2. Tools from Diophantine approximation

Henceforth, we frequently use the Vinogradov notation $f \ll_{\kappa_1,\kappa_2,\ldots,\kappa_n} g$ which means that $|f/g|$ is less than some positive constant depending only on $\kappa_1,\kappa_2,\ldots$ and $\kappa_n$ , where we simply write $f \ll g$ if the implied constant is absolutely finite. Note that the implied constant of each Vinogradov notation appearing below is effectively computable unless otherwise referred to.

Let $P[\mathcal A]$ denote the greatest prime factor of a nonzero integer $\mathcal A$ , with the convention that $P[1]=P[-1]=1$ .

Let $f(t_1,\ldots,t_n)$ be a real-valued function defined on an unbounded region in $\mathbb R^n$ . We write

\begin{equation*}\lim_{\max\{\,|t_1|,\ldots,|t_n|\,\} \to \infty} f(t_1,\ldots,t_n) = \infty, \ \ \text{effectively},\end{equation*}

if, for any $M \gt 0$ , there exists some number T which depends only on M and is effectively computable such that $f(t_1,\ldots,t_n) \gt M$ whenever $\max\{\,|t_1|,\ldots,|t_n|\,\} \gt T$ .

The following is a direct consequence of [ Reference ShoreyShTi , corollary 8·1].

Proposition 2·1 Let m and n be positive integers with $m \gt 1$ and $n \gt 1$ such that $\max\{m,n\} \ge 3.$ Let $\mathcal X$ and $\mathcal Y$ be variables as a pair of nonzero relatively prime integers. Then

\begin{equation*}\lim_{\max\{\,|{\mathcal X}|,\,|{\mathcal Y}|\,\} \to \infty} P[{\mathcal X}^m+{\mathcal Y}^n] = \infty, \ \ \text{effectively}.\end{equation*}

For any positive integer $M \gt 1$ , we let $\nu_M(\mathcal A)$ denote the M-adic valuation of a nonzero integer $\mathcal A$ , namely, the highest exponent e such that $M^e$ divides $\mathcal A$ . Further, for any nonzero rational number x, we set $\nu_M(x)\,{:\!=}\,\nu_M(p)-\nu_M(q)$ , where p and q are relatively prime integers such that $x=p/q$ .

For any algebraic number $\gamma$ , we define the absolute logarithmic height ${\rm h}(\gamma)$ of $\gamma$ as follows:

\begin{equation*}{\rm h}(\gamma) =\frac{1}{[\mathbb{Q}(\gamma):\mathbb{Q}]}\,\Bigl(\, \log c_0 \,+ \,\sum\, \log\, \max \{\,1,\,| \gamma' |\,\} \,\Bigl),\end{equation*}

where $c_0 \gt 0$ is the leading coefficient of the minimal polynomial of $\gamma$ over $\mathbb Z$ , and the sum extends over all conjugates $\gamma'$ of $\gamma$ in the field of complex numbers.

The following is a simple consequence of Bugeaud [ Reference BugeaudBu2 , theorem 1·14] with the choice $(\mu,c_2(\mu))=(4,53.6)$ .

Proposition 2·2 Let M be an integer with $M \gt 1.$ Let $\alpha_1$ and $\alpha_2$ be nonzero rational numbers such that $\nu_q(\alpha_1)=0$ and $\nu_q(\alpha_2)=0$ for any prime factor q of M. Assume that $\alpha_1$ and $\alpha_2$ are multiplicatively independent. Let g be a positive integer with $\gcd({\rm g},M)=1$ such that

\begin{gather*} \nu_{q}( {\alpha_1}^{{\rm g}}-1 ) \ge \nu_{q}(M), \nu_{q}( {\alpha_2}^{{\rm g}}-1 ) \ge 1\end{gather*}

for any prime factor q of M. If M is even, then further assume that

\begin{equation*} \nu_{2}( {\alpha_j}^{{\rm g}}-1 ) \ge 2 \quad (j=1,2).\end{equation*}

Let $H_1$ and $H_2$ be positive numbers such that

\begin{equation*}H_j \ge \max \{ \,{\rm h}(\alpha_j),\,\log M\, \} \quad (j=1,2).\end{equation*}

Then, for any positive integers $b_1$ and $b_2$ with $\gcd(b_1,b_2,M)=1,$

\begin{equation*}\nu_M( {\alpha_1}^{b_1} - {\alpha_2}^{b_2}) \leq \frac{53.6 \, {\rm g}\, H_1 H_2}{\log^4 M}\, \bigl(\max \{\, \log b^{\ast}+\log \log M+0.64,\,4\log M\, \}\bigl)^2\end{equation*}

with $b^{\ast}=b_1/H_2+b_2/H_1.$

For a number field $\mathbb K$ and a prime ideal $\pi$ in $\mathbb K$ , we let $\nu_{\pi}(\alpha)$ denote the exponent of $\pi$ in the prime ideal factorisation of the fractional ideal generated by a nonzero element $\alpha$ in $\mathbb K$ .

The following is a simple consequence of Bugeaud and Laurent [ Reference Bugeaud and LaurentBuLa , héorème 4] with the choice of $(\mu,\nu)=(8,10)$ and $A_j\,(j=1,2)$ chosen so $H_j=(D \log A_j) /f_\pi$ .

Proposition 2·3 Let $\mathbb K$ be a number field. Let $\pi$ be a prime ideal in $\mathbb K,$ and p the rational prime lying above $\pi.$ Let $\alpha_1$ and $\alpha_2$ be nonzero elements in $\mathbb K$ such that the fractional ideal generated by $\alpha_1 \alpha_2$ is not divisible by $\pi.$ Assume that $\alpha_1$ and $\alpha_2$ are multiplicatively independent. Let g be a positive integer such that

\begin{equation*}\nu_{\pi}({\alpha_j}^{\rm g} - 1) \ge 1 \quad (j=1,2).\end{equation*}

Let $H_1$ and $H_2$ be positive numbers such that

\begin{equation*}\quad H_j \ge \max \biggl\{ \frac{D}{f_{\pi}}\,{\rm h}(\alpha_j), \, \log p \biggl\} \quad (j=1,2),\end{equation*}

where $D=[\mathbb Q(\alpha_1,\alpha_2):\mathbb Q]$ and $f_{\pi}$ is the inertia index of $\pi.$ Then, for any positive integers $b_1$ and $b_2,$

\begin{align*}\nu_{\pi}({\alpha_1}^{b_1}-{\alpha_2}^{b_2}) &\leq\frac{27.3 \, D^2 p\, {\rm g} \, H_1 H_2}{{f_{\pi}}^2(p-1)(\log p)^4}\\&\quad{}\times \Bigl(\max \Bigl\{ \log b^{\ast}+\log \log p+0.4,\,\textstyle{\frac{8 f_{\pi}}{D}}\log p,\,10 \Bigl\} \Bigl)^2\end{align*}

with $b^{\ast}=b_1/H_2+b_2/H_1.$

The following is a particular case of a generalisation of Thue–Siegel–Roth theorem due to Ridout [ Reference RidoutRid ].

Proposition 2·4 Let $\xi$ be a nonzero algebraic number. Let $\mathcal S$ be a non-empty set of finitely many primes. Then, for every positive number $\varepsilon,$ there exists some positive constant C depending only on $\xi,\mathcal S$ and $\varepsilon$ such that the inequality

\begin{equation*}\left|\xi - \frac{p}{q}\right| \gt \frac{C}{q^{1+\varepsilon}}\end{equation*}

holds for all integers p and q with $q \gt 0$ and $p/q \ne \xi$ such that every prime factor of q belongs to $\mathcal S.$

For any nonzero $x \in \mathbb Q$ , we write $|x|_p=1/p^{\nu_{p}(x)}$ for each prime p. The product formula says that $\prod_{\mu \in \mathbb P \,\cup \,\{\infty\}} |x|_{\mu}=1$ , where $|x|_\infty=|x|$ and $\mathbb P$ denotes the set of all prime numbers.

The following is a simplified form of Schmidt Subspace theorem (cf. [ Reference SchmidtSc , Reference SchmidtSc2 ]).

Proposition 2·5 Let N be a positive integer. Let $\mathcal S$ be a finite set of primes. For each $\mu \in \mathcal S \cup \{\infty\},$ let $L_{\mu,1},\ldots,L_{\mu,N}$ be N linear forms in N variables with rational coefficients, and assume that they are linearly independent. Then, for every positive number $\varepsilon$ with $\varepsilon \lt 1,$ there exist finitely many proper subspaces of the $\mathbb Q$ vector space $\mathbb Q^N$ such that any nonzero element $\boldsymbol{x}=(x_1,\ldots,x_N)$ in $\mathbb Q^N$ with $x_i \in \mathbb Z$ for any i satisfying the inequality

\begin{equation*}\prod_{\mu \in \mathcal S \cup \{\infty\}} \,\prod_{i=1}^{N}\,|L_{\mu,i}(\boldsymbol{x})|_\mu\leq \frac{1}{(\,\max\{\,|x_i|\,:\, i=1,\ldots,N\,\}\,)^\varepsilon}\end{equation*}

belongs to one of those subspaces.

3. Reducing to weak forms

For a positive integer M, we define $e_M(\mathcal A)$ to be the extended multiplicative order modulo M of a nonzero integer $\mathcal A$ being coprime to M, namely, $e_M(\mathcal A)$ equals the least positive integer e such that $\mathcal A^e$ is congruent to 1 or $-1$ modulo M. The extended multiplicative orders have properties similar to those of the usual multiplicative orders (cf. [ Reference MiyazakiMiyPin2 , lemma 2·1]). In particular, $e_M(\mathcal A)$ is a divisor of $\varphi(M)$ .

The following lemma found in [ Reference MiyazakiMiyPin2 , lemma 2·3] tells us that proving Conjecture 1·3 is reduced to studying a specialisation of it.

Lemma 3·1 Let d be a positive divisor of c with $d \gt 2.$ Define positive integers A and B by

\begin{equation*}A=a^{\,e_d(a)/g}, \quad B=b^{\,e_d(b)/g},\end{equation*}

where $g=\gcd(\,e_d(a),e_d(b)\,).$ Then the following hold.

1. (i) The number of solutions (x,y,z) to equation (1·5) equals that to the equation

   \begin{equation*}A^X+B^Y=c^Z\end{equation*}
   in positive integers X,Y and Z. Further, its correspondence is given by
   \begin{equation*}X=\frac{x}{e_{d}(A)/g}, \ Y=\frac{y}{e_{d}(B)/g}, \ Z=z.\end{equation*}

2. (ii) $e_{d}(A)=e_{d}(B).$

3. (iii) (a,b,c) belongs to set (1·7) if and only if (A,B,c) belongs to the same set.

Based on this lemma, in all situations in the remaining except for the first part of the next section, for some proper divisor d of c, in equation (1·5), we actually consider (A,B,c,X,Y,z) instead of (a,b,c,x,y,z). In particular, we consider equation (1·5) for the triples (a,b,c) such that $e_{d}(a)=e_{d}(b)$ . It is enough for our purposes to do so, because the following straightforward properties hold.

\begin{gather*}\max\{a,b\} \gt \mathcal C \ \Rightarrow \ \max\{A,B\} \gt \mathcal C\,;\\X,Y \ll_{c}1 \ \Rightarrow \ x,y \ll_{c}1\,;\\(A,B,13) \text{ belongs to set} {(1\!\cdot\!7)} \ \Leftrightarrow \ \{a,b\}=\{3,10\},\end{gather*}

where $\mathcal C$ is any positive number. This restriction is basically not essential in the study of Conjecture 1·3, but it will turn out that it is helpful in the proof of Theorem 1·9 (cf. Section 9).

4. Cases for general values of c

The numbers $\mathcal C_i\,(i=1,2,\ldots,5)$ and ${\mathcal C_5}'$ appearing in this or the next section are some positive constants which depend only on c and are effectively computable.

In this section, we give several finiteness results on the solutions to equation (1·5) for general values of c. In equation (1·5), it is clear that the implication

\begin{equation*}z \ll_c 1 \ \Rightarrow \ \max\{a,b\} \ll_c 1\end{equation*}

holds, and this will be frequently and implicitly used below.

Write $x=p x_0,\,y=p y_0$ . If $p=2$ , then c is even, so that both a,b are odd, leading to $c^z=(a^{x_0})^2+(b^{y_0})^2 \equiv 2 \pmod{4}$ , whereby $z=1$ . For odd p, one considers the greatest prime factor of each of both sides of equation (1·5) to see that

\begin{equation*}P[{\mathcal X}^m+{\mathcal Y}^n]=P[c],\end{equation*}

where $(m,n)\,{:\!=}\,(p,p)$ and $({\mathcal X},{\mathcal Y})\,{:\!=}\,(a^{x_0},b^{y_0})$ . For each possible p, we shall apply Proposition 2·1. Since $P[c] \ll_c 1$ , it turns out that $\mathcal X,\mathcal Y \ll_{c} 1$ , so that $a,b \ll_c 1$ .

Since Conjecture 1·11 is true for $c=2$ by Scott [ Reference ScottSco , theorem 6; $p=2$ ] (or [ Reference MiyazakiMiyPin2 , corollary 1]), we may assume that c is not a power of 2. Thus there is some divisor $c_1$ of c with $c_1 \gt 2$ such that

(4·1) \begin{equation}\gcd(\,c_1,\varphi(c_1)\,)=1.\end{equation}

Note that $c_1$ is odd.

