1 Introduction
Let
$\mathbb {N}$
be the set of all positive integers. Let a, b, c be fixed coprime positive integers with
$\min \{a,b,c\}>1$
. We assume without loss of generality that a, b and c are not perfect powers. The purely exponential Diophantine equation
$$ \begin{align} a^{x}+b^{y}=c^{z},\quad x,y,z \in \mathbb{N} \end{align} $$
has been studied deeply (see [Reference Le, Scott and Styer17] for a survey of the results). In 1933, Mahler [Reference Mahler18] used his p-adic analogue of the Diophantine approximation method of Thue–Siegel to prove that (1.1) has only finitely many solutions
$(x,y,z)$
, but his method is ineffective. Let
$N(a,b,c)$
denote the number of solutions
$(x,y,z)$
of (1.1). An effective upper bound for
$N(a,b,c)$
was first given by Gel’fond [Reference Gel’fond7], using his new method in transcendental number theory. Subsequently, as a straightforward consequence of an upper bound for the number of solutions of binary S-unit equations due to Beukers and Schlickewei [Reference Beukers and Schlickewei2], the bound was improved to
$N(a,b,c)\le 2^{36}$
. More accurate upper bounds for
$N(a,b,c)$
have been obtained under certain conditions:
-
(i) if
$2 \nmid c$
, then
$N(a,b,c) \le 2$
(Scott and Styer [Reference Scott and Styer21]); -
(ii) if
$\max \{a,b,c\}>5\times 10^{27}$
, then
$N(a,b,c) \le 3$
(Hu and Le [Reference Hu and Le10]); -
(iii) if
$2 \mid c$
and
$\max \{a,b,c\}>10^{62}$
, then
$N(a,b,c) \le 2$
(Hu and Le [Reference Hu and Le11]); -
(iv) if
$2\mid c$
and
$a<b<10^{62}$
, then
$N(a,b,c) \le 2$
, except for
$N(3,5,2)=3$
(Miyazaki and Pink [Reference Miyazaki and Pink19]).
Nevertheless, deeper problems about the number of solutions of (1.1) remain unresolved. Let
$N^{\prime }(a,b,c)$
denote the number of solutions
$(x,y,z)$
of (1.1) with
$\min \{x,y,z\}>1$
. In this paper, we will discuss the following conjecture.
Conjecture 1.1 (Terai–Jeśmanowicz conjecture).
For any triple
$(a,b,c)$
of positive integers with
$\min\{a,b,c\}>1$
, we have
$N^{\prime }(a,b,c)\le 1$
.
This conjecture contains the famous Jeśmanowicz conjecture concerning Pythagorean triples (see [Reference Jeśmanowicz12]) and its original form was put forward by Terai [Reference Terai22]. It is related to the generalised Fermat conjecture (see Problems B19 and D2 of [Reference Guy8]) and seems very difficult. In 2015, Hu and Le [Reference Hu and Le9] gave a general criterion to judge whether Conjecture 1.1 is true, but this criterion is difficult to apply because it involves some unsolved problems such as the existence of Wieferich primes (see Problem A3 of [Reference Guy8]).
We now discuss Conjecture 1.1 for
$2 \in \{a,b,c\}$
using a different approach from the one in [Reference Hu and Le9]. First, by means of the results of [Reference Le14, Reference Scott20], we can directly prove the following result.
Theorem 1.2. If
$c=2$
or a, b and c are distinct primes, then Conjecture 1.1 is true.
For any odd positive integer r with
$r>1$
, we define
$f(r)=(-1)^{(r-1)/2}$
and
$2^{g(r)}\,\|\, r-(-1)^{(r-1)/2}$
. Obviously,
$f(r) \in \{-1,1\}$
and
$g(r)$
is a positive integer with
$g(r) \ge 2$
. Using a combination of various methods including Baker’s method and known results on exponential Diophantine equations, we prove the following result.
Theorem 1.3. If
$a=2$
and either
$f(b)\ne f(c)$
, or
$f(b)=f(c)$
and
$g(b)\ne g(c)$
, then Conjecture 1.1 is true.
2 Preliminaries
For any positive integer s, let
${\mathop{\mathrm{ord}}\nolimits}_{2}(s)$
denote the order of
$2$
in s, namely,
$2^{{\mathop{\mathrm{ord}}\nolimits}_{2}(s)}\,\|\,s$
.
