Proof. Since c is an odd prime, using the notation of [Reference Scott and Styer22], we see that for any given parity class of (1.1), there is only one ideal factorisation in $C_D$ . Therefore, by Lemma 2 of [Reference Scott and Styer22], we obtain the lemma immediately.

Proof. This lemma is a special case of Corollary 1.7 of [Reference Bauer and Bennett1] with $y=2$ .

Proof. This follows easily from a conjecture of Pillai [Reference Pillai18] first proven by Stroeker and Tijdeman [Reference Stroeker, Tijdeman, Lenstra and Tijdeman23] using lower bonds on linear forms in logarithms. It is also an immediate consequence of the elementary corollary to Theorem 4 of [Reference Scott19].

Proof. If $a \equiv 2 \bmod 3$ , $b \not \equiv 0 \bmod 3$ and c is an odd prime, then the parity of y determines the parity class. Since Lemma 2.1 shows that there is at most one solution per parity class, we see that $N(a,b,c)>1$ implies that there are exactly two solutions $(x_1, y_1, z_1)$ and $(x_2,y_2,z_2)$ and $ 2 \nmid y_1 - y_2$ .

Proof of Theorem 1.3.

Assume $N(a,b,c)>1$ with $a \equiv 2 \bmod 3$ , $b \not \equiv 0\, \mod 3$ and $c>3$ an odd prime. By Lemma 2.12, there must be exactly two solutions $(x_1, y_1, z_1)$ and $(x_2,y_2,z_2)$ , with $2 \nmid y_1 - y_2$ . Take $y_1$ even. Then, since $c>3$ , consideration modulo 3 shows that $2 \mid x_1$ , $2 \mid y_1$ and $2 \nmid z_1$ , which requires $c \equiv 2 \bmod 3$ and $c \equiv 1 \bmod 4$ . Thus, $c \equiv 5 \bmod 12$ .

Proof of Theorem 1.4.

Assume that a, b and c are distinct primes with $a<b$ . If $N(a,b,c)>1$ , we can take $a=2$ ( $c \ne 2$ by Lemma 2.7, except for $(a,b,c)=(3,5,2)$ and $(3,13,2)$ , given as exceptions in the statement of the theorem).

Additionally, we can also take $b \ne 3$ : using Lemma 2.2, we find that if the equation $2^x+3^y=c^z$ has two solutions $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ , we either must have $c=5$ with $\{z_1,z_2 \}=\{ 1,2 \}$ or we must have $z_1=z_2=1$ , in which case we have $3^{y_1}-2^{x_2} = 3^{y_2}-2^{x_1} = d$ , where $d=1$ , $-5$ or $-13$ by Lemma 2.11; recalling Lemma 1.2 and using the solutions given in cases (x) and (iii) in Conjecture 1.1, we see that $d = -5$ corresponds to $c=35$ , while using case (xii) with case (iv) shows that $d=-13$ corresponds to $c=259$ ; since neither of these values of c is prime, we must have $d=1$ , and it is a familiar elementary result that we must have $(x_1, y_1, x_2, y_2) = (1,2,3,1)$ giving $c=11$ .

Thus we find that the only cases with $a=2$ , $b=3$ and $N(a,b,c)>1$ are $(a,b,c) = (2,3,5)$ and $(2,3,11)$ , given as exceptions in the statement of the theorem.

Now assume $a=2$ and $c=3$ , and assume (1.1) has two solutions $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ . By Lemma 2.12, we can assume $2 \mid y_1$ . If $z_1$ is even, then $3^{z_1/2} - b^{y_1/2} =~2$ and $3^{z_1/2}+b^{y_1/2} = 2^{x_1-1}$ , so that $ 3^{z_1/2} = 2^{x_1 -2} +1$ ; it is a familiar elementary result that we must have either $(x_1, z_1) = (3,2)$ (giving $b^{y_1/2} = 1$ ) or $(x_1, z_1) = (5,4)$ (giving $b^{y_1/2} = 7$ ). Since $b>1$ , we find that $(a,b,c) = (2,7,3)$ is the only possibility when $z_1$ is even.