Lemma 4·2 Let (x,y,z) be a solution to equation $(1.5)$ such that $\gcd(x,y,c_1)=1.$ Then

\begin{equation*}z \ll \max\{1,F_1\} \cdot e \,\log a \,\log b,\end{equation*}

where

\begin{equation*}F_1=F_1(c,c_1)=\max\biggr\{1,\,\frac{\log^2 (\log c\,\log c_1)}{ \log^2 {c_1} }\biggr\} \cdot \frac{1}{\nu_{c_1}(c)\,\log^2 {c_1}},\end{equation*}

and $e=e_{c_1}(a) \cdot e_{c_1}(b) \cdot \mathrm{lcm}(\,e_{c_1}(a),e_{c_1}(b)\,).$

Proof. By equation (1·5),

\begin{equation*}\nu_{c_1}(a^{2x}-b^{2y}) \ge \nu_{c_1}(a^x+b^y) = \nu_{c_1}(c^z) \ge \nu_{c_1}(c)\cdot z.\end{equation*}

To find an upper bound for the left-most side above, we shall apply Proposition 2·2 for

\begin{equation*}M\,{:\!=}\,c_1, \ (\alpha_1,\alpha_2)\,{:\!=}\,(a,b), \ (b_1,b_2)\,{:\!=}\,(2x,2y).\end{equation*}

Since, for each $h \in \{a,b\}$ , one finds that $h^{2\,e_{c_1}(h)} \equiv (h^{e_{c_1}(h)})^2 \equiv (\pm1)^2 \equiv 1 \pmod{c_1}$ , and that $\gcd(\,2\,e_{c_1}(h),c_1\,)=\gcd(\,e_{c_1}(h),c_1\,)=1$ by (4·1) with $e_{c_1}(h) \mid \varphi(c_1)$ , one may set ${\rm g}\,{:\!=}\,2\,\mathrm{lcm}(\,e_{c_1}(a),e_{c_1}(b)\,)$ . Observe that $\gcd(b_1,b_2,M)=\gcd(2x,2y,c_1)=\gcd(x,y,c_1)=1$ by assumption. Putting

\begin{equation*}H_1\,{:\!=}\,\log a', \ H_2\,{:\!=}\,\log b',\end{equation*}

where $a'=\max\{a,c_1\}$ and $b'=\max\{b,c_1\}$ , one obtains

\begin{equation*}\nu_{c_1}(a^{2x}-b^{2y}) \ll \frac{\mathrm{lcm}(\,e_{c_1}(a),e_{c_1}(b)\,)\,\log a'\,\log b'}{\log^4 {c_1}} \cdot \mathcal B^2,\end{equation*}

where

\begin{equation*}\mathcal B=\max \biggr\{ \log \biggr( \frac{2x}{\log b'}+\frac{2y}{\log a'} \biggr)+\log \log c_1, \ \log c_1 \biggr\}.\end{equation*}

By the trivial inequalities $x \lt {(\log c)}/{(\log a)}\,z,\, y \lt {(\log c)}/{(\log b)}\,z$ from equation (1·5), observe that

\begin{equation*}\log \biggr( \frac{2x}{\log b'}+\frac{2y}{\log a'} \biggr)\leq \log \biggr( \frac{2x}{\log b}+\frac{2y}{\log a} \biggr) \lt \log \biggr( \frac{4\log c}{\log a\,\log b}\,z \biggr).\end{equation*}

This together with the two bounds for $\nu_{c_1}(a^{2x}-b^{2y})$ implies

\begin{equation*}T \ll \frac{1}{\nu_{c_1}(c)} \cdot \frac{\log a'}{\log a} \cdot \frac{\log b'}{\log b} \cdot \frac{\mathrm{lcm}(\,e_{c_1}(a),e_{c_1}(b)\,)}{\log^4 {c_1}} \cdot {\mathcal B'}^2,\end{equation*}

where

\begin{equation*}T\,{:\!=}\,\frac{4\,z}{\,\log a\,\log b\,}, \quad \mathcal B'\,{:\!=}\,\log \,\max \bigr\{ \,(\log c\,\log c_1)\,T ,\,c_1\,\bigr\}.\end{equation*}

For each $h \in \{a,b\}$ , since ${c_1} \lt h^{2\,e_{c_1}(h)}$ as $h^{2\,e_{c_1}(h)} \equiv 1 \pmod{c_1}$ , it follows that

\begin{equation*}\frac{\log h'}{\log h} = \max\biggl\{1,\,\frac{\log c_1}{\log h}\biggr\} \leq 2\,e_{c_1}(h),\end{equation*}

so that

\begin{equation*}T \ll \frac{e}{\nu_{c_1}(c)\,\log^4 {c_1}} \cdot {\mathcal B'}^2.\end{equation*}

If $(\log c\,\log c_1)\,T \leq c_1$ , then $\mathcal B' =\log c_1$ , so that

(4·2) \begin{equation}T \ll \frac{e}{\nu_{c_1}(c)\,\log^2 {c_1}}.\end{equation}

Finally, suppose that $(\log c\,\log c_1)\,T \gt c_1$ . Then

\begin{equation*}T \ll \frac{e}{\nu_{c_1}(c)\,\log^4 {c_1}} \cdot \log^2 \bigr(\,(\log c\,\log c_1)\,T\,\bigr).\end{equation*}

If $\log c\,\log c_1 \gt T$ , then

(4·3) \begin{equation}T \ll \frac{\log^2 (\log c\,\log c_1)}{ \log^2{c_1} }\cdot \frac{e}{\nu_{c_1}(c)\,\log^2 {c_1}}.\end{equation}

While if $\log c\,\log c_1 \leq T$ , then putting

\begin{equation*}R\,{:\!=}\,\frac{e}{\nu_{c_1}(c)\,\log^4 {c_1}},\end{equation*}

one has

(4·4) \begin{equation}\quad \frac{T}{\log^2 T} \lt u\,R\end{equation}

for some constant u with $1 \leq u \ll 1$ . If $R \lt 3$ , then inequality (4·4) implies that

(4·5) \begin{equation}T \ll 1.\end{equation}

In the remaining parts, we assume that $R \ge 3$ . If $T \ge 4u R \log^2 R\,( \gt \exp(2))$ , then replacing the number T in (4·4) by $4u R \log^2 R$ implies that

\begin{equation*}\frac{R}{\log^2 R} \lt 4u,\end{equation*}

so that $R \ll 1$ , leading to the same as (4·5) by (4·4). If $T \lt 4u R \log^2 R$ , then

\begin{equation*}T \ll R \log^2 R=\frac{e}{\nu_{c_1}(c)\,\log^4 {c_1}} \cdot \log^2 \biggr( \frac{e}{\nu_{c_1}(c)\,\log^4 {c_1}} \biggr).\end{equation*}

Since $e \leq \varphi(c_1)^4 \lt {c_1}^4$ , the above implies the same as (4·2). The combination of (4·2), (4·3), (4·5) completes the proof.

From now on, suppose that equation (1·5) has two solutions, say (x,y,z) and (X,Y,Z), with $(x,y,z) \ne (X,Y,Z)$ . We have

\begin{eqnarray*}& a^x+b^y=c^z,\\& a^X+b^Y=c^Z.\end{eqnarray*}

There is no loss of generality in assuming that $z \leq Z$ . The above two equations are referred to as the 1st and 2nd equation, respectively.

In what follows, we put

\begin{equation*}\ \ \Delta\,{:\!=}\,|xY-Xy| \ \ ( \gt 0).\end{equation*}

Note that $\Delta$ is nonzero in general (cf. [ Reference HuHuLe , lemma 3·3]).

It holds trivially that

(4·6) \begin{equation}x \lt \frac{\log c}{\log a}\,z, \ \ y \lt \frac{\log c}{\log b}\,z, \ \ X \lt \frac{\log c}{\log a}\,Z, \ \ Y \lt \frac{\log c}{\log b}\,Z.\end{equation}

Based on the contents of the previous section, from now on, we assume that $e_{c_1}(a)=e_{c_1}(b)$ (however, this can allow a or b to be a perfect power). We let $E_1$ be its common value, namely,

\begin{equation*}E_1\,{:\!=}\, e_{c_1}(a)=e_{c_1}(b).\end{equation*}

Recall that there are some properties on $E_1$ similar to those of the usual multiplicative orders modulo $c_1$ , in particular, $E_1$ is a divisor of $\varphi(c_1)$ .

For each $h \in \{a,b\}$ , since $c_1 \gt 2$ , we can uniquely define $\delta_h \in \{1,-1\}$ by the following congruence:

(4·7) \begin{equation}h^{E_1} \equiv \delta_h \mod{c_1}.\end{equation}

The following lemma is a well-known fact in p-adic calculations.

Lemma 4·3 Let p be a prime number. Let U and V be relatively prime nonzero integers. Assume that

\begin{equation*}\begin{cases}\, U \equiv V \pmod{p} & \text{if $p \ne 2$},\\\, U \equiv V \pmod{4} & \text{if $p=2$}.\end{cases}\end{equation*}

Then

\begin{equation*}\nu_p(U^N-V^N)=\nu_p(U-V)+\nu_p(N)\end{equation*}

for any positive integer N.

Let p be any prime factor of $c_1$ . Note that p is odd. It follows from (4·8) that

\begin{equation*}\nu_p(h^{\Delta}-{\delta_h}^{\Delta/E_1}) \ge \nu_p(c_1) \cdot z.\end{equation*}

From congruence (4·7) observe that $h^{E_1} \equiv \delta_h \pmod{p}$ . To calculate the left-hand side of the above displayed inequality, lemma 4·3 is applied for $(U,V)=(h^{E_1},\delta_h)$ and $N=\Delta/E_1$ . It turns out that

\begin{equation*}\nu_p(h^{E_1}-\delta_h)+\nu_p(\Delta/E_1) \ge \nu_p(c_1) \cdot z.\end{equation*}

Since p is chosen arbitrarily, it follows that $(h^{E_1}-\delta_h) \cdot \Delta/E_1$ is divisible by ${c_1}^z$ , and the assertion holds.

Lemma 4·5 If $\max\{a,b\} \gt \mathcal C_1,$ then

\begin{equation*}\Delta \ll F_2\,z,\end{equation*}

where $F_2=F_2(c,c_1,E_1)=\max\{1,F_1\} \cdot{E_1}^3 \log^2 c.$

Proof. Since $\Delta=|x Y-X y| \lt \max\{x Y,X y\}$ , it follows from inequalities (4.6) that

(4·9) \begin{equation} \Delta \lt \frac{\log^2 c}{\log a\,\log b}\,z\,Z.\end{equation}

This together with the upper bound for Z obtained from lemmas 4·1 and 4.2 on the solution (X,Y,Z) implies the assertion.

The above lemma in particular says that the magnitude of $\Delta$ as a function of z is ‘very small’ relative to that of ${c_1}^z$ appearing in the divisibility relation of lemma 4·4 (iii). This observation is the most important one in our previous work [ Reference MiyazakiMiyPin2 ], and also for the present one. Based on this, we put

\begin{equation*}\mathcal D_1\,{:\!=}\,\frac{{c_1}^z}{\Delta'},\end{equation*}

where

\begin{equation*}\Delta'\,{:\!=}\,\gcd(\Delta/E_1,{c_1}^z).\end{equation*}

It turns out from lemma 4·4 (iii) that

(4·10) \begin{equation} h^{E_1} \equiv \delta_h \mod{\mathcal D_1}\end{equation}

for each $h \in \{a,b\}$ .

In what follows, we frequently use the following inequalities:

\begin{equation*}c_1 \ll_{c} 1, \ \ E_1 \ll_{c} 1.\end{equation*}

Proof. We may assume that $\max\{a,b\} \gt \mathcal C_1$ . lemma 4·5 gives that $\Delta' \leq \Delta/E_1 \ll (F_2/E_1) z$ , so that

\begin{align*}\\[-2pc]\Delta' \lt \eta \,z\end{align*}

for some $\eta \gt 0$ which depends only on c and is effectively computable. Then

\begin{equation*}\log \mathcal D_1 = z \log c_1 - \log \Delta' \gt z \log c_1 - \log (\eta z),\end{equation*}

so that

\begin{equation*}\frac{\log \mathcal D_1}{z} \gt \log c_1 - \frac{\log (\eta z)}{z} \gg \log c_1,\end{equation*}

whenever z exceeds some constant depending only on $\eta$ . This completes the proof.

The following lemma is the most important step in the proofs of our results.

Lemma 4·7 If $\max\{a,b\} \gt \mathcal C_3,$ then

\begin{equation*}z\,Z \ll \frac{1}{\nu_{c_1}(c) \log^2 c_1} \, {E_1}^3\,\log a\,\log b.\end{equation*}

First, from the 2nd equation, observe that

\begin{equation*}\nu_{\mathcal D_1}(a^{2X}-b^{2Y}) \ge \nu_{\mathcal D_1}(a^X+b^Y)=\nu_{\mathcal D_1}(c^Z) \ge \nu_{{c_1}^z}(c^Z) \ge \nu_{c_1}(c) \cdot \lfloor Z/z \rfloor.\end{equation*}

To find an upper bound for the left-most side above, we shall apply Proposition 2·2 with the same values of $\alpha_1,\alpha_2,b_1,b_2$ as those in the proof of lemma 4·2 (on the solution (X,Y,Z)). In this case, we set $M\,{:\!=}\,\mathcal D_1$ , and all required conditions for this choice are satisfied by condition (4·1), congruence (4·10) and lemma 4·1. Then

\begin{equation*}\nu_{\mathcal D_1}(a^{2X}-b^{2Y}) \ll \frac{E_1 \log a'\, \log b'}{\log^4 \mathcal D_1} \cdot \mathcal B^2,\end{equation*}

where $a'=\max\{a,\mathcal D_1\},\, b'=\max\{b,\mathcal D_1\}$ and

\begin{equation*}\mathcal B=\max \biggr\{ \log \biggr( \frac{2X}{\log b'}+\frac{2Y}{\log a'} \biggr)+\log \log \mathcal D_1, \ \log \mathcal D_1 \biggr\}.\end{equation*}

Since $\lfloor Z/z \rfloor \gg Z/z$ , and

\begin{equation*}\log \biggr( \frac{2X}{\log b'}+\frac{2Y}{\log a'} \biggr) \lt \log \biggr( \frac{4\log c}{\log a\,\log b}\,Z \biggr),\end{equation*}

the two bounds for $\nu_{\mathcal D_1}(a^{2X}-b^{2Y})$ together imply

\begin{equation*}T \ll \frac{1}{\nu_{c_1}(c)} \cdot \frac{\log a'}{\log a} \cdot \frac{\log b'}{\log b} \cdot \frac{E_1\,z^2}{\log^4 \mathcal D_1} \cdot {\mathcal B'}^2,\end{equation*}

where

\begin{align*}\\[-2pc]T\,{:\!=}\,\frac{4\,z\,Z}{\,\log a\,\log b\,}, \quad \mathcal B'\,{:\!=}\,\log\,\max \bigr\{\, (\log c)(\log \mathcal D_1)\,T/z, \,\mathcal D_1\,\bigr\}.\end{align*}