Lemma 2.1. For any positive integers r and s such that
$r>1$
,
$2 \nmid r$
and
$2\mid s$
, we have
$$ \begin{align} {\mathop{\mathrm{ord}}\nolimits}_{2}(r^{s}-1)=g(r)+{\mathop{\mathrm{ord}}\nolimits}_{2}(s). \end{align} $$
Proof. Since
$2\mid s$
and
$r=2^{g(r)}r_{1}+f(r)$
, where
$r_{1}$
is an odd positive integer,
$$ \begin{align} r^{s}-1&=(2^{g(r)}r_{1}+f(r))^{s}-1\notag\\ &=((f(r))^{s}-1)+2^{g(r)}r_{1}s(f(r))^{s-1}+\sum_{i=2}^{s}\binom{s}{i}(2^{g(r)}r_{1})^{i}(f(r))^{s-i}\notag\\ &=2^{g(r)}r_{1}sf(r)+\sum_{i=2}^{s}\binom{s}{i}(2^{g(r)}r_{1}f(r))^{i}. \end{align} $$
Further, since
$g(r) \ge 2$
and
$2\nmid r_{1}f(r)$
, we see that
$2^{g(r)+{\mathop{\mathrm{ord}}\nolimits}_{2}(s)}\,\|\; 2^{g(r)}r_{1}sf(r)$
and
$$ \begin{align*} \binom{s}{i}(2^{g(r)}r_{1}f(r))^{i} \equiv 2^{g(r)}s\binom{s-1}{i-1}\frac{2^{g(r)(i-1)}}{i}(r_{1}f(r))^{i} \equiv 0 \pmod{2^{g(r)+{\mathop{\mathrm{ord}}\nolimits}_{2}(s)+1}},\quad i \ge 2. \end{align*} $$
For any real number
$\alpha $
, let
$\log \alpha $
denote the natural logarithm of
$\alpha $
.
Lemma 2.2. Let
$\alpha _{1}$
,
$\alpha _{2}$
,
$\beta _{1}$
,
$\beta _{2}$
be positive integers with
$\min \{\alpha _{1},\alpha _{2}\}>1$
. Further, let
$\Lambda =\beta _{1} \log \alpha _{1}-\beta _{2}\log \alpha _{2}$
. If
$\Lambda \ne 0$
, then
$$ \begin{align*} \log|\Lambda|>-25.2(\log \alpha_{1})(\log \alpha_{2})\bigg(\max\bigg\{10,0.38+\log \bigg(\frac{\beta_{1}}{\log \alpha_{2}}+\frac{\beta_{2}}{\log \alpha_{1}}\bigg)\bigg\}\bigg)^{2}. \end{align*} $$
Proof. This is the special case of [Reference Laurent13, Corollary 2] for
$m=10$
.
Lemma 2.3. Let
$\alpha _{1}$
,
$\alpha _{2}$
be odd integers with
$\min \{|\alpha _{1}|,|\alpha _{2}|\}\ge 3$
and let
$\beta _{1}$
,
$\beta _{2}$
be positive integers. Further, let
$\Lambda ^{\prime }=\alpha _{1}^{\,\beta _{1}}-\alpha _{2}^{\,\beta _{2}}$
. If
$\Lambda ^{\prime }\ne 0$
and
$\alpha _{1}\equiv \alpha _{2}\equiv 1 \pmod {4}$
, then
$$ \begin{align*} {\mathop{\mathrm{ord}}\nolimits}_{2}(|\Lambda^{\prime}|)&<19.55(\log|\alpha_{1}|)(\log|\alpha_{2}|)\\ &\quad \times \bigg(\max\bigg\{12\log 2,0.\,4+\log(2\log 2)+\log\bigg(\frac{\beta_{1}}{\log|\alpha_{2}|}+\frac{\beta_{2}}{\log|\alpha_{1}|}\bigg)\bigg\}\bigg)^{2}. \end{align*} $$
Proof. This is the special case of [Reference Bugeaud4, Theorem 2] for
$p=2$
,
$y_{1}=y_{2}=1$
,
$\alpha _{1}\equiv \alpha _{2}\equiv 1 \pmod 4$
,
$g=1$
and
$E=2$
.
Lemma 2.4 [Reference Cohn6, Reference Le16].
The equation
$$ \begin{align*} X^{2}+2^{m}=Y^{n},\quad X,Y,m,n \in \mathbb{N},\, \gcd(X,Y)=1,~n>2 \end{align*} $$
has only the solutions
$(X,Y,m,n)=(5,3,1,3), (7,3,5,4)$
and
$(11,5,2,3)$
.
Lemma 2.5 [Reference Bennett and Skinner1, Theorem 8.4].
The equation
$$ \begin{align*} X^{2}-2^{m}=Y^{n},\quad X,Y,m,n \in \mathbb{N},\, \gcd(X,Y)=1,~m>1,n>2 \end{align*} $$
has only the solution
$(X,Y,m,n)=(71,17,7,3)$
.
Lemma 2.6 [Reference Cao5].
If
$a=2$
and b and c are distinct odd primes with
$\max \{b,c\}<100$
, then
$N^{\prime }(a,b,c)\le 1$
.
Lemma 2.7 [Reference Scott and Styer21].
If
$2 \nmid c$
, then
$N(a,b,c) \le 2$
.
Lemma 2.8 [Reference Scott20, Theorem 6].
$N^{\prime }(a,b,2)\le 1$
.
According to Theorems 1, 2 and 3 of [Reference Le14] and the proof of them, we can obtain the following lemma.
Lemma 2.9. Let p and q be fixed odd primes with
$p \ne q$
.
-
(i) The equation
(2.3)has at most one solution
$$ \begin{align} 2^{x}+p^{y}=q^{z},\quad x,y,z \in \mathbb{N} \end{align} $$
$(x,y,z)$
with
$2\mid y$
and this solution has
$z=1$
, except for
$(p,q,x,y,z)=(3,5,4,2,2),~(5,3,1,2,3),~(7,3,5,2,4)$
and
$(11,5,2,2,3)$
.
