Resolution of the equation $(3^{x_1}-1)(3^{x_2}-1)=(5^{y_1}-1)(5^{y_2}-1)$

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.