Note that $\log \mathcal D_1 \gg (\log c_1)\,z$ by lemma 4·6. Since $\mathcal D_1 \lt a^{2E_1},b^{2E_1}$ by congruence (4·10), one has

\begin{align*}\\[-2pc] T \ll \frac{{E_1}^3 z^2}{\nu_{c_1}(c)\,\log^4 \mathcal D_1} \cdot {\mathcal B'}^2.\end{align*}

If $(\log c)(\log \mathcal D_1)\,T/z \leq \mathcal D_1$ , then ${\mathcal B}'=\log {\mathcal D}_1$ , so that

\begin{equation*}T \ll \frac{{E_1}^3 z^2}{\nu_{c_1}(c)\,\log^2 \mathcal D_1} \ll \frac{{E_1}^3}{\nu_{c_1}(c) \log^2 c_1},\end{equation*}

which gives the asserted inequality. Finally, suppose that $(\log c)(\log \mathcal D_1)\,T/z \gt \mathcal D_1$ . Then

\begin{equation*}\frac{\mathcal D_1}{(\log c)\log \mathcal D_1} \lt T/z \ll \frac{{E_1}^3\,z}{\nu_{c_1}(c)\,\log^4 \mathcal D_1} \cdot \log^2\bigr(\,(\log c)(\log \mathcal D_1)\,T/z \,\big).\end{equation*}

If $(\log c)\log \mathcal D_1 \gt T/z$ , then

\begin{equation*}\frac{\mathcal D_1}{\log \mathcal D_1} \ll_c T/z \ll_c \log \mathcal D_1,\end{equation*}

so that $\mathcal D_1 \ll_c 1$ , whereby $z \ll \log \mathcal D_1 \ll_c 1$ . While if $(\log c)\log \mathcal D_1 \leq T/z$ , then

\begin{equation*}\frac{\mathcal D_1}{\log \mathcal D_1} \ll_c T/z \ll_c \frac{z}{\log^4 \mathcal D_1} \cdot \log^2(T/z).\end{equation*}

Since $z \ll \log \mathcal D_1$ , it is not hard to see from the right-hand inequality above that $T/z \ll_c 1$ , so that the left-hand one yields $\mathcal D_1 \ll_c 1$ , and $z \ll_c 1$ . This completes the proof.

Lemma 4·8 If $\max\{a,b\} \gt \mathcal C_3,$ then

\begin{equation*}\max\{x,y,X,Y\} \ll {F_3}^2{E_1}^6 \ll_{c} 1,\end{equation*}

where $F_3=F_3(c,c_1)=({1}/{\nu_{c_1}(c)}) ({\log^2 c}/{\log^2 c_1}).$

Proof. Lemma 4·7 together with inequalities (4.6) yields

\begin{align*}z\,Z \ll F_3 {E_1}^3 \cdot \frac{\log a}{\log c} \cdot \frac{\log b}{\log c}& \lt F_3 {E_1}^3 \cdot \min \biggr\{\dfrac{z}{x},\dfrac{Z}{X} \biggr\} \cdot \min \biggr\{\dfrac{z}{y},\dfrac{Z}{Y} \biggr\}\\& \leq F_3 {E_1}^3 \cdot \min \biggr\{\dfrac{z^2}{xy},\dfrac{Z^2}{XY} \biggr\}.\end{align*}

This implies that

\begin{equation*}X Y \ll F_3 {E_1}^3 \cdot Z/z \ll F_3 {E_1}^3 \cdot F_3 {E_1}^3/(x y),\end{equation*}

leading to the assertion.

A bright idea of Luca, appearing in the proof of [ Reference LucaLu2 , lemma 7] over the ring of Gaussian integer, is used in the proof of the following lemma.

Suppose that $x=y=2$ . We shall observe that this leads to $z \ll_c 1$ . Note that $\Delta\,=2|X-Y|$ is a positive even integer. As seen in the proof of lemma 4·1, it suffices to consider when c is odd. Below we shall argue over the ring of Gaussian integers. A usual factorisation argument on the 1st equation over ${\mathbb{Z}}[i]$ shows that there is some element $\beta$ with $\beta\bar{\beta}=c$ such that $a+b i$ equals $\beta^z$ times a unit. Note that $\beta$ is not a unit and is coprime to $\bar{\beta}$ . This implies that

\begin{equation*}a=\dfrac{\epsilon}{2}\,(\,\beta^z +\delta \,{\bar{\beta}}^z\,)\end{equation*}

for some unit $\epsilon$ and $\delta=\pm 1$ . Observe that $a^{2\Delta} \equiv 1 \pmod{c^z}$ from lemma 4·4 (i), and that

\begin{equation*}a^4=\dfrac{1}{2^4}\,(\,\beta^z +\delta\,{\bar{\beta}}^z\,)^4.\end{equation*}

These together yield

\begin{equation*}(\,\beta^z+\delta\,{\bar{\beta}}^z\,)^{2\Delta} \equiv 2^{2\Delta} \mod{c^z}.\end{equation*}

One reduces this modulo $\bar{\beta}^z$ to obtain $\beta^{2\Delta z} \equiv 2^{2\Delta} \pmod{\bar{\beta}^z}$ . Taking a prime element $\pi$ dividing $\bar{\beta}$ , one has

\begin{equation*}\nu_{\pi} (\beta^{2 \Delta z}-2^{2\Delta}) \ge z.\end{equation*}

To find an upper bound for the left-hand side above, we shall apply Proposition 2·3 for $(\alpha_1,\alpha_2)\,{:\!=}\,(\beta,2),\,(b_1,b_2)\,{:\!=}\,(2\Delta z,2\Delta)$ . Note that the corresponding p is some odd prime factor $(\!\ge 5)$ of c and $(D, f_{\pi})=(2,1)$ . One may set ${\rm g}\,{:\!=}\,\pi\bar{\pi}-1=p-1$ . Further, one may set $H_1\,{:\!=}\,\log c$ as ${\rm h}(\beta)={1}/{2}\log c$ , and $H_2\,{:\!=}\,\log p$ . Therefore,

\begin{equation*}\nu_{\pi} (\beta^{2\Delta z}-2^{2\Delta}) \ll \frac{(p-1) \log c}{\log^3 p} \,\log^2 \max \{\mathcal B, p\},\end{equation*}

where

\begin{equation*} \quad \mathcal B= \Delta z \cdot \left( 2 + \frac{2\log p}{z \log c} \right).\end{equation*}

To sum up, the two bounds for $\nu_{\pi} (\beta^{2\Delta z}-2^{2\Delta})$ together imply

\begin{equation*}z \ll_c \,\log^2 (2\Delta z).\end{equation*}

Since $\Delta=2|X-Y| \ll_c 1$ , the above implies that $z \ll_c 1$ .

1. (ii) Similarly to the proof of (i), Proposition 2·1 is applied for the 2nd equation, and it suffices to show $(X,Y) \ne (2,2)$ . In the case where $X=Y=2$ , from the 2nd equation, one can find that

   \begin{equation*}z \ll_{c} \,\log^2 (2\Delta Z)\end{equation*}
   similarly to the proof of (i). Since $Z \ll_{c} z$ , the above implies that $z \ll_c 1$ . This completes the proof.

The above lemma brings the following consequence to (a more general version of) Terai’s conjecture (cf. [ Reference TeraiTe ]) involving the condition $\min\{x,y,z\} \gt 1$ . His conjecture is a special case of Conjecture 1·11 and it asserts that if there is some solution to equation (1·5) in positive integers x,y and z all greater than 1, then $N(a,b,c) \leq 1,$ except for specific cases.

The following proposition is a particular case of [ Reference Bugeaud and LaurentBuLu , theorem 4·2].

Proposition 4·13 Let m and n be positive integers with $m \gt n.$ Let q be a positive integer with $q \gt 1.$ Then there are only finitely many solutions to the equation

(4·11) \begin{equation} \mathcal Z^m - \mathcal Z^n= q^{y_1} - q^{y_2}\end{equation}

in positive integers $\mathcal Z, y_1$ and $y_2$ with $\mathcal Z \gt 1$ and $\gcd(\mathcal Z,q)=1.$

The proof of the above proposition largely relies on Propositions 2·4 and 2·5. Their ineffective nature makes the proof of lemma 4·10 below provide no effective way to quantify the size of $\mathcal C$ in terms of c. This means that the proofs of Theorems 1.1 and 1.2 ultimately rely on the Subspace Theorem. It is worth noting that Subspace Theorem is required only for the case where $m \ge 3,\,n=1$ and q is composite.

For the sake of completeness, and for later discussions on the possibility of making the results introduced in Section 1, except Theorem 1·15 effective (cf. Section 11·2), at the end of this section we give a proof of Proposition 4·13 following [ Reference LucaLu , Reference BugeaudBuLu ], with some modifications and simpler treatments.

Proof. Lemma 4·8 says that $x,y,X,Y \ll_c 1$ . If $x=X$ , then eliminating the terms $a^x$ from the 1st and 2nd equations leads to

\begin{equation*}b^Y - b^y = c^Z - c^z.\end{equation*}

For each possible pair (y, Y), Proposition 4·13 is applied for $(m,n,q)\,{:\!=}\,(Y,y,c)$ and $(\mathcal Z,y_1,y_2)\,{:\!=}\,(b,Z,z)$ to see in particular that z has to be less than some constant depending only on c. Similarly the same conclusion holds when $y=Y$ . This completes the proof.

We are now in position to prove Theorem 1·1.

Proof of Theorem 1. Let $a \gt 1$ be a fixed positive integer. For some pair (b,c) with $\gcd(a,b)=1$ , suppose that equation (1·2) has two solutions. Write

\begin{equation*}c+b^y=a^x, \quad c+b^Y=a^X.\end{equation*}

Since there is no loss of generality in assuming by [ Reference LeVequeLeV , theorem 6] that $c \gt 1$ , lemma 4·10 is applied to the triple (c,b,a) to imply that both b,c have to be less than some constant depending only on a. This completes the proof.

The following lemma will not be actually used in the proofs of the results of this paper (cf. Section 11·1).

Lemma 4·15 If $\max\{a,b\} \gt \mathcal C_5,$ then

\begin{equation*}\max\{x,y\} \leq \biggl \lfloor \frac{E_1 \log c}{\log c_1}\biggl \rfloor.\end{equation*}

Proof. Set $m\,{:\!=}\,\min\{a,b\}$ . By lemma 4·4 (iii), it follows that ${c_1}^z$ divides W, where $W=(m^{E_1}-\delta_m) \cdot \Delta/E_1.$ Note that $W \ge {c_1}^z$ . We shall estimate the size of W. Since there is no loss of generality in assuming by lemma 4·8 that $\Delta \lt xyXY \ll {F_3}^2 {E_1}^6$ , it follows that

\begin{equation*}W \leq (m^{E_1}+1) \cdot \Delta/E_1 \ll m^{E_1} \cdot {F_3}^2 {E_1}^5.\end{equation*}

Since $m=\min\{a,b\} \lt c^{\,z/\max\{x,y\}}$ by the first two inequalities in (4·6), one has

\begin{equation*}W \ll \big(c^{ \,E_1/\max\{x,y\} }\bigr)^z \cdot {F_3}^2 {E_1}^5.\end{equation*}

Since ${c_1}^z \leq W$ , it follows that

(4·12) \begin{equation} {L_1}^z \ll {F_3}^2 {E_1}^5 \ll_c 1,\end{equation}

where $L_1=c_1/\,c^{\,E_1/\max\{x,y\}}$ .

Put $l={E_1\log c}/{\log c_1}$ . Then $L_1 \gt 1$ if and only if $\max\{x,y\} \gt \lfloor l \rfloor$ . Indeed,

\begin{equation*}L_1 \gt 1 \,\Leftrightarrow\, c_1 \gt c^{\,E_1/\max\{x,y\}} \,\Leftrightarrow\, \max\{x,y\} \gt l.\end{equation*}

It follows from (4·12) that if $\max\{x,y\} \gt \lfloor l \rfloor$ , then

\begin{equation*}z \ll_c \frac{1}{\log L_1} \leq \frac{1}{\log (\,c_1\,/\,c^{\,E_1/(\lfloor l \rfloor+1)}\,)} \ll_c 1.\end{equation*}

This completes the proof.

A summary of this section is as follows.

1. (i) There exists some constant $K_1$ which depends only on c and is effectively computable such that if $\max\{a,b\} \gt K_1,$ then

   \begin{alignat*}{3}&\min\{x,y\}=1, &&\max\{x,y\} \leq \displaystyle \biggl \lfloor \frac{E \log c}{\log c'}\biggl \rfloor, \\&\min\{X,Y\}=1, &\quad &\max\{X,Y\} \lt \,\frac{\mathcal C_0\,E^6 \log^4 c}{\nu_{c'}(c)^2 (\log c')^4\max\{x,y\}},\end{alignat*}
   where $\mathcal C_0$ is some positive absolute constantFootnote ² being effectively computable.

2. (ii) There exists some constant $K_2$ depending only on c such that if $\max\{a,b\} \gt K_2,$ then $(x,Y)=(1,1)$ or $(y,X)=(1,1).$

We finish this section by introducing a proof of Proposition 4·13, where the implied constant of each Vinogradov notation depends only on the parameters of equation (4·11) (namely, m,n and q) and is not always effectively computable.

Proof of Proposition 4.1. Let $(\mathcal Z, y_1,y_2)$ be a solution to equation (4·11). Clearly, $y_1 \gt y_2$ . We show that $\mathcal Z, y_1 \ll 1$ in any case. Note that $\mathcal Z, y_1 \ll 1$ if $\mathcal Z \ll 1$ or $y_1 \ll 1$ .