-
(ii) Equation (2.3) has at most one solution
$(x,y,z)$
with
$2 \mid x$
and
$2\nmid y$
. -
(iii) Equation (2.3) has at most one solution
$(x,y,z)$
with
$2 \nmid xy$
.
Remark 2.10. The reference [Reference Le14] is written in Chinese and the proof of the theorems mentioned is rather complicated. In the present case, this lemma can be easily obtained using the tools in [Reference Bilu, Hanrot and Voutier3, Reference Le15, Reference Voutier24].
3 Further lemmas on (1.1) for
$a=2$
Let
$a=2$
and let b and c be fixed coprime odd positive integers with
$\min \{b,c\}\ge 3$
. In this section, we give some results on the solutions
$(x,y,z)$
of the equation
$$ \begin{align} 2^{x}+b^{y}=c^{z},\quad x,y,z\in \mathbb{N},\ \min\{x,y,z\}>1. \end{align} $$
Lemma 3.1.
Proof. Let
$(x,y,z)$
be a solution of (3.1) with
$2 \nmid yz$
.
(i) Note that
$f(b) \ne f(c)$
is equivalent to
$b \not \equiv c \pmod 4$
. Since
$2 \nmid yz$
, by (3.1), we have
$2^{x}=c^{z}-b^{y} \equiv 2 \pmod 4$
. This means
$x=1$
, which contradicts
$x>1$
. Therefore, we obtain the conclusion (i).
(ii) Since
$2 \nmid bc$
, we may write
$$ \begin{align} b=2^{g(b)}b_{1}+f(b), \quad c=2^{g(c)}c_{1}+f(c),\quad b_{1},c_{1}\in \mathbb{N},~2\nmid b_{1}c_{1}. \end{align} $$
Assume that
$f(b)=f(c)$
. Since
$2 \nmid yz$
and
$f(b)=f(c)\in \{-1,1\}$
, we see from (3.1) and (3.2) that
$$ \begin{align} 2^{x}&=(2^{g(c)}c_{1}+f(c))^{z}-(2^{g(b)}b_{1}+f(b))^{y}\notag\\ &=\sum_{i=1}^{\max\{y,z\}}\bigg(\binom{z}{i}(2^{g(c)}c_{1})^{i}(f(c))^{z-i}-\binom{y}{i}(2^{g(b)}b_{1})^{i}(f(b))^{y-i}\bigg)\notag\\ &=(2^{g(c)}c_{1}z-2^{g(b)}b_{1}y)+\sum_{i=2}^{\max\{y,z\}}\bigg(\binom{z}{i}(2^{g(c)}c_{1})^{i}-\binom{y}{i}(2^{g(b)}b_{1})^{i}\bigg)(f(c))^{i+1}. \end{align} $$
When
$g(b)\ne g(c)$
, since
$2 \nmid b_{1}c_{1}yz$
, we have
$$ \begin{align*} 2^{\min\{g(b),g(c)\}}\,\|\; 2^{g(c)}c_{1}z-2^{g(b)}b_{1}y \end{align*} $$
and
$$ \begin{align*} \sum_{i=2}^{\max\{y,z\}}\bigg(\binom{z}{i}(2^{g(c)}c_{1})^{i}-\binom{y}{i}(2^{g(b)}b_{1})^{i}\bigg)(f(c))^{i+1} \equiv 0 \pmod{2^{\min\{g(b),g(c)\}+1}}. \end{align*} $$
Since
$\min \{g(b),g(c)\}\ge 2$
, we see from (3.3) that
$x=\min \{g(b),g(c)\}$
and the conclusion (ii) is obtained.
