On the diophantine equation $(x^m + 1)(x^n + 1) = y²$
Acta Arithmetica, Tome 82 (1997) no. 1, pp. 17-26
Cet article a éte moissonné depuis la source Institute of Mathematics Polish Academy of Sciences
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).
@article{10_4064_aa_82_1_17_26,
author = {Maohua Le},
title = {On the diophantine equation $(x^m + 1)(x^n + 1) = y{\texttwosuperior}$},
journal = {Acta Arithmetica},
pages = {17--26},
year = {1997},
volume = {82},
number = {1},
doi = {10.4064/aa-82-1-17-26},
language = {fr},
url = {http://geodesic.mathdoc.fr/articles/10.4064/aa-82-1-17-26/}
}
Maohua Le. On the diophantine equation $(x^m + 1)(x^n + 1) = y²$. Acta Arithmetica, Tome 82 (1997) no. 1, pp. 17-26. doi: 10.4064/aa-82-1-17-26
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