Firstly, observe that

(4·13) \begin{align}q^{y_1/m} \ll \mathcal Z \ll q^{y_1/m},\end{align}

(4·14) \begin{align}q^{y_2} \mid (\mathcal Z^{m-n}-1).\end{align}

Define an integer M by

\begin{equation*}M\,{:\!=}\,q^{y_1} - \mathcal Z^m.\end{equation*}

Note that M is nonzero as $\mathcal Z \gt 1$ and $\gcd(\mathcal Z,q)=1$ by assumption. We are interested in the size of M. Since $M= q^{y_2} - \mathcal Z^n$ by equation (4·11), it holds trivially that

\begin{equation*}|M| \lt \max\{q^{y_2}, \mathcal Z^n\}.\end{equation*}

On the other hand,

\begin{align*}|M|&=|(q^{y_1/m})^m - \mathcal Z^m|\\[6pt]&=|q^{y_1/m} - \mathcal Z| \cdot (\,(q^{y_1/m})^{m-1}+ (q^{y_1/m})^{m-2}\mathcal Z +\cdots+\mathcal Z^{m-1}\,)\\[6pt]&\ge |q^{y_1/m} - \mathcal Z| \cdot m\,\min\{q^{y_1/m},\mathcal Z\}^{m-1}.\end{align*}

These two bounds for M together lead to

\begin{equation*}|q^{y_1/m} - \mathcal Z| \lt \frac{\max\{q^{y_2}, \mathcal Z^n\}}{m\, \min\{q^{y_1/m},\mathcal Z\}^{m-1}}.\end{equation*}

Note that the left-hand side above does not vanish.

Write

\begin{equation*}y_1=m Q_1+R_1\end{equation*}

for integers $Q_1,R_1$ with $Q_1 \ge 0$ and $0 \leq R_1 \lt m$ . Dividing both sides of the previous displayed inequality by $q^{Q_1}$ gives

\begin{equation*}\left|q^{R_1/m} - \frac{\mathcal Z}{q^{Q_1}}\right| \lt \frac{\max\{q^{y_2}, \mathcal Z^n\}}{m\,q^{Q_1} \min\{q^{y_1/m},\mathcal Z\}^{m-1}}.\end{equation*}

Since $q^{y_1/m} \ll \mathcal Z \ll q^{y_1/m}$ by (4·13), it follows that

(4·15) \begin{equation} \left|\xi - \frac{\mathcal Z}{q^{Q_1}}\right| \ll \frac{\max\{q^{y_2}, q^{n\,y_1/m}\}} {q^{Q_1+(m-1) y_1/m}},\end{equation}

where $\xi$ is an algebraic number given by $\xi=q^{R_1/m}$ .

Since $\mathcal Z \ll q^{y_1/m}$ , and $q^{y_2} \ll \mathcal Z^{m-n}$ by (4·14), one has

\begin{equation*}q^{y_2} \ll q^{(m-n) y_1/m}.\end{equation*}

It follows from (4·15) that

(4·16) \begin{equation} \begin{split}\left|\xi - \frac{\mathcal Z}{q^{Q_1}}\right| &\ll \frac{q^{\,\max\{m-n,\,n\}\,y_1/m }} {q^{Q_1+(m-1)y_1/m}}=\frac{1} {q^{Q_1+Q_2}},\end{split}\end{equation}

where

\begin{align*}Q_2&=(\,m-1-\max\{m-n,n\}\,) \cdot y_1/m\\&=\min\{n-1,m-n-1\} \cdot (Q_1+R_1/m).\end{align*}

Let $\varepsilon\,{:\!=}\,Q_2 / Q_1$ . Then (4·16) gives

\begin{equation*}\left|\xi - \frac{\mathcal Z}{q^{Q_1}}\right|\ll \frac{1} {(q^{Q_1})^{1+\varepsilon}}.\end{equation*}

This is a restricted rational approximation to $\xi$ . If $\varepsilon \gg 1$ , then Proposition 2·4 with $\mathcal S$ the set of all prime factors of q is applied to see that

\begin{equation*}\left|\xi - \frac{\mathcal Z}{q^{Q_1}}\right| \gt \frac{C}{(q^{Q_1})^{1+\varepsilon/2}}\end{equation*}

for some constant C depending only on $\xi,\mathcal S$ and $\varepsilon$ . The two bounds for $|\xi - \mathcal Z/q^{Q_1}|$ together imply that $q^{Q_1} \ll 1$ , in particular, $y_1 \ll m Q_1 \ll 1$ . By these observations, we shall next estimate the size of $\varepsilon$ . Observe that

\begin{align*}\varepsilon=Q_2 / Q_1&=\min\{n-1,\,m-n-1\} \cdot (\,1+R_1/(m Q_1)\,)\\& \ge \min\{n-1,\,m-n-1\}.\end{align*}

Therefore, it suffices to consider when $n=1$ or $m-n=1$ .

Suppose that $n=m-1$ . It follows from [ Reference MiyazakiMiyPin , lemma 9·2(i)] that $\mathcal Z \gg q^{3y_2}$ if $q^{y_2} \gt n$ . Thus, in any case,

(4·17) \begin{equation} q^{y_2} \ll \mathcal Z^{1/3}.\end{equation}

Multiplying both sides of equation (4·11) by $m^m$ leads to

(4·18) \begin{equation} m^m q^{y_1} - (m \mathcal Z-1)^m = m^m q^{y_2} - G(\mathcal Z),\end{equation}

where G is a polynomial with integer coefficients of degree $m-2$ given by

\begin{align*}G(\mathcal Z)&=(m \mathcal Z-1)^m - m^m(\mathcal Z^m-\mathcal Z^{m-1})\\&=\binom{m}{2}(m \mathcal Z)^{m-2} - \binom{m}{3}(m \mathcal Z)^{m-3} + \cdots + (-1)^m.\end{align*}

Note that the left-hand side of (4·18) does not vanish (as it is not divisible by m). Similarly to a previous case, putting $\xi\,{:\!=}\,m q^{R_1/m}$ , one finds from (4·18) with (4·17) that

\begin{align*}\left| \xi - \frac{m \mathcal Z -1}{q^{Q_1}}\right| & \lt \frac{\max\{m^m q^{y_2},G(\mathcal Z)\}} {m\,q^{Q_1} \min\{m q^{y_1/m},m\mathcal Z-1\}^{m-1}}\\[3pt]&\ll \frac{\max\{q^{y_2},\mathcal Z^{m-2}\}}{q^{Q_1} \min\{q^{y_1/m},\mathcal Z\}^{m-1}}\\[3pt]&\ll \frac{\mathcal Z^{\,\max\{1/3,\,m-2\}}}{q^{Q_1} (q^{y_1/m})^{m-1}}\\[3pt]&\ll \frac{q^{\,\max\{1/3,\,m-2\} \, y_1/m}}{q^{Q_1+(m-1)y_1/m}} = \frac{1}{q^{Q_1+Q_2}},\end{align*}

where $Q_2= \min\{m-4/3,1\} \cdot y_1/m$ . Since $m \ge 2$ and $y_1 / m \ge Q_1$ , it follows that $Q_2 \ge (2/3)Q_1$ , so that

\begin{equation*}\left|\xi - \frac{m \mathcal Z -1}{q^{Q_1}}\right| \ll \frac{1} {(q^{Q_1})^{5/3}}.\end{equation*}

Similarly to a previous case, by considering the above restricted rational approximation to $\xi$ , it turns out from applying Proposition 2·4 that $Q_1 \ll 1$ , and $y_1 \ll 1$ . Therefore, it remains to consider when $n=1$ with $m \ge 3$ .

In the remaining part, we suppose that $m \ge 3$ and $n=1$ . By divisibility relation (4·14),

\begin{equation*}q^{y_2} \ \Big| \ (\mathcal Z-1) \cdot \frac{\mathcal Z^{m-1}-1}{\mathcal Z-1}.\end{equation*}

Since the greatest common divisor of $\mathcal Z-1$ and ${\mathcal Z^{m-1}-1}/{\mathcal Z-1}$ is a divisor of $m-1$ , the above relation implies that

(4·19) \begin{equation} \mathcal Z-1={q_2}^{y_2}\,u, \ \ \frac{\mathcal Z^{m-1}-1}{\mathcal Z-1}={q_3}^{y_2}\,v\end{equation}

for some positive integers $q_2,q_3$ with $q_2 q_3=q$ , and some positive rational numbers u,v each of whose denominator is a divisor of $m-1$ . Let $\delta \gt 0$ be any fixed constant satisfying $\delta \lt {2(m-2)}/{(m-1)(m-2)+2}\,( \lt 1)$ . We distinguish two cases according to whether $\max\{u,v\} \ge \mathcal Z^\delta$ or not.

Assume that

(4·20) \begin{equation} u \ge \mathcal Z^\delta \text{ or } v \ge \mathcal Z^\delta.\end{equation}

Then

\begin{equation*}q^{y_2} = {q_2}^{y_2} \cdot {q_3}^{y_2} = \frac{\mathcal Z^{m-1}-1}{u v} \lt \frac{\mathcal Z^{m-1}}{\max\{u,v\}}\leq \mathcal Z^{m-1-\delta}.\end{equation*}

This leads to

\begin{equation*}| M |=| \mathcal Z^1 - q^{y_2} | \ll \mathcal Z^{\,\max\{1,\,m-1-\delta\}}=\mathcal Z^{m-1-\delta}\ll (q^{y_1/m})^{m-1-\delta}.\end{equation*}

Similarly to previous cases, the above implies that

\begin{equation*}\left|q^{R_1/m} - \frac{\mathcal Z}{q^{Q_1}} \right|\ll \frac{(q^{y_1/m})^{m-1-\delta}}{q^{Q_1+(m-1)\,y_1/m}}=\frac{1}{q^{Q_1+\delta(y_1/m)}}\leq \frac{1}{(q^{Q_1})^{1+\delta}}.\end{equation*}

Thus, Proposition 2·4 is applied for the above restricted rational approximation to $q^{R_1/m}$ to find that the solutions to equation (4·11) with property (4·20) are only finitely many.

From now on, we consider only the solutions with the property that

(4·21) \begin{equation} u \lt \mathcal Z^\delta \text{ and } v \lt \mathcal Z^\delta.\end{equation}

We shall observe that $\mathcal Z \ll 1$ . Write D for the least common multiple of the denominators of u and v. We know that D is composed of primes less than m. Observe that

\begin{equation*}\frac{\mathcal Z^{m-1}-1}{\mathcal Z-1}=m-1+\sum_{k=2}^{m-1}\binom{m-1}{k}(\mathcal Z-1)^{k-1}.\end{equation*}

With the notation in (4·19), one multiplies both sides above by $D^{m-2}$ and rearranges terms to find that

(4·22) \begin{equation} (m-1) D^{m-2} = {q_3}^{y_2}v D^{m-2} - \sum_{k=2}^{m-1}\binom{m-1}{k}\, {q_2}^{(k-1)y_2}\,u^{k-1} D^{m-2}.\end{equation}

To apply Proposition 2·5 to the above relation, set $N\,{:\!=}\,m-1\,(\ge 2)$ , and let $\mathcal S$ be the set of all primes which divide q or are less than m. For each $p \in \mathcal S$ , define N linear forms $L_{p,i}=L_{p,i}(X_1,...,X_N) \,(i=1,...,N)$ by $L_{p,i}\,{:\!=}\,X_i$ for any i. Further, define N linear forms $L_{\infty,i} =L_{\infty,i}(X_1,...,X_N) \,(i=1,...,N)$ by $L_{\infty,i}\,{:\!=}\,X_i$ for $i \gt 1$ , and

\begin{equation*}L_{\infty,1}\,{:\!=}\, X_1 - \sum_{k=2}^{N}\binom{N}{k} X_k.\end{equation*}

Note that the defined forms depend only on N, and that $L_{\mu,1},...,L_{\mu,N}$ are linearly independent for each $\mu \in \mathcal S \cup\{\infty\}$ .

We shall estimate the left-hand side of the inequality in Proposition 2·5 at the points $\boldsymbol{x}=(x_1,...,x_N) \in {\mathbb Q}^N$ given by

\begin{align*}&x_1\,{:\!=}\,{q_3}^{y_2} v D^{N-1} \ =\frac{\mathcal Z^N-1}{\mathcal Z-1}\,D^{N-1}, \\[3pt] &x_k\,{:\!=}\, {q_2}^{(k-1)y_2}\,u^{k-1} D^{N-1} \ =(\mathcal Z-1)^{k-1} D^{N-1} \ \text{for $k \ge 2$}.\end{align*}

Clearly, $x_i \in \mathbb Z$ for any i, and $\max\{|x_i|:i=1,...,N\}=x_1 \ll \mathcal Z^{N-1}$ . Note that $q_2,q_3,D$ are composed of primes in $\mathcal S$ . By the product formula, one finds that

\begin{align*}\prod_{\mu \in \mathcal S} \, |L_{\mu,1}(\boldsymbol{x})|_\mu=\prod_{\mu \in \mathcal S} \, |x_1|_\mu&=\prod_{\mu \in \mathcal S} \, |{q_3}^{y_2}v D^{N-1}|_\mu\\[6pt]&=\prod_{\mu \in \mathcal S} \, |{q_3}^{y_2}D^{N-2}|_\mu\,\cdot\, \prod_{\mu \in \mathcal S} \, |v D|_\mu\\[6pt]&=\frac{1}{|{q_3}^{y_2}D^{N-2}|_\infty}\cdot\prod_{\mu \in \mathcal S} \, |v D|_\mu\leq \frac{1}{{q_3}^{y_2}D^{N-2}},\end{align*}

and $|L_{\infty,1}(\boldsymbol{x})|_\infty= N D^{N-1}$ by (4·22). For $k \ge 2$ ,

\begin{align*}\prod_{\mu \in \mathcal S \cup\{\infty\}} |L_{\mu,k}(\boldsymbol{x})|_\mu&=\prod_{\mu \in \mathcal S \cup\{\infty\}} |x_k|_\mu=\prod_{\mu \in \mathcal S \cup\{\infty\}} \,\bigr|\,{q_2}^{(k-1)y_2}\,u^{k-1} D^{N-1}\,\bigr|_{\mu}\\[6pt]&=\prod_{\mu \in \mathcal S \cup\{\infty\}} |\,{\rm num}\,(u)^{k-1}\,|_\mu \ \cdot\prod_{\mu \in \mathcal S \cup\{\infty\}} \bigr|\,{q_2}^{(k-1)y_2}\,D^{N-1}/{\rm den}\,(u)^{k-1}\,\bigr|_\mu\\[6pt]&=\Biggr(\prod_{\mu \in \mathcal S \cup\{\infty\}} |{\rm num}\,(u)|_\mu\Biggr)^{k-1} \cdot \,1\\[6pt]&\leq \bigl(\,|{\rm num}\,(u)|_{\infty}\,\bigl)^{k-1}= {\rm num}\,(u)^{k-1},\end{align*}

where ${\rm num}\,(u),{\rm den}\,(u)$ are the numerator and denominator of u, respectively. Since ${\rm den}\,(u) \ll 1$ , and ${q_3}^{y_2}\,v \gg \mathcal Z^{N-1}$ by (4·19), it follows from property (4·21) that

\begin{alignat*}{3}\prod_{\mu \in \mathcal S \cup\{\infty\}}\,\prod_{i=1}^{N} \, |L_{\mu,i}(\boldsymbol{x})|_\mu &\leq \frac{{\rm num}\,(u)^{1+2+\cdots+N-1} \cdot N D^{N-1}}{{q_3}^{y_2} D^{N-2}}&\\&\ll \frac{ {\rm num}\,(u)^{\frac{N(N-1)}{2}} }{{q_3}^{y_2}}&\\& \ll \frac{ u^{\frac{N(N-1)}{2}}\,v}{ \mathcal Z^{N-1} }&\\& \ll \frac{ \mathcal Z^{(N^2/2-N/2+1)\delta} }{ \mathcal Z^{N-1} }=\frac{ 1 }{ \mathcal Z^{N-1-(N^2/2-N/2+1)\delta} }=\frac{ 1 }{ (\mathcal Z^{N-1})^{\varepsilon} },\end{alignat*}

where $\varepsilon=1-\frac{N(N-1)+2}{2(N-1)}\,\delta$ . To sum up,

\begin{equation*}\prod_{\mu \in \mathcal S \cup\{\infty\}} \,\prod_{i=1}^{N} \, |L_{\mu,i}(\boldsymbol{x})|_\mu\ll \frac{1}{ (\,\max\{\,|x_i|:i=1,...,N\,\}\,)^{\varepsilon} },\end{equation*}

