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August 2020 Resolution of the equation $(3^{x_1}-1)(3^{x_2}-1)=(5^{y_1}-1)(5^{y_2}-1)$
Kálmán Liptai, László Németh, Gökhan Soydan, László Szalay
Rocky Mountain J. Math. 50(4): 1425-1433 (August 2020). DOI: 10.1216/rmj.2020.50.1425

## Abstract

Consider the diophantine equation $\left({3}^{{x}_{1}}-1\right)\left({3}^{{x}_{2}}-1\right)=\left({5}^{{y}_{1}}-1\right)\left({5}^{{y}_{2}}-1\right)$ in positive integers ${x}_{1}\le {x}_{2}$ and ${y}_{1}\le {y}_{2}$. Each side of the equation is a product of two terms of a given binary recurrence. We prove that the only solution to the title equation is $\left({x}_{1},{x}_{2},{y}_{1},{y}_{2}\right)=\left(1,2,1,1\right)$. The main novelty of our result is that we allow products of two terms on both sides.

## Citation

Kálmán Liptai. László Németh. Gökhan Soydan. László Szalay. "Resolution of the equation $(3^{x_1}-1)(3^{x_2}-1)=(5^{y_1}-1)(5^{y_2}-1)$." Rocky Mountain J. Math. 50 (4) 1425 - 1433, August 2020. https://doi.org/10.1216/rmj.2020.50.1425

## Information

Received: 16 December 2019; Revised: 21 January 2020; Accepted: 21 January 2020; Published: August 2020
First available in Project Euclid: 29 September 2020

zbMATH: 07261872
MathSciNet: MR4154815
Digital Object Identifier: 10.1216/rmj.2020.50.1425

Subjects:
Primary: 11D61
Secondary: 11B37

Rights: Copyright © 2020 Rocky Mountain Mathematics Consortium

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