Lemma 3.2. All solutions
$(x,y,z)$
of (3.1) with
$\max \{b,c\}> 100$
satisfy
$$ \begin{align} x<16460(\log b)(\log c),~y<14261\log c,~z<14261\log b\quad \text{if \,}2^{x}<c^{0.8z}, \end{align} $$
and
$$ \begin{align} x<1784(\log b)(\log c),~y<1236\log c,~z<1236\log b\quad \text{if \,}2^{x}>c^{0.8z}. \end{align} $$
Proof. We first consider the case
$2^{x}<c^{0.8z}$
. By (3.1) with
$\max \{b,c\}> 100$
, we have
$b^{y}>2^{x}$
and
$$ \begin{align} z \log c &=\log(b^{y}+2^{x})=y\log b+\log \bigg(1+\frac{2^{x}}{b^{y}}\bigg)\notag\\ &<y \log b + \frac{2^{x}}{b^{y}}=y\log b+\frac{2^{x+1}}{2b^{y}}<y\log b+\frac{2c^{0.8z}}{c^{z}} =y\log b+\frac{2}{c^{z/5}}. \end{align} $$
Let
$(\alpha _{1},\alpha _{2},\beta _{1},\beta _{2})=(c,b,z,y)$
and
$\Lambda =z\log c - y\log b$
. By (3.6),
$0<\Lambda <2/c^{z/5}$
and
$$ \begin{align} \log2-\log|\Lambda|>\frac{z}{5}\log c. \end{align} $$
Since
$\min \{b,c\}\ge 3$
, using Lemma 2.2, we have
$$ \begin{align} \log|\Lambda|>-25.2(\log c)(\log b)\bigg(\max\bigg\{10,0.38+\log\bigg(\frac{z}{\log b}+\frac{y}{\log c}\bigg)\bigg\}\bigg)^{2}. \end{align} $$
When
$10 \ge 0.38+\log (z/\!\log b+y/\!\log c)$
, by (3.7) and (3.8), we have
$$ \begin{align*} \log 2+2520(\log c)(\log b)>\frac{z}{5}\log c, \end{align*} $$
which gives
$$ \begin{align} z<12601\log b. \end{align} $$
When
$10<0.38+\log (z/\!\log b+y/\!\log c)$
, by (3.6), (3.7) and (3.8), we have
$$ \begin{align*} &\log 2 +25.2(\log c)(\log b)\bigg(0.38+\log\bigg(\frac{2z}{\log b}\bigg)\bigg)^{2}\\ &\quad>\log 2+25.2(\log c)(\log b)\bigg(0.38+\log \bigg(\frac{z}{\log b}+\frac{y}{\log c}\bigg)\bigg)^{2} >\frac{z}{5}\log c, \end{align*} $$
which gives
$$ \begin{align} \frac{5\log 2}{(\log b)(\log c)}+126\bigg(0.38+\log 2+\log \bigg(\frac{z}{\log b}\bigg)\bigg)^{2}>\frac{z}{\log b}. \end{align} $$
Since
$(\log b)(\log c)> (\log 3)(\log 100)$
, we can calculate from (3.10) that z satisfies
$$ \begin{align} z<14261\log b. \end{align} $$
Therefore, since
$y \log b<z \log c$
and
$x \log 2<(4z \log c)/5$
, by (3.9) and (3.11), we obtain (3.4).
Next, we consider the case
$2^{x}>c^{0.8z}$
. Let
$(\alpha _{1},\alpha _{2},\beta _{1},\beta _{2})=(cf(c),bf(b),z,y)$
and
$\Lambda ^{\prime }=(cf(c))^{z}-(bf(b))^{y}$
. Since
$x \ge 2$
, by (3.1) and (3.2), we have
$|\Lambda ^{\prime }|=2^{x}$
, that is,
${\mathop{\mathrm{ord}}\nolimits}_{2}(|\Lambda ^{\prime }|)=x$
. Since
$\min \{|cf(c)|,|b(f(b)|\}=\min \{b,c\}\ge 3$
and
$cf(c)\equiv bf(b)\equiv 1 \pmod 4$
, by Lemma 2.3,
$$ \begin{align} x<19.55(\log c)(\log b) \bigg(\max\bigg\{12\log 2,0.4+\log(2\log2)+\log\bigg(\frac{z}{\log b}+\frac{y}{\log c}\bigg)\bigg\}\bigg)^{2}. \end{align} $$
However, since
$2^{x}>c^{0.8z}$
,
$$ \begin{align} x>\frac{0.8z\log c}{\log 2}. \end{align} $$
The combination of (3.12) and (3.13) yields
$$ \begin{align} z <16.94(\log b) \bigg(\max\bigg\{12\log 2,0.4+\log(2\log 2)+\log\bigg(\frac{z}{\log b}+\frac{y}{\log c}\bigg)\bigg\}\bigg)^{2}. \end{align} $$
When
$12\log 2 \ge 0.4+\log (2\log 2)+\log (z/\!\log b+y/\!\log c)$
, by (3.14), we get
$z<1172\log b$
. When
$12\log 2<0.4+\log (2\log 2)+\log (z/\!\log b+y/\!\log c)$
, by (3.6) and (3.14), we have
$$ \begin{align*} \frac{z}{\log b}&<16.94\bigg(0.4+\log(2\log2)+\log\bigg(\frac{z}{\log b}+\frac{y}{\log c}\bigg)\bigg)^{2}\\ &<16.94\bigg(0.4+\log(2\log2)+\log\bigg(\frac{2z}{\log b}\bigg)\bigg)^{2}, \end{align*} $$
whence
$$ \begin{align} z<1236\log b. \end{align} $$
Hence, if
$2^{x}>c^{0.8z}$
, then all the solutions
$(x,y,z)$
of (3.1) satisfy (3.15). Therefore, since
$y \log b<z \log c$
and
$x \log 2<z \log c$
, by (3.15), we obtain (3.5).
Lemma 3.3. Assume that
$x=\min \{g(b),g(c)\}\le 23$
. Then, all solutions
$(x,y,z)$
of (3.1) satisfy
$$ \begin{align*} y<2530\log c, \quad z<2530\log b. \end{align*} $$
Proof. The proof proceeds along the same lines as that of the first half of Lemma 3.2. Since
$2^{x} \le 2^{g(b)}\le b+1 < b^{y}$
, we see from (3.1) that
$$ \begin{align} 0<\Lambda:=z\log c-y\log b=\log\bigg(1+\frac{2^{x}}{b^{y}}\bigg)<\frac{2^{x}}{b^{y}}<\frac{2^{x+1}}{c^{z}}, \end{align} $$
which, together with the assumption
$x \le 23$
, implies that
$$ \begin{align} 24\log 2 - \log|\Lambda|> z \log c. \end{align} $$
We know by Lemma 2.2 that (3.8) holds.