Since $0 \lt \varepsilon \lt 1$ by the choice of $\delta$ , Proposition 2·5 is applied to imply that there are only finitely many proper subspaces of $\mathbb Q^N$ such that each of the points $\boldsymbol{x}$ under consideration belongs to one of them. This means that for each such $\boldsymbol{x}=(x_1,...,x_N)$ the linear relation $\sum_{k=1}^{N}R_k x_k =0$ holds for some N-tuple $(R_1,...,R_N)$ of integers belonging to a finite subset of ${\mathbb Z}^N \backslash \{\textbf{0}\}$ depending only on N. Thus

\begin{equation*}R_1 w_1 + \sum_{k=2}^{N} R_k {w_2}^{k-1} =0,\end{equation*}

where $w_1={{\mathcal Z}^N-1}/{\mathcal Z-1}, \,w_2=\mathcal Z-1$ . Observe that $w_2 \mid R_1 N$ as $w_2 \mid R_1 w_1$ with $\gcd(w_1,w_2) \mid N$ . If $R_1 \ne 0$ , then $w_2 \leq R_1 N \ll 1$ . If $R_1=0$ , then the non-trivial algebraic equation $\sum_{k=2}^{N} R_k {w_2}^{k-1}=0$ for $w_2$ holds, in particular, $w_2 \ll_{R_2,...,R_N} 1 \ll 1$ . To sum up, $\mathcal Z=w_2+1 \ll 1$ , and the solutions to equation (4·11) with property (4·21) are only finitely many. This completes the proof.

5. Cases for prime values of c

Throughout this section, we assume that c is an odd prime, and prove stronger results than some of the lemmas in the previous section. In particular, lemma 4·11 is substantially improved. Note that in (4·1) we can take $c_1$ as c itself. Further, we simply write $E_1=E$ .

From now on, similarly to the previous section, we assume that equation (1·5) has two solutions (x,y,z) and (X,Y,Z) with $z \leq Z$ .

The following is a direct consequence of Scott [ Reference ScottSco , lemma 6].

Proof. Suppose that $\Delta$ is even, that is,

\begin{equation*}x Y \equiv X y \mod{2}.\end{equation*}

On the other hand, by Proposition 5·2, one may assume that

\begin{equation*}x \not\equiv X \mod{2} \quad \text{or} \quad y \not\equiv Y \mod{2}.\end{equation*}

These congruence conditions together imply that x,y are even or X,Y are even. Thus lemma 4·9 gives $a,b \leq \mathcal C_4$ . The second assertion follows from lemma 4·4 (ii).

The following lemma for the case where $E=3$ will be mainly used in the proofs of Theorems 1.2 and 1.3.

Proof. The proof proceeds along the similar lines to that of lemma 4·11. By lemma 4·4 (iii), it follows that $c^z$ divides $(m^E-\delta_m) \cdot \Delta/E$ , where $m=\min\{a,b\}$ . Since c is a prime, $m \not\equiv \delta_m \pmod{c}$ as $E \gt 1$ , and E is odd, one finds that $c^z$ divides W, where

\begin{equation*}W\,{:\!=}\,\frac{m^E-\delta_m}{m-\delta_m} \cdot \gcd(\Delta/E,c^z).\end{equation*}

We shall estimate the size of W. Since one may assume by lemma 4·8 that $\Delta \lt xyXY \ll E^6$ , it follows that

\begin{equation*}W \leq (m^{E-1}+m^{E-2}+\cdots+1) \cdot \Delta/E \ll m^{E-1} \cdot E^5.\end{equation*}

Since $m \lt c^{\,z/\max\{x,y\}}$ , it follows that

(5·1) \begin{equation}W \ll c^{\,z(E-1)/\max\{x,y\}} \cdot E^5.\end{equation}

If $\max\{x,y\} \ge E$ , then, since $c^z \leq W \ll c^{\,z(E-1)/E} \cdot E^5$ by (5·1), one has $c^{z/E} \ll E^5$ , so that $c^z \ll_{c} 1$ , whereby $z \ll_c 1$ .

Finally, suppose that $\max\{x,y\}=E-1$ . It suffices to show that $z \ll_{c} 1$ . By (5·1),

\begin{equation*} W/c^z \ll E^5 \end{equation*}

with $W \equiv 0 \pmod{c^z}$ . Thus

\begin{equation*} W =K \cdot c^z \end{equation*}

for some positive integer K satisfying $K \ll E^5 \ll_{c} 1$ . Since $\gcd(\Delta/E,c^z)=c^n$ with some integer $n \ge 0$ being absolutely finite and effectively computable as $\Delta \ll E^6 \lt c^6$ , one concludes from the definition of W that

(5·2) \begin{equation}\frac{m^E-\delta_m}{m-\delta_m}=K \cdot c^{z-n}.\end{equation}

This implies that the integer

\begin{equation*} m^{E-1}+\delta_m m^{E-2}+{\delta_m}^2\,m^{E-3}+\cdots +{\delta_m}^{E-1} \left(=\frac{m^E-\delta_m}{m-\delta_m}\right)\end{equation*}

is composed of only the primes dividing K and c. In particular,

\begin{equation*}P[\,g(m)\,] \ll_c 1,\end{equation*}

where g is a polynomial with integer coefficients given by $g(t)=t^{E-1}+\delta_m t^{E-2}+{\delta_m}^{2}\,t^{E-3}+\cdots +{\delta_m}^{E-1}$ . Since g has no multiple roots of degree $E-1 \ge 2$ , it is known for each possible E that

\begin{equation*}\lim_{j \to \infty} P[\,g(j)\,]=\infty, \ \ \text{effectively}\end{equation*}

(cf. [ Reference ShoreyShTi , theorem 7·1]). Therefore, $m \ll_{c,E} 1$ . Since $z-n \ll_{m,E} 1$ by (5·2), one has $z \ll_{c} 1$ .

A summary of this section is as follows.

6. Proof of Theorem 1·2

Let c be any prime of the form $c=2^r \cdot 3 +1$ with a positive integer r. Assume that $\max\{a,b\} \gt \mathcal C_4,{\mathcal C_5}'$ . Suppose that equation (1·5) has two solutions (x,y,z),(X,Y,Z) with $z \leq Z$ . Note that $E \gt 1$ by [ Reference MiyazakiMiyPin2 , corollary 1]. Since $E \mid \Delta$ by lemma 4·4 (ii), and $\Delta$ is odd by lemma 5·5, it follows that E is odd. Now, Lemmas 5·3 and 4.9 (ii) say that

\begin{gather*}x \leq E-2 \text{ and } y \leq E-2;\\X=1 \text{ or } Y=1,\end{gather*}

respectively. Further, one can show that $E=3$ as follows. Recall that $E \mid \varphi(c)$ . Since $\varphi(c)=c-1=2^r \cdot 3$ , it follows that $E=3$ . Therefore, $x=y=1$ . lemma 4·10 says that $a,b \leq \mathcal C$ . This completes the proof of Theorem 2.

7. Towards an effective version of Theorem 1·2

Based on the proof of Theorem 1·2 with Proposition 5·2, we can find the following proposition towards making Theorem 1·2 effective.

Let b be a positive integer with $b \gt 1$ and $\gcd(b,c)=1.$ Let $Y_0$ be a positive even integer with $4 \leq Y_0 \ll_c 1.$ Then the equation

(7·1) \begin{equation}c^Z-c^z=b^{Y_0}-b\end{equation}

has no solution in positive integers z and Z such that $c^z \gt b,$ whenever b exceeds some constant which depends only on $c,\epsilon$ and $k_0$ and is effectively computable.

As seen in (1·8) there are many values of c for which both conditions (i) and (ii) of the above proposition hold. Among those values, we could only find that condition (iii) also holds for $c=13$ with $(\epsilon,k_0)=(0.53,3)$ , and this fact follows from a striking result of Bauer and Bennett [ Reference Bauer and BennettBaBe , corollary 1·6; $y=13$ ] based on the hypergeometric method. Another key idea for proving Proposition 7·1 is to find a non-trivial lower bound for the ratio of z to Z in equation (7·1). Note that exceptional cases in Theorem 1·2 for $c=13$ can be effectively determined by Proposition 7·1 together with the proof of Theorem 1·2, and those cases will be handled completely in the next section.

Below, in order to prove Proposition 7·1, we shall prepare a few lemmas, where, to follow the arguments in previous sections, in equation (7·1) we define a positive integer a by

\begin{equation*}a+b=c^z, \ \ a+b^{Y_0}=c^Z.\end{equation*}

It suffices to consider when $a \gt 1$ by [ Reference LeVequeLeV , theorem 6]. The above two equations are in this section also referred to as the 1st and 2nd equation, respectively. Note that $\Delta=Y_0-1$ . Further, similarly to the proof of Theorem 1·2, one may assume $e_c(b)=E$ , where $E=3$ .

First, suppose that Z is even. From the 2nd equation, it follows that a is divisible by the positive integer $ c^{Z/2}+b^{Y_0/2}$ . In particular, $a \ge c^{Z/2}+b^{Y_0/2} \gt c^{Z/2}$ . Since $a \lt c^z$ , one has $Z \lt 2z$ .

Next, suppose that Z is odd. From the 2nd equation,

\begin{equation*}a=c^Z-b^{Y_0}=\bigr(c^{\frac{Z-1}{2}}\sqrt{c}+b^{\frac{Y_0}{2}}\big)\big(c^{\frac{Z-1}{2}}\sqrt{c}-b^{\frac{Y_0}{2}}\bigr).\end{equation*}

Dividing both left-most and right-most sides above by $(c^{\frac{Z}{2}}+b^{\frac{Y_0}{2}})\,c^{\frac{Z-1}{2}}$ yields

\begin{equation*}\frac{a}{(c^{\frac{Z}{2}}+b^{\frac{Y_0}{2}})\,c^{\frac{Z-1}{2}} }=\sqrt{c}-\frac{b^{\frac{Y_0}{2}}}{c^{\frac{Z-1}{2}}}.\end{equation*}

Since $a \lt c^z$ , it follows that

\begin{equation*} 0 \lt \sqrt{c}-\frac{b^{\frac{Y_0}{2}}}{c^{\frac{Z-1}{2}}} \lt \frac{c^z}{c^{\frac{Z}{2}} \cdot c^{\frac{Z-1}{2}}},\end{equation*}

which leads to a restricted rational approximation to $\sqrt{c}$ , as follows:

\begin{equation*}\left|\sqrt{c}-\frac{b^{\frac{Y_0}{2}}}{c^{\frac{Z-1}{2}}}\right| \lt \frac{1}{c^{\,Z-\frac{1}{2}-z}}.\end{equation*}

On the other hand, if ${Z-1}/{2} \ge k_0$ , then property (iii) of c for $(P,Q,k)\,{:\!=}\,(b^{\frac{Y_0}{2}},c^{\frac{Z-1}{2}},\frac{Z-1}{2})$ is used to give an opposite restriction to the above approximation, as follows:

\begin{equation*}\left|\sqrt{c}-\frac{b^{\frac{Y_0}{2}}}{c^{\frac{Z-1}{2}}}\right| \gt \frac{1}{c^{\frac{Z-1}{2} \cdot (1+\epsilon)}}.\end{equation*}

These inequalities together lead to

\begin{equation*}Z-\frac{1}{2}-z \lt \frac{Z-1}{2} \cdot (1+\epsilon),\end{equation*}

so that

\begin{equation*}\frac{1-\epsilon}{2} \cdot Z \lt z-\frac{\epsilon}{2}.\end{equation*}

Since $\epsilon \lt 1$ , this implies the assertion.

