Publication date: 15 December 2024
Volume: Vol 59
Issue: Svezak 2
Pages: 299-312
DOI: 10.3336/gm.59.2.03
Izvorni znanstveni članak
https://doi.org/10.3336/gm.59.2.03
On a conjecture concerning the number of solutions to \(a^x+b^y=c^z\), II
Maohua Le
; Institute of Mathematics, Lingnan Normal College, Zhanjiang 524048, Guangdong, China
Reese Scott
; Somerville, MA, USA
Robert Styer
; Department of Mathematics, Villanova University, Villanova, PA, USA
Let \(a\), \(b\), \(c\) be distinct primes with \(a\lt b\). Let \(N(a,b,c)\) denote the number of positive integer solutions \((x,y,z)\) of the equation \(a^x + b^y = c^z\). In a previous paper [16] it was shown that if \((a,b,c)\) is a triple of distinct primes for which \(N(a,b,c)\gt 1\) 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: \(a=2\), \(b \equiv 1 \bmod 48\), \(c \equiv 17 \mod 48\), \(b \gt 10^9\), \(c \gt 10^{18}\); 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 \(1 = z_1 \lt z_2\) and either \(x_1 \ge 28\) or \(x_2 \ge 88\). These results support a conjecture put forward in [28] and improve results in [16].
Ternary purely exponential Diophantine equation, upper bound for number of solutions
325172
26.12.2024.
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