When
$10 \ge 0.38+\log (z/\!\log b+y/\!\log c)$
, by (3.8) and (3.17),
$$ \begin{align*} 24\log 2+2520(\log c)(\log b)> z \log c, \end{align*} $$
whence
$$ \begin{align} z<2530\log b. \end{align} $$
When
$10<0.38+\log (z/\!\log b+y/\!\log c)$
, by (3.8) and (3.17),
$$ \begin{align*} 24\log 2+25.2(\log c)(\log b)\bigg(0.38+\log \bigg(\frac{2z}{\log b}\bigg)\bigg)^{2}> z \log c, \end{align*} $$
which together with
$(\log b)(\log c) \ge (\log 3)(\log 5)$
yields
$$ \begin{align} z<1879\log b. \end{align} $$
The inequalities in the lemma now follow from (3.18), (3.19) and
$y \log b < z \log c$
.
Lemma 3.4. If
$(x,y,z)$
is a solution of (3.1) with
$2 \nmid yz$
, then
$\gcd (y,z)=1$
.
Proof. Let
$d=\gcd (y,z)$
. Then we have
$$ \begin{align} y=dY, \quad z=dZ,\quad Y,Z \in \mathbb{N}. \end{align} $$
$$ \begin{align} 2^{x}=c^{z}-b^{y}=(c^{Z})^{d}-(b^{Y})^{d}=(c^{Z}-b^{Y})(c^{Z(d-1)}+\cdots+b^{Y(d-1)}). \end{align} $$
Since
$2 \nmid yz$
, we have
$2 \nmid d$
. Since
$2 \nmid bc$
, if
$d>1$
, then
$c^{Z(d-1)}+\cdots +b^{Y(d-1)}$
is an odd positive integer greater than
$1$
contradicting (3.21). So we must have
$d=1$
.
By Lemmas 2.4 and 2.5, we can directly obtain the next two lemmas.
Lemma 3.5. Equation (3.1) has only the solution
$(b,c,x,y,z)=(11,5,2,2,3)$
satisfying
$2 \mid y$
and
$2 \nmid z$
.
Lemma 3.6. Equation (3.1) has only the solution
$(b,c,x,y,z)=(17,71,7,3,2)$
satisfying
$2 \nmid y$
and
$2 \mid z$
.
Lemma 3.7. Equation (3.1) has only the solutions
$$ \begin{align} (b,c,x,y,z)=(7,3,5,2,4) \end{align} $$
and
$$ \begin{align} (b,c,x,y,z)=(2^{t}-1,2^{t}+1,t+2,2,2),\quad t \in \mathbb{N},~t \ge 2 \end{align} $$
satisfying
$2 \mid y$
and
$2 \mid z$
.
Proof. Let
$(x,y,z)$
be a solution of (3.1) with
$2 \mid y$
and
$2 \mid z$
. Then,
$2^{x}=c^{z}-b^{y}=(c^{z/2}+b^{y/2})(c^{z/2}-b^{y/2})$
. Further, since
$\gcd (c^{z/2}+b^{y/2},c^{z/2}-b^{y/2})=2$
,
$$ \begin{align*} c^{z/2}+b^{y/2}=2^{x-1}, \quad c^{z/2}-b^{y/2}=2, \end{align*} $$
which gives
$$ \begin{align} c^{z/2}=2^{x-2}+1, \quad b^{y/2}=2^{x-2}-1. \end{align} $$
Since
$b>1$
, we see from the second equality of (3.24) that
$y/2$
is odd. If
${y/2>1}$
, then
$2^{x-2}=b^{y/2}+1=(b+1)(b^{y/2-1}-b^{y/2-2}+\cdots -b+1)$
, where
$b^{y/2-1}-b^{y/2-2}+\cdots -b+1$
is an odd positive integer greater than
$1$
, a contradiction. So we have
$$ \begin{align} \frac{y}{2}=1, \quad b=2^{x-2}-1,\quad x \ge 4. \end{align} $$
Similarly, if
$z/2$
is odd, then from the first equality of (3.24),
$$ \begin{align} \frac{z}{2}=1, \quad c=2^{x-2}+1. \end{align} $$
Hence, by (3.25) and (3.26), we obtain (3.23).
If
$z/2$
is even, then
$2^{x-2}=c^{z/2}-1=(c^{z/4}+1)(c^{z/4}-1)$
, whence
$$ \begin{align} c^{z/4}+1=2^{x-3}, \quad c^{z/4}-1=2. \end{align} $$
Therefore, by (3.25) and (3.27), we obtain (3.22). The lemma is proved.
Here and below, we assume that
$(x_{1},y_{1},z_{1})$
and
$(x_{2},y_{2},z_{2})$
are two distinct solutions of (3.1). We can further assume without loss of generality that
$x_{1} \le x_{2}$
.