Additionally, if $2 \nmid z_1$ when $a=2$ and $c=3$ , then, since $2 \mid y_1$ , we have $x_1 = 1$ , and we can factor in $\mathbb {Q} (\sqrt {-2})$ :

$$ \begin{align*} \pm b^{y_1/2} \pm \sqrt{-2} = (1+\sqrt{-2})^{z_1}, \end{align*} $$

where the ‘ $\pm $ ’ are independent. Since clearly $z_1>1$ , we can use Lemma 2.10 to see that $z_1 =3$ and $(a,b,c)=(2,5,3)$ .

Thus, recalling Lemma 1.2, we find that the only cases with $a=2$ , $c=3$ and $N(a,b,c)>1$ are $(a,b,c)=(2,7,3)$ and $(2,5,3)$ , given as exceptions in the statement of the theorem.

So, excluding the six exceptions given in the statement of the theorem, we can assume $a=2$ , $b \ne 3$ and $c \ne 3$ when $N(a,b,c)>1$ for a triple of primes $(a,b,c)$ . Now, Theorem 1.4 follows immediately from Theorem 1.3, except for the conclusion $z_1=1$ . To see that $z_1=1$ , note that, since $2 \nmid z_1$ , it suffices to use Lemma 2.4, noting $(a,b,c) = (2,5,3)$ and $(2,7,3)$ have been excepted, and handling $(2,11,5)$ using Lemma 2.1 with consideration modulo 3 and modulo 5.

Proof of Theorem 1.5.

Assume a, b and c are distinct primes with $a<b$ , $(a,b,c) \ne (2,3,5)$ , $(2,3,11)$ , $(2,5,3)$ , $(2,7,3)$ , $(3,5,2)$ or $(3,13,2)$ , and $N(a,b,c)> 1$ . From Theorem 1.4,

(4.1) $$ \begin{align} c \equiv 5 \bmod 12 \end{align} $$

and

(4.2) $$ \begin{align} 2^{x_1} + b^{y_1} = c, \quad 2 \mid x_1, 2 \mid y_1. \end{align} $$

Assume first

(4.3) $$ \begin{align} b \equiv 1 \bmod 3. \end{align} $$

We have

(4.4) $$ \begin{align} 2^{x_2} + b^{y_2} = c^{z_2}, \quad 2 \mid x_2, 2 \nmid y_2, 2 \nmid z_2. \end{align} $$

Since $2 \mid x_2$ and $2 \nmid y_2$ , by (4.1) and (4.4), we get $ b \equiv b^{y_2} \equiv c^{z_2} - 2^{x_2} \equiv 1-0 \equiv 1 \bmod 4$ . Hence, by (4.3),

(4.5) $$ \begin{align} b \equiv 1 \bmod 12. \end{align} $$

If $c \equiv 5 \bmod 8$ and $b \equiv 1 \bmod 8$ , then from (4.2) and (4.4), we get $x_1=2$ and $x_2=2$ , respectively. This implies that (2.2) has two solutions $(m,n)= (1, y_1)$ and $(z_2, y_2)$ for $(A,B) = (c,b)$ and $k=2$ . By Lemma 2.6, this is impossible. Hence, by (4.1) and (4.5), $(b,c) \not \equiv (1,5) \bmod 24$ and

(4.6) $$ \begin{align} (b,c) \equiv (1,17), (13,5) {\textrm{ or }} (13,17) \bmod 24. \end{align} $$

Now assume $ b \equiv 2 \bmod 3$ . Then we have (4.1), (4.2) and

(4.7) $$ \begin{align} 2^{x_2} + b^{y_2} &= c^{z_2}, \quad 2 \nmid x_2, 2 \nmid y_2, \end{align} $$

(4.8) $$ \begin{align} b & \equiv 2 \bmod 3, \end{align} $$

and

(4.9) $$ \begin{align} c^{z_2} \equiv 1 \bmod 3. \end{align} $$

Further, by (4.1) and (4.9),

(4.10) $$ \begin{align} 2 \mid z_2. \end{align} $$

When $x_2>1$ and $y_2>1$ , we see from (4.7) and (4.10) that (2.1) has a solution $(X,Y,m,n)=(c^{z_2/2}, b, x_2, y_2)$ . Hence, by Lemma 2.5, $(b,c) = (17,71)$ , for which a solution $(x_1, y_1, z_1)$ is impossible since $z_1=1$ . So we have either $x_2=1$ or $y_2=1$ .

