On the Diophantine equation \(x^2 + p^{2k} = y^n\)
The Diophantine equation \[ x^2+D=y^n, \] where \(D, n, x, y\) are positive integers and \(n\geq 3\), has a very long and rich history. In 1850, Lebesgue was the first to obtain a non-trivial result. He proved that the above equation has no solutions when \(C=1\) [Nouv. Ann. Math. 9, 178--181 (1850)]. Since then, the equation was solved for several values of the parameter \(D\) in the range \(1\leq D\leq 100\). Recently, several authors became interested in the case when \(D\) is positive and only the prime factors of \(D\) are specified, for example, the case when \(D=p^k\), where \(p\) is a prime number. See the brief survey done by \textit{F. S. Abu Muriefah} and \textit{Y. Bugeaud} [Rev. Colomb. Math. 40, No. 1, 31--37 (2006; Zbl 1189.11019)]. In this paper, the authors extend earlier results obtained by \textit{I. Pink} [Publ. Math. 70, No. 1--2, 149--166 (2006; Zbl 1121.11028)]. They consider the case \(D=p^{2k}\), where \(p\geq 2\) is a prime number. So for \(2\leq p\leq 100\), they determine all solutions \((x, y, p, n, k)\) of the Diophantine equation \[ x^2+p^{2k}=y^n,\; \text{where}\; x\geq 1,\; y\geq 1,\; n\geq 3\; \text{prime},\; k\geq 0,\; \text{and}\; \gcd(x, y)=1. \] Moreover, they deduce that the equation has no solution \((x, y, p, n, k)\) with \(x\geq 1,\; y\geq 1,\; n\geq 5\; \text{prime},\; k\geq 0,\; \text{and}\; \gcd(x, y)=1\). The proofs of these results are based on Pink's result cited above and a deep result of \textit{Y. Bilu}, \textit{G. Hanrot} and \textit{P. M. Voutier} on primitive divisors of Lehmer numbers [see J. Reine Angew. Math. 539, 75--122 (2001; Zbl 0995.11010)].
scientific article; zbMATH DE number 5500892