Proof. Let (z,Z) be a solution to equation (7·1). Similarly to the proof of Theorem 1·2, one finds that $b^2+\delta b+1$ is divisible by $c^z/\gcd(\Delta/E,c^z)$ for some $\delta=\pm1$ . In particular,

\begin{equation*}b^2+\delta b+1 \ge \frac{c^z}{\gcd(\Delta/E,c^z)}.\end{equation*}

Since $\Delta=Y_0-1$ , and $Y_0 \ll_c 1$ by assumption, it follows that

\begin{equation*}2b^2 \gt b^2+\delta b+1 \ge \frac{c^z}{\Delta/E} \gg_{c} c^z,\end{equation*}

so that $b \gg_c c^{z/2}$ . This together with the trivial inequality $b \lt c^{Z/Y_0}$ leads to

\begin{equation*}c^{z/2} \ll_c c^{Z/Y_0}.\end{equation*}

Since $b \lt c^Z \ll_{c,k_0} 1$ if $Z \lt 2k_0+1$ , one may assume by lemma 7·6 that $Z \lt {2}/{1-\epsilon}\,z$ . Thus the above displayed inequality implies that

(7·2) \begin{equation}c^{\,\mathcal E z} \ll_c 1,\end{equation}

where

\begin{equation*}\mathcal E\,{:\!=}\,\frac{1}{2} - \frac{2}{(1-\epsilon)Y_0}.\end{equation*}

From now on, suppose that $Y_0 \gt E+1\,(=4)$ . Since $Y_0$ is even and $Y_0=\Delta+1$ with $\Delta \equiv 0 \pmod{E}$ , it follows that $Y_0 \ge 3E+1=10$ , whereby

\begin{equation*}\mathcal E \ge \frac{1}{2} - \frac{1}{5(1-\epsilon)} \gt 0,\end{equation*}

where the last inequality follows from assumption $\epsilon \lt 0.6$ . To sum up, putting $\epsilon'\,{:\!=}\,{1}/{2} - {1}/{(5(1-\epsilon))}$ , one has $c^{\,\epsilon'z} \ll_c 1$ by (7·2), so that $b \lt c^z \ll_{c,\epsilon} 1$ . This completes the proof.

Proof. We shall preserve the letter $Y_0$ . Observe from the 1st and 2nd equations that $c^{Y_0 z}=(a+b)^{Y_0} \gt a+b^{Y_0}=c^Z$ , so that $Y_0 z \gt Z$ . Further, one reduces the 1st and 2nd equations modulo a to see that

\begin{equation*}c^{Y_0 z-Z} \equiv 1 \mod{a}.\end{equation*}

In particular, $c^{Y_0 z-Z} \gt a$ . We distinguish two cases.

Suppose that $a \leq c^{z-1}$ . Since $c^Z \gt b^{Y_0}$ and $b=c^z-a \ge (1-1/c)\,c^z$ , it follows that

\begin{equation*}c^Z \gt (\,(1-1/c)\,c^z\,)^{Y_0}=\frac{1}{\bigr(\frac{c}{c-1}\bigr)^{Y_0}}\,c^{Y_0 z},\end{equation*}

which implies that

\begin{equation*}Y_0 \gt \frac{\log c}{\log(1+\frac{1}{c-1})} \cdot (Y_0 z-Z).\end{equation*}

Since $c \ge 4$ and $Y_0 z-Z \ge 1$ , the above gives $Y_0 \gt 4.81\ldots$ , a contradiction.

If $a \gt c^{z-1}$ , then $c^{Y_0 z-Z} \gt a \gt c^{z-1}$ , so that $Y_0 z-Z \ge z$ , i.e., $Z \leq (Y_0-1)z=3z$ .

Proof. We have $b(b-1)(b^2+b+1)=c^z(c^{Z-z}-1)$ . Since $b-1$ is not divisible by c (as $e_c(b)=E \gt 1$ ) and c is a prime, it follows that $b^2+b+1 \equiv 0 \pmod{c^z}$ , so that

(7·3) \begin{gather}b^2+b+1=Kc^z,\end{gather}

(7·4) \begin{gather}(b^2-b)K=c^{Z-z}-1\end{gather}

for some positive odd integer K. The key idea is to derive a non-trivial lower bound for K. For this we follow an idea used in the proofs of [ Reference MiyazakiMiyPin , lemma 9·2] and [ Reference MiyazakiMiyPin2 , lemma 5·6]. Note that $Z \ge 2z$ as $b^2 \gt c^z/2$ by (7·3), and $c^Z \gt b^{Y_0}=b^4$ by the 2nd equation with $c \gt 4$ .

Since, by (7·3),

\begin{equation*}b^2-b \equiv -2b-1 \mod{c^z},\end{equation*}

one finds from (7·4) that

\begin{equation*}(2b+1)K \equiv 1 \mod{c^z}.\end{equation*}

Squaring both sides above shows

\begin{equation*}(2b+1)^2 K^2 \equiv 1 \mod{c^z}.\end{equation*}

Since

\begin{equation*}(2b+1)^2 \equiv 4b^2+4b+1 \equiv -3 \mod{c^z},\end{equation*}

it follows that

\begin{equation*}3 K^2 \equiv -1 \mod{c^z}.\end{equation*}

This implies the following lower bound for K:

\begin{equation*}K \ge \sqrt{\frac{4c^z-1}{3}}=\sqrt{\frac{4-1/c^z}{3}} \cdot c^{z/2}.\end{equation*}

Then

\begin{equation*}2 b^2 \gt b^2+b+1=K c^z \ge \sqrt{\frac{4-1/c^z}{3}} \cdot c^{3z/2},\end{equation*}

so that

\begin{equation*}b^4 \gt \frac{1-1/(4c^z)}{3} \cdot c^{3 z}.\end{equation*}

Since $c^Z \gt b^4$ with $c \ge 4$ , it follows that

\begin{equation*}c^{Z-3z} \gt \frac{1-1/(4c^z)}{3} \gt c^{-1}.\end{equation*}

This leads to the assertion.

Proof. For $Y_0=4$ , suppose that equation (7·1) has a solution (z,Z). Lemmas 7.3 and 7.4 together say that $Z=3z$ . Reducing the 1st and 2nd equations modulo a implies that $b \equiv 1 \pmod{a}$ , in particular, $b \gt a$ . Thus,

\begin{equation*}c^z=a+b \lt 2b \lt 2 c^{Z/Y_0}=2 c^{3z/4},\end{equation*}

so that $c^z \lt 16$ . Now brute force computation suffices to verify that equation (7·1): $c^{3z}-c^z=b^4-b$ does not hold in any possible case.

Proposition 7·1 follows from the combination of lemmas 7·2 and 7·5.

8. Case where $c=13$ with $\Delta$ odd

In this and the next section, we put $c=13$ . We may assume that $e_{c}(a)=e_{c}(b) \gt 1$ , and write $E=e_{c}(a)=e_{c}(b)$ . Observe that

\begin{equation*}e_{c}(h)=3 \ \Leftrightarrow \ h \equiv 3,4,9,10 \pmod{c}. \end{equation*}

It suffices to consider when $\max\{a,b\} \gt c$ (because Conjecture 1·11 is proved to be true for each pair (a,b) with $\max\{a,b\} \lt c$ (cf. [ Reference StyerSty ])).

Let x,y,z,X,Y,Z be positive integers such that

(8·1) \begin{eqnarray} &a^x+b^y=c^z, \end{eqnarray}

(8·2) \begin{eqnarray} &a^X+b^Y=c^Z. \end{eqnarray}

As in previous sections, we put $\Delta,\Delta',\mathcal D$ as follows:

\begin{equation*}\Delta=|xY-yX|, \ \ \Delta'=\gcd(\Delta/E,c^{\,\min\{z,Z\}}), \ \ \mathcal D=c^{\,\min\{z,Z\}}/\Delta'.\end{equation*}

Since c is a prime, there exists some nonnegative integer n’ with $n' \leq \min\{z,Z\}$ such that

\begin{equation*}\Delta'=c^{n'}, \ \ \mathcal D=c^{\,\min\{z,Z\}-n'}.\end{equation*}

Further, for each $h \in \{a,b\}$ , we have

(8·3) \begin{equation} h^{E-1}+\delta_h h^{E-2}+{\delta_h}^2\,h^{E-3}+\cdots +{\delta_h}^{E-1} \equiv 0 \mod{\mathcal D},\end{equation}

where $\delta_h \in \{1,-1\}$ is defined by the congruence $h^E \equiv \delta_h \pmod{c}$ . This in particular implies that

\begin{equation*}m \gt (\mathcal D /2)^{\frac{1}{E-1}},\end{equation*}

where $m\,{:\!=}\,\min\{a,b\}$ .

In the remainder of this section, we assume that $z \leq Z$ . Equations (8·1) and (8·2) are often referred to as the 1st and 2nd equation, respectively.

8·1. Finding explicit bounds for solutions

Here we shall attempt to find sharp upper bounds for the solutions such that $\Delta$ is odd. The aim is to show the following lemma.

Lemma 8·1. Suppose that $\Delta$ is odd. Then the following hold.

1. (i) $Z \lt K_1\,\log a\,\log b,$ where

   \begin{equation*} K_1=\begin{cases}\,2367 & \text{if $m \lt c,$} \\\,843 & \text{if $m \gt c$}.\end{cases} \end{equation*}

2. (ii) $\Delta \lt K_1(\log^2 c)\,z.$

3. (iii) If $z \ge 9,$ then $z\,Z \leq K_2\,\log a\,\log b,$ where $K_2=10459.$

4. (iv) $\Delta \leq K_3,$ where

   \begin{equation*} K_3=\begin{cases}\,77862 & \text{if $m \lt c$ and $z \leq 8,$} \\\,44368 & \text{if $m \gt c$ and $z \leq 8,$} \\\,68809 & \text{if $z \ge 9.$}\end{cases} \end{equation*}
   Further, $n' \leq 3.$

5. (v) If $\max\{x,y\} \ge 3,$ then $z \leq 9.$

6. (vi) If $\max\{x,y\}=2,$ then

   \begin{equation*} z \leq \begin{cases}\,23650 & \text{if $n'=0$}, \\\,23651 & \text{if $n'=1$}, \\\,47322 & \text{if $n'=2$}, \\\,70986 & \text{if $n'=3$}.\end{cases} \end{equation*}

7. (vii) $x \gt 1$ or $y \gt 1.$

Before going to the proof of the above lemma, we make some remarks on its last two statements. For (vi), while we follow the proof of lemma 5·5 for $E=3$ , which relies on an explicit version of the size of greatest prime factor of the values of polynomials with integer coefficients at integer points, we adopt another way using the non-Archimedean analogue to Baker’s method over the ring of Eisenstein integers in a non-artificial sense, which is regarded as an analogue to a bright idea of Luca [ Reference LucaLu ] over the ring of Gaussian integers. For (vii), in order to show that $\min\{X,Y\}=1$ , we rely on modular approach recently established by Bennett and Siksek [ Reference BennettBeSi ] instead of a strategy used in the proof of lemma 4·9.

Proof of Lemma 8.1. Note that $E=3$ , and that

\begin{equation*}a \equiv 3,4,9,10 \pmod{c}, \ \ b \equiv 3,4,9,10 \pmod{c}.\end{equation*}

In particular, either $m \in \{3,4,9,10\}$ or $m \gt c$ .

1. (i) We follow the proof of lemma 4·2 on solution (X,Y,Z). Then

   \begin{equation*}Z \leq \nu_{c}(a^{2X}-b^{2Y}) \leq \frac{53.6\cdot 2E\cdot \log a'\,\log b'}{\log^4 c} \cdot {\mathcal B'}^2,\end{equation*}
   where $a'=\max\{a,c\},\,b'=\max\{b,c\}$ ,
   \begin{equation*}\mathcal{B'}=\log\,\max \bigr\{\,4\exp(0\cdot 64)(\log^2 c)\,T,\,c^4 \,\bigr\}\end{equation*}
   with $T= {Z}/{(\log a\,\log b)}$ . Thus
   \begin{align*}\\[-2pc] T \lt f\cdot \frac{53.6\cdot 2E}{\log^4 c} \cdot \mathcal{B'}^2,\end{align*}
   where $f= {(\log a')}/{(\log a)}\cdot {(\log b')}/{(\log b)}$ . Since $\max\{a,b\} \gt c$ , one has
   \begin{align*}f=\frac{\log\,\max \{m,c\}}{\log m}=\begin{cases}\,\frac{\log c}{\log m} & \text{if $m \lt c,$} \\\,1 & \text{if $m \gt c.$}\end{cases} \end{align*}
   These together imply that $T \lt K_1$ .

2. (ii) This follows from inequality (4·9) together with (i).

3. (iii) Firstly we follow the proof of lemma 4·6. Since $\Delta' \leq \Delta/E \lt K_1/E\cdot(\log^2 c)\,z$ by (ii), one has

   \begin{align*}\\[-2pc]\frac{\log \mathcal D}{z} \gt \log c-\frac{\log\bigr( K_1/E \cdot (\log^2{c}) \, z \bigr)}{z}.\end{align*}
   Assume that $z \ge 9$ . Then the above displayed inequality implies
   (8·4) \begin{equation}\log{\mathcal D} \gt \zeta \,z,\end{equation}
   where
   \begin{align*}\\[-2pc] \zeta=\begin{cases}\,1.37 & \text{if $m \lt c,$} \\\,1.48 & \text{if $m \gt c.$}\end{cases}\end{align*}
   Since $\mathcal D \gt \exp(\zeta\,z) \ge \exp(1.37\cdot9) \gt 2.2 \cdot 10^5$ , it follows that $m=\min\{a,b\} \gt (\mathcal D /2)^{\frac{1}{E-1}} \gt 13\,(=c)$ . Thus,
   (8·5) \begin{gather}\mathcal D \gt \exp(\zeta\,z) \gt 6 \cdot 10^5,\end{gather}
   (8·6) \begin{gather}m \gt (\mathcal D /2)^{\frac{1}{E-1}} \gt (\,\exp(\zeta \,z) /2\,)^{\frac{1}{E-1}}\end{gather}
   with $\zeta=1.48$ .