Lemma 3.8. We have
$2 \nmid y_{1}y_{2}z_{1}z_{2}$
.
Proof. By Lemmas 3.5, 3.6 and 3.7, if
$2 \mid y_{1}y_{2}z_{1}z_{2}$
, then
$$ \begin{align} (b,c)\in \{(11,5),(17,71),(7,3),(2^{t}-1,2^{t}+1)\},\quad t \in \mathbb{N},~t \ge 2. \end{align} $$
However, by Lemma 2.6, we can eliminate the cases
$(b,c)=(11,5)$
,
$(17,71)$
and
$(7,3)$
. Alternatively, by Lemma 2.7, if
$(a,b,c)=(2,2^{t}-1,2^{t}+1)$
, then (1.1) has only two solutions
$(x,y,z)=(1,1,1)$
and
$(t+2,2,2)$
. Therefore, we can eliminate the cases
$(b,c)=(2^{t}-1,2^{t}+1) (t=2,3,\ldots )$
in (3.28). Thus, the lemma is proved.
Lemma 3.9. We have
$y_{1}z_{2} \ne y_{2}z_{1}$
.
Proof. By Lemmas 3.4 and 3.8,
$\gcd (y_{1},z_{1})=\gcd (y_{2},z_{2})=1$
. Hence, if
$y_{1}z_{2}=y_{2}z_{1}$
, then
$y_{1}\mid y_{2}$
and
$y_{2}\mid y_{1}$
. This implies that
$y_{1}=y_{2}$
,
$z_{1}=z_{2}$
and
$(x_{1},y_{1},z_{1})=(x_{2},y_{2},z_{2})$
, a contradiction. The lemma is proved.
Lemma 3.10. If
$\max \{b,c\}>8\times 10^{6}$
, then
$2^{x_{1}}<c^{0.8z_{1}}$
.
Proof. By (3.1),
$b^{y_{1}}\equiv c^{z_{1}}\pmod {2^{x_{1}}}$
and
$b^{y_{2}}\equiv c^{z_{2}}\pmod {2^{x_{2}}}$
. Since
$x_{1} \le x_{2}$
, we get
$b^{y_{1}y_{2}}\equiv c^{z_{1}y_{2}}\equiv c^{z_{2}y_{1}}\pmod {2^{x_{1}}}$
and
$c^{z_{1}z_{2}}\equiv b^{y_{1}z_{2}}\equiv b^{y_{2}z_{1}}\pmod {2^{x_{1}}}$
. Consequently,
$b^{|y_{1}z_{2}-y_{2}z_{1}|}\equiv c^{|y_{1}z_{2}-y_{2}z_{1}|}\equiv 1 \pmod {2^{x_{1}}}$
. Let
$m =\max \{b,c\}$
. We have
$$ \begin{align} m^{|y_{1}z_{2}-y_{2}z_{1}|}\equiv 1 \pmod{2^{x_{1}}}. \end{align} $$
By Lemmas 3.8 and 3.9,
$|y_{1}z_{2}-y_{2}z_{1}|$
is an even positive integer. Since
$2 \nmid m$
, by Lemma 2.1,
$$ \begin{align} 2^{g(m)+{\mathop{\mathrm{ord}}\nolimits}_{2}|y_{1}z_{2}-y_{2}z_{1}|}\,\|\; m^{|y_{1}z_{2}-y_{2}z_{1}|}-1. \end{align} $$
$$ \begin{align} 2^{g(m)+{\mathop{\mathrm{ord}}\nolimits}_{2}|y_{1}z_{2}-y_{2}z_{1}|} \ge 2^{x_{1}}. \end{align} $$
Further, since
$2^{g(m)}\le m+1$
and
$2^{{\mathop{\mathrm{ord}}\nolimits}_{2}|y_{1}z_{2}-y_{2}z_{1}|}\le |y_{1}z_{2}-y_{2}z_{1}|$
, by (3.31),
$$ \begin{align} (m+1)|y_{1}z_{2}-y_{2}z_{1}|\ge 2^{x_{1}}. \end{align} $$
Furthermore, by Lemma 3.2, if
$2^{x_{1}}>c^{0.8z_{1}}$
, then
$$ \begin{align} |y_{1}z_{2}-y_{2}z_{1}| <\max\{y_{1}z_{2},y_{2}z_{1}\}<14261\times 1236\,(\log b)(\log c) <(4199\log m)^{2}. \end{align} $$
$$ \begin{align} (4199\log m)^{2}(m+1)>2^{x_{1}}. \end{align} $$
Recall that
$c^{z_{1}}>b^{y_{1}}$
,
$2\nmid y_{1}z_{1}$
and
$\min \{y_{1},z_{1}\}\ge 3$
. We have
$c^{z_{1}} \ge m^{3}$
. Therefore, if
$2^{x_{1}}>c^{0.8z_{1}}$
, then from (3.34), we get
$$ \begin{align*} (4199\log m)^{2}(m+1)>m^{2.4}, \end{align*} $$
whence
$m < 8\times 10^{6}$
. Thus, if
$m>8\times 10^{6}$
, then
$2^{x_{1}}<c^{0.8z_{1}}$
.
