## On the Diophantine equation $$x^2 + p^{2k} = y^n$$.(English)Zbl 1175.11018

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 F. S. Abu Muriefah and Y. Bugeaud [Rev. Colomb. Math. 40, No. 1, 31–37 (2006; Zbl 1189.11019)].
In this paper, the authors extend earlier results obtained by 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 Y. Bilu, G. Hanrot and P. M. Voutier on primitive divisors of Lehmer numbers [see J. Reine Angew. Math. 539, 75–122 (2001; Zbl 0995.11010)].

### MSC:

 11D41 Higher degree equations; Fermat’s equation 11D61 Exponential Diophantine equations

### Keywords:

exponential Diophantine equations; primitive divisors

### Citations:

Zbl 0995.11010; Zbl 1121.11028; Zbl 1189.11019
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