Next, we closely follow the proof of lemma 4·7. Then

\begin{equation*}Z/z-1 \lt \nu_{\mathcal D}(a^{2X}-b^{2Y}) \leq \frac{53.6\cdot 2E\cdot \log a'\,\log b'}{\log^4 \mathcal D} \cdot {\mathcal B'}^2,\end{equation*}

where $a'\,{:\!=}\,\max\{a,\mathcal D\},\, b'\,{:\!=}\,\max\{b,\mathcal D\}$ ,

\begin{equation*}\mathcal{B'}\,{:\!=}\,\log\,\max \bigr\{ \,4\,{\rm e}^{0.64}(\log c)(\log \mathcal D)\,T/z, \, {\mathcal D}^4\, \bigr\}\end{equation*}

with ${\rm e}=\exp(1)$ and $T\,{:\!=}\,{(z\,Z)}/{(\log a\,\log b)}$ . These together imply

\begin{equation*}T \lt \frac{53.611\cdot 2E\cdot f\,z^2}{\log^4 \mathcal D}\cdot \mathcal{B'}^2,\end{equation*}

where $f\,{:\!=}\,{(\log a')}/{(\log a)}\cdot {(\log b')}/{(\log b)}$ . Observe that

\begin{equation*}f \leq \max\biggr\{1,\,\frac{\log^2 \mathcal D}{\log^2 m}\biggr\}.\end{equation*}

If $4\,{\rm e}^{0.64}(\log c)(\log \mathcal D)\,T/z \gt \mathcal D^4$ , then

\begin{equation*}\frac{\mathcal D^4}{4\,{\rm e}^{0.64}(\log c)\log \mathcal D} \lt T/z \lt \frac{53.611\cdot 2E\cdot f\,z}{\log^4 \mathcal D}\,\log^2 \bigr(\,4\,{\rm e}^{0.64}(\log c)(\log \mathcal D)\,T/z\,\bigr).\end{equation*}

Since $\mathcal D^4 \gt (\,4\,{\rm e}^{0.64}(\log c)\log \mathcal D\,)^9$ by (8·5), it follows from the left-hand inequality displayed above, that

\begin{equation*}(T/z)^{1/8} \ge 4\,{\rm e}^{0.64}(\log c)\log \mathcal D.\end{equation*}

In particular, $T/z \gt 2 \cdot 10^{19}$ . Therefore, by the second one,

\begin{equation*}T/z \lt \frac{53.611\cdot 2E\cdot f\,z}{\log^4 \mathcal D}\, \log^2 \bigr(\,(T/z)^{9/8}\bigr).\end{equation*}

Since $z \ge 9$ , $m \gt (\mathcal D /2)^{\frac{1}{E-1}}$ by (8·6), and $\log \mathcal D \gt \zeta\,z$ by (8·4), it follows that

\begin{align*}\frac{T/z}{\log^2 \bigr(\,(T/z)^{9/8}\bigr)}& \lt 53.611\cdot 2E\cdot \max\biggr\{\frac{z}{\log^4 \mathcal D},\,\frac{z}{\log^2 m\,\log^2 \mathcal D}\biggr\}\\& \lt 53.611\cdot 2E\cdot \max\biggr\{\frac{1}{\zeta^4 z^3},\,\frac{1}{\zeta^2(\log^2 m)\,z}\biggr\}\ \lt \ 0.41,\end{align*}

which is a contradiction as $T/z$ is large.

Finally, if $4\,{\rm e}^{0.64}(\log c)(\log \mathcal D)\,T/z \leq \mathcal D^4$ , then

\begin{align*}T& \lt \frac{53.611\cdot 2E\cdot f\,z^2 \cdot 4^2}{\log^2 \mathcal D}\\& \lt 53.611\cdot 2E\cdot \max\biggr\{\frac{z^2}{\log^2 \mathcal D},\frac{z^2}{\log^2 m}\biggr\} \cdot 4^2\\& \lt 53.611\cdot 2E\cdot \max\biggr\{\frac{1}{\zeta^2},\frac{z^2}{\log^2 m}\biggr\} \cdot 4^2\ \leq \ K_2.\end{align*}

1. (iv) Firstly, since $\Delta \lt K_1(\log^2 c)\,z$ by (ii), if $z \leq 8$ , then

   (8·7) \begin{equation} \Delta \leq\begin{cases}\,124579 & \text{if $m \lt c,$} \\\,44368 & \text{if $m \gt c.$}\end{cases}\end{equation}
   If $z \ge 9$ , then, since $zZ \leq K_2\,\log a\,\log b$ by (iii), it follows that $\Delta \lt {(z\,Z)}/ {(\log a \, \log b)} \log^2 c \leq K_2\log^2 c \lt 68810$ . Thus the inequality $\Delta \leq 124579$ always holds. Also, $n' \leq 4$ , as $c^{n'}=\Delta' \leq \Delta/E \lt 41527 \lt c^5$ .

We shall reduce the bound for $\Delta$ in (8·7) when $m \lt c$ . By congruence (8·3) with $h=m$ ,

(8·8) \begin{equation} c^{z-n'} \mid (m^2+\delta_m \,m+1).\end{equation}

If $m \lt c$ , then $\nu_c(m^2+\delta_m \,m+1) \leq 1$ , so that $c^{z-n'} \mid c,$ whereby $z \leq 1+n' \leq 5$ . Then $\Delta \lt 2367 \cdot (\log^2 c) \cdot 5 \lt 77863$ . To sum up, the asserted bounds for $\Delta$ hold. Further observe that $n' \leq 3$ .

1. (v) We basically follow the proof of lemma 5·5 for the case where $\max\{x,y\} \ge E\,(=3)$ . Relation (8.8) implies that $c^{z-n'} \leq m^2+m+1$ . Since $n' \leq 3$ by (iv), and $m \lt c^{\,z/\max\{x,y\}} \leq c^{z/3}$ by assumption that $\max\{x,y\} \ge 3$ , it follows that $c^{z-3} \lt c^{2z/3}+c^{z/3}+1$ , leading to $z \leq 9$ .

2. (vi) Firstly, we follow the proof of lemma 5·5 for the case where $\max\{x,y\}=E-1\,(=2)$ . By (8·8), there is a positive integer K such that

   (8·9) \begin{equation} m^2+\delta_m \,m+1=K c^{z-n'}.\end{equation}
   Since $a \ne b$ , and $\min\{x,y\}=1$ as $\Delta$ is odd, one finds that
   \begin{equation*}K c^{z-n'}=m^2+\delta_m \,m+1 \leq m^2+m+1 \leq m^2+\max\{a,b\} \leq c^z,\end{equation*}
   so that
   (8·10) \begin{equation} K \leq c^{n'}.\end{equation}

Below, we shall argue over the ring of Eisenstein integers. Let ${\mathcal O}$ be the ring of integers of $\mathbb{Q}(\omega)$ , where $\omega=\exp(2\pi i/3)$ . One factorises the left-hand side of (8·9) to see that

(8·11) \begin{equation} f \cdot \bar{f}=K' \cdot c^{z_1},\end{equation}

where

\begin{equation*}f=m -\delta_m\,\omega, \ \ \bar{f}=m -\delta_m\,\bar{\omega},\ \ K'=K/c^{\,\nu_c(K)}, \ \ z_1=z-n'+\nu_c(K).\end{equation*}

Note that $z_1 \gt 0$ . Observe that

(8·12) \begin{equation} f - \bar{f} = -\delta_m\,\omega (1-\omega).\end{equation}

In particular, $f-\bar{f}$ is not divisible by any rational prime. From this, it is not hard to see that any rational prime dividing the right-hand side of (8·11) has not to be inert, namely, it has to be factorised into the product of two prime elements. This implies that

\begin{equation*}f= K_0 \cdot \pi^{z_1}\end{equation*}

for some $K_0 \in {\mathcal O}$ with $K_0\, \overline{K_0}=K'$ , and $\pi$ is a prime element with $\pi \bar{\pi}=c$ . Note that $\pi \ne \bar{\pi}$ (as $c \ne 3$ ). Equality (8·12) becomes

\begin{equation*}K_0 \cdot \pi^{z_1}-\overline{K_0} \cdot {\bar{\pi}}^{\,z_1}= -\delta_m\,\omega (1-\omega).\end{equation*}

Reducing this modulo $\pi^{z_1}$ gives

\begin{equation*}\overline{K_0} \cdot {\bar{\pi}}^{\,z_1} -\delta_m\,\omega (1-\omega) \equiv 0 \mod{\pi^{z_1}}.\end{equation*}

Putting

\begin{equation*}\alpha_1\,{:\!=}\,\bar{\pi}, \ \ \alpha_2\,{:\!=}\,\delta_m\,\omega (1-\omega)/\,\overline{K_0},\end{equation*}

one has

\begin{equation*}\nu_{\pi} ({\alpha_1}^{z_1} - {\alpha_2}) \ge z_1.\end{equation*}

Now we shall apply Proposition 2·3 to find an upper bound for the left-hand side above. For this, set $(b_1,b_2)\,{:\!=}\,(z_1,1)$ . Observe that $p=c,\,D=2,\,f_{\pi}=1$ . It is easy to see that $\alpha_1,\alpha_2$ are multiplicatively independent from the fact that $\pi$ divides neither $K_0$ nor $1-\omega$ . One may choose ${\rm g}\,{:\!=}\,\pi \cdot \bar{\pi}-1=c-1$ . Since ${\rm h}(\alpha_1)={1}/{2}\log c$ , one may take $H_1\,{:\!=}\,\log c$ . Further, observe that ${\rm h}(\alpha_2)={\rm h}({\alpha_2}')$ , where ${\alpha_2}'=(1-\omega)/\,\overline{K_0}$ , and that the minimal polynomial $g_2(t)$ of ${\alpha_2}'$ over $\mathbb Z$ is given by

\begin{equation*}g_2(t)=\begin{cases}K' t^2 - 3u t +3 & \text{if $3 \nmid K'$},\\(K'/3)\,t^2 - u t +1 & \text{if $3 \mid K'$},\end{cases}\end{equation*}

where u is the rational integer such that ${\alpha_2}'=u+v\,\omega$ with some $v \in \mathbb Z$ . Since $|{\alpha_2}'|=\sqrt{3/K'}$ , it turns out that

\begin{align*}{\rm h}( {\alpha_2}')&\leq\frac{1}{2}\,\Bigr(\!\log K' + 2\,\log\,\max \{\,1,|{\alpha_2}'|\,\}\Bigr)=\frac{1}{2}\log\, \max \{K',3\}.\end{align*}

Since $K' \not\equiv 0 \pmod{c}$ and $K'$ is a divisor of the left-hand side of (8·9), it follows from inequality (8·10) that

\begin{equation*}K' \leq K_u,\end{equation*}

where $K_u =c^{n'}$ . Therefore,

\begin{equation*}\frac{D}{f_\pi}\,{\rm h}(\alpha_2)=2\,{\rm h}({\alpha_2}') \leq \log K_u,\end{equation*}

thereby one may take $H_2\,{:\!=}\,\log\,\max\{K_u,c\}$ . Then

\begin{equation*}\nu_{\pi} ({\alpha_1}^{z_1} - {\alpha_2}) \lt \frac{27.3 \cdot 2^2\,c\,H_2}{\log^3 c} \cdot \mathcal B^2\end{equation*}

with $\mathcal B=\max\{\, \log (z_1/H_2+1/\log c)+\log \log c+0.4,\,4\log c \,\}$ . Finally, the two bounds for $\nu_{\pi} ({\alpha_1}^{z_1} - {\alpha_2})$ together lead to the bounds for $z_1\,(=z-n')$ , implying the assertion.

1. (vii) Suppose that $x=y=1$ . Since $\{a,b\} \ne \{3,10\}$ as $\max\{a,b\} \gt c=13$ , by Proposition 5·2, there is no loss of generality in assuming that X is odd and Y is even. equation (8·2) is rewritten as

   \begin{equation*}(b^{Y/2})^2-13^Z=(-a)^X.\end{equation*}
   We can directly apply the following result of Bennett and Siksek which is a consequence of [ Reference BennettBeSi , theorems 2 and 5; $q=13$ ].

Proposition 8·2. All solutions to the equation

\begin{equation*}S^2 - 13^k = T^n\end{equation*}

in integers S,T,k and n with $S \gt 0, \gcd(S,13)=1, k \ge 1, n \ge 3$ are given by $(S,T,k,n)=(16,3,1,5),(14,3,2,3)$ and $(499,12,2,5).$

It turns out that $X=1$ , so that

(8·13) \begin{equation} c^Z-c^z=b^Y-b\end{equation}

with $c^z \gt b$ . Since $Y-1=\Delta \leq K_3$ by (iv), and $\Delta \equiv 0 \pmod{E}$ with $\Delta$ odd, it holds that

(8·14) \begin{equation} Y \leq K_3+1, \ \ Y \equiv 4 \pmod 6.\end{equation}

Thus equation (8·13) is the same as (7·1), and we can use the arguments in Section 7 for $c=13$ with $(\epsilon,k_0)=(0.53,3)$ .

First, lemma 7·6 says that $Y \gt 4$ . Next, we shall apply lemma 7·6. If $Z \ge 7$ , then $Z \ge 2k_0+1\,(=7)$ , so that

\begin{equation*}Z \lt \frac{200}{47}\,z.\end{equation*}

On the other hand, from the proof of lemma 7·6,

\begin{equation*}c^{z-n'}=\frac{c^z}{\Delta'} \lt 2 b^2 \lt 2 c^{2Z/Y}.\end{equation*}

These together imply

\begin{equation*}c^{\,(1-\frac{400}{47Y})\,z-n'} \lt 2.\end{equation*}

Since $n' \leq 3$ by (iv), and $Y \ge 10$ by (8·14), one concludes from the above displayed inequality that $z \leq 21$ . Thus $Z \lt \frac{200}{47}\,z \lt 90$ . To sum up, one always has $Z \lt 90$ . Now brute force computation suffices to verify that equation (8·13) does not hold in any possible case.

8·2. Sieving the remaining cases

Here we sieve the remaining cases in lemma 8·1 (v, vi). Based on restrictions provided by lemma 8·1 and its proof, we check whether the system of 1st and 2nd equations holds, according to assumptions $z \leq 9$ in (v) and $z \leq 70986$ in (vi), respectively. For this purpose, it suffices to consider only the case where $x \ge y$ by symmetry of a and b.

We know from congruence (8·3) with $h=a$ there is a positive integer $t=t(a)$ such that $a^2+\delta_a \, a+1=tc^{z-n'}$ , whence

(8·15) \begin{equation}(2h+\delta_a)^2=4tc^{z-n'}-3.\end{equation}

This implies that

(8·16) \begin{equation}t \not\equiv 0 \pmod{2}, \ \ t \not\equiv 0 \pmod{9}, \ \ p \not\equiv 2 \pmod{3}\end{equation}

for any odd prime factor p of t. Further, since $a \lt c^{z/x} \leq c^{z/2}$ as $x=\max\{x,y\} \ge 2$ , and $t=(a^2+\delta_a \,a+1)/c^{z-n'} \leq (a^2+a+1)/c^{z-n'}$ , observe that

(8·17) \begin{gather}t \lt (1+1/c^{z/2}+1/c^z)\,c^{n'}.\end{gather}

8·2·1. Case (v)

We perform the algorithm consisting of the following three steps. The computation time was a few seconds.

