\theta^n=(3+2\sqrt{2}\,)^{n-2t}\ge 3+2\sqrt{2}. \end{align} $$

Similarly, if $\lambda =-1$ , then

$$ \begin{align*} \theta^n=\bigg(\frac{x+\sqrt{2}}{x-\sqrt{2}}\bigg)(3+2\sqrt{2}\,)^{n-2t}>(3+2\sqrt{2}\,)^{n-2t}\ge 3+2\sqrt{2}. \end{align*} $$

This implies that (4.19) is still true for $\lambda =-1$ .

Taking logarithms on both sides of (4.19),

(4.20) $$ \begin{align} n\log \theta>\log \bigg(\frac{3+2\sqrt{2}}{1.01}\bigg)>1.75. \end{align} $$

Further, since $\log (1+\alpha )<\alpha $ for any $\alpha>0$ , by (4.18),

(4.21) $$ \begin{align} \log \theta=\log \bigg(1+\frac{2}{2^{q/2}-1}\bigg)<\frac{2}{2^{q/2}-1}. \end{align} $$

Therefore, by (4.11), (4.20) and (4.21),

(4.22) $$ \begin{align} n>\tfrac12(1.75(2^{q/2}-1))>0.87(2^{61/2}-1)>10^9. \end{align} $$

However, by Lemma 4.1, (4.22) is impossible. The lemma is proved.

Proof of Theorem 1.2.

Assume that $(x,m,n)$ is a solution of (1.1) for ${p \equiv 7 \pmod 8}$ . Since

(4.23) $$ \begin{align} p^n\equiv \begin{cases}1&\pmod8\quad\text{if }2\mid n,\\ 7&\pmod8\quad\text{if }2 \nmid n,\end{cases} \end{align} $$

by (3.1) and (4.23), the solution $(x,m,n)$ satisfies either $2 \mid n$ or $m=1$ and $2\nmid n$ .

We first consider the case that p is a Mersenne prime. Then, p can be expressed as (1.5), where q is an odd prime. By Lemma 2.12, (1.1) has only one solution (1.6) with $2 \mid n$ . However, by Lemma 4.5, if the solution $(x,m,n)$ satisfies $m=1$ and $2 \nmid n$ , then $n=1$ . Hence, by (1.1) and (1.5),

(4.24) $$ \begin{align} x^2-2^q=1. \end{align} $$

Applying Lemma 2.9 to (4.24), we can only obtain $x=3$ , $q=3$ and

(4.25) $$ \begin{align} p=7,\quad (x,m,n)=(3,1,1). \end{align} $$

From the combination of (1.6) and (4.25), if $p=7$ , then (1.1) has exactly two solutions (1.4); otherwise, (1.1) has only one solution (1.6). It follows that the theorem holds when p is a Mersenne prime.

Next, we consider the case where p is not a Mersenne prime. Then, by Lemma 2.12, (1.1) has no solutions $(x,m,n)$ with $2 \mid n$ . Hence, the solution $(x,m,n)$ must satisfy $2 \nmid n$ . Further, by Lemma 2.10, we have $m=1$ . Therefore, from the uniqueness of the fundamental solution $(U_1,V_1)$ of (2.15), by Lemma 2.4(i), we find from Lemma 4.4 that (1.1) has at most one solution $(x,m,n)$ for any fixed p in this case. Further, by Lemmas 2.2, 2.4, 4.4 and 4.3, $p>1.5\times 10^{12}$ and $p<C$ if $n>1$ . However, if $n=1$ , then from (4.2), we obtain only the solution (1.7). From this analysis, the theorem is proved.

Proof. When D is an exceptional case of type (I) with $D \ne 17$ ,

$$ \begin{align*} D=(2^r-1)^2-2^{r+2}=(2^r+1)^2-2^{r+3}\quad \mbox{with } r \in {\mathbb{N}},\;r \ge 4, \end{align*} $$

whence we get the following factorisations of D in positive integers:

(5.2) $$ \begin{align} D=\begin{cases}(2^r+2^{r/2+1}-1)(2^r-2^{r/2+1}-1)&\text{if }2 \mid r ,\\ (2^r+2^{(r+3)/2}+1)(2^r-2^{(r+3)/2}+1)&\text{if }2\nmid r. \end{cases} \end{align} $$

Since $r \ge 4$ , we have $2^r+2^{r/2+1}-1>2^r-2^{r/2+1}-1>1$ and also $2^r+2^{(r+3)/2}+1> 2^r-2^{(r+3)/2}+1>1$ . Hence, we see from (5.2) that D is not an odd prime.

When D is an exceptional case of type (II),

(5.3) $$ \begin{align} D=\tfrac19(2^{2r+1}+2^{r+3}-1)(2^{2r+1}-2^{r+3}-1). \end{align} $$

Since $r \ge 3$ , we have $2^{2r+1}+2^{r+3}-1>2^{2r+1}-2^{r+2}-1 \ge 95$ and

(5.4) $$ \begin{align} 0\equiv \begin{cases}2^{2r+1}+2^{r+3}-1&\pmod9\quad\text{if }r \equiv 0,2,4\quad\pmod6, \\ 2^{2r+1}-2^{r+3}-1&\pmod9\quad\text{if }r\equiv 1,3,5\quad\pmod6. \end{cases} \end{align} $$

We see from (5.3) and (5.4) that D is not an odd prime.

When D is an exceptional case of type (III),

(5.5) $$ \begin{align} D= \begin{cases}(2^{r_2}-2^{r_1}+2^{r_1/2}-1)(2^{r_2}-2^{r_1}-2^{r_1/2+1}-1)&\text{if }2 \mid r_1 ,\\ (2^{r_2}+2^{r_2/2+1}-2^{r_1}-1)(2^{r_2}-2^{r_2/2+1}-2^{r_1}-1)&\text{if }2 \mid r_2, \\ (2^{r_2}+2^{r_1}+2^{(r_2+r_1)/2+1}-1)(2^{r_2}+2^{r_1}-2^{(r_2+r_1)/2+1}-1)&\text{if }2 \nmid r_1r_2. \end{cases} \end{align} $$

Since $r_2>r_1+1>2$ , we see from (5.5) that D is not an odd prime. The lemma is proved.

Proof of Theorem 1.3.

Since $p \equiv 1\pmod 8$ , by Lemma 2.12, (1.1) has no solutions $(x,m,n)$ with $2 \mid n$ . So we have $2 \nmid n$ . Further, since $p \equiv 1 \pmod 8$ , we see from (1.1) that $m \ge 3$ . Therefore, since $p \ne 17$ , by Lemma 2.10, the solutions $(x,m,n)$ of (1.1) must satisfy $n=1$ . This implies that there is a one-to-one correspondence $(x,m,n)=(y,z+2,1)$ between the solutions $(x,m,n)$ of (1.1) and the solutions $(y,z)$ of the equation

(5.6) $$ \begin{align} y^2-p=2^{z+2},\quad y,z\in {\mathbb{N}}. \end{align} $$

In particular, if $p \equiv 1\pmod 8$ and $p \ne 17$ , then the numbers of solutions to these two equations are the same.

Since $p \equiv 1\pmod 8$ , by Lemma 2.3, (2.3) has solutions $(u',v')$ . Since p is an odd prime with $p \ne 17$ , by Lemma 5.2, p is nonexceptional for (5.6). Therefore, by Lemma 5.1, (5.6) has at most two solutions $(y,z)$ . The theorem is proved.