## On the diophantine equation $(x^m + 1)(x^n + 1) = y²$

### Volume 82 / 1997

Acta Arithmetica 82 (1997), 17-26
DOI: 10.4064/aa-82-1-17-26

#### Abstract

1. Introduction. Let ℤ, ℕ, ℚ be the sets of integers, positive integers and rational numbers respectively. In [7], Ribenboim proved that the equation (1) $(x^m + 1)(x^n + 1) = y²$, x,y,m,n ∈ ℕ, x > 1, n > m ≥ 1, has no solution (x,y,m,n) with 2|x and (1) has only finitely many solutions (x,y,m,n) with 2∤x. Moreover, all solutions of (1) with 2∤x satisfy max(x,m,n) < C, where C is an effectively computable constant. In this paper we completely determine all solutions of (1) as follows. Theorem. Equation (1) has only the solution (x,y,m,n)=(7,20,1,2).