4 Proof of Theorem 1.2
Obviously, by Lemma 2.8, the theorem holds for
$c=2$
. Moreover, in case a, b and c are distinct primes, we only have to consider
$(a,b,c)=(2,p,q)$
, where p and q are odd primes with
$p \ne q$
. Then, (1.1) can be rewritten as (2.3). Further, by Lemma 2.6, the theorem holds for
$(p,q)=(3,5),~(5,3),~(7,3)$
and
$(11,5)$
.
We now assume that
$N^{\prime }(2,p,q)>1$
. It follows that (2.3) has two solutions
$(x_{1},y_{1},z_{1})$
and
$(x_{2},y_{2},z_{2})$
with
$\min \{x_{j},y_{j},z_{j}\}>1$
for
$j=1,2$
. Since we have excluded the cases
$(p,q)=(3,5),~(5,3),~(7,3)$
and
$(11,5)$
, by Lemma 2.9, we can assume without loss of generality that
$2 \mid x_{1}$
,
$2\nmid x_{2}$
and
$2\nmid y_{1}y_{2}$
. Then
$$ \begin{align} 2^{x_{1}}+p^{y_{1}}=q^{z_{1}},\quad 2^{x_{2}}+p^{y_{2}}=q^{z_{2}}. \end{align} $$
If
$p\ne 3$
, since
$2\nmid y_{1}y_{2}$
and
$p^{y_{1}}\equiv p^{y_{2}}\equiv p \pmod 3$
, then from (4.1),
$$ \begin{align} q^{z_{1}}\equiv 1+p \pmod3,\quad q^{z_{2}}\equiv 2+p\pmod3. \end{align} $$
However, since
$p \ne 3$
,
$1+p\not \equiv 2+p\pmod 3$
and
$3 \mid (1+p)(2+p)$
, (4.2) is false.
If
$p=3$
, by (4.1), then we have
$q^{z_{1}}\equiv 1 \pmod 3$
and
$q^{z_{2}} \equiv 2 \pmod 3$
, whence
$q \equiv 2 \pmod 3$
,
$2\mid z_{1}$
and
$2 \nmid z_{2}$
. Hence, by the first equality of (4.1),
$$ \begin{align} q^{z_{1}/2}+2^{x_{1}/2}=3^{y_{1}},\quad q^{z_{1}/2}-2^{x_{1}/2}=1. \end{align} $$
Eliminating
$q^{z_{1}/2}$
from (4.3), we have
$$ \begin{align} 2^{x_{1}/2+1}=3^{y_{1}}-1. \end{align} $$
However, since
$x_{1}/2+1 \ge 2$
and
$2\nmid y_{1}$
, we get from (4.4) that
$0 \equiv 2^{x_{1}/2+1}\equiv 3^{y_{1}}-1\equiv 3-1\equiv 2 \pmod 4$
, a contradiction. Thus, the theorem is proved.
5 Proof of Theorem 1.3
To show Theorem 1.3, we need the following lemma.
Lemma 5.1. If
$a=2$
,
$f(b)=f(c)$
,
$g(b)\ne g(c)$
and
$\max \{b,c\}>8.4\times 10^{6}$
, then Conjecture 1.1 is true.
Proof. Assume that
$a=2$
,
$f(b)=f(c)$
,
$g(b)\ne g(c)$
,
$\max \{b,c\}>8.4\times 10^{6}$
and
$N^{\prime }(a,b,c)>1$
. Then, by the conclusions of Lemma 3.8 and of Lemma 3.1(ii), (3.1) has two distinct solutions
$(x_{1},y_{1},z_{1})$
and
$(x_{2},y_{2},z_{2})$
with
$$ \begin{align} x_{1}=x_{2}. \end{align} $$
Since
$(x_{1},y_{1},z_{1})\ne (x_{2},y_{2},z_{2})$
, by (5.1), we may assume without loss of generality that
$$ \begin{align} y_{1}<y_{2},\quad z_{1}<z_{2}. \end{align} $$
Let
$$ \begin{align} \Lambda_{j}=z_{j}\log c-y_{j}\log b,\quad j=1,2. \end{align} $$
We see from (3.6) and (5.3) that
$$ \begin{align} 0<\Lambda_{j}=\log\bigg(1+\frac{2^{x_{j}}}{b^{y_{j}}}\bigg),\quad j=1,2. \end{align} $$
Further, by (5.1), (5.2) and (5.4),
$$ \begin{align} 0<\Lambda_{2}<\Lambda_{1}=\log\bigg(1+\frac{2^{x_{1}}}{b^{y_{1}}}\bigg). \end{align} $$
Furthermore, by Lemma 3.10, we have
$2^{x_{1}}<c^{0.8z_{1}}$
. Hence, by (3.6) and (5.5),
$$ \begin{align} 0<\Lambda_{2}<\Lambda_{1}<\frac{2}{c^{z_{1}/5}}. \end{align} $$
However, by Lemmas 3.8 and 3.9,
$|y_{1}z_{2}-y_{2}z_{1}|$
is an even positive integer. So,
$$ \begin{align} |y_{1}z_{2}-y_{2}z_{1}| \ge 2. \end{align} $$
By (5.3),