Step 1. Find all possible numbers $z,n',t.$

We generate a list including all possible values of $z,n'$ and t satisfying (8·15), (8·16) and (8·17). The fact that $4tc^{z-n'}-3$ is a square following from (8·15) is a key sieving relation. It turns out that the resulting list, say list1, contains 114 elements.

begin

$z_u\,{:\!=}\,9$

for $n'\,{:\!=}\,0$ to 3 do

for $z\,{:\!=}\,\max\{1,n'\}$ to $z_u$ do

if $z \ge 7$ then $t_u\,{:\!=}\,c^{n'}$ ; else $t_u\,{:\!=}\, \lfloor (1+1/c^{z/2}+1/c^z)\,c^{n'}\rfloor$

for $t\,{:\!=}\,1$ to $t_u$ do

Sieve with (8·16)

if $4tc^{z-n'}-3$ is a square then

Put [ $z,n',t$ ] into list1

end

Step 2. Find all possible numbers $a,b,x,y,z,n'.$

We generate a list including all possible values of a, b, x, y, z and $n'$ by using the 1st equation. It turns out that the resulting list, say list2, contains 108 elements.

begin

$x_l\,{:\!=}\,3$

for each element [ $z,n',t$ ] in list1 do

$\mathcal D\,{:\!=}\,c^{z-n'}$ ; $A\,{:\!=}\,\sqrt{4t\mathcal D-3}$

$m_l\,{:\!=}\,\max\bigr\{ \,3, \bigr\lceil \,(\,\mathcal D/2\,)^{\frac{1}{E-1}} \,\bigr\rceil \, \bigr\}$

for $\delta_a$ in $[-1,1]$ do

$a\,{:\!=}\,(A-\delta_a)/2$

if $a \ge m_l$ and $a \leq \lfloor c^{z/x_l} \rfloor$ and $a \equiv 3,4,9,10 \pmod{c}$

and $a^2+\delta_a \, a+1 \equiv 0 \pmod{c^{z-n'}}$ then

for $x\,{:\!=}\,x_l$ to $\bigr\lfloor \frac{\log c}{\log a}\,z \bigr\rfloor$ do

for $y\,{:\!=}\,1$ to x do

if $c^z-a^x$ is a y th power then

$b\,{:\!=}\,(c^z-a^x)^{1/y}$

if $b \ge m_l$ and $b \equiv 3,4,9,10 \pmod{c}$

and $b^3 \equiv \pm1 \pmod{c^{z-n'}}$

and $b \not\equiv a \pmod{2}$ and $\gcd(b,a)=1$ then

Put [a,b,x,y,z, $n'$ ] into list2

end

Step 3. Find all possible numbers $a,b,x,y,z,X,Y,Z.$

We search all possible values of a,b,x,y,z,X,Y and Z by using the 2nd equation. For this, we use lemma 8·1 (i, ii, iv) and the latter two inequalities in (4·6), and distinguish two cases according to whether $\Delta=xY-Xy$ or $Xy-xY$ . It turns out that the output is the empty list.

begin

for each element [a,b,x,y,z, $n'$ ] in list2 do

$X_u\,{:\!=}\,\lfloor K_1 \log b\,\log c \rfloor$ ; $Y_u\,{:\!=}\,\lfloor K_1 \log a\,\log c \rfloor$ ;

$\Delta_u\,{:\!=}\,\min\bigr\{\,\lfloor K_1 (\log^2 c)\,z \rfloor,\, K_3 \,\bigr\}$

(Case where $\Delta=xY-Xy$ )

for $X\,{:\!=}\,1$ to $\min\bigr\{X_u, \bigr\lfloor (x Y_u - E c^{n'}) / y \bigr\rfloor\,\bigr\}$ do

for $k\,{:\!=}\,1$ to $\bigr\lfloor\,\min\{ \Delta_u, \,x \cdot Y_u-y \cdot X\}/(E\, c^{n'}) \bigr\rfloor$ by 2 do

$xY\,{:\!=}\,y \cdot X+k\cdot E \cdot c^{n'}$

if $xY \equiv 0 \pmod{x}$ then

$Y\,{:\!=}\,xY$ div x

if $a^X+b^Y$ is a power of c then

$Z\,{:\!=}\,\log(a^X+b^Y) / \log c$

Print [a,b,x,y,z,X,Y,Z]

(Case where $\Delta=Xy-xY$ )

for $Y\,{:\!=}\,1$ to $\min\bigr\{Y_u, \bigr\lfloor (X_u y - E c^{n'}) / x \bigr\rfloor\,\bigr\}$ do

for $k\,{:\!=}\,1$ to $\bigr\lfloor\,\min\{ \Delta_u, \,y \cdot X_u-x \cdot Y\}/(E\,c^{n'}) \ \bigr\rfloor$ by 2 do

$yX\,{:\!=}\,x \cdot Y+k\cdot Ec^{n'}$

if $yX \equiv 0 \pmod{y}$ then

$X\,{:\!=}\,yX$ div y

if $a^X+b^Y$ is a power of c then

$Z\,{:\!=}\,\log(a^X+b^Y) / \log c$

Print [a,b,x,y,z,X,Y,Z]

end

8·2·2. Case (vi)

This case is dealt with by performing the same algorithm as in Section 8·2·1 by resetting the value of x as $x\,{:\!=}\,2$ and resetting the value of $z_u$ as $z_u\,{:\!=}\,23650, 23651, 47322$ or 70986, according to whether $n'=0,1,2$ or 3, respectively. The computation time was less than 1 hour. We mention that it turned out from Step 1, which was the most time-consuming part, that none of the numbers $4tc^{z-n'}-3$ is a square for $z \ge 8$ .

This finishes studying system of equations (8·1) and (8·2) for $c=13$ with $\Delta$ odd.

9. Case where $c=13$ with $\Delta$ even

With the same notation as in the previous section, we consider the system of equations (8·1) and (8·2). Suppose that $\Delta$ is even. We shall observe that this leads to a contradiction.

From the proof of lemma 5·5, either x and y are even, or X and Y are even. Since the following argument hold in either case, we will assume that both X and Y are even. Write $X=2X'$ and $Y=2Y'$ . equation (8·2) becomes

(9·1) \begin{eqnarray} a^{2X'}+b^{2Y'}=c^Z.\end{eqnarray}

Below, we follow the first part of the proof of [ Reference MiyazakiMiyPin2 , theorem 3].

We can write

\begin{equation*}c=\beta \cdot \bar{\beta},\end{equation*}

where $\beta=2+3i$ . By a usual factorisation argument to equation (9·1) over $\mathbb Z[i]$ (see the proof of [ Reference MiyazakiMiyPin2 , lemma 7·6(i)]), one can show that

(9·2) \begin{equation}\{a^{X'},b^{Y'}\}=\{a(Z),b(Z)\},\end{equation}

where

\begin{equation*}a(Z)=\dfrac{1}{2}\,\bigr|\beta^Z+(-\bar{\beta})^Z\bigr|, \quad b(Z)=\dfrac{1}{2}\,\bigr|\beta^Z-(-\bar{\beta})^Z\bigr|.\end{equation*}

Lemma 9·1: The following hold:

\begin{equation*}\bigr(\,e_{c}(\,a(Z)\,),\,e_{c}(\,b(Z)\,)\,\bigr)=\,\begin{cases}\,(3,6) & \text{if $Z \equiv 1,3 \pmod{6}$},\\\,(2,1) & \text{if $Z \equiv 2 \pmod{6}$},\\\,(6,3) & \text{if $Z \equiv 0,4 \pmod{6}$},\\\,(1,2) & \text{if $Z \equiv 5 \pmod{6}$}.\end{cases}\end{equation*}

Proof. We show the assertion only for the value of $e_{c}(\,a(Z)\,)$ . It is easy to verify the assertion for $Z=1,2,...,6$ . Thus it suffices to show that

(9·3) \begin{equation} a(Z+6) \equiv \pm a(Z) \mod{c} \ \ \ (Z=1,2,\ldots).\end{equation}

For this, one can use some congruence properties of the Lucas sequence of the first and second kinds corresponding to the pair $(\beta,\bar{\beta})$ . Define two integer sequences $\{U_{n}\}_{n=1}^{\infty}$ and $\{V_{n}\}_{n=1}^{\infty}$ as follows:

\begin{equation*}U_{n}\,{:\!=}\, \frac{\beta^n-{\bar{\beta}}^n}{\beta-\bar{\beta}}, V_{n}\,{:\!=}\, \beta^n+{\bar{\beta}}^n.\end{equation*}

With this notation, since $\beta-\bar{\beta}=6i$ , one may write

\begin{equation*}2a(Z) = \begin{cases}\,\bigr|\beta^Z-{\bar{\beta}}^Z\bigr|=6\,|\,U_{Z}\,| &\text{if }{\textit{Z} \, \text{is odd},}\\\,\bigr|\beta^Z+{\bar{\beta}}^Z\bigr|=|\,V_{Z}\,| &\text{if} \, {\textit{Z} \, \text{is} \, \text{even}.}\end{cases}\end{equation*}

It is known that

\begin{equation*}U_{n} \equiv V_{n-1} \mod{c}, \quad V_{n} \equiv P^n \mod{c} \quad (n=1,2,\ldots),\end{equation*}

where $P=\beta+\bar{\beta}=4$ (cf. [ Reference RibenboimRi2 , (IV.12)]). These together imply that

\begin{equation*}\begin{cases}a(Z) \equiv \pm 4^{Z+1} \!\!\! \pmod{c} &\text{if} {\textit{Z} \, \text{is odd},}\\a(Z) \equiv \pm 7 \cdot 4^Z \!\!\! \pmod{c} &\text{if} {\textit{Z} \, \text{is even}.}\end{cases}\end{equation*}

This together with the fact that $e_{c}(4)=3$ implies congruence (9·3).

Since $e_{c}(a)=e_{c}(b)$ , and

\begin{equation*}e_{c}(a^{X'})=\frac{e_{c}(a)}{\gcd(\,e_{c}(a),X'\,)}, \ \ e_{c}(b^{Y'})=\frac{e_{c}(b)}{\gcd(\,e_{c}(b),Y'\,)}\end{equation*}

(cf. [ Reference MiyazakiMiyPin2 , lemma 2·1(iii)]), it follows from lemma 9·1 that at least one of X’ or Y’ is even. Then a simple application of the works [ Reference BennettBeElNg , Reference BruinBr , Reference EllenbergEl ] on the generalised Fermat equation of signature (2,4,r) with $r \ge 4$ to equation (9·1) yields $Z \leq 3$ . We shall apply relation (9·2) for $Z=1,2,3$ . It turns out that if $Z \ne 3$ , then both $a^{X'}$ and $b^{Y'}$ are not perfect powers, in particular $X'=Y'=1$ . Further, $a \in \{3,9\}$ and $b=46$ for $Z=3$ . In any case, one observes that $e_{c}(a) \ne e_{c}(b)$ , contradicting the premise.

The contents of this and the previous sections together complete the proof of Theorem 1·3.

10. Application of abc-conjecture

Here we give some applications of the abc-conjecture to Conjecture 1·11, which strengthen many of the lemmas in Sections 4 and 5. We begin by quoting the following result of Bugeaud and Luca [ Reference Bugeaud and LaurentBuLu ].

Proposition 10·1 ([BuLu, theorem 6·1]) Under the truth of the abc-conjecture, there are only finitely many solutions to the equation

\begin{equation*}\mathcal A^{x_1} - \mathcal A^{x_2} = \mathcal B^{y_1} - \mathcal B^{y_2}\end{equation*}

in positive integers $\mathcal A,\mathcal B,x_1,x_2, y_1$ and $y_2$ such that $\mathcal A \gt 1, \mathcal B \gt 1, x_1 \ne x_2$ and $\mathcal A^{x_1} \ne \mathcal B^{y_1}.$

It should be noted that the conclusion of the above proposition ensures that the number of exceptional triples (a,b,c) for Conjecture 1·3 is finite.

Supposing that equation (1·5) has two solutions (x,y,z), (X,Y,Z) with $z \leq Z$ , the following two lemmas are proved by using Proposition 10·1. We emphasise that any Vinogradov notation appearing in their proofs is not always effective on its implied constant.

Proof. (i) The proof proceeds similarly to that of lemma 4·10. If $z=Z$ , then

\begin{equation*}a^X - a^x = b^y - b^Y.\end{equation*}

Proposition 10·1 for $(\mathcal A,\mathcal B)\,{:\!=}\,(a,b),\, (x_1,x_2, y_1, y_2)\,{:\!=}\,(X,x,y,Y)$ is applied to see in particular that $a,b,x,y \ll 1$ , so that $c \leq c^z=a^x+b^y \ll 1$ . Similarly, the same conclusion holds when $x=X$ or $y=Y$ .

(ii) Let $\varepsilon \gt 0$ be any constant. A direct application of the abc-conjecture to the 2nd equation implies

\begin{equation*}c^Z \ll_{\varepsilon} (abc)^{1+\varepsilon}.\end{equation*}

Since $a \lt c^p$ and $b \lt c^q$ , where $p=\min\{z/x,Z/X\}$ and $q=\min\{z/y,Z/Y\}$ , it follows that $c^Z \ll_{\varepsilon} (c^{\,p+q+1})^{1+\varepsilon}$ , so that

(10·1) \begin{equation} c^{\,\mathcal E} \ll_{\varepsilon} 1,\end{equation}

where

\begin{equation*}\mathcal E\,{:\!=}\,Z-(p+q)(1+\varepsilon) \,-1-\varepsilon.\end{equation*}

In particular, $\mathcal E \ll_{\varepsilon} 1$ .

Suppose that $x \gt 1,\,y \gt 1$ . Since $p \leq z/x$ and $q \leq z/y$ , and one may assume by (i) that $Z \ge z+1$ , it follows that