The diophantine equation \(x^ 2+119 = 15\cdot 2^ n\) has exactly six solutions
The author considers the diophantine equation \(x^2 + D = A \cdot 2^n\), \(x,D, A,n\) positive integers, \((A,2D) = 1\). If \(A = 1\) and \(D \neq 7\), \textit{F. Beukers} [Acta Arith. 38, 389--410 (1981; Zbl 0371.10014)] proved that there are at most four solutions \(x,n\). As the (historically famous) case \(D = 7\), \(A = 1\) shows, it is possible that there are five solutions. The author shows that if \(D = 119\), \(A = 15\) there are exactly six solutions, which he can give explicitly. The proof is based on ideas of \textit{P. Bundschuh} [Semin. Math. Sci. 12, 31--40 (1988; Zbl 0682.10014)] and consideration of congruence properties of recurrent series given by the solutions of \(15y^2 - x^2 = 119\) (if \(n\) is even), or \(30y^2 - x^2 = 119\) (if \(n\) is odd). In the same way, the author is able to proof that the diophantine equation \(x^2 + 391 = 35 \cdot 2^n\) has exactly five, explicitly given solutions.
congruence properties of recurrent series
scientific article; zbMATH DE number 894685