$$ \begin{align} |y_{1}z_{2}-y_{2}z_{1}|&=\frac{1}{\log c}|y_{1}(z_{2}\log c)-y_{2}(z_{1}\log c)|\notag\\ &=\frac{1}{\log c}|y_{1}(y_{2}\log b+\Lambda_{2})-y_{2}(y_{1}\log b+\Lambda_{1})|\notag\\ &=\frac{1}{\log c}|y_{1}\Lambda_{2}-y_{2}\Lambda_{1}|. \end{align} $$
Since
$y_{1} \Lambda _{2}>0$
and
$y_{2}\Lambda _{1}>0$
by (5.4), we see from (5.8) that
$$ \begin{align} |y_{1}z_{2}-y_{2}z_{1}|<\max\bigg\{\frac{y_{1}\Lambda_{2}}{\log c},\frac{y_{2}\Lambda_{1}}{\log c}\bigg\}. \end{align} $$
Further, by (5.2) and Lemma 3.2,
$$ \begin{align} \frac{y_{1}}{\log c}<\frac{y_{2}}{\log c}<14261. \end{align} $$
Hence, by (5.6), (5.7), (5.9) and (5.10),
$$ \begin{align*} 2 \le |y_{1}z_{2}-y_{2}z_{1}|<\max\{14261\Lambda_{2},14261\Lambda_{1}\}=14261\Lambda_{1}<\frac{2\times 14261}{c^{z_{1}/5}}, \end{align*} $$
whence we obtain
$$ \begin{align} c^{z_{1}/5}<14261. \end{align} $$
However, since
$\max \{b,c\}>8.4\times 10^{6}$
and
$c^{z_{1}}\ge (\max \{b,c\})^{3}$
, (5.11) is false. Thus, we have
$N^{\prime }(a,b,c)\le 1$
if
$a=2$
,
$f(b)=f(c)$
,
$g(b)\ne g(c)$
and
$\max \{b,c\}>8.4\times 10^{6}$
.
We are now ready to prove Theorem 1.3.
Proof of Theorem 1.3.
Obviously, by the conclusion of Lemma 3.1(i), Conjecture 1.1 is true if
$a=2$
and
$f(b)\ne f(c)$
. We now assume that
$a=2$
,
$f(b)=f(c)$
,
$g(b)\ne g(c)$
and
$N^{\prime }(a,b,c)>1$
. Moreover, by Lemma 5.1, we may assume that
$$ \begin{align} \max\{b,c\} < 8.4\times 10^{6}. \end{align} $$
Then, by the conclusion of Lemma 3.1(ii), (3.1) has two distinct solutions
$(x_{1},y_{1},z_{1})$
and
$(x_{2},y_{2},z_{2})$
with
$$ \begin{align} x_{1}=x_{2}=\min\{g(b),g(c)\}, \end{align} $$
and we may assume that
$y_{1}<y_{2}$
and
$z_{1}<z_{2}$
. Further, by (5.12), we have
$2^{g(b)}\le b+1 \le 8.4\times 10^{6}$
, which together with (5.13) implies that
$$ \begin{align} x_{1}=x_{2}=\min\{g(b),g(c)\}\le 23. \end{align} $$
It follows from Lemma 3.3 that
$$ \begin{align} \frac{y_{1}}{\log c}<\frac{y_{2}}{\log c}<2530, \quad \frac{z_{1}}{\log b}<\frac{z_{2}}{\log b}<2530. \end{align} $$
Furthermore, by (3.16) and (5.14),
$$ \begin{align} 0<\Lambda_{1}:=z_{1}\log c-y_{1}\log b<\frac{2^{x_{1}+1}}{c^{z_{1}}}\le \frac{2^{24}}{c^{z_{1}}}. \end{align} $$
Thus, by the same argument as the proof of Lemma 5.1, we see from (5.15) and (5.16) that
$$ \begin{align*} 2 \le|y_{1}z_{2}-y_{2}z_{1}|<2530\Lambda_{1}<\frac{2^{24}\times 2530}{c^{z_{1}}}, \end{align*} $$
whence we obtain
$$ \begin{align*} b^{y_{1}}<c^{z_{1}}<2^{23}\times 2530 < 2.123\times 10^{10}. \end{align*} $$
Consequently, it only remains to show that (3.1) has no solutions if
$$ \begin{align*} 3 \le b \le 2767,~3\le c \le 2767,~2\le x_{1} \le 23,~3\le y_{1} \le 21,~3\le z_{1} \le 21 \end{align*} $$
with
$2 \nmid bcy_{1}z_{1}$
(by Lemma 3.8). We checked that the above claim is true by a simple program in PARI/GP [23] with precision
$100$
. Indeed, the result showed that for any c,
$x_{1}$
,
$y_{1}$
,
$z_{1}$
in the above ranges, the fractional part of
$(c^{z_{1}}-2^{x_{1}})^{1/y_{1}}$
is greater than
$10^{-6}$
. The computation time was within
$1$
minute. Thus, the theorem is proved.
Acknowledgement
The authors thank the referee for reading the draft very carefully and putting forward some valuable suggestions.
