Resolution of the equation $(3^{x_1}-1)(3^{x_2}-1)=(5^{y_1}-1)(5^{y_2}-1)$
Resolution of the equation $(3^{x_1}-1)(3^{x_2}-1)=(5^{y_1}-1)(5^{y_2}-1)$
Consider the diophantine equation (3x1−1)(3x2−1)=(5y1−1)(5y2−1) in positive integers x1≤x2 and y1≤y2. 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 (x1,x2,y1,y2)=(1,2,1,1). The main novelty of our result is that we allow products …