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Let      D D       be a positive integer which is square free, and let      p p       be a prime. In this note we prove that if      p = 2 p = 2       and      D &gt; exp ⁡  64 D &gt; \exp 64       , then the equation      x 2 p −  D y 2 = 1 {x^{2p}} - D{y^2} = 1       has at most one positive integer solution      ( x , y ) \left ( {x,y} \right )      ; if      p &gt; 2 p &gt; 2       and      D &gt; exp ⁡  exp ⁡  exp ⁡  exp ⁡  10 D &gt; \exp \exp \exp \exp 10      , then every positive integer solution      ( x , y ) \left ( {x,y} \right )       can be expressed as      x p + y D = ε  1 m {x^p} + y\sqrt D = \varepsilon _1^m      , where      m m       is a positive integer with      2 ∤  m , ε  1 2\nmid m,{\varepsilon _1}       is the fundamental solution of Pell’s equation      u 2 −  D v 2 = 1 {u^2} - D{v^2} = 1      .

A note on the Diophantine equation 𝑥^{2𝑝}-𝐷𝑦²=1