If $y_2=1$ , then from (4.7) and (4.10),

(4.11) $$ \begin{align} 2^{x_2} + b = c^{z_2}, \quad 2 \nmid x_2, 2 \mid z_2. \end{align} $$

Note that $x_2>1$ since $c^{z_2}>c$ .

Apply Lemma 2.3 to (4.11) to obtain

(4.12) $$ \begin{align} b = ( c^{z_2/2} )^2 - 2^{x_2}> 2^{0.26 x_2}. \end{align} $$

Further, by (4.2), (4.11) and (4.12),

(4.13) $$ \begin{align} 2^{x_2} + b = c^{z_2} \ge c^2 = (2^{x_1} + b^{y_1})^2> b^{2y_1} \ge b^4 > 2^{1.04 x_2}, \end{align} $$

whence we obtain

(4.14) $$ \begin{align} b> 2^{x_2} (2^{0.04 x_2} -1). \end{align} $$

However, by (4.2), we have $b < \sqrt {c}$ . By Lemma 2.9, we can assume $\max \{ b,c\}> 100$ , so by (4.11), $2^{x_2} = c^{z_2} - b> c^2 - \sqrt {c} \ge 101^2 - \sqrt {101} > 10190$ , whence we get $x_2 \ge 15$ . So we have $2^{0.04 x_2} - 1 \ge 2^{0.6}-1> 1/2$ . By (4.14),

(4.15) $$ \begin{align} b> 2^{x_2-1}. \end{align} $$

Therefore, by (4.11), (4.13) and (4.15), we have $3b> 2^{x_2} + b = c^{z_2} \ge c^2 > b^4$ , which is a contradiction. So we obtain $y_2>1$ . This implies

(4.16) $$ \begin{align} x_2=1. \end{align} $$

By (4.16), (4.1), (4.7) and (4.8),

$$ \begin{align*} b = 2^r h - 1, \quad c = 2^s k + 1, \quad r>1, s>1, 2 \nmid h, 2 \nmid k, 3 \mid h. \end{align*} $$

Recalling (4.16), (4.7) and (4.10), we have

(4.17) $$ \begin{align} 2^{x_2} + b^{y_2} = 2 + (2^r h - 1)^{y_2} = 2^r h_1 + 1 = c^{z_2} = 2^{s + v_2(z_2)} k_1 + 1,\quad 2 \nmid h_1, 2 \nmid k_1, 3 \mid h_1, \end{align} $$

where $v_2(n)$ is the greatest integer t such that $2^t \mid n$ . From (4.17), we see that $r>s$ , so that in (4.2), we must have $x_1 = s$ .

Now we apply Lemma 2.13. Clearly $2< {c^{z_2/2}}/{4}$ (recall Lemma 2.9), so that applying Lemma 2.13 to (4.2), we find that we must have

$$ \begin{align*}2^{x_1} = 2^s \ge \frac{c^{1/2}}{4}> \frac{b^{y_1/2}}{4} \ge \frac{b}{4} \ge \frac{3 \cdot 2^r - 1 }{4} \ge \frac{3 \cdot 2^{s+1} -1}{4} = 3 \cdot 2^{s-1} - \frac{1}{4} \end{align*} $$

so that

$$ \begin{align*} 2 \ge 3 - \frac{1}{2^{s+1}}, \end{align*} $$

which is false for all positive s.

Thus, we have $b \not \equiv 2 \bmod 3$ . So $b \equiv 1 \bmod 3$ and (4.6) holds.

Proof of Corollary 1.6.

First, $N(a,b,c)=2$ follows from Theorem 1.4. If $z_1=z_2$ , then, recalling Lemma 2.7 and noting that $c \ne 35$ , $133$ or $259$ , we must have $b^{y_2} - 2^{x_1} = b^{y_1} - 2^{x_2}> 0$ , so that Lemma 2.1 gives $2 \nmid x_1 - x_2$ . Consideration modulo 3 shows that this requires $2 \nmid y_1 - y_2$ and $b \equiv 2 \bmod 3$ , contradicting Theorem 1.5. So $z_1 \ne z_2$ , and by Theorem 1.4, $z_2>1$ .