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 orcid.org/0000-0003-4570-4581 ; Department of Mathematics, Villanova University, Villanova, PA, USA

Preuzmi JATS datoteku

Sažetak

Let $a$, $b$, $c$ be distinct primes with $a<b$. 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 $N(a,b,c)>1$ 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: $a=2$, $b\equiv 1mod48$, $c\equiv 17\mathrm{mod}48$, $b>{10}^9$, $c>{10}^{18}$; 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 $n^t\equiv 1modp$); 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 [28] and improve results in [16].

Ključne riječi

Ternary purely exponential Diophantine equation, upper bound for number of solutions

Hrčak ID:

325172

URI

https://hrcak.srce.hr/325172

Datum izdavanja:

11.4.2025.

Posjeta: 158 *