On a conjecture concerning the number of solutions to $a^x+b^y=c^z$, II

TL;DR

This paper investigates the solutions to the exponential Diophantine equation involving three primes, eliminating possible cases and establishing severe restrictions on the solutions, thereby supporting a longstanding conjecture.

Contribution

It refines the classification of solutions to $a^x + b^y = c^z$ for prime triples, eliminating two cases and detailing restrictions on the remaining case.

Findings

Eliminates two of the three possible cases for solutions.

Establishes specific congruence and divisibility conditions on primes.

Shows the number of solutions is severely limited under certain conditions.

Abstract

Let $a$, $b$, $c$ be distinct primes with $a<b$. Let $S(a,b,c)$ denote the number of positive integer solutions $(x,y,z)$ of the equation $a^x + b^y = c^z$. In a previous paper \cite{LeSt} it was shown that if $(a,b,c)$ is a triple of distinct primes for which $S(a,b,c)>1$ and $(a,b,c)$ is not one of the six known such triples then $(a,b,c)$ must be one of three cases. In the present paper, we eliminate two of these cases (using the special properties of certain continued fractions for one of these cases, and using a result of Dirichlet on quartic residues for the other). Then we show that the single remaining case requires severe restrictions, including the following: $a=2$, $b \equiv 1 \bmod 48$, $c \equiv 17 \mod 48$, $b > 10^9$, $c > 10^{18}$; at least one of the multiplicative orders $u_c(b)$ or $u_b(c)$ must be odd (where $u_p(n)$ is the least integer $t$ such that $n^t \equiv 1 \bmod p$); 2 must be an octic residue modulo $c$ except for one specific case; $2 \mid v_2(b-1) \le v_2(c-1)$ (where $v_2(n)$ satisfies $2^{v_2(n)} \parallel n$); there must be exactly two solutions $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ with $1 = z_1 < z_2$ and either $x_1 \ge 28$ or $x_2 \ge 88$. These results support a conjecture put forward in \cite{ScSt6} and improve results in \